CLP-4 Vector Calculus [4]

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CLP-4 VECTOR CALCULUS (FELDMAN, RECHNITZER, AND YEAGER)

Joel Feldman, Andrew Rechnitzer and Elyse Yeager University of British Columbia

University of British Columbia CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer and Elyse Yeager

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TABLE OF CONTENTS Colophon Licensing Preface Feedback about the text

1: Curves 1.1: Derivatives, Velocity, Etc. 1.2: Reparametrization 1.3: Curvature 1.4: Curves in Three Dimensions 1.5: A Compendium of Curve Formula 1.6: Integrating Along a Curve 1.7: Sliding on a Curve 1.8: Optional — Polar Coordinates 1.9: Optional — Central Forces 1.10: Optional — Planetary Motion 1.11: Optional — The Astroid 1.12: Optional — Parametrizing Circles

2: Vector Fields 2.1: Definitions and First Examples 2.2: Optional — Field Lines 2.3: Conservative Vector Fields 2.4: Line Integrals 2.5: Optional — The Pendulum

3: Surface Integrals 3.1: Parametrized Surfaces 3.2: Tangent Planes 3.3: Surface Integrals 3.4: Interpretation of Flux Integrals 3.5: Orientation of Surfaces

4: Integral Theorems 4.1: Gradient, Divergence and Curl 4.2: The Divergence Theorem 4.3: Green's Theorem 4.4: Stokes' Theorem 4.5: Optional — Which Vector Fields Obey ∇ × F = 0 4.6: Really Optional — More Interpretation of Div and Curl 4.7: Optional — A Generalized Stokes' Theorem

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5: True/False and Other Short Questions 5.2: Exercises

Appendices A: Appendices A.1: Trigonometry A.2: Powers and Logarithms A.3: Table of Derivatives A.4: Table of Integrals A.5: Table of Taylor Expansions A.6: 3d Coordinate Systems A.7: ISO Coordinate System Notation A.8: Conic Sections and Quadric Surfaces A.9: Review of Linear Ordinary Differential Equations B: Hints for Exercises C: Answers to Exercises D: Solutions to Exercises

Index Glossary Detailed Licensing

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Colophon Cover Design Nick Loewen — licensed under the CC-BY-NC-SA 4.0 License. Source files A link to the source files for this document can be found at the CLP textbook website. The sources are licensed under the CC-BY-NC-SA 4.0 License. Edition CLP4 Vector Calculus: May 2021 Website CLP-4 ©2016 – 2021 Joel Feldman, Andrew Rechnitzer, Elyse Yeager This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. You can view a copy of the license here.

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Licensing A detailed breakdown of this resource's licensing can be found in Back Matter/Detailed Licensing.

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Preface This text is a merger of the CLP Vector Calculus textbook and problembook. It is, at the time that we write this, still a work in progress; some bits and pieces around the edges still need polish. Consequently we recommend to the student that they still consult text webpage for links to the errata — especially if they think there might be a typo or error. We also request that you send us an email at [email protected] Additionally, if you are not a student at UBC and using these texts please send us an email (again using the feedback button) — we'd love to hear from you. Joel Feldman, Andrew Rechnitzer and Elyse Yeager

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Feedback about the text The CLP-4 Vector Calculus text is still undergoing testing and changes. Because of this we request that if you find a problem or error in the text then: 1. Please check the errata list that can be found at the text webpage. 2. Is the problem in the online version or the PDF version or both? 3. Note the URL of the online version and the page number in the PDF 4. Send an email to [email protected] . Please be sure to include a description of the error the URL of the page, if found in the online edition and if the problem also exists in the PDF, then the page number in the PDF and the compile date on the front page of PDF.

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CHAPTER OVERVIEW 1: Curves We are now going to study vector-valued functions of one real variable. That is, we are going to study functions that assign to each real number t (typically in some interval) a vector 1 r (t). For example ⇀

⇀

r (t) = (x(t), y(t), z(t))

might be the position of a particle at time t. As t varies,

⇀

r (t)

sweeps out a curve.

While in some applications t will indeed be “time”, it does not have to be. It can be simply a parameter that is used to label the different points on the curve that r (t) sweeps out. We then say that r (t) provides a parameterization of the curve. ⇀

⇀

 Example 1.0.1. Parametrization of x

2

+ y

2

2

= a

While we will often use t as the parameter in a parametrized curve r (t), there is no need to call it t. Sometimes it is natural to use a different name for the parameter. For example, consider the circle x + y = a . It is natural to use the angle θ in the sketch below to label the point (a cos θ , a sin θ) on the circle. ⇀

2

2

2

That is, ⇀

r (θ) = (a cos θ , a sin θ)

is a parametrization of the circle x traces out the full circle.

2

+y

2

2

=a .

0 ≤ θ < 2π

Just looking at the figure above, it is clear that, as θ runs from

0

to 2π,

⇀

r (θ)

However beware that just knowing that r (t) lies on a specified curve does not guarantee that, as t varies, r (t) covers the entire curve. For example, as t runs over the whole real line, arctan(t) runs over the interval (−1, 1). For all t, ⇀

⇀

2

π

2

⇀

r (t) = (x(t), y(t)) = a (

− −−−−−−−−−−−− − 4 2 arctan (t) )

arctan(t) , √ 1 −

π

is well-defined and obeys x(t)

2

2

+ y(t)

2

=a .

But this

 Example 1.0.2. Parametrization of (x − h)

2

⇀

r (t)

π

2

does not cover the entire circle because y(t) is always positive. 2

+ (y − k)

2

= a

We can tweak the parametrization of Example 1.0.1 to get a parametrization of the circle of radius a that is centred on (h, k). One way to do so is to redraw the sketch of Example 1.0.1 with the circle translated so that its centre is at (h, k).

1

We see from the sketch that ⇀

r (θ) = (h + a cos θ , k + a sin θ)

is a parametrization of the circle (x − h)

2

2

0 ≤ θ < 2π

2

+ (y − k)

=a .

A second way to come up with this parametrization is to observe that we can turn the trig identity cos equation (x − h) + (y − k) = a of the circle by

2

2

2

multiplying the trig identity by a to get (a cos t) + (a sin t) = a and then setting  a cos t = x − h  and  a sin t = y − k , which turns (a cos t) + (a sin t) 2

2

2

 Example 1.0.3. Parametrization of

2

x

+

2

a

2

y

2

2

a

+

 

Setting   cos t = (

and   sin t = 1

x a

)

3

y b

 

turns cos

2

and   sin t = (

 

y a

1

)

3

 

and of x

2/3

2

y

2

x

like we did in the second part of the last example. a

= 1

b

We can build parametrizations of the curves

x

into the

t =1

2

2

Setting   cos t =

2

t + sin

2

and x

2/3

=1

2

b

2

t + sin

turns cos

2

t =1

+ y

+y

into 2

t + sin

2

2/3

2/3

2

x

2

a

t =1

+

2

=a

into (x − h)

2

2

+ (y − k)

2

=a .

2/3

= a

from the trig identity

2/3

=a

2

cos

2

t + sin

t = 1,

2

y

2

= 1.

b

into

2/3

x

2/3

+

a

2/3

y

2/3

= 1.

a

So ⇀

r (t) = (a cos t , b sin t)

⇀

3

r (t) = (a cos

give parametrizations of

a2

2

y

2

x

+

2

b

=1

and

2/3

x

+y

2/3

3

t , a sin

2/3

=a

,

0 ≤ t < 2π t)

0 ≤ t < 2π

respectively. To see that running

t

from

0

to

2π

runs

⇀

r (t)

once around the curve, look at the figures below.

The curve x + y = a is called an astroid. From its equation, we would expect its sketch to look like a deformed circle. But it is probably not so obvious that it would have the pointy bits of the right hand figure. We will not explain here why they arise. The astroid is studied in some detail in Example 1.1.9. In particular, the above sketch is carefully developed there. 2/3

2/3

2/3

 Example 1.0.4. Parametrization of e

y

2

= 1 + x

A very easy method that can often create parametrizations for a curve is to use x or y as a parameter. Because we can solve e = 1 +x for y as a function of x, namely y = ln (1 + x ), we can use x as the parameter simply by setting t = x. This gives the parametrization y

2

2

2

⇀

2

r (t) = (t , ln(1 + t ))

 Example 1.0.5. Parametrization of x

2

+ y

2

2

= a ,

−∞ < t < ∞

again

It is also quite common that one can use either x or y to parametrize part of, but all of, a curve. A simple example is the circle x + y = a . For each −a < x < a, there are two points on the circle with that value of x. So one cannot use x to parametrize the whole circle. Similarly, for each −a < y < a, there are two points on the circle with that value of y. So one cannot use y to parametrize the whole circle. On the other hand 2

2

2

− − − − − − 2 2 r (t) = (t , √ a − t )

−a < t < a

− − − − − − 2 2 r (t) = (t , −√ a − t )

−a < t < a

⇀

⇀

provide parametrizations of the top half and bottom half, respectively, of the circle using x as the parameter, and − − − − − − 2 2 r (t) = (√ a − t , t)

−a < t < a

− − − − − − 2 2 r (t) = ( − √ a − t , t)

−a < t < a

⇀

⇀

provide parametrizations of the right half and left half, respectively, of the circle using y as the parameter.

 Example 1.0.6. Unparametrization of

⇀

r (t) = (cos t, 7 − t)

In this example, we will undo the parametrization question. We may rewrite the parametrization as

and find the Cartesian equation of the curve in

⇀

r (t) = (cos t, 7 − t)

x = cos t y = 7 −t

Note that we can eliminate the parameter t simply by using the second equation to solve for t = 7 − y. Substituting this into the first equation gives us the Cartesian equation

t

as a function of

y.

Namely

x = cos(7 − y)

Curves often arise as the intersection of two surfaces. For example, the intersection of the ellipsoid paraboloid z = x + 2y is the blue curve in the figure below. 2

2

x

y

+

2

2

+

z

2

3

=1

with the

2

One way to parametrize such curves is to choose one of the three coordinates x, y, z as the parameter, and solve the two given equations for the remaining two coordinates, as functions of the parameter. Here are two examples.

 Example 1.0.7 The set of all (x, y, z) obeying 3

x

−e

2

x

3y

−e

y

=0

+z = 0

is a curve. We can choose to use y as the parameter and think of 3

x

2

x

=e +z = e

3y

y

as a system of two equations for the two unknowns x and z, with y being treated as a given constant, rather than as an unknown. We can now solve the first equation for x, substitute the result into the second equation, and finally solve for z. 3

x

2

x

=e +z = e

3y

⟹

x =e

y

y

⟹

e

2y

+z = e

y

So ⇀

r (y) = (e

y

, y , e

3

y

−e

2y

)

⟹

z =e

y

−e

2y

is a parametrization for the given curve.

 Example 1.0.8 The previous example was rigged so that it was easy to solve for x and z as functions of y. In practice it is not always easy, or even possible, to do so. A more realistic example is the set of all (x, y, z) obeying 2

x

y

2

+

z

2

2

3 −

y

2

=1 3

2

x

which is the blue curve in the figure above. Substituting x

2

+ + 2y

2

= z − 2y

z

=z

2

+z+

2

(from the second equation) into the first equation gives

2

=1 3

or, completing the square, 3 −

y

2

1 +

2

3 (z +

3

2

) 2

7 = 4

If, for example, we are interested in points (x, y, z) on the curve with y ≥ 0, this can be solved to give y as a function of z. − −−−−−−−−−−−− − 2 2 3 14 (z + ) − 9 2 12

y =√

Then x

2

= z − 2y

2

also gives x as a function of z. If x ≥ 0, − −−−−−−−−−−−−−−− − 2 4 3 14 (z + ) + 9 2 6

x = √ z−

− −−−−−−−−−− − 4 4 1 2 − z − z 3 9 3

=√

The other signs of x and y can be gotten by using the appropriate square roots. In this example, satisfies the two original equations, if and only if all of (±x, ±y, z) are also on the curve. 1. We are going to use boldface letters, like instead.

⇀

r,

(x, y, z)

is on the curve, i.e.

to designate vectors. When writing by hand, it is clearer to use arrows, like r ,⃗ 

1.1: Derivatives, Velocity, Etc. 1.2: Reparametrization 1.3: Curvature 1.4: Curves in Three Dimensions 1.5: A Compendium of Curve Formula 1.6: Integrating Along a Curve 1.7: Sliding on a Curve 1.8: Optional — Polar Coordinates 1.9: Optional — Central Forces 1.10: Optional — Planetary Motion 1.11: Optional — The Astroid 1.12: Optional — Parametrizing Circles

This page titled 1: Curves is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

4

1.1: Derivatives, Velocity, Etc. This being a Calculus text, one of our main operations is differentiation. We are now interested in parametrizations easy and natural to extend our definition of derivative to r (t) as follows.

⇀

r (t).

It is very

⇀

 Definition 1.1.1 The derivative of the vector valued function

⇀

r (t)

is defined to be ⇀

⇀

r (t) =

(t) = lim dt

when the limit exists. In particular, if

⇀

r (t + h) − r (t)

dr

⇀′

h

h→0

then

⇀

r (t) = (x(t) , y(t) , z(t)), ⇀′

′

′

′

r (t) = (x (t) , y (t) , z (t))

That is, to differentiate a vector valued function of t, just differentiate each of its components. And of course differentiation interacts with arithmetic operations, like addition, in the obvious way. Only a little more thought is required to see that differentiation interacts quite nicely with dot and cross products too. Here are some examples.

 Example 1.1.2 Let 2

a(t) = t

b(t) = e

4

6

^ ı ı +t

−t

^ ȷ ȷ +t

^ ı ı +e

−3t

^ k

^ ȷ ȷ +e

−5t

^ k

2

γ(t) = t

s(t) = sin t

We are about to compute some derivatives. To make it easier to follow what is going on, we'll use some colour. When we apply the product rule d

′

′

[f (t) g(t)] = f (t) g(t) + f (t) g (t) dt

we'll use blue to highlight the factors f

′

(t)

and g

′

(t).

Here we go. 2

γ(t) b(t) = t e

−t

2

^ ı ı +t e

−3t

2

^ ȷ ȷ +t e

−5t

^ k

gives d [γ(t)b(t)] = [2te

−t

2

−t e

−t

]^ ı ı + [2te

−3t

2

−3 t e

−3t

]^ ȷ ȷ + [2te

−5t

2

−5 t e

−5t

^ ]k

dt = 2t{ e ′

−t

^ ı ı +e

−3t

^ ȷ ȷ +e

−5t

2 −t −3t −5t ^ ^ ^ ^ k} + t { − e ı ı − 3e ȷ ȷ − 5e k}

′

= γ (t)b(t) + γ(t)b (t)

and 2

a(t) ⋅ b(t) = t e

−t

4

+t e

−3t

6

+t e

−5t

gives

1.1.1

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d [a(t) ⋅ b(t)]

= [2te

−t

2

−t e

−t

3

] + [4 t e

−3t

4

−3 t e

−3t

5

] + [6 t e

−5t

6

−5 t e

−5t

]

dt = [2te

−t

3

+ 4t e 3

+ {t

5

+ 6t e 5

= {2t ^ ı ı + 4t 2

−3t

^ k} ⋅ { e

^ ȷ ȷ + 6t 4

6

^ ı ı +t

−5t

^ ȷ ȷ +t

′

2

] + [−t e −t

^ ı ı +e

^ k} ⋅ { − e

−t

−3t

−t

4

−3 t e

^ ȷ ȷ +e

^ ı ı − 3e

−3t

−5t

−3t

6

−5 t e

−5t

]

^ k}

^ ȷ ȷ − 5e

−5t

^ k}

′

= a (t) ⋅ b(t) + a(t) ⋅ b (t)

and ^ ı ı

^ ȷ ȷ

^ k

a(t) × b(t) = det ⎢ t2

4

6

⎡

⎣

e

4

= ^ ı ı (t e

−t

t

t

⎥

−3t

−5t

⎦

e

−5t

⎤

6

−t e

e

−3t

2

)−^ ȷ ȷ (t e

−5t

6

−t e

2 −3t 4 −t ^ ) + k(t e −t e )

−t

gives d [a(t) × b(t)] dt 3

=  ^ ı ı ( 4 t e

−5t

4

+^ ı ı (−5 t e

−5t

3

= {2t ^ ı ı + 4t 2

+ {t

5

   −  6 t e 6

+3 t e

−3t

5

^ ȷ ȷ + 6t 4

^ ı ı +t

−3t

)  −  ^ ȷ ȷ ( 2te 2

)−^ ȷ ȷ (−5 t e

−5t

−5t

5

  −  6 t e 6

+t e

−t

−t

−3t 3 −t ^ ) + k( 2te   −  4 t e )

2 −3t 4 −t ^ ) + k(−3 t e +t e )

−t −3t −5t ^ ^ ^ ^ k} × { e ı ı +e ȷ ȷ +e k} 6

^ ȷ ȷ +t

′

−t −3t −5t ^ ^ ^ ^ k} × { − e ı ı − 3e ȷ ȷ − 5e k} ′

= a (t) × b(t) + a(t) × b (t)

and 2

a(s(t)) = (sin t)

4

^ ı ı + (sin t)

6

^ ȷ ȷ + (sin t)

d ⟹

[a(s(t))]

3

= 2(sin t) cos t ^ ı ı + 4(sin t)

^ k 5

cos t ^ ȷ ȷ + 6(sin t)

^ cos t k

dt 3 5^ = {2(sin t) ^ ı ı + 4(sin t) ^ ȷ ȷ + 6(sin t) k} cos t ′

′

= a (s(t)) s (t)

Of course these examples extend to general (differentiable) a(t),

b(t), γ(t)

and s(t) and give us (most of) the following theorem.

 Theorem 1.1.3. Arithmetic of differentiation Let be vector valued differentiable functions of t ∈ R that take values in R and α, β ∈ R be constants and γ(t) and s(t) be real valued differentiable functions of t ∈ R n

a(t), b(t)

Then d

′

(a)

′

[α a(t) + β b(t)] = α a (t) + β b (t)

(linear combination)

dt d (b)

′

′

[γ(t)b(t)] = γ (t)b(t) + γ(t)b (t)

(multiplication by scalar function)

dt d

′

(c)

′

[a(t) ⋅ b(t)] = a (t) ⋅ b(t) + a(t) ⋅ b (t)

(dot product)

dt d

′

′

(d)

[a(t) × b(t)] = a (t) × b(t) + a(t) × b (t)  

(cross product)

dt d (e)

′

′

[a(s(t))] = a (s(t)) s (t)

(composition)

dt

1.1.2

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Let's think about the geometric significance of ⇀′

along the curve. The derivative r head at r (t + h) and tail at r (t). ⇀

(t)

⇀′

r (t).

is the limit of

In particular, let's think about the relationship between

⇀ ⇀ r (t+h)− r (t)

as

h

h → 0.

The numerator,

⇀

⇀

⇀′

r (t + h) − r (t),

r (t)

and distances

is the vector with

⇀

When h is very small this vector has the essentially the same direction as the tangent vector to the curve at r (t) and has length being essentially the length of the part of the curve between r (t) and r (t + h). ⇀

⇀

⇀

Taking the limit as h → 0 yields that ⇀′

r (t)

is a tangent vector to the curve at

⇀

r (t)

that points in the direction of increasing t and

if s(t) is the length of the part of the curve between

and

⇀

r (0)

⇀

r (t),

then

ds dt

⇀

(t) = ∣ ∣

dr dt

(t)∣ ∣.

This is worth stating formally.

 Lemma 1.1.4 Let

⇀

r (t)

be a parametrized curve.

^ 1. Denote by T (t) the unit tangent vector to the curve at

⇀

r (t)

pointing in the direction of increasing t. If

⇀′

r (t) ≠ 0

then

⇀′

r (t)

^ T (t) =

⇀′

| r (t)|

2. Denote by s(t) the length of the part of the curve between

⇀

r (0)

and

⇀

r (t).

Then

⇀

ds (t) = dt

∣ ∣

dr dt T

s(T ) − s(T0 ) = ∫ T0

∣ (t) ∣ ⇀

∣ dr ∣ ∣ (t)∣ dt ∣ dt ∣

3. In particular, if the parameter happens to be arc length, i.e. if t = s, so that

ds = 1, ds

then

⇀

∣ dr ∣ ∣ (s)∣ = 1 ∣ ds ∣

′

^ (s) = ⇀ T r (s)

As an application, we have the

 Lemma 1.1.5 If

⇀

r (t) = (x(t) , y(t) , z(t))

is the position of a particle at time t, then

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⇀

position at time t = r (t) = (x(t) , y(t) , z(t)) ⇀′

⇀

′

′

ds

′

velocity at time t = v (t) = r (t) = (x (t) , y (t) , z (t)) =

^ (t) T (t)

dt ds speed at time t =

−−−−−−−−−−−−−−−−− −

⇀′

⇀

′

2

(t) = | v (t)| = | r (t)| = √ (x (t)

′

2

+ y (t)

′

2

+ z (t)

dt ⇀′′

acceleration at time t = a(t) = r

⇀′

′′

(t) = v (t) = (x (t) , y

′′

(t) , z

′′

(t))

and the distance travelled between times T and T is 0

T

⇀

∣ ∣

s(T ) − s(T0 ) = ∫

dr

T0

Note that the velocity

⇀′

⇀

v (t) = r (t)

dt

T

−−−−−−−−−−−−−−−−− − ′

∣ (t) dt = ∫ ∣

2

√ (x (t)

′

2

+ y (t)

′

2

+ z (t)

dt

T0

is a vector quantity while the speed

ds

⇀′

(t) = | r (t)| dt

is a scalar quantity.

 Example 1.1.6. Circumference of a circle In general it can be quite difficult to compute arc lengths. So, as an easy warmup example, we will compute the circumference of the circle x + y = a . We'll also find a unit tangent to the circle at any point on the circle. We'll use the parametrization 2

2

2

⇀

r (θ) = (a cos θ , a sin θ)

0 ≤ θ ≤ 2π

of Example 1.0.1. Using Lemma 1.1.4, but with the parameter t renamed to θ ⇀′

r (θ) = a( − sin θ , cos θ) ⇀′

r (θ)

^ T (θ) =

= ( − sin θ , cos θ)

⇀′

| r (θ)| ds dθ

⇀′

(θ) = ∣ ∣ r (θ)∣ ∣ =a Θ

s(Θ) − s(0) = ∫

⇀′

∣ ∣ r (θ)∣ ∣ dθ = aΘ

0

As 1 s(Θ) is the arc length of the part of the circle with 0 ≤ θ ≤ Θ, the circumference of the whole circle is s(2π) = 2πa

which is reassuring, since this formula has been known 2 for thousands of years.

The formula s(Θ) − s(0) = aΘ also makes sense — the part of the circle with circle, and so should have length × 2πa. Also note that

0 ≤θ ≤Θ

is the fraction

Θ 2π

of the whole

Θ

2π

⇀

^ (θ) = (a cos θ , a sin θ) ⋅ ( − sin θ , cos θ) = 0 r (θ) ⋅ T

so that the tangent to the circle at any point is perpendicular to the radius vector of the circle at that point. This is another geometric fact that has been known 3 for thousands of years.

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 Example 1.1.7. Arc length of a helix Consider the curve ^ r (t) = 6 sin(2t) ^ ı ı + 6 cos(2t)^ ȷ ȷ + 5tk

⇀

^ where the standard basis vectors ^ ı ı = (1, 0, 0), ^ ȷ ȷ = (0, 1, 0) and k = (0, 0, 1). We'll first sketch it, by observing that

and y(t) = 6 cos(2t) obey x(t) + y(t) = 36 sin lie on the cylinder x + y = 36 and as t increases, (x(t), y(t)) runs clockwise around the circle x + y linearly. 2

x(t) = 6 sin(2t)

2

2

2

2

(2t) + 36 cos (2t) = 36.

So all points of the curve

2

2

2

= 36

and at the same time z(t) = 5t just increases

Our curve is the helix

^ We have marked three points of the curve on the above sketch. The first has t = 0 and is 0 ^ ı ı + 6^ ȷ ȷ + 0 k. The second has t = ^ ^ ^ (t) and is 0 ^ ı ı − 6^ ȷ ȷ + k, and the third has t = π and is 0 ^ ı ı + 6^ ȷ ȷ + 5π k. We'll now use Lemma 1.1.4 to find a unit tangent T to the curve at r (t) and also the arclength of the part of curve between t = 0 and t = π. π 2

5π 2

⇀

^ r (t) = 6 sin(2t) ^ ı ı + 6 cos(2t)^ ȷ ȷ + 5tk

⇀

⇀′

^ r (t) = 12 cos(2t) ^ ı ı − 12 sin(2t)^ ȷ ȷ + 5k

ds dt

− −−−−−−−−−−−−−−−−−−−−−− −

⇀′

2

(t) = ∣ ∣ r (t)∣ ∣ = √ 12

2

2

cos (2t) + 12

2

2

sin (2t) + 5

− − − − − − − 2 2 = √ 12 + 5

= 13 ⇀′

r (t)

^ T (t) =

12 =

⇀′

| r (t))|

cos(2t) ^ ı ı −

13

12 13

sin(2t)^ ȷ ȷ +

5

^ k

13

π ⇀′

∣ ∣ r (t)∣ ∣ dt = 13π

s(π) − s(0) = ∫ 0

 Example 1.1.8. Velocity and acceleration Imagine that, at time t, a particle is at t

⇀

r (t) = [h + a cos(2π

)] ^ ı ı + [k + a sin(2π

T

t

)] ^ ȷ ȷ

T

As | r (t) − h ^ ı ı −k ^ ȷ ȷ | = a, the particle is running around the circle of radius a centred on (h, k). When t increases by T , the argument, 2π , of cos(2π ) and sin(2π ) increases by exactly 2π and the particle runs exactly once around the circle. In particular, it travels a distance 2πa. So it is moving at speed . According to Lemma 1.1.5, it has ⇀

t

t

t

T

T

T

2πa T

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2πa

⇀′

velocity  = r (t) = −

t

ds speed =

)^ ı ı +

sin(2π T

2πa

T

t cos(2π

T

)^ ȷ ȷ

T

2πa

⇀′

(t) = | r (t)| = dt

T 2

⇀′′

acceleration = r

4π a (t) = −

2

T 4π =− T

2

t cos(2π

4π a )^ ı ı −

T

T

2

t sin(2π

)^ ȷ ȷ T

2 ⇀

2

[ r (t) − h ^ ı ı −k ^ ȷ ȷ]

Here are some observations. ⇀′

The velocity r (t) has dot product zero with r (t) − h ^ ı ı −k ^ ȷ ȷ , which is the radius vector from the centre of the circle to the particle. So the velocity is perpendicular to the radius vector, and hence parallel to the tangent vector of the circle at ⇀

⇀

r (t).

The speed given by Lemma 1.1.5 is exactly the speed we found above, just before we started applying Lemma 1.1.5. The acceleration r (t) points in the direction opposite to the radius vector. ⇀′′

 Example 1.1.9. Perimeter of the astroid In this example, we find the perimeter of the astroid 4 2/3

x

+y

2/3

2/3

=a

A geometric construction of this curve, as well as a derivation of its equation is given in the optional section 1.11 later. We'll start by finding a convenient parametrization. To do so, notice that x + y = a looks somewhat like the equation of the circle x + y = a . The standard parametrization of the circle, namely x = a cos t, y = a sin t works because of the elementary trig identity 2/3

2

2

cos

t + sin

2/3

2/3

2

2/3

2/3

=a

2

cos

t

and y(t)

2/3

as desired. But of course its easy to arrange that: just solve x(t) y(t) =a sin t for y(t), namely y(t) = a sin t. x(t)

2/3

+ y(t)

2/3

=a

2/3

2/3

=a

2

sin

then the same elementary trig identity will give

t,

,

2/3

2/3

2

t = 1.

If we can arrange that x(t) 2/3

2

2

2/3

=a

2

cos

t

for x(t), namely x(t) = a cos

3

t,

and solve

3

Our parametrization is ⇀

3

r (t) = a cos

3

t ^ ı ı + a sin

t^ ȷ ȷ

By Lemma 1.1.4

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⇀

3

r (t) = a cos

3

t ^ ı ı + a sin

⇀′

2

2

t ^ ı ı + 3a sin

r (t) = −3a sin t cos

ds dt

t^ ȷ ȷ t cos t ^ ȷ ȷ

− −−−−−−−−−−−−−−−−−−−−−−− − ⇀′ √ 9 a2 sin2 t cos4 t + 9 a2 sin4 t cos2 t (t) = ∣ ∣ r (t)∣ ∣ = −−−−−−−−−−−−−−−−−−− − 2

= 3a√ sin

2

t cos

2

t(cos

2

t + sin

t)

= 3a∣ ∣ sin t cos t∣ ∣ ⇀′

^ T (t) =

r (t)

sin t cos t =

⇀′

| r (t))|

 ( − cos t ^ ı ı + sin t ^ ȷ ȷ)

| sin t cos t|

= sgn(sin t cos t) ( − cos t ^ ı ı + sin t ^ ȷ ȷ)

Here sgn(sin t cos t) means “the sign of sin t cos t ”, i.e +1 when sin t cos t > 0 and −1 when sin t cos t < 0. So ^ (t) T

={

1

if  sin t > 0,   cos t > 0 or  sin t < 0,   cos t < 0

−1

if  sin t > 0,   cos t < 0 or  sin t < 0,   cos t > 0

1

if 0 < t
0. Let the positive z axis point vertically upwards, as usual. When is the particle moving upwards, and when is it moving downwards? Is it moving faster at time t = 1 or at time t = 3?

 7 Below is the graph of the parametrized function Indicate on the graph s(t + h) − s(t) and

⇀

r (t).

⇀

Let s(t) be the arclength along the curve from ⇀

r (t + h) − r (t).

⇀

r (0)

to

⇀

r (t).

Are the quantities scalars or vectors?

 8 What is the relationship between velocity and speed in a vector-valued function of time?

 9✳ ⇀

Let

⇀

r (t)

be a vector valued function. Let

⇀′

⇀′′

r , r

, and

⇀′′′

r

denote

dr

2⇀

,

d

dt d

⇀

⇀′

r

2

dt

⇀′′

[( r × r ) ⋅ r

and

3⇀

d

r

3

dt

,

respectively. Express

]

dt

in terms of

⇀

⇀′

r, r

,

⇀′′

r

, and

1. ( r × r ) ⋅ r 2. ( r × r ) ⋅ r + ( r × r 3. ( r × r ) ⋅ r 4. 0 5. None of the above. ⇀′ ⇀′ ⇀

⇀′′ ⇀′′

⇀′

⇀′′′

r

.

Select the correct answer.

⇀′′′

⇀

⇀

⇀′

⇀′′′

)⋅ r

⇀′′′

 10 Show that, if the position and velocity vectors of a moving particle are always perpendicular, then the path of the particle lies on a sphere.

Stage 2  11 ✳ Find the speed of a particle with the given position function – 5t −5t ^ r (t) = 5 √2 t ^ ı ı +e ^ ȷ ȷ −e k

⇀

Select the correct answer: 1. | v (t)| = (e ⇀

5t

+e

−5t

)

1.1.11

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− −−−−−−−−−− −

2. | v (t)| = √10 + 5e + 5e − −−−−−−−−−−− − 3. | v (t)| = √10 + e + e 4. | v (t)| = 5(e + e ) 5. | v (t)| = 5(e + e ) ⇀

t

⇀

10t

⇀

5t

⇀

t

−t

−10t

−5t

−t

 12 Find the velocity, speed and acceleration at time t of the particle whose position is ⇀

^ r (t) = a cos t ^ ı ı + a sin t ^ ȷ ȷ + ct k

Describe the path of the particle.

 13 ✳ 1. Let ⇀

2

r (t) = (t , 3,

1

3

t )

3

Find the unit tangent vector to this parametrized curve at t = 1, pointing in the direction of increasing t. 2. Find the arc length of the curve from (a) between the points (0, 3, 0) and (1, 3, − ). 1 3

 14 Using Lemma 1.1.4, find the arclength of

− −

⇀

r (t) = (t, √

3 2

2

3

t ,t )

from t = 0 to t = 1.

 15 Find the length of the parametric curve x = a cos t sin t

between t = 0 and t = T

2

y = a sin

t

z = bt

> 0.

 16 A particle's position at time t is given by at time t?

⇀

r (t) = (t + sin t, cos t)

7.

What is the magnitude of the acceleration of the particle

 17 ✳ A curve in R is given by the vector equation 3

3

⇀

r (t) = (2t cos t, 2t sin t,

t

3

)

1. Find the length of the curve between t = 0 and t = 2. 2. Find the parametric equations of the tangent line to the curve at t = π.

 18 ✳ Let

⇀

r (t) = (3 cos t, 3 sin t, 4t)

be the position vector of a particle as a function of time t ≥ 0.

1. Find the velocity of the particle as a function of time t. 2. Find the arclength of its path between t = 1 and t = 2.

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 19 The plane  z = 2x + 3y  intersects the cylinder  x

2

+y

2

= 9 

in an ellipse.

1. Find a parametrization of the ellipse. 2. Express the circumference of this ellipse as an integral. You need not evaluate the integral 8.

 20 ✳ Consider the curve 1

⇀

r (t) =

1

3

cos

t ^ ı ı +

3

3

sin

3

t^ ȷ ȷ + sin

^ tk

3

1. Compute the arc length of the curve from t = 0 to t = . 2. Compute the arc length of the curve from t = 0 to t = π. π 2

 21 ✳ Let

⇀

r (t) = (

1 3

3

t ,

1 2

2

t ,

1 2

t), t ≥ 0.

Compute s(t ), the arclength of the curve at time t.

 22 ✳ Find the arc length of the curve of m, a, and b.

⇀

m

r (t) = (t

m

, t

3m/2

, t

)

for 0 ≤ a ≤ t ≤ b, and where

m > 0.

Express your result in terms

 23 Let C be the part of the curve of intersection of the parabolic cylinder

x =y

2

and the hyperbolic paraboloid

3z = 2xy

with

y ≥ 0.

1. Write a vector parametric equation for C using x as the parameter. 2. Find the length of the part of C between the origin and the point (9, 3, 18). 3. A particle moves along C in the direction for which x is increasing. If the particle moves with constant speed 9, find its velocity vector when it is at the point (1, 1, ). 4. Find the acceleration vector of the particle of part (c) when it is at the point (1, 1, ). 2 3

2 3

 24 If a particle has constant mass m, position

⇀

r,

and is moving with velocity

For a particle with mass m = 1 and position function

⇀

r = (sin t, cos t, t),

⇀

v,

then its angular momentum is L = m( r × v ). ⇀

⇀

∣ dL ∣ ∣. ∣ dt ∣

find ∣

Stage 3  25 ✳ A particle moves along the curve C of intersection of the surfaces z particle is at (1, 3, 6) its velocity v and acceleration a are given by

2

= 12y

and 18x = yz in the upward direction. When the

⇀

^ v =6 ^ ı ı + 12 ^ ȷ ȷ + 12 k

⇀

1. Write a vector parametric equation for C using u = 2. Find the length of C from (0, 0, 0) to (1, 3, 6).

z 6

^ a = 27 ^ ı ı + 30 ^ ȷ ȷ +6 k

as a parameter.

3. If u = u(t) is the parameter value for the particle's position at time t, find

1.1.13

du dt

when the particle is at (1, 3, 6).

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4. Find

2

d u 2

dt

when the particle is at (1, 3, 6).

 26 ✳ A particle of mass m = 1 has position

⇀ r0

=

^ k

1 2

and velocity v

⇀ 0

π

=

2

^ ı ı

2

at time 0. It moves under a force

⇀

F(t) = −3t ^ ı ı + sin t ^ ȷ ȷ + 2e

2t

^ k.

1. Determine the position r (t) of the particle depending on t. 2. At what time after time t = 0 does the particle cross the plane x = 0 for the first time? 3. What is the velocity of the particle when it crosses the plane x = 0 in part (b)? ⇀

 27 ✳ Let C be the curve of intersection of the surfaces that

dx > 0. dt

2

y =x

and z =

2 3

3

x .

A particle moves along

The particle is at (0, 0, 0) at time t = 0 and is at (3, 9, 18) at time t =

7 2

C

with constant speed such

.

1. Find the length of the part of C between (0, 0, 0) and (3, 9, 18). 2. Find the constant speed of the particle. 3. Find the velocity of the particle when it is at (1, 1, ). 4. Find the acceleration of the particle when it is at (1, 1, ). 2 3

2 3

 28 A camera mounted to a pole can swivel around in a full circle. It is tracking an object whose position at time t seconds is x(t) metres east of the pole, and y(t) metres north of the pole. In order to always be pointing directly at the object, how fast should the camera be programmed to rotate at time t? (Give your answer in terms of x(t) and y(t) and their derivatives, in the units rad/sec.)

 29 A pipe of radius 3 follows the path of the curve

⇀

r (t) = (

2 √2 3

3/2

t

,

1 2

2

t , t + 2),

for 0 ≤ t ≤ 10.

What is the volume inside the pipe? What is the surface area of the pipe?

 30 A wire of total length 1000cm is formed into a flexible coil that is a circular helix. If there are 10 turns to each centimetre of height and the radius of the helix is 3 cm, how tall is the coil?

 31 A projectile falling under the influence of gravity and slowed by air resistance proportional to its speed has position satisfying 2

d r 2

⇀

^ = −gk − α

dr dt

dt ⇀

where α is a positive constant. If

⇀

⇀

r = r0

and

dr dt

⇀

= v0

at time t = 0, find

⇀

r (t).

1. You might guess that Θ is a capital Greek theta. You'd be right. 2. The earliest known written approximations of π, in Egypt and Babylon, date from 1900--1600BC. The first recorded algorithm for rigorously evaluating π was developed by Archimedes around 250 BC. The first use of the symbol π, for the ratio between the circumference of a circle and its diameter, in print was in 1706 by William Jones.

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3. It is Proposition 18 in Book 3 of Euclid's Elements. It was published around 300BC. 4. Astroid should not be confused with asteroid, though both words derive from the Greek word for star. 5. Like a cross-walk sign. 6. For your velocity to jump discontinuously, your acceleration has to be infinite, which requires an infinite force. You might not look so healthy afterwards 7. The particle traces out a cycloid — see Question 1.1.1.4 8. The indefinite integral involved is one of a class of integrals called elliptic integrals because of their connections to arc lengths of ellipses. In general, elliptic integrals cannot be expressed in terms of elementary functions. You can easily find discussions of elliptic integrals using your favourite search engine. This page titled 1.1: Derivatives, Velocity, Etc. is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.1.15

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1.2: Reparametrization There are invariably many ways to parametrize a given curve. Kind of trivially, one can always replace t by, for example, 3u. But there are also more substantial ways to reparametrize curves. It often pays to tailor the parametrization used to the application of interest. For example, we shall see in the next couple of sections that many curve formulae simplify a lot when arc length is used as the parameter.

 Example 1.2.1 Here are three different parametrizations of the semi-circle x

2

+y

2

2

=r ,

y ≥ 0.

The first uses the polar angle θ as the parameter. We have already seen, in Example 1.0.1, the parametrization

⇀ r 1 (θ)

= (r cos θ , r sin θ)

0 ≤θ ≤π

The second uses x as the parameter. Just solving x so gives the parametrization

2

+y

2

2

=r ,

y ≥0

− − − − − − 2 2 −x

for y as a function of x, gives y(x) = √r

and

− − − − − − 2 2 r 2 (x) = (x , √ r − x )

⇀

−r ≤ x ≤ r

The third uses arc length from (r, 0) as the parameter. We have seen, in Example 1.1.6, that the arc length from (r, 0) to r (θ) is just s = rθ. So the point on the semicircle that is arc length s away from (r, 0) is ⇀ 1

⇀

s

⇀

r 3 (s) = r 1 (

) r

s

s

= (r cos

, r sin r

) r

0 ≤ s ≤ πr

We shall see that, for some purposes, it is convenient to use parametrization by arc length. Here is a messier example in which we reparametrize a curve so as to use the arc length as the parameter.

 Example 1.2.2 We saw in Example 1.1.9, that, as t runs from 0 to , r (t) = a cos t ^ ı ı + a sin t ^ ȷ ȷ runs from (a, 0) to (0, a) along the astroid x + y = a . Suppose that we want a new parametrization R(s) chosen so that, as s runs from 0 to some appropriate value, R(s) runs from (a, 0) to (0, a) along x + y = a , with s being the arc length from (a, 0) to R(s) along x + y = a . π

⇀

3

3

2

2/3

2/3

2/3

2/3

2/3

2/3

2/3

2/3

2/3

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We saw, in Example 1.1.9, that, for 0 ≤ t ≤

π 2

ds ,

3a

=

2

dt t

sin(2t)

3a

s(t) = ∫

′

so that the arclength from (a, 0) =

π

2

2

to

⇀

r (t)

is

[1 − cos(2t)] 4

which runs from 0, at t = 0, to , at t = . This is relatively clean and we can invert value, T (s), of t that corresponds to any given 0 ≤ s ≤ is determined by 3a

r (0)

3a

′

sin(2 t ) dt = 2

0

⇀

s(t)

to find

t

as a function of s. The

3a 2

3a s =

1 [1 − cos (2T (s))]

⟺

T (s) =

4s arccos (1 −

4

2

) 3a

and ⇀

3

R(s) = r (T (s)) = a cos

We can simplify s =

3a 4

3

cos

[1 − cos (2T (s))]

(T (s))^ ȷ ȷ

by just using trig identities to convert the into cos (T (s))'s and sin (T (s))'s.

(T (s))

and

3

(T (s)) ^ ı ı + a sin

3

sin

(T (s))

3a s =

3a [1 − cos (2T (s))]

=

2

[1 − (2 cos

4

cos (2T (s))

in

(T (s) − 1)]

4 2

⟺

cos

2s (T (s)) = 1 − 3a

3a s =

3a [1 − cos (2T (s))]

=

2

[1 − (1 − 2 sin

4

(T (s))]

4 2

⟺

sin

2s (T (s)) = 3a

Consequently the desired parametrization is 3/2

2s R(s) = a [1 −

]

^ ı ı + a[

3a

3/2

2s ]

3a

^ ȷ ȷ

0 ≤s ≤

3a

2

which is remarkably simple.

Exercises Stage 1  1 t

A curve

⇀

r (s)

is parametrized in terms of arclength. What is ∫

⇀′

| r (s)| ds

when t ≥ 1?

1

 2 The function ⇀

s+1

r (s) = sin(

)^ ı ı + cos(

2

s+1 2

1.2.2

)^ ȷ ȷ +

– √3

^ (s + 1)k

2

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is parametrized in terms of arclength, starting from the point P . What is P ?

 3 A curve R = a(t) is reparametrized in terms of arclength as R = b(s) = a(t(s)). Of the following options, which best describes the relationship between the vectors a (t ) and b (s ), where t(s ) = t ? ′

′

0

0

0

0

You may assume a (t) and b (s) exist and are nonzero for all t, s ≥ 0. ′

′

1. they are parallel and point in the same direction 2. they are parallel and point in opposite directions 3. they are perpendicular 4. they have the same magnitude 5. they are equal

Stage 2  4✳ 1. Let ⇀

3

r (t) = (2 sin

3

t, 2 cos

t, 3 sin t cos t)

Find the unit tangent vector to this parametrized curve at t = π/3, pointing in the direction of increasing t. 2. Reparametrize the vector function r (t) from (a) with respect to arc length measured from the point t = 0 in the direction of increasing t. ⇀

 5✳ This problem is about the logarithmic spiral in the plane ⇀

t

r (t) = e (cos t, sin t),

t ∈ R

1. Find the arc length of the piece of this spiral which is contained in the unit circle. 2. Reparametrize the logarithmic spiral with respect to arc length, measured from t = −∞.

Stage 3  6 Define 1

⇀

r (t) = (

for 0 ≤ t. Reparametrize the function using interpretation of the new parameter z?

arctan t

− − − − − √ 1 + t2

,

z = arctan t,

− − − − − − √ 1 + t−2

, arctan t)

and describe the curve it defines. What is the geometric

 7 Reparametrize the function

⇀

r (t) = (

1 2

2

t ,

1 3

3

t )

in terms of arclength from t = −1.

This page titled 1.2: Reparametrization is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.2.3

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1.3: Curvature So far, when we have wanted to approximate a complicated curve by a simple curve near some point, we drew the tangent line to the curve at the point. That's pretty crude. In particular tangent lines are straight — they don't curve. We will get a much better idea of what the complicated curve looks like if we approximate it, locally, by a very simple “curvy curve” rather than by a straight line. Probably the simplest “curvy curve” is a circle 1 and that's what we'll use.

 Definition 1.3.1 1. The circle which best approximates a given curve near a given point is called the circle of curvature or the osculating circle 2 at the point. 2. The radius of the circle of curvature is called the radius of curvature at the point and is normally denoted ρ. 3. The curvature at the point is κ = . 4. The centre of the circle of curvature is called centre of curvature at the point. 1

ρ

These definitions are illustrated in the figure below. It shows (part of) the osculating circle at the point P . The point C is the centre of curvature.

Note that when the curvature κ is large, the radius of curvature ρ is small and we have a very curvy curve. On the other hand when the curvature κ is small, the radius of curvature ρ is large and our curve is almost straight. In particular, straight lines have curvature exactly zero. We are now going to determine how to find the circle of curvature, starting by figuring out what its radius should be. We'll first look at curves 3 that lie in the xy-plane and then move on to curves in 3d. Consider the black curve in the figure below.

That figure also contains a (portion of a) red circle that fits the curve really well between the two radial lines that are (a very small) angle dθ apart. So the arclength ds of the part of the black curve between the two radial lines, should be (essentially) the same as the arc length of the circle between the two radial lines, which is ρ |dθ|, where ρ is the radius of the circle. (We put in absolute values to take into account the possibility that dθ could be negative.) Thus ds = ρ |dθ|. When dθ is a macroscopic angle, this is of course an approximation. But in the limit as dθ → 0, we should end up with ∣ ds ∣ ρ =∣ ∣ ∣ dθ ∣

We now have a formula for the radius of curvature, but not in a very convenient form, because to evaluate it we would need to know the arc length along the curve as a function of the angle θ in the rightmost figure below. We'll now spend some time developing more convenient formulae for ρ. First consider the three figures below. They all show the same curve as in the last figure. The leftmost figure just shows the curve of interest, which is the black curve, and the (blue) point of interest on the black curve. We want to find the curvature at that point. The middle figure shows the same curve and point of interest and also shows the red circle of curvature (i.e. best fitting circle) for the black curve at the blue dot.

1.3.1

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The red dot is the centre of curvature. The rightmost figure shows the same black curve, blue point of interest and red circle of curvature (at least part of it) somewhat enlarged. The angle θ is the angle between ^ ı ı and the radius vector from the red dot (the centre of curvature) to the blue dot (the point of interest). ^ is the tangent vector to the black curve at the blue dot. T ^ ^ The angle ϕ is the angle between ^ ı ı and T. The vector T is also tangent to the red circle. As the tangent and radius vectors for circles are perpendicular to each other 4, we have that ϕ = θ +

and hence ρ = ∣∣

π 2

ds dϕ

We are now in a position to develop a bunch of formulae for the radius of curvature convenient than κ = ∣∣

ds dϕ

−1

∣ ∣

.

∣ ∣

ρ

too.

and the curvature

κ =

1 ρ

,

that are more

These formulae will use the

 Definition 1.3.2 If

⇀

r (t)

is a parametrized curve, then ⇀

dr

⇀

v (t) =

(t)

is the velocity vector at

(t)

is the acceleration vector at

⇀

r (t)

dt

2⇀

a(t) =

d

r

2

dt

⇀

r (t)

is the unit tangent vector to the curve at ^ N(t) is the unit normal vector to the curve at κ(t) is the curvature at r (t) ρ(t) is the radius of curvature at r (t) ^ T(t)

that points in the direction of increasing t. r (t) that points toward the centre of curvature.

⇀

r (t)

⇀

⇀

⇀

 Theorem 1.3.3 ^ 1. Given 5 s(ϕ), i.e. if we know the arc length along the curve as a function of the angle 6 ϕ = ∡( ^ ı ı , T), then

∣ ds ∣ ρ =∣ ∣ ∣ dϕ ∣

2. Given

⇀

r (s),

∣ ds ∣ κ =∣ ∣ ∣ dϕ ∣

−1

∣ dϕ ∣ κ =∣ ∣ ∣ ds ∣

i.e. if we have a parametrization of the curve in terms of arc length, then ^ dT

^ (s) = κ(s) N(s)

ds ^ (s) is the unit normal vector to the curve at r (s) that points toward the centre of curvature. where N 3. Given r (t), i.e. if we have a general parametrized curve, then ⇀

⇀

^ dT

ds =κ

dt

4. Given (x(t) ,

y(t)),

^ N

⇀

ds

v (t) =

dt

2

d s

^ (t) T(t)

a(t) =

dt

2

^ T + κ(

ds

2

^ ) N

dt

dt

(for curves in the xy-plane) ∣ ∣ ⇀ v (t) × a(t) κ =∣ ds 3 ∣ ( ) ∣ dt

∣

∣ ∣

∣ ∣ = ∣ ∣

dx dt dx

[(

1.3.2

2

dt

− dt dy

2

) dt

dy

2

d y

+(

2

d x 2

dt

∣ ∣ 3/2

2

) ] dt

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5. Given y(x), (again for curves in the xy-plane) 2

d y

∣ ∣

dx2

κ =

∣ ∣ 3/2

dy [1 + (

2

) ] dx

Proof (a) Given s(ϕ), then ρ =

As we are assuming that 0 < ρ = ∣∣

ds dϕ

∣ < ∞, ∣

ds

∣ ∣

dϕ

∣ ∣

κ =

∣ ∣

ds dϕ

−1

∣ ∣

the inverse function theorem says that we can invert the function s(ϕ) (at least

locally) to get ϕ as a function of s, and that κ =

(b) Given

⇀

r (s),

^ then, by Lemma 1.1.4.c, T (s) =

⇀′

∣ ∣

dϕ ds

∣ ∣

is a unit tangent to the curve at

r (s)

^ dT

⇀

r (s)

and

^ dT dϕ =

ds

Now up to a sign

dϕ

dϕ

ds

^ ^ is κ, and just because ϕ = ∡( ^ ıı , T), with T a unit vector,

ds ^ T = cos ϕ ^ ıı + sin ϕ ^ ȷ ȷ ^ dT ⟹

= − sin ϕ ^ ıı + cos ϕ ^ ȷ ȷ

dϕ

So

^ dT

^ is a unit vector that is perpendicular 7 to T , and hence to the curve at

⇀

r (s),

and

dϕ ^ dT

^ (s) = κ(s) N(s)

ds ^ with N (s) a unit normal vector to the curve at toward the centre of curvature.

⇀

r (s).

In fact,

^ N(s)

is the unit normal vector to the curve at

⇀

r (s)

that points

To see that, look at the figures below 8, and note that substituting the sign information from each figure into (∗) gives († ). For example,

1.3.3

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consider the figure on the lower left. In that figure, ^ ^ the x component of T is negative (T is leftward pointing in the figure),

which makes cos ϕ negative (see (∗∗)), which makes the y component of ^ dT

so so

dϕ

^ dT

^ dT

negative (see (∗∗) again),

dϕ

is downward pointing,

^ = −N

dϕ

(the centre of curvature is the red dot above the curve) and

as s increases (i.e. as you move in the direction of the arrow on the curve), ϕ decreases (on the far right hand part of the curve ϕ ≈ So by (∗),

3π 2

,

^ dT

^ dT dϕ =

ds

dϕ

r (t),

and κ = ∣∣

dϕ ds

∣ ∣ =−

dϕ . ds

^ ^ = ( − N)(−κ) = κ N.

ds

Note that if κ(s) = 0, then fitting circle”. ⇀

0. This is, of course, the circle of radius a centred on the origin. As ⇀

⇀

dr

v (t) =

(t) = −a sin t ^ ı ı + a cos t ^ ȷ ȷ

dt

ds ⟹

⇀

(t) = | v (t)| = a dt

we have that the unit tangent vector ⇀

v (t)

⇀

T(t) =

⇀

= − sin t ^ ı ı + cos t ^ ȷ ȷ

| v (t)|

Note, as a check, that this is indeed a vector of length one and is perpendicular to the radius vector (as expected — the curve is a circle). As ^ dT (t) = − cos t ^ ı ı − sin t ^ ȷ ȷ dt

we have that

1.3.5

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^ dT (t) dt

^ N(t) = ∣ ∣

∣ ∣ κ(t) =

= − cos t ^ ı ı − sin t ^ ȷ ȷ

^ dT

(t)∣ ∣

dt ^ dT

(t)∣ ∣

dt

1 =

ds

a (t)

dt 1 ρ(t) =

=a κ(t)

Now look at the figure.

To get to the centre of curvature we should start from r (t) and walk a distance ρ(t), which after all is the radius of curvature, ^ in the direction N (T ), which is pointing towards the centre of curvature. So the centre of curvature is ⇀

⇀

^ r (t) + ρ(t)N(t) = [a cos t ^ ı ı + a sin t ^ ȷ ȷ ] + a[ − cos t ^ ı ı − sin t ^ ȷ ȷ] = 0

⇀

This makes perfectly good sense — the radius of curvature is the radius of the original circle and the centre of curvature is the centre of the original circle. One alternative calculation of the curvature, using x(t) = a cos t, ∣ ∣

dx dt

κ(t) =

dy

2

d y

(t)

2

dt

(t) − dy

2

(t))

+(

dt

is

2

(t) dt

dx [(

y(t) = a sin t,

d x 2

dt

(t)∣ ∣

3/2

2

(t)) ] dt

∣ ∣ − a sin t( − a sin t) − a cos t( − a cos t)∣ ∣ = 2

[( − a sin t)

2

3/2

+ (a cos t) ]

1 = a − − − − − − 2 2 −x

Another alternative calculation of the curvature, using y(x) = √a x

′

y (x) = −

− − − − − − =− y(x) √ a2 − x2 ′

y

′′

(for the part of the circle with y > 0 ),

x

2

y(x) − x y (x) (x) = −

y(x ) =−

2

2

a =−

3

y(x)

2

+x

y(x)

3

y(x)

is 2

a

2

∣ ∣ κ(x) =

d y 2

dx

dy [1 + (

(x)∣ ∣

2

3

a

y(x)

= 2

=

3/2

(x)) ]

[1 +

dx

3/2

x2 2

]

2

[y(x )

2

3/2

+x ]

y(x)

1 = a

1.3.6

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 Example 1.3.5 As a more computationally involved example, we'll analyze ⇀

r (t)

= ( cos t + t sin t) ^ ı ı + ( sin t − t cos t)^ ȷ ȷ

t >0

⇀

v (t) = t cos t ^ ı ı + t sin t ^ ȷ ȷ a(t)

We can read off from

⇀

v (t),

= ( cos t − t sin t) ^ ı ı + ( sin t + t cos t)^ ȷ ȷ

that ds

⇀

(t) = | v (t)| = t dt 2

d s 2

(t) = 1

dt

⇀

v (t)

⇀

T(t) =

= cos t ^ ı ı + sin t ^ ȷ ȷ

⇀

| v (t)|

Next, from a(t), we read off that a(t)

= ( cos t − t sin t) ^ ıı + ( sin t + t cos t)^ ȷ ȷ 2

d s a(t) =

^ (t) T(t) + κ(t)(

2

ds

and

2

^ (t)) N(t)

dt

dt

(by Theorem 1.3.3 .c) 2 ^ = cos t ^ ıı + sin t ^ ȷ ȷ + t κ(t)N(t)

⟹

so that t

2

κ(t)

2 ^ t κ(t)N(t) = −t sin t ^ ıı + t cos t ^ ȷ ȷ

is the length of −t sin t ^ ı ı + t cos t ^ ȷ ȷ , which is t. Thus 1 κ(t) =

and

^ N(t) =

−t sin t ^ ı ı + t cos t ^ ȷ ȷ 2

t

= − sin t ^ ı ı + cos t ^ ȷ ȷ

t κ(t)

As an alternative calculation of the curvature, we have ⇀

| v (t) × a(t)| κ(t) = ds (

(t))3

dt ∣ ı ı + t sin t ^ ȷ ȷ ] × [( cos t − t sin t) ^ ı ı + ( sin t + t cos t)^ ȷ ȷ ]∣ ∣[t cos t ^ ∣ = ds (

3

(t)) dt

^ ∣ ∣[t cos t( sin t + t cos t) − t sin t( cos t − t sin t)] k∣ ∣ = ds (

(t))3

dt 2^ | t k|

=

3

t

1 = t

It pays to think before you calculate!

Exercises Stage 1 There are a lot of constants in this chapter that might be new to you. They can take a little getting used to. Questions 1.3.1.1-1.3.1.5 provide practice working with and interpreting these constants and their relations to each other.

1.3.7

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 1 Sketch the curve

⇀

r (t) = (3 sin t, 3 cos t).

^ ^ and N At the point (0, 3), label T . Give the values of κ and ρ at this point as well.

 2 Consider the circle ^ and N (s).

⇀

r (t) = (3 sin t, 3 cos t).

^ ^ Find T (t) and T(s). Then, use parts (b) and (c) of Theorem 1.3.3 to find

^ N(t)

 3 The functon r (t) = (t cos t, t sin t), calculation), predict lim κ(t). ⇀

t ≥ 0,

defines a spiral centred at the origin. Using only geometric intuition (no

t→∞

 4 Let

⇀

t

r (t) = (e , 3t, sin t).

What is

ds ? dt

 5 In Question 1.2.1.5 of Section 1.2, we found that the spiral ⇀

t

r (t) = e (cos t, sin t)

parametrized in terms of arclength is s R(s) =

Find

^ dT ds

and

^ dT dt

– √2

s (cos ( ln (

– √2

s )) , sin ( ln (

– √2

))) .

for this curve.

 6 In this exercise, we make more precise the sense in which the osculating circle is the circle which best approximates a plane curve at a point. By translating and rotating our coordinate system, we can always arrange that the point is (0, 0) and that the curve is y = f (x) with f (0) = 0 and f (0) > 0. (We are assuming that the curvature at the point is nonzero.) Let y = g(x) be the bottom half of the circle of radius r which is centred at (0, r). ′

′′

Show that if f (x) and g(x) have the same second order Taylor approximation at y = f (x) at x = 0.

x = 0,

then

r

is the radius of curvature of

Stage 2  7 Given a curve

⇀

t

2

r (t) = (e , t

+ t),

compute the following quantities:

1. v (t) 2. a(t) ⇀

3.

ds dt

^ 4. T (t) 5. κ(t)

1.3.8

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 8 Find the curvature κ(t) of

⇀

r (t) = (cos t + sin t, sin t − cos t).

 9 Find the minimum and maximum values for the curvature of the ellipse x(t) = a cos t,

y(t) = b sin t.

Here a > b > 0.

 10 ✳ 1. Find the curvature of y = e at (0, 1). 2. Find the equation of the circle best fitting y = e at (0, 1). x

x

 11 ✳ Consider the motion of a thumbtack stuck in the tread of a tire which is on a bicycle moving at constant speed. This motion is given by the parametrized curve ⇀

r (t) = (t − sin t , 1 − cos t)

with t > 0. 1. Sketch the curve in the xy-plane for 0 < t < 4π. 2. Find and simplify the formula for the curvature κ(t). 3. Find the radius of curvature of the osculating circle to r (t) at t = π. 4. Find the equation of the osculating circle to r (t) at t = π. ⇀

⇀

Stage 3  12 Find the curvature κ as a function of arclength s (measured from (0, 0)) for the curve θ

x(θ) = ∫ 0

1 cos (

θ 2

π t )dt

y(θ) = ∫

2

0

1 sin (

2

π t )dt 2

 13 ✳ 3

Let C be the curve in R given by the graph of the function y = . Let κ(x) be the curvature of C at the point (x, x /3). Find all points where κ(x) attains its maximal values, or else explain why such points do not exist. What are the limits of κ(x) as x → ∞ and x → −∞? 2

x

3

3

1. Circles are good for studying “curvature”, because, unlike parabolas for example, the rate at which a circle curves is uniform over the entire circle. 2. “Osculare” is the Latin verb “to kiss”. The German mathematician Gottfried Wilhelm (von) Leibniz (1646--1716) named the circle the “circulus osculans”. 3. We'll also assume that the curves of interest are smooth, with no cusps for example, and not straight, so that the radius of curvature 0 < ρ < ∞. 4. We saw that in Example 1.1.6. 5. The equation s = s(ϕ) is called the “intrinsic equation of the curve”. ^ ^ 6. The notation ∡( ^ ı ı , T) means “the angle between ^ ı ı and T ”. ^ and ϕ and think about what the sketch says about 7. Think about why this should be the case. In particular, sketch T

^ dT . dϕ

8. In each of the four figures, the arrow on the curve specifies the direction of increasing arc length s and the red dot is the centre of curvature for the curve at the blue dot.

1.3.9

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This page titled 1.3: Curvature is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.3.10

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1.4: Curves in Three Dimensions So far, we have developed formulae for the curvature, unit tangent vector, etc., at a point r (t) on a curve that lies in the xy-plane. We now extend our discussion to curves in R . Fix any t. For t very close to t, r (t ), will, by the Taylor expansion to second order, be very close to r (t) + r (t) (t − t) + r (t) (t − t) , so that r (t ) almost lies in the plane through r (t) that is determined by the two vectors r (t) and r (t). Thus, if we restrict our attention to a very small part of the curve near the point of interest r (t), the curve will, to a very good approximation lie in some plane. So we can still define, for example, the osculating circle to the curve at r (t) to be the circle in that plane that fits the curve best near r (t). And we still have the formulae 1 ⇀

3

⇀′

⇀

′

1 ⇀′

′

⇀

′

⇀

2

′

⇀

′

2

⇀′

⇀′

⇀

⇀

⇀

⇀

dr

⇀

ds

v =

^ dT

^ T

= dt

dt

^ = κN

ds ^ dT

ds =κ

dt

^ N

dt 2⇀

d a =

2

r

d s =

2

dt v × a = κ(

2

ds

^ ) N

dt

dt

ds

⇀

^ T + κ(

2

3

^ ×N ^ ) T

dt

The only 2 difference is that

^ v , a, T

⇀

^ and N are now three component vectors rather than two component vectors.

^ (s) and N ^ (s) are mutually perpendicular If we are lucky and our curve happens to lie completely in a single plane, the vectors T ^ ^ ^ (s) × N(s) is a unit vector that is perpendicular to the unit vectors that lie in the same plane, so that their cross product B(s) = T ^ plane. By continuity, B(s) has to be a constant vector, i.e. be independent of s. ^ (s) is not constant, then our curve doesn't lie in a single plane, and we can use the derivative If, on the other hand, B ^ dB

d =

ds

^ ^ ×N (T ) =

ds

^ dT

^ ^ × ×N+T

ds

^ = T×

^ dN ds

^ dN

^ dT (since 

ds

^  is parallel to N)

ds

as a measure of how badly the curve fails to lie in a plane, i.e. how much the plane that fits the curve best near The cross product in because

^ dB

^ = T×

ds

^ dN ds

implies that

^ | B| = 1

⟹

twists as s increases,

⇀

r (s)

^ dB ds

is perpendicular to

^ ^ 1 = B⋅B

d ⟹

0 =

So

(s) ds

In addition,

^ ^ ^ [B ⋅ B] = 2 B ⋅

ds ^ dB

^ T.

^ dB ds

must be perpendicular to

^ B

^ dB ds

^ must be parallel to N (s).

 Definition 1.4.1 ^ ^ ^ ^ 1. The binormal vector at r (s) is B (s) = T(s) × N(s). The normal vector N(s) is sometimes called the unit principal normal vector to distinguish it from the binormal vector. 2. We define the torsion τ (s) by ⇀

^ dB

^ (s) = −τ (s)N(s)

ds

The negative sign is included so that τ (s) > 0 indicates “right handed twisting”. There will be an explanation of what this means in Example 1.4.4 below.

1.4.1

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3. The osculating plane at r (s) (the plane that fits the curve best at ^ B(s). The equation of the plane is ⇀

⇀

r (s)

) is the plane through

⇀

r (s)

with normal vector

⇀ ^ B(s) ⋅ {(x, y, z) − r (s)} = 0

^ ^ ^ For each s, T (s), N(s) and B(s) are mutually perpendicular unit vectors. They form an orthonormal basis for R , just as ^ ı ı, ^ ȷ ȷ 3 ^ ^ ^ ^ ^ and k form an orthonormal basis for R . Furthermore both (T(s) , N(s) , B(s)) and ( ^ ı ı , ^ ȷ ȷ , k) are “right handed triples” , ^ ^ ^ ^ meaning that B (s) = T(s) × N(s) and k = ^ ı ı ×^ ȷ ȷ. 3

3

We have already computed

^ dT ds

and

^ dB

It is now an easy matter to compute

. ds

^ dN

d =

ds

^ ^ (B(s) × T(s))

ds ^ ^ ^ ^ (s) + B = −τ (s)N(s) × T (s) × (κ(s)N(s)) ^ ^ = τ (s)B(s) − κ(s)T(s)

^ ^ ^ ^ ^ ^ To see that N (s) × T(s) = −B(s) and B(s) × N(s) = −T(s), just look at the right hand figure above.

Now suppose that we have a curve that is parametrized by t rather than s. How do we find the torsion τ ? The most obvious method is to recall that v × a = κ( ⇀

ds

ds

3

^ ^ ) T × N = κ(

dt

3

^ and that B (t) is a unit vector. So

^ ) B

dt

⇀

v (t) × a(t)

^ (t) = B

⇀

| v (t) × a(t)|

Having found B(t) we can differentiate it and use

^ dB

^ (s) (s) = −τ (s)N

and the chain rule to give

ds dB

dB ds =

dt ds

from which we can read off τ , provided we know

ds = −τ

ds

dt

^ B

dt

^ and N .

dt

There is another, often more efficient, method to find the torsion τ that uses da

2

d

d s

= dt

( dt

2

^ T + κ(

=

dt

3

dt

2

^ ) N)

dt

3

d s

ds

2

^ T+

d s 2

dt

ds

^ κN +

dt

d

ds (κ(

dt

2

^ ) )N + κ(

dt

ds

3

^ ^ ) (τ B − κ T)

dt

While this looks a little complicated, notice that, with just one exception, namely κ(

ds

3

^ ) τ (s)B(s),

dt

^ ^ ^ side is either in the direction T or in the direction N and so is perpendicular to B . So, dotting with

⇀

da

( v × a) ⋅

2

ds

=κ ( dt

6

)

⇀

2

τ = | v × a|

⇀

every term on the right hand ds

v × a = κ( dt

3

^ ) B

gives

τ

dt

and hence

1.4.2

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da

⇀

( v × a) ⋅ dt

τ =

⇀

2

| v × a|

If the curvature 4 κ(s) > 0 and the torsion τ (s) are known, then the system of equations 5

 Equation 1.4.2. Frenet–Serret Formulae ^ dT (s) =

^ κ(s) N(s)

(s) =

^ ^ τ (s) B(s) − κ(s) T(s)

ds ^ dN ds ^ dB

^ (s) = −τ (s) N(s)

ds

is a first order linear system of ordinary differential equations ^ ⎡ T(s) ⎤ d ds

⎡

0

κ(s)

⎢ ^ ⎥ = ⎢ −κ(s) ⎢ N(s) ⎥ ⎣

⎣ ^ ⎦ B(s) ^ for the 9 component vector valued function (T (s) ,

0

^ ⎥ τ (s) ⎥ ⎢ N (s) ⎥ ⎢

0

0

^ (s) T ⎤ ⎤⎡

−τ (s)

0

⎦

⎣ ^ ⎦ B(s)

^ ^ N(s) , B(s)).

Any first order linear initial value problem d x(s) = M (s)x(s)

x(0) = x0

ds

where x is an n -component vector and M (s) is an n × n matrix with continuous entries, has exactly one solution. If n = 1, so that x(s) and M (s) are just functions, this is easy to see. Just let M(s) be the antiderivative of M (s) that obeys M(0) = 0. Then d x(s) = M (s)x(s)

⟺

e

d

−M(s)

ds

x(s) − M (s)e

−M(s)

x(s) = 0

ds d ⟺

(e

−M(s)

x(s)) = 0

ds

by the product rule. So e x(s) is a constant independent of s. In particular e x(s) = e x(0) = x x(s) = x e . This argument can be generalized to any natural number n. But that is beyond the scope of this book. −M(s)

−M(s)

−M(0)

0

so that

M(s)

0

Since the Frenet-Serret formulae constitute a first order system of ordinary differential equations for the vector ^ ^ ^ (T(s) , N(s) , B(s)) and since any first order linear initial value problem has a exactly one solution, ^ ^ ^ the vector valued function (T (s) , N(s) , B(s)) is determined by the functions κ(s) and τ (s) (assuming that they are ^ ^ ^ continuous) together with the initial condition (T (0) , N(0) , B(0)). ⇀

^ Furthermore, once you know T (s), then

⇀

r (s)

is determined by

⇀

r (0)

and

dr

^ (s) = T(s).

ds

^ ^ ^ So any smooth curve r (s) is completely determined by r (0), (T (0) , N(0) , B(0)), κ(s) and τ (s). That is, up to translations (you can move between any two possible choices of r (0) by a translation) and rotations (you can ^ ^ ^ move between any two possible choices of (T (0) , N(0) , B(0)) by a rotation) a curve is completely determined by the curvature κ(s) > 0 and the torsion τ (s). This result is called “The fundamental theorem of space curves”. ⇀

⇀

⇀

 Theorem 1.4.3. The Fundamental Theorem of Space Curves Let κ(s) > 0 and τ (s) be continuous. Then up to translations and rotations, there is a unique curve with curvature torsion τ (s).

1.4.3

κ(s)

and

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 Example 1.4.4. Right circular helix The right circular helix is the curve ^ r (t) = a cos t ^ ı ı + a sin t ^ ȷ ȷ + bt k

⇀

with a, b > 0 as in the figure on the left below.

Here is why it is called a right helix rather than a left helix. If the helix is the thread of a bolt that you are screwing into a nut, and you turn the bolt in the direction of the (curled) fingers of your right hand (as in the figure 6 on the right above), then it moves in the direction of your thumb (as in the long straight arrow of the figure on the right above). To determine the curvature and torsion of this curve we compute ^ v (t) = −a sin t ^ ı ı + a cos t ^ ȷ ȷ +b k

⇀

a(t) = −a cos t ^ ı ı − a sin t ^ ȷ ȷ da

(t) = a sin t ^ ı ı − a cos t ^ ȷ ȷ

dt

From

we read off

⇀

v (t)

ds

− − − − − − 2 2 = √a + b

dt ^ (t) = − T

From a =

2

d s 2

dt

^ T + κ(

ds

a

a

− − − − − − √ a2 + b2

2

sin t ^ ı ı +

cos t ^ ȷ ȷ +

− − − − − − √ a2 + b2

^ k

we read off that

2 2 ^ ^ ) N = κ(a + b )N,

dt

a κ(t) =

b

− − − − − − √ a2 + b2

^ N(t) = − cos t ^ ı ı − sin t ^ ȷ ȷ

a2 + b2

From

v (t) × a(t)

| v (t) × a(t)|

2

= det ⎢ −a sin t ⎣

2

⇀

2

2

=a b

+a

ı ı − ab cos t ^ ȷ ȷ +a b ⎥ = ab sin t ^

a cos t

−a cos t 4

^ k⎤

^ ȷ ȷ

^ ı ı

⎡ ⇀

−a sin t

2

2

= a (a

0

^ k

⎦

2

+b )

we read off ⇀

^ (t) = B

v (t) × a(t)

b =

⇀

| v (t) × a(t)|

b

− − − − − − √ a2 + b2

sin t ^ ı ı −

a

− − − − − − √ a2 + b2

cos t ^ ȷ ȷ +

− − − − − − √ a2 + b2

^ k

and da

⇀

( v × a) ⋅

2

dt

τ (t) = ⇀

2

| v × a|

Note that, for the right handed helix, τ

> 0.

a b =

2

2

a (a

b 2

+b )

=

2

a

2

+b

Finally the centre of curvature is

1.4.4

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⇀

1

r (t) +

2

^ N(t)

a

2

+b

= (a −

2

) cos t ^ ı ı + (a −

a

κ(t) 2

b =−

a

2

+b

^ ) sin t ^ ı ı + bt k

a 2

cos t ^ ı ı −

a

b

^ sin t ^ ı ı + bt k

a

which is another helix. In the figure below, the red curve is the original helix and the blue curve is the helix traced by the centre of curvature.

Exercises Stage 1  1 ^ ^ ^ In the sketch below of a three-dimensional curve and its osculating circle at a point, label T and N . Will B be pointing out of the paper towards the reader, or into the paper away from the reader?

 2 In the formula ds

⇀

⇀′

(t) = | v (t)| = | r (t)| dt

does s stand for speed, or for arclength?

 3 Which curve (or curves) below have positive torsion, which have negative torsion, and which have zero torsion? The arrows indicate the direction of increasing t.

1.4.5

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 4 Consider a curve that is parametrized by arc length s. 1. Show that if the curve has curvature κ(s) = 0 for all s, then the curve is a straight line. 2. Show that if the curve has curvature κ(s) > 0 and torsion τ (s) = 0 for all s, then the curve lies in a plane. 3. Show that if the curve has curvature κ(s) = κ , a strictly positive constant, and torsion τ (s) = 0 for all s, then the curve is a circle. 0

 5✳ The surface z = x

2

+y

2

is sliced by the plane x = y. The resulting curve is oriented from (0, 0, 0) to (1, 1, 2).

1. Sketch the curve from (0, 0, 0) to (1, 1, 2). ^ ^ ^ 2. Sketch T , N and B at ( , , ). 3. Find the torsion at ( , , ). 1

1

1

2

2

2

1

1

1

2

2

2

Stage 2  6✳ Let C be the space curve ⇀

t

r (t) = (e − e

⇀′

−t

t

) ^ ı ı + (e + e

−t

^ )^ ȷ ȷ + 2t k

⇀′′

1. Find r , r and the curvature of C . 2. Find the length of the curve between

⇀

r (0)

and

⇀

r (1).

 7 Find the torsion of

⇀

2

3

r (t) = (t, t , t )

at the point (2, 4, 8).

 8 Find the unit tangent, unit normal and binormal vectors and the curvature and torsion of the curve 2

⇀

r (t) = t ^ ı ı +

t

2

3

^ ȷ ȷ +

t

^ k

3

 9 For some constant c, define r (t) = (t , t, e ). For which value(s) of c is equation for the plane containing the osculating circle to the curve at t = 5. ⇀

3

ct

1.4.6

τ (5) = 0?

For each of those values of

c,

find an

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 10 ✳ 1. Consider the parametrized space curve ⇀

2

3

r (t) = (t , t, t )

Find an equation for the plane passing through (1, 1, 1) with normal vector tangent to 2. Find the curvature of the curve from (a) as a function of the parameter t.

⇀

r

at that point.

 11 ✳ Let C be the osculating circle to the helix

⇀

r (t) = ( cos t , sin t , t)

at the point where t = π/6. Find:

1. the radius of curvature of C 2. the center of C 3. the unit normal to the plane of C

 12 ✳ 1. Consider the parametrized space curve ⇀

2

r (t) = (cos(t), sin(t), t )

Find a parametric form for the tangent line at the point corresponding to t = π. 2. Find the tangential component a (t) of acceleration, as a function of t, for the parametrized space curve T

⇀

r (t).

 13 ✳ Suppose,

in

terms

of

the

time

⇀

parameter 2

r (t) = (sin t − t cos t) ^ ı ı + (cos t + t sin t) ^ ȷ ȷ +t

^ k,

t

,

a

particle

moves

along

the

path

1 ≤ t < ∞.

1. Find the speed of the particle at time t. 2. Find the tangential component of acceleration at time t. 3. Find the normal component of acceleration at time t. 4. Find the curvature of the path at time t.

 14 ✳ Assume the paraboloid z = x + y viewed from the positive z -axis. 2

2

and the plane

2x + z = 8

intersect in a curve

C. C

is traversed counter-clockwise if

1. Parametrize the curve C . ^ ^ ^ 2. Find the unit tangent vector T , the principal normal vector N, the binormal vector B and the curvature κ all at the point (2, 0, 4).

 15 ✳ Consider the curve C given by ⇀

1

r (t) =

3

t 3

1 ^ ı ı +

– √2

^ 1. Find the unit tangent T (t) as a function of t. 2. Find the curvature κ(t) as a function of t. ^ 3. Determine the principal normal vector N at the point (

2

t

8 3

^ ^ ȷ ȷ + t k,

−∞ < t < ∞.

– , 2 √2, 2).

1.4.7

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 16 ✳ Suppose the curve C is the intersection of the cylinder x

2

+y

2

=1

with the plane x + y + z = 1.

1. Find a parameterization of C . 2. Determine the curvature of C . 3. Find the points at which the curvature is maximum and determine the value of the curvature at these points.

 17 ✳ Let ⇀

2

r (t) = t

^ ^ ı ı + 2t ^ ȷ ȷ + ln t k

^ ^ Compute the unit tangent and unit normal vectors T (t) and N. Compute the curvature κ(t). Simplify whenever possible!

 18 ✳ 1. Find the length of the curve

2

⇀

r (t) = (1,

t

2

3

,

t

)

3

for 0 ≤ t ≤ 1.

^ 2. Find the principal unit normal vector to r (t) = cos(t) ^ ı ı + sin(t) ^ ȷ ȷ + t k at t = π/4. ^ 3. Find the curvature of r (t) = cos(t) ^ ı ı + sin(t) ^ ȷ ȷ + t k at t = π/4. ^ N

⇀

⇀

 19 ✳ A particle moves along a curve with position vector given by ⇀

2

r (t) = (t + 2 , 1 − t , t /2)

for −∞ < t < ∞. 1. Find the velocity as a function of t. 2. Find the speed as a function of t. 3. Find the acceleration as a function of t. 4. Find the curvature as a function of t. 5. Recall that the decomposition of the acceleration into tangential and normal components is given by the formula 2

⇀′′

r

d s (t) =

2

^ T(t) + κ(t)(

ds

2

^ ) N(t)

dt

dt

^ Use this formula and your answers to the previous parts of this question to find N (t), the principal unit normal vector, as a function of t. 6. Find an equation for the osculating plane (the plane which best fits the curve) at the point corresponding to t = 0. 7. Find the centre of the osculating circle at the point corresponding to t = 0.

 20 ✳ Consider the curve C given by 3

⇀

2

t

r (t) =

t ^ ı ı +

3

– √2

^ ^ ȷ ȷ +t k

−∞ < t < ∞

^ 1. Find the unit tangent T (t) as a function of t. 2. Find the curvature κ(t) as a function of t. 3. Evaluate κ(t) at t = 0. ^ 4. Determine the principal normal vector N (t) at t = 0. ^ 5. Compute the binormal vector B(t) at t = 0.

1.4.8

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 21 ✳ A curve in R is given by 3

⇀

2

r (t) = (t

3

, t , t ).

1. Find the parametric equations of the tangent line to the curve at the point (1, −1, −1). 2. Find an equation for the osculating plane of the curve at the point (1, 1, 1).

 22 ✳ A curve in R is given by 3

⇀

2

r (t) = (sin t − t cos t) ^ ı ı + (cos t + t sin t) ^ ȷ ȷ +t

1. Find the length of the curve r (t) from r (0) = (0, 1, 0) to 2. Find the curvature of the curve at time t > 0. ⇀

⇀

^ k,

⇀

0 ≤t 0, the hill would have to pull on you to ^ < 0, this happens whenever keep you on hill. It can't, so you become airborne. Since ^ȷȷ ⋅ N

W >0

⟺

⇀ 2

mκ| v |

^ + mg^ ȷ ȷ ⋅N >0

⟺

⇀

− − − − − − − g ^ |^ ȷ ȷ ⋅ N| κ

| v| > √

^ If the hill is concave upward as in the figure on the right above, then N points upward and the hill is allowed to have W ^ ^ (which corresponds to the normal force W N pushing upward). Since ^ȷȷ ⋅ N > 0 we always have ^ W = mκ| v | + mg^ ȷ ȷ ⋅ N > 0. You never become airborne. On the other hand your knees may complain.

≥0

⇀ 2

The Skate Boarder So far, Equations 1.7.1 and 1.7.2 apply to any stiff frictionless “wire”. We now specialize to the special case of a skateboarder inside a circular culvert of radius a. Let's put the bottom of the circle at the origin (0, 0), so that the centre of the circle is at (0, a).

In this case the curvature is  κ =

1 a

 

^ and ^ȷȷ ⋅ N = cos ϕ =

a−y a

so 1.7.1 and 1.7.2 simplify to

1.7.2

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−−−−−−−−− − − −−−−−−−−− − 2 E (E − mgy) = √ 2g( − y) m mg

⇀

| v| = √

2 W

=

mg (E − mgy) +

a

3mg (a − y) =

2 (

a

a

a E+

3mg

− y) 3

Imagine now that you start at the bottom of the culvert, that is at y = 0, with energy E > 0. As time progresses, y increases and consequently | v | and W both decrease, as, of course, they should. This continues until one of the following three things happen. ⇀

1. | v | hits 0, in which case you stop rising and start descending. The speed | v | is zero when y = y ⇀

⇀

=

S

E mg

.

(The subscript “S ”

stands for “stop”.) Physicists say that when you reach y all of your kinetic energy ( m| v | ) has been converted into potential energy (mgy ). 2. W hits zero. When you get higher than this, W becomes negative and the culvert would have to pull on you to keep your feet on the culvert. As the culvert can only push on you, you become airborne. The normal force W is zero when y =y = + . (The subscript “A ” stands for “airborne”.) 1

S

A

2

E

a

3

mg

3

⇀ 2

2

3. y hits 2a. This is the summit of the culvert. You descend on the other side. Which case actually happens is determined by the relative sizes of y

S,

Comparing y

S

Comparing y

A

Comparing y

A

= = =

2

E

3

mg

2

E

3

mg

2

E

3

mg

+ + +

1

E

3

mg

and y

A

and a =

a 3

2 3

and 2a =

a 3

=

2

E

3

mg

a+ 5 3

a 3

a+

+

3

3

,

S

A

,

 yA

we see that y

we see that y

, a

a

we see that y

and 2a.

≤ yA

≤a ⟺

A

⟺ E mg

≤ 2a ⟺

E mg

≤ a.

≤ a. E mg

≤

5 2

a.

So the conclusions are: If  0 ≤

E mg

≤ a 

then  0 ≤ y

S

≤ yA ≤ a .

In this case you just oscillate between heights 0 and y

S

≤a

in the bottom half of

the culvert, as in the figure on the left below. If  a ≤ ≤ a  then  a ≤ y ≤ y , 2a . In this case you make it more than half way to the top. But you become airborne at y =y which is somewhere between the half way mark y = a and the top y = 2a. At this point our model breaks down because you are no longer in contact with the culvert. You just freely follow a parabolic arc until you crash back into the culvert, as in the figure in the centre below. If   a <   then  2a < y < y  . In this case you successfully go all the way around the culvert, looping the loop, as in the E

5

mg

2

A

S

A

5

E

2

mg

A

S

figure on the right below. Note that, as top, i.e. to reach height 2a.

E mg

>

5 2

a > 2a,

this requires significantly more energy than that required to reach the

Exercises Stage 1 You may assume the acceleration due to gravity is g = 9.8 m/s . You may also assume that the systems described function as they do in the book: so tracks are frictionless, etc., unless otherwise mentioned. 2

 1 The figure below represents a bead sliding down a wire. Sketch vectors representing the normal force the wire exerts on the bead, and the force of gravity.

1.7.3

https://math.libretexts.org/@go/page/92309

Assume the top of the page is “straight up.”

 2 In the definition E =

1 2

⇀ 2

m| v |

+ mgy,

⇀

v

is the derivative of position with respect to what quantity?

 3 A bead slides down a wire with the shape shown below, x < 0.

Let W \hN be the normal force exerted by the wire when the bead is at position x. Note W

> 0.

Is

dW dx

positive or negative?

 4 A skateboarder is rolling on a frictionless, very tall parabolic ramp with cross-section described by y = x . Given a boarder of mass m with system energy E, what is the highest elevation the skater reaches? How does this compare to a circular culvert? 2

Stage 2  5 A skateboarder of mass 100 kg is freely rolling in a frictionless circular culvert of radius 5 m. If the skateboarder oscillates between vertical heights of 0 and 3 m, what is the energy E of the system?

 6 A skateboarder is rolling on a frictionless circular culvert of radius 5 m. What should their speed be when they're at the bottom of the culvert (y = 0 ) for them to make it all the way around?

 7 A ball of mass 1 kg rolls down a track with the shape

⇀

r (θ) = (3 cos θ, 5 sin θ, 4 + 4 cos θ)

measured in metres, and the z axis is vertical (so the force due to gravity is When θ = π/4, the particle has instantaneous velocity time? Give your answer as a vector.

⇀

| v (t)| = 5

1.7.4

^ −mgk.

for

0 ≤θ ≤

π 2

.

Coordinates are

)

m/s. What is the normal force exerted by the track at that

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 8 A bead of mass kg slides down a wire in the shape of the curve r (θ) = (sin θ, sin θ − θ), θ ≥ 0, with coordinates measured in metres. The bead will break off the wire when the wire exerts a force of 100 N on the bead. 1

⇀

9.8

If the bead breaks off the wire at θ =

13π 3

,

how fast is the bead moving at that point?

 9 A skier is gliding down a hill. The hill can be described as r (t) = (ln t, 1 − t), kilometres. How fast would the skier have to be moving in order to catch air? ⇀

1/e ≤ t ≤ e,

with coordinates measured in

Stage 3  10 A wire follows the arclength-parametrized path r (s) = (x(s), y(s)). A bead, equipped with a jet pack, slides down the wire. ^ The jet pack can exert a variable force in a direction tangent to the wire, U T . Assuming the bead slides with constant speed ⇀

⇀

⇀

∣ dr ∣ ∣ dr ∣ ∣ ∣ =c∣ ∣ = c, ∣ dt ∣ ∣ ds ∣

find a simplified equation for U , the signed magnitude of the force exerted by the jet pack.

Let the acceleration due to gravity be g, and let the mass of the bead with its jet pack be m. Give U as a function of s. Remark: most beads this author has seen did not have jet packs. However, in modelling a frictionful 4 system, friction acts as a force that is directly opposing the direction of motion — much like our jet pack.

 11 ^ parallel to the direction of motion. A snowmachine is cautiously descending a hill in low gear. Its engine provides a force M T The engine provides whatever force is necessary to keep the snowmachine moving at a constant speed, | v |. Its treads do not slip. ⇀

1. Give a formula for M in terms of the mass m of the snowmachine, the acceleration due to gravity g, and the tangent vector ^ to the hill. T ^ 2. Let T point in the downhill direction. Do you expect M to be positive or negative as the snowmachine moves downhill? 3. Find M for the hill of shape y = 1 + cos x (measured in metres) at the point x = for a snowmachine of mass 200 kg. 3π 4

 12 A skateboarder rolls along a culvert with elliptical cross-section described by ⇀

r (θ) = (4 cos θ, 3(1 + sin θ)), 0 ≤ θ ≤ 2π,

with coordinates measured in metres. 1. Give the height y (in terms of m, g, and E ) where the skater's speed is zero. 2. Write an equation relating E, m, g, and y , where y is the y -value where the skater would become airborne, i.e. where W = 0. (You do not have to solve for y explicitly.) 3. Suppose the skater has speed 11 m/s at the bottom of the culvert. Which of the following describes their journey: they make it all the way around; they roll back and forth in the bottom half; or they make it onto the ceiling, then fall off? S

A

A

A

 13 A frictionless roller-coaster track has the form of one turn of the circular helix with parametrization  (a cos θ, a sin θ, bθ). A car leaves the point where  θ = 2π  with zero velocity and moves under gravity to the point where  θ = 0. By Newton's law of motion, the position r (t) of the car at time t obeys ⇀

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⇀′′

mr

^ ⇀ ^ (t) = N( r (t)) − mgk

^ ^ Here m is the mass of the car, g is a constant, −mgk is the force due to gravity and N ( r (t)) is the force that the roller-coaster ^ ( r (t)) is always perpendicular to track applies to the car to keep the car on the track. Since the track is frictionless, N ⇀

⇀

⇀

⇀

dr

v (t) =

(t). dt

1. Prove that E(t) =

1 2

⇀

2

m| v (t)|

⇀ ^ + mg r (t) ⋅ k

is a constant, independent of t. (This is called “conservation of energy”.) ⇀ 2

2. Prove that the speed | v | at the point θ obeys  | v | 3. Find the time it takes to reach θ = 0. ⇀

= 2gb(2π − θ).

1. We are mathematicians — we like idealized situations. 2. This force is required to keep the bead from either passing through the wire or flying off the wire. 3. We assume that you are going downhill and that the curvature κ > 0. 4. Frictionated? Frictiony? Befrictioned? This page titled 1.7: Sliding on a Curve is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.7.6

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1.8: Optional — Polar Coordinates So far we have always written vectors in two dimensions in terms of the basis vectors ^ ı ı and ^ ȷ ȷ . This is not always convenient. For ^ example, when working in polar coordinates it is often convenient to use basis vectors ^ r(θ), θ (θ) which depend on the value of the ^ current polar coordinate θ — though one usually just writes ^ r, θ , suppressing the dependence on θ from the notation. When one is at the point with polar coordinates (r, θ), these basis vectors are defined by

 Equation 1.8.1 ^ r(θ) = cos θ ^ ı ı + sin θ ^ ȷ ȷ ^ θ (θ) = − sin θ ^ ı ı + cos θ ^ ȷ ȷ

Note that this basis has two very nice properties. ^ 1. |^ r(θ)| = | θ (θ)| = 1,

2.

^ dr

^ (θ) = θ (θ),

dθ

That

(orthonormality)

(θ) = −^ r(θ)

dθ

^ dr (θ) dθ

^ dθ

^ ^ r(θ) ⊥ θ (θ)

is some scalar multiple of θ^(θ) follows just from the fact that |^ r(θ)| = 1. |^ r(θ)| = 1

⟹

^ r(θ) ⋅ ^ r(θ) = 1

⟹

^ r(θ) ⋅

dr ^

1

^ dr ⟹

d

(θ) = dθ

(^ r(θ) ⋅ ^ r(θ)) = 0 2 dt

(θ) ⊥ ^ r(θ) ⟹

dθ

Similarly, that

^ dθ (θ) dθ

^ dr

^ (θ) ∥ θ (θ)

dθ

^ is some scalar multiple of ^ r(θ) follows just from the fact that | θ (θ)| = 1.

 Lemma 1.8.2 If we parametrize a curve by giving its polar coordinates 1 (r(t) ,

θ(t)),

then

1. the position vector of the point at time t is ⇀

r (t) = r(t) ^ r(θ(t))

2. and the velocity vector of the point at time t is dr

⇀

dθ

v (t) =

(t) ^ r(θ(t)) + r(t)  dt

^ (t) θ (θ(t))

dt

3. and the acceleration vector of the point at time t is 2

d r a(t) = [

2

dt

dθ (t)−r(t)(

2

(t)) ] ^ rr(θ(t))

dt 2

d θ + [r(t) 

2

dr (t)+2

dt

dθ (t)

dt

^ (t)] θ (θ(t))

dt

It is standard to suppress the arguments t and θ(t) and write, for example,

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dr

⇀

 ^ r + r 

v = dt

dθ

^  θ

dt

But it is important to remember that the arguments really are there. Proof The vector from the origin to the point whose polar coordinates are (r, θ) is the polar coordinates at time t,

⇀

r =r^ r(θ).

So if we parametrize a curve by giving

⇀

r (t) = r(t) ^ r(θ(t)) dr

⇀

(t) ^ r(θ(t)) + r(t) 

v (t) = dt dr =

(t) ^ r(θ(t)) + r(t) 

2

2

 ^ r+

 

2

d r 2

^ dr

dr dt

dt =[

dθ

(t) dt

^ (t) θ (θ(t))

dt

d r =

dθ (θ(t)) 

dθ

dt

a(t)

^ dr

dθ

dr

 

+

dθ

dθ − r (

dt

dt

2

^  θ + r 

dt

) ]^ r + [r 

d θ 2

^  θ + r (

d θ

dr +2

2

dθ  

dt

dt

dθ

2

^ dθ

)   dt

dt

2

2

dt

dt

dθ  

dθ

^ ]θ

dt

 Example 1.8.3 As an example, consider a bead that is sliding on a frictionless rod that has one end fixed at the origin and that is rotating about the origin at a constant Ω rad/sec.

Because the rod is frictionless, it is incapable of applying to the bead any force parallel to the rod. So under Newton's law, the radial 2 component of the acceleration of the particle is exactly zero. So, if the polar coordinates of the bead at time t are (r(t), θ(t)), then, by Lemma 1.8.2.c, ⇀

ma = F,

2

d r dt

As the rod is rotating at Ω rad/sec,

dθ =Ω

2

dθ − r (

2

)

=0

dt

and

dt 2

d r 2

2

− Ω  r = 0

dt

The general solution to this constant coefficient second order ordinary differential equation is 3 r(t) = Ae

Ω t

+ Be

−Ω t

where A and B are arbitrary constants that are determined by initial conditions. Just as an example, if r(0) = 1 and r (0) = 0, then A + B = 1 and AΩ − BΩ = 0, so that A = B = and ′

1 2

1 r(t) =

(e

Ω t

+e

−Ω t

)

2

If, again for example, θ(0) = 0, then θ(t) = Ωt and the bead follows the polar coordinate curve 1 r(θ) =

(e

θ

+e

−θ

)

2

Observe that r(θ) is 1 when θ = 0, increases as θ increases, and tends to ∞ as θ → +∞. The curve is a spiral.

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 Example 1.8.4. Conic sections in polar coordinates In this example, we derive the equation of a general conic section in polar coordinates. A conic section is the intersection of a plane with a cone. This is illustrated in the figures below.

For our current purposes, it is convenient to use the equivalent 4 (and often used) definition that a conic section is the set of points P in the xy-plane whose distance from a fixed point F (called the focus of the conic) is a constant multiple ε ≥ 0 (called the eccentricity of the conic) of the distance from P to a fixed line L (called the directrix of the conic). Choose a coordinate system with the focus

F

of the conic being the origin and with the directrix

L

being

x =p

for some

p > 0.

If P has polar coordinates (r, θ), then P has x-coordinate r cos θ. The point Q on the line L in the figure above has xcoordinate p. So the distance from P to L, which is also the distance from P to Q, is p − r cos θ. The distance from P to F is r. We require that the distance from P to F is ε times the distance from P to L. So εp r = ε(p − r cos θ) ⟺

r = 1 + ε cos θ

The numerator εp is usually renamed to ℓ giving the equation ℓ r = 1 + ε cos θ

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 Example 1.8.5. Conic sections in polar coordinates, again We'll now take the equation r = for a conic section in polar coordinates, from the last example, and convert it to the more familiar Cartesian coordinates. Just by the definition of polar coordinates ℓ

1+ε cos θ

r(1 + ε cos θ) = ℓ

⟺

r = ℓ − εx

⟺

x

⟺

(1 − ε )x

2

+y

2

2

=ℓ

2

2

2

2

− 2εℓx + ε x

+ 2εℓx + y

2

2

=ℓ

(C)

Now consider separately four different cases, depending on the value of ε ≥ 0. If ε = 0, (C) reduces to 2

x

+y

2

2

=ℓ

which is of course a circle of radius ℓ. If 0 < ε < 1, completing the square in (C) gives 2

2

εℓ

2

(1 − ε )(x +

)

1 − ε2

+y

2

2

=ℓ

2

2

ε ℓ +

1 − ε2

ℓ =

1 − ε2

which is equivalent to (x +

2

εℓ 1−ε

2

)

y +

2

ℓ

2

2

M

=

=1

ℓ

1−ε

(1−ε )

and is of course an ellipse with semi-major axis r

2

2

ℓ 1−ε

2

2

and semi-minor axis r

m

=

ℓ √1−ε2

.

If ε = 1, (C) reduces to y

2

2

=ℓ

− 2ℓx

which is of course a parabola. If ε > 1, the same computation as in the 0 < ε < 1 case gives (x −

2

εℓ 2

ε −1

)

2

ℓ

y −

2

2

=1

ℓ 2

( ε2 −1 )

1.8.4

2

ε −1

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and is of course a hyperbola.

Exercises Stage 1  1 Consider the points (x1 , y1 ) = (3, 0)

(x2 , y2 ) = (1, 1)

(x4 , y4 ) = (−1, 1)

(x5 , y5 ) = (−2, 0)

(x3 , y3 ) = (0, 1)

For each 1 ≤ i ≤ 5, sketch, in the xy-plane, the point (x , y ) and find the polar coordinates r and θ , with 0 ≤ θ i

i

i

i

i

< 2π,

for the point (x

i,

yi ).

 2 1. Find all pairs (r, θ) such that (−2, 0) = (r cos θ , r sin θ)

2. Find all pairs (r, θ) such that (1, 1) = (r cos θ , r sin θ)

3. Find all pairs (r, θ) such that (−1, −1) = (r cos θ , r sin θ)

 3 Consider the points (x1 , y1 ) = (3, 0)

(x2 , y2 ) = (1, 1)

(x4 , y4 ) = (−1, 1)

(x5 , y5 ) = (−2, 0)

(x3 , y3 ) = (0, 1)

Also define, for each angle θ, the vectors ^r (θ) = cos θ  ^ e ı ı + sin θ ^ ȷ ȷ

^θ (θ) = − sin θ  ^ e ı ı + cos θ ^ ȷ ȷ

^ (θ) and e ^ (θ) and the angle between the vectors e ^ (θ) and e ^ 1. Determine, for each angle θ, the lengths of the vectors e ^ ^ (θ) × e ^ (θ) (viewing e ^ (θ) and e ^ (θ) as vectors in three dimensions with zero k Compute e components). ^ (θ ) and e ^ (θ ). In your sketch of the 2. For each 1 ≤ i ≤ 5, sketch, in the xy-plane, the point (x , y ) and the vectors e ^ (θ ) and e ^ (θ ) at (x , y ). vectors, place the tails of the vectors e r

r

θ

r

θ

r

θ (θ).

θ

i

r

i

θ

i

i

i

r

i

θ

i

i

 4✳ Match the following equations with the corresponding pictures. Cartesian coordinates are

(x, y)

and polar coordinates are

(r, θ).

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(a)

r = 2 + sin(4θ)

(b)

r = 1 + 2 sin(4θ)

(c)

r =1

(d)

r = 2 cos(θ),   −

(f)

r =θ

(e)

r =e

θ/10

+e

−θ/10

(A)

(B)

(C)

(D)

(E)

(F)

π 2

≤θ ≤

π 2

Stage 2  5 Recall that a point with polar coordinates r and θ has x = r cos θ and y = r sin θ. Let curve in polar coordinates. Find the curvature of this curve at a general point θ.

r = f (θ)

be the equation of a plane

 6 Find the curvature of the cardioid r = a(1 − cos θ). 1. As usual r is the distance from the origin to the point and θ is angle between the x-axis and the vector from the origin to the point. The symbols r, θ are the standard mathematics symbols for the polar coordinates. Appendix A.7 gives another set of symbols that is commonly used in the physical sciences and engineering. 2. The θ^ component of the acceleration just tells us how much normal force the rod is applying to the bead to keep it on the rod. 3. A review of the technique used to find this solution is given in Appendix A.9. In any event, it is easy to check that r(t) = Ae + Be really does obey − Ω  r = 0. Ω t

−Ω t

2

d r

2

2

dt

4. It is outside our scope to prove this equivalence.

This page titled 1.8: Optional — Polar Coordinates is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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1.9: Optional — Central Forces One of the great triumphs of Newtonian mechanics was the explanation of Kepler's laws 1, which said 1. The planets trace out ellipses about the sun as focus. 2. The radius vector r sweeps out equal areas in equal times. 3. The square of the period of each planet is proportional to the cube of the major axis of the planet's orbit. ⇀

Newton showed that all of these behaviours follow from the assumption that the acceleration a(t) of each planet obeys the law of ⇀

motion ma = F where m is the mass of the planet and GM m

⇀

F =−

⇀

r

r3

is the “gravitational force” applied on the planet by the sun. Here G is a constant 2, called the “gravitational constant” or the “universal gravitational constant”, M is the mass of the sun, r is the vector from the sun to the planet and r = | r |. ⇀

⇀

In this section, we'll show that some of these properties follow from the weaker assumption that the acceleration a(t) of each planet obeys the law of motion ma = F with F being a central force. That is, the assumption that F is parallel to r . The verification that the other properties follow from the specific form of the gravitational force, proportional to r , will be delayed until the optional §1.10. ⇀

⇀

⇀

⇀

−2

So, in this section, we assume that we have a parametrized curve

⇀

r (t)

2⇀

d m

r

2

⇀

and that this curve obeys ⇀

(t) = F( r (t))

dt

where, for all

⇀

⇀

3

⇀

r ∈ R , F( r )

is parallel to

We shall show that

⇀

r.

1. The path r (t) lies in a plane through the origin and that 2. the radius vector r sweeps out equal areas in equal times. ⇀

⇀

We'll start by trying to guess what the plane is. Pretend that we know that

⇀

r (t)

lies in a fixed plane through the origin. Then

⇀

dr

⇀

v (t) =

(t) dt

⇀

⇀

r (t) × v (t)

lies in the same plane and

⇀

is perpendicular to the plane. If our path really does lie in a fixed plane,

⇀

r (t) × v (t)

cannot change direction — it must always be parallel to the normal vector to the plane. So let's define ⇀

⇀

Ω(t) = r (t) × v (t)

and check how it depends on time. By the product rule, dΩ

d  (t)

=

dt

⇀

⇀

⇀

⇀

⇀

( r (t) × v (t)) = v (t) × v (t) + r (t) × a(t) dt 1

=

⇀

⇀

⇀

r (t) × F( r (t))

m ⇀

= 0

because As

⇀

⇀

r (t)

(A)

and F( r (t)) are parallel. So Ω(t) is 3 in fact independent of t. It is a constant vector that we'll just denote Ω. ⇀

⇀

⇀

r (t) × v (t) = Ω,

we have that

⇀

r (t)

is always perpendicular to Ω and ⇀

r (t) ⋅ Ω = 0

⇀

If Ω ≠ 0 , this is exactly the statement that r (t) always lies in the plane through the origin with normal vector Ω. If Ω = 0 , then r (t) is always parallel to v (t) and there is some function α(t) such that ⇀

⇀

⇀

⇀

⇀

dr

⇀

⇀

(t) = v (t) = α(t) r (t) dt

This is a first order, linear, ordinary differential equation that we can solve by using an integrating factor. Set t

β(t) = ∫

α(t) dt

0

1.9.1

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Then ⇀

⇀

dr

⇀

(t) = α(t) r (t)

⟺

e

−β(t)

dt

dr

(t) − α(t)e

−β(t) ⇀

r (t) = 0

dt d  ⟺

[e

−β(t) ⇀

r (t)] = 0

dt ⟺ ⟺

so that

⇀

r (t)

e

−β(t) ⇀

⇀

r (t) = r (0)

⇀

r (t) = e

β(t) ⇀

r (0)

lies on a line through the origin. This makes sense — the particle is always moving parallel to its radius vector.

This completes the verification that

⇀

r (t)

lies in a plane through the origin.

Now we show that the radius vector r (t) sweeps out equal areas in equal times. In other words, we now verify that the rate at which r (t) sweeps out area is independent of time. To do so we rewrite the statement that | r (t) × v (t)∣∣ is constant in polar coordinates. Writing r (t) = r(t)^ r(θ(t)) and then applying Lemma 1.8.2.b gives that ⇀

⇀

⇀

⇀

⇀

⇀

⇀

constant = ∣ ∣ r × v∣ ∣

∣ = r^ r ×( ∣

2

dθ (t) dt

 ^ r + r 

dt

dθ dt

2 ^ ∣  θ ) = r ∣

dθ dt

^ |^ r ×^ r| = 0,  | ^ r × θ| = 1

since

is constant. It now suffices to observe that r(t)

dr

is exactly twice the rate at which

⇀

r (t)

sweeps out area. To see this, just look

at the figure below. The shaded area is essentially a wedge of a circular disk of radius r. (If r(t) were independent of t, it would be exactly a wedge of a circular disk.) Its area is the fraction of the area of the full disk, which is dθ

2π

dθ

2

 π r 2π

1 =

2

r

dθ

2

Exercises Stage 3  1✳ Let

^ r (t) = x(t) ^ ı ı + y(t) ^ ȷ ȷ + z(t) k

⇀

differential equation

2⇀

d

r

2

dt

⇀

= f (r) r

be the position of a particle at time t . Suppose the motion of the particle satisfies the

where r = | r | . ⇀

1. Suppose f (r) is an arbitrary function of r. Prove or disprove each of the following statements. 1. The motion of the particle is planar. 2. The path of the particle sweeps out equal areas in equal times. 2. Find all forms of f (r) for which the motion of the particle always lies on a straight line. 3. Give a specific form of f (r) for which the motion of the particle could lie on an ellipse.

 2✳ An object moves along a curve in the xy-plane having polar equation r = central force so that the object has no transverse acceleration.

1 θ+α

(where α is a constant) under the influence of a

1. Verify that r θ˙ = h remains constant as the object moves. 2. Express the magnitude of the acceleration of the object as a function of r and h. 2

1.9.2

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1. The German astronomer Johannes Kepler (1571–1630) developed these laws during the course of an attempt to relate the five extraterrestrial planets then known to the five Platonic solids. He based the laws on a great number of careful measurements made by the Danish Astronomer Tycho Brahe (1546–1601). Then Isaac Newton (English, 1642–1727) provided the explanation in 1687. Kepler also wrote a paper entitled “On the Six-Cornered Snowflake”. Tycho Brahe lost his nose in a sword duel and wore a prosthetic nose from then on. The story is that Brahe died from a burst bladder that resulted from his refusing to leave the dinner table before his host. 2. Its value is about 6.67408 × 10 m kg sec . 3. Physicists call m Ω(t) the angular momentum at time t and refer to (A) as (an example of) conservation of angular momentum. Conservation of angular momentum is exploited in gyro-compasses and by ice skaters (to spin faster/slower). −11

3

−1

−2

This page titled 1.9: Optional — Central Forces is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.9.3

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1.10: Optional — Planetary Motion We now return to the claim, made in §1.9 on central forces, that if 2⇀

d

r

GM =−

2

⇀

r (t)

GM

⇀

r =−

3

dt

obeys Newton's inverse square law

2

r

^ r

r

then the curve obeys Kepler's laws 1. r (t) runs over an ellipse having one focus at the origin and 2. r (t) sweeps out equal areas in equal times and 3. the square of the period is proportional to the cube of the major axis of the ellipse. ⇀ ⇀

We just showed, in §1.9, that the fact that − r is parallel to r implies that out equal area in equal times. We now verify the remaining Kepler laws. GM ⇀

⇀

⇀

r (t)

3

r

lies in a plane through the origin and sweeps

We start by just rewriting Newton's laws above in polar coordinates. We saw in Lemma 1.8.2.c, that if we write then 2⇀

d

2

r

d r =(

2

dt

dθ − r (

2

2

dt

dt

GM =−

2

) )^ r + (r 

r =−

dθ  

dt

r (t) = r(t) ^ r(t),

^ )θ

dt

^ r

2

r

dr +2

2

dt

GM

⇀

3

d θ

⇀

r

^ The ^ r and θ components of this equation are 2

d r

dθ − r (

2

2

GM

)

=−

2

dt

dt

2

d θ r 

dr

dθ

+2

2

r

 

dt

dt

=0 dt

The second of these two equations only tells us that d 

2

dt

2

dθ

{r

d θ } = r {r 

dt

which we already knew. Substituting

dr +2

2

dθ

h

=

2

r

dt

dθ  

dt

dt

} =0

2

⟹

dθ

r

dt

= h,

a constant

dt

into the first equation gives

 Equation 1.10.1 2

2

d r 2

dt

h −

GM =−

3

r

2

r

This equations contains a lot of 's. So let's set u = . Furthermore, for the first of Kepler's laws, we really want r as a function of θ rather than t. So let's make u a function of θ and write 1

1

r

r

1 r(t) = u(θ(t))

Then dr

1 (t) = −

dt 2

2

dθ (θ(t))

dθ

u

2

d r dt

du

2

d u (t) = −h

2

dθ

du (t) = −h

dt dθ

(θ(t))

dθ (θ(t))

since 

dθ

h =

dt

2

2

= hu

r

2

2

2

(t) = −h u(θ(t)) dt

d u 2

(θ(t))

dθ

and our equation becomes

1.10.1

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 Equation 1.10.2 2

2

2

−h u

2

d u dθ2

2

3

d u

2

−h u

= −GM u

or

dθ2

GM +u =

h2

This is a second order, linear, ordinary differential equation with constant coefficients. Recall 1 that the general solution of such an equation is the sum of a “particular solution” (i.e. any one solution, which in this case we can take to be the constant function ) GM 2

plus the general solution of the homogeneous equation u

′

h

which one often writes as

+ u = 0,

A cos θ + B sin θ

with A and B arbitrary constants. In this particular application it is more convenient to write the solution in a different, standard but less commonly used, form. Namely, we can use the triangle

to write A = C cos α and B = C sin α so that the general solution of the homogeneous equation u

′

+u = 0

becomes

C cos α cos θ + C sin α sin θ = C cos(θ − α)

with C and α being arbitrary constants. So the general solution to 1.10.2 is GM u(θ) =

+ C cos(θ − α)

2

h

and the general solution to 1.10.1 is 1 r(t) =

GM 2

+ C cos(θ(t) − α)

h

The angle α just shifts the zero point of our coordinate θ. By rotating our coordinate system by α, we can arrange that α = 0 and then 1 r(t) =

GM 2

2

ℓ with

ℓ =

1 + ε cos θ

+ C cos(θ(t))

2

h

=

Ch ,  ε =

GM

GM

h

As we saw in Example 1.8.4, this is exactly the equation of a conic section with eccentricity ε. That leaves only the last of Kepler's laws, relating the period to the semi-major axis. As we are talking about planets, whose orbits remain bounded, our conic section must be a circle or ellipse, rather than a parabola or hyperbola. Looking back at Example 1.8.5, we see that the semi-major and semi-minor axes of our ellipse are ℓ a = 1 −ε

ℓ b =

2

− − − − − √ 1 − ε2

The period T of our orbit is just the length of time it takes the radius vector πab.

As the rate at which the radius vector is sweeping out area is T

2

πab =( h/2

2

2

)

2

2

2

4π a b =

2

h

1 2

2

dθ

r

2

=

r (t)

h

dt 2

4π a b =

⇀

4π

2

we have

2

= GM ℓ

,

to sweep out the area of the ellipse 2 , which is

2

3

 a GM

b since ℓ =

a

1. See Appendix A.9. 2. You probably computed the area of an ellipse in first year calculus. If not, you should be able to do it now in a few lines. This page titled 1.10: Optional — Planetary Motion is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a

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detailed edit history is available upon request.

1.10.3

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1.11: Optional — The Astroid Imagine a ball of radius a/4 rolling around the inside of a circle of radius a. The curve traced by a point circle (that's the blue curve in the figures below) is called an astroid 1. We shall find its equation.

P

painted on the inner

Define the angles θ and ϕ as in the figure in the left below.

That is the vector from the centre, O, of the circle of radius a to the centre, Q, of the ball of radius a/4 is the vector from the centre, Q, of the ball of radius a/4 to the point P is a( cos ϕ, − sin ϕ)

3 4

a( cos θ, sin θ)

and

1 4

As θ runs from 0 to , the point of contact between the two circles travels through one quarter of the circumference of the circle of radius a, which is a distance (2πa), which, in turn, is exactly the circumference of the inner circle. Hence if ϕ = 0 for θ = 0 (i.e. if P starts on the x-axis), then for θ = , P is back in contact with the big circle at the north pole of both the inner and outer circles. That is, ϕ = when θ = . (See the figure on the right above.) So ϕ = 3θ and P has coordinates π 2

1 4

π 2

3π

π

2

2

3

1 a( cos θ, sin θ) +

4

a a( cos ϕ, − sin ϕ) =

(3 cos θ + cos 3θ, 3 sin θ − sin 3θ)

4

4

As, recalling your double angle, or even better your triple angle, trig identities, cos 3θ = cos θ cos 2θ − sin θ sin 2θ 2

2

= cos θ[ cos

2

θ − sin

2

θ] − 2 sin 2

= cos θ[ cos

θ − 3 sin

θ cos θ

θ]

sin 3θ = sin θ cos 2θ + cos θ sin 2θ 2

= sin θ[ cos

2

2

θ − sin

θ] + 2 sin θ cos

2

2

= sin θ[3 cos

θ − sin

θ

θ]

we have

1.11.1

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3 cos θ + cos 3θ

2

3

= 4 cos 3 sin θ − sin 3θ

2

= cos θ[3 + cos

θ − 3 sin

2

= 4 sin

= cos θ[3 + cos

2

θ − 3(1 − cos

θ)]

θ

= sin θ[3 − 3 cos 3

2

θ]

2

θ + sin

2

θ]

= sin θ[3 − 3(1 − sin

2

θ) + sin

θ]

θ

and the coordinates of P simplify to 3

x(θ) = a cos

Oof! As  x

2/3

+y

2/3

2/3

=a

2

cos

2/3

θ+a

2

sin

θ ,

3

θ

y(θ) = a sin

θ

the path traced by P obeys the equation 2/3

x

+y

2/3

2/3

=a

which is surprisingly simple, considering what we went through to get here. There remains the danger that there could exist points (x, y) obeying the equation x + y = a that are not of the form x = a cos θ,  y = a sin θ for any θ. That is, there is a danger that the parametrized curve x = a cos θ,  y = a sin θ covers only a portion of x + y = a . We now show that the parametrized curve x = a cos θ,  y = a sin θ in fact covers all of x +y =a as θ runs from 0 to 2π. 2/3

3

2/3

2/3

2/3

2/3

2/3 3

2/3

3

3

3

2/3

First, observe that x 0 ≤x ≤a and so

2/3

2/3

2/3

3

2/3

2 3 − = (√x ) ≥ 0

−a ≤ x ≤ a.

2

and y = (√y) ≥ 0. Hence, if (x, y) obeys x + y = a As θ runs from 0 to 2π, a cos θ takes all values between −a and 2/3

2/3

3

2/3

2/3

3

possible values of x. For each x ∈ [−a, a], y takes two values, namely ±[a the two corresponding values of y are precisely a sin θ and −a sin θ = a sin

2/3

3

3

0

3

0

2/3

−x

3/2

]

.

If

3

x = a cos

a

then necessarily and hence takes all

,

3

θ0 = a cos (2π − θ0 ),

(2π − θ0 ).

1. The name “astroid” comes from the Greek word “aster”, meaning star, with the suffix “oid” meaning “having the shape of”. The curve was first discussed by Johann Bernoulli in 1691–92. This page titled 1.11: Optional — The Astroid is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.11.2

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1.12: Optional — Parametrizing Circles We now discuss a simple strategy for parametrizing circles in three dimensions, starting with the circle in the radius ρ and is centred on the origin. This is easy to parametrize:

-plane that has

xy

⇀

r (t) = ρ cos t ^ ı ı + ρ sin t ^ ȷ ȷ

0 ≤ t < 2π

Now let's move the circle so that its centre is at some general point c. To parametrize this new circle, which still has radius ρ and which is still parallel to the xy-plane, we just translate by c:

⇀

r (t) = c + ρ cos t ^ ı ı + ρ sin t ^ ȷ ȷ

0 ≤ t < 2π

Finally, let's consider a circle in general position. The secret to parametrizing a general circle is to replace vectors ^ ı ı and ^ ȷ ȷ which ′

^ ı ı

and

^ ȷ ȷ

by two new

′

1. are unit vectors, 2. are parallel to the plane of the desired circle and 3. are mutually perpendicular.

′

⇀

′

r (t) = c + ρ cos t ^ ı ı + ρ sin t ^ ȷ ȷ

0 ≤ t < 2π

To check that this is correct, observe that ⇀

r (t) − c

⇀

is parallel to the plane of the desired circle because both ^ ı ı and ^ ȷ ȷ are parallel to the plane of the desired circle and ′

′

′

′

r (t) − c = ρ cos t ^ ı ı + ρ sin t ^ ȷ ȷ

⇀

r (t) − c

is of length ρ for all t because

1.12.1

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2

⇀

⇀

| r (t) − c |

⇀

= ( r (t) − c ) ⋅ ( r (t) − c ) ′

′

′

′

= (ρ cos t ^ ı ı + ρ sin t ^ ȷ ȷ ) ⋅ (ρ cos t ^ ı ı + ρ sin t ^ ȷ ȷ ) 2

=ρ

2

cos

2

2

= ρ (cos

since ^ ı ı

′

′

′

′

′

2

2

t + sin

′

sin

′

′

′

t ^ ȷ ȷ ⋅^ ȷ ȷ + 2ρ cos t sin t  ^ ı ı ⋅^ ȷ ȷ

2

t) = ρ

(^ ı ı and ^ ȷ ȷ are both unit vectors) and ^ ı ı

′

⋅ ^ ı ı =^ ȷ ȷ ⋅^ ȷ ȷ =1

2

t  ^ ı ı ⋅ ^ ı ı +ρ

′

′

′

′

⋅^ ȷ ȷ =0

(^ ı ı and ^ ȷ ȷ are perpendicular). ′

′

To find such a parametrization in practice, we need to find the centre c of the circle, the radius ρ of the circle and two mutually perpendicular unit vectors, ^ ı ı and ^ ȷ ȷ , in the plane of the circle. It is often easy to find at least one point p on the circle. Then we ′

can take

′

^ ı ı =

′

′ ′ ^ ^ ȷ ȷ =k ×^ ı ı .

p−c |p−c|

.

′

It is also often easy to find a unit vector,

′

^ k ,

that is normal to the plane of the circle. Then we can choose

We'll illustrate this now.

 Example 1.12.1 Let C be the intersection of the sphere x

2

+y

2

+z

2

and the plane z = y.

=4

The intersection of any plane with any sphere is a circle. The plane in question passes through the centre of the sphere, so C has the same centre and same radius as the sphere. So C has radius 2 and centre (0, 0, 0). Notice that the point (2, 0, 0) satisfies both x + y + z = 4 and z = y and so is on C . We may choose ^ ı ı to be the unit vector in the direction from the centre (0, 0, 0) of the circle towards (2, 0, 0). Namely ^ ı ı = (1, 0, 0). 2

2

′

2

′

⇀

Since the plane of the circle is z − y = 0, the vector ∇(z − y) = (0, −1, 1) is perpendicular to the plane of C . So we may ′

^ take k

=

1

(0, −1, 1).

√2

Then ^ȷȷ

′

′

′ ^ =k ×^ ı ı =

1 √2

(0, −1, 1) × (1, 0, 0) =

Substituting in c = (0, 0, 0), ρ = 2,

′

^ ı ı = (1, 0, 0)

1 √2

(0, 1, 1).

and ^ȷȷ

′

=

1 √2

(0, 1, 1)

⇀

gives

r (t) = 2 cos t (1, 0, 0) + 2 sin t

sin t = 2( cos t,

– , √2

1 – (0, 1, 1) √2

sin t – ) √2

0 ≤ t < 2π

To check this, note that

– – x = 2 cos t, y = √2 sin t, z = √2 sin t

satisfies both x

2

+y

2

+z

2

=4

and z = y.

 Example 1.12.2 Let C be the circle that passes through the three points (3, 0, 0), (0, 3, 0) and (0, 0, 3). All three points obey x + y + z = 3. So the circle lies in the plane x + y + z = 3. We guess, by symmetry, or by looking at the figure below, that the centre of the circle is at the centre of mass of the three points, which is [(3, 0, 0) + (0, 3, 0) + (0, 0, 3)] = (1, 1, 1). We must check this and can do so by checking that (1, 1, 1) is equidistant from the three points: 1 3

– ∣ ∣(3, 0, 0) − (1, 1, 1)∣ ∣ =∣ ∣(2, −1, −1)∣ ∣ = √6 – ∣ ∣ =∣ ∣(0, 3, 0) − (1, 1, 1)∣ ∣(−1, 2, −1)∣ ∣ = √6 – ∣ ∣ =∣ ∣(0, 0, 3) − (1, 1, 1)∣ ∣(−1, −1, 2)∣ ∣ = √6

1.12.2

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This tells us both that (1, 1, 1) is indeed the centre (as only the centre is equidistant from any three distinct points on a – circle) and that the radius of C is √6. We may choose ^ ı ı to be the unit vector in the direction from the centre (1, 1, 1) of the circle towards (3, 0, 0). Namely ′

′

^ ı ı =

1 √6

(2, −1, −1). ⇀

Since the plane of the circle is x + y + z = 3, the vector ∇(x + y + z) = (1, 1, 1) is perpendicular to the plane of C . So ′

^ we may take k

=

1 √3

(1, 1, 1).

Then ′

^ ȷ ȷ

1

′

′ ^ =k ×^ ı ı =

1

− − √18

(1, 1, 1) × (2, −1, −1) =

− − √18

(0, 3, −3)

1 =

– √2 –

Substituting in \vc = (1, 1, 1), ρ = √6,

(0, 1, −1)

′

^ ı ı =

1 √6

(2, −1, −1)

– r (t) = (1, 1, 1) + √6 cos t

⇀

1 – √6

′

and ^ȷȷ

=

1 √2

(0, 1, −1)

– (2, −1, −1) + √6 sin t

gives

1 – √2

(0, 1, −1)

– – = (1 + 2 cos t, 1 − cos t + √3 sin t, 1 − cos t − √3 sin t)

To check this, note that sin

2π 3

=

√3 2

and sin

4π 3

⇀

r (0) = (3, 0, 0),

=−

√3 2

⇀

r(

2π 3

) = (0, 3, 0)

and

⇀

r(

4π 3

) = (0, 0, 3)

since

cos

2π 3

= cos

4π 3

=−

1 2

,

.

This page titled 1.12: Optional — Parametrizing Circles is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.12.3

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CHAPTER OVERVIEW 2: Vector Fields 2.1: Definitions and First Examples 2.2: Optional — Field Lines 2.3: Conservative Vector Fields 2.4: Line Integrals 2.5: Optional — The Pendulum Thumbnail: A unit sphere with surface vectors ( CC BY-SA 3.0 Unported; Cronholm144 via Wikipedia) This page titled 2: Vector Fields is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1

2.1: Definitions and First Examples In the last chapter, we studied vector valued functions of a single variable, like, for example, the velocity v (t) of a particle at time t. Suppose however that we are interested in a fluid. There is a, possibly different, velocity at each point in the fluid. So the velocity of a fluid is really a vector valued function of several variables. Such a function is called a vector field. ⇀

 Definition 2.1.1 1. A vector field in the plane is a rule which assigns to each point (x, y) in a subset, D, of the xy-plane, a two component vector v (x, y). 2. A vector field in space is a rule which assigns to each point (x, y, z) in a subset of R , a three component vector v (x, y, z). ⇀

3

⇀

Here are two typical applications that naturally involve vector fields. If v (x, y, z) is the velocity of a moving fluid at position (x, y, z), then v is called a velocity field. If F(x, y, z) is the force at position (x, y, z), then F is called 1 a force field. ⇀

⇀

⇀

⇀

 Example 2.1.2. The Point Source Imagine The whole world is filled with an incompressible fluid. Call it water. Somehow you find a way to produce still more water at the origin. Say you create 4πm litres per second. This forces the water to flow outward. Let's suppose that it flows symmetrically outward from the origin. Let's find the resulting vector field

⇀

v (x, y, z).

As the flow is to be symmetric, the velocity of the water at the point (x, y, z)

has to be pointing radially outward from the origin. That is, the direction of the velocity vector radial vector ^ r(x, y, z) =

⇀

v (x, y, z)

has to be the unit

^ x^ ı ı + y^ ȷ ȷ + zk − −−−−−−−− − √ x2 + y 2 + z 2

The magnitude of the velocity, i.e. the speed | v (x, y, z)| of the water, has to depend only on the distance from the origin. That is, the speed can only be some function of ⇀

− −−−−−−−− − 2

r(x, y, z) = √ x

+y

2

+z

2

Thus the velocity field is of the form ⇀

v (x, y, z) = v(r(x, y, z)) ^ r(x, y, z)

We just have to determine the function v(r). Fix any r > 0 and concentrate on the sphere x red in the figure below.

2

+y

2

+z

2

2

=r .

It is sketched in

During a very short time interval dt seconds, 4πm dt litres of water is created at the origin (which is the red dot). As the water is incompressible, 4πm dt litres of water must exit through the sphere during the same time interval to make room for the newly created water. But, at the surface of the sphere the water is flowing radially outward with speed v(r). So during the time interval in question the water near the surface of the sphere moves outward a distance v(r) dt, and in particular the water that was in the thin − −−−−−−−− − spherical shell   r − v(r) dt ≤ √x + y + z ≤ r   at the beginning of the time interval exits through the sphere 2

2

2

2.1.1

https://math.libretexts.org/@go/page/91898

− −−−−−−−− − √x2 + y 2 + z 2 = r

during the time interval. The shell is sketched in gray in the figure above. The volume of water in the gray shell is essentially the surface area of the shell, which is 4π r , times the thickness of the shell, which is v(r) dt. So, equating the volume of water created inside the sphere with the volume of water that exited the sphere, 2

4πm

2

4πm dt = (4π r )(v(r) dt) ⟹

v(r) =

2

m =

4πr

2

r

Thus our vector field is m

⇀

v (x, y, z) =

r(x, y, z)2

^ r(x, y, z)

If the world were two, rather than three dimensional 2, and the source created 2πm litres per second, the same argument leads to 2πm 2πm dt = (2πr)(v(r) dt) ⟹

v(r) =

m =

2πr

r

and to the vector field − −− −− −

m

⇀

v (x, y) =

2

^ r(x, y)

r(x, y) = √ x

+y

2

x^ ı ı + y^ ȷ ȷ ^ r(x, y) =

r(x, y)

To get a mental image of what this field looks like, imagine sketching, for each point tail at (x, y). Note that the vector

m r(x,y)

− −− −− − 2 2 √x + y

(x, y),

the vector

m r(x,y)

^ r(x, y)

with its

^ r(x, y)

points radially outward and has length which m

r(x,y)

depends only on r = |(x, y)| and is very long when (x, y) is near the origin and decreases in length like as r = |(x, y)| increases. 1 r

Here is a sketch of a bunch of such vectors.

Figure 2.1.3.

Note that as |(x, y)| → 0, the magnitude of the velocity that we are producing water at a single point (the origin).

⇀

| v (x, y)| → ∞.

This is a consequence of our idealized assumption

 Example 2.1.4. The Vortex In this example, we sketch the vector field ⇀

v (x, y) = Ω( − y ^ ı ı + x^ ȷ ȷ)

2.1.2

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where Ω is just a strictly positive constant. We give an efficient procedure for getting a rough sketch, which still provides a pretty realistic picture of the vector field, and which also generalises to other vector fields. First concentrate on the horizontal component ^ ı ı ⋅ v (x, y) of the vector field and determine in which part of the xy-plane it is zero, in which part it is positive and in which part it is negative. ⇀

⎧=0

if y = 0

^ ı ı ⋅ v (x, y) = −Ωy   ⎨ < 0 ⎩ >0

if y > 0

⇀

if y < 0

Next repeat with the vertical component. ⎧=0

if x = 0

^ ȷ ȷ ⋅ v (x, y) = Ωx   ⎨ < 0 ⎩ >0

if x < 0

⇀

if x > 0

This naturally divides the xy-plane into nine parts according to whether each of the components is positive, 0 or negative — ⇀

^ ı ı ⋅ v >0 ⇀

^ ı ı ⋅ v >0 ⇀

^ ı ı ⋅ v >0 ⇀

^ ı ı ⋅ v =0

and ^ȷȷ ⋅ v > 0 and ^ȷȷ ⋅ v = 0 and ^ȷȷ ⋅ v < 0 and ^ȷȷ ⋅ v > 0 ⇀

⇀ ⇀

⇀

in { (x, y) ∈ R in { (x, y) ∈ R in { (x, y) ∈ R in { (x, y) ∈ R

2

 ∣ ∣ y < 0,  x > 0 }

2

 ∣ ∣ y < 0,  x = 0 }

2

2

 ∣ ∣ y < 0,  x < 0 }  ∣ ∣ y = 0,  x > 0 }

and so on Now think of

⇀

v (x, y)

as being the velocity at (x, y) of a flowing fluid.

Look at the first bullet point above. It says that in the first of the nine parts, namely { (x, y) ∈ R  ∣∣ y < 0,  x > 0 }, which is the fourth quadrant, the horizontal component ^ ı ı ⋅ v > 0 signifying that the fluid is flowing rightwards. Indicate this in the sketch by drawing a rightward pointing horizontal arrow at some generic point in the middle of the fourth quadrant. (It's the blue arrow in the figure below.) The vertical component ^ȷȷ ⋅ v > 0 signifying that the fluid is also moving upwards. Indicate this in the sketch by drawing an upward pointing vertical arrow at the same generic point in the fourth quadrant. (It's the red arrow in the figure below.) 2

⇀

⇀

Next, look at the second bullet point above. It says that on the second of the nine parts, namely { (x, y) ∈ R  ∣ ı ı ⋅ v > 0, signifying ∣ y < 0,  x = 0 }, which is the bottom half of the y -axis, the horizontal component ^ that the fluid is moving rightwards. Indicate this in the sketch by drawing a rightward pointing horizontal arrow at some generic point in the middle of the bottom half of the y -axis. (It's the second blue arrow in the figure below.) The vertical component ^ȷȷ ⋅ v = 0 signifying that the fluid has no vertical motion at all. Indicate this in the sketch by not drawing any vertical arrow on the bottom half of the y -axis. 2

⇀

⇀

and so on

2.1.3

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By the time we have looked at all nine regions we will have built up the following sketch.

Figure 2.1.5.

From this sketch we see that, for example, in the first quadrant, the fluid is moving upwards and to the left and the fluid crosses the x-axis vertically (so that close to the x-axis, the arrows will be almost vertical) and the fluid crosses the y -axis horizontally (so that close to the y -axis, the arrows will be almost horizontal) and there is one point, namely (0, 0), where the vector field is exactly zero. It's the black dot in the centre of the figure above. Furthermore v (x, y) = Ω(−y ^ ı ı + x^ ȷ ȷ ) is smaller when (x, y) is closer to (0, 0) and v (x, y) is larger when (x, y) is farther from (0, 0), ⇀

⇀

Putting all of this accumulated wisdom together, we come up with this better sketch of the vector field.

Figure 2.1.6.

This shows the field swirling around the origin in a counterclockwise direction. Hence the name “vortex”.

 Example 2.1.7. The Undamped Nonlinear Pendulum In this example, we illustrate another way in which vector fields arise. Model a pendulum by a mass m that is connected to a hinge by an idealized rod that is massless 3 and of fixed length ℓ. Denote by θ the angle

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between the rod and vertical. The forces acting on the mass are gravity and the tension in the rod, whose magnitude, τ , automatically adjusts itself so that the distance between the mass and the hinge is fixed at ℓ. In the optional 4 Section 2.5, we show that the angle θ(t) obeys the second order nonlinear 5 differential equation 2

d θ

g +

2

sin θ = 0 ℓ

dt

It is often much more convenient to deal with first order, rather than second order, differential equations. The second order pendulum equation above may be reformulated 6 as a system of first order ordinary differential equations, by the simple expedient of defining ′

x(t) = θ(t)

y(t) = θ (t)

So x(t) is the angle at time t and y(t) is the angular velocity at time t. Then, ′

′

′

′′

x (t) = θ (t) = y(t) g

y (t) =  θ (t) = −

sin x(t) ℓ

Usually, one does not write in the (t) dependence explicitly. ′

x y

′

=y g =−

sin x ℓ

The right hand sides form the vector field g

⇀

v ((x, y)) = (y , −

sin x) ℓ

We can sketch this vector field, just as we sketched the vector field of Example 2.1.4. Noting that the horizontal component ⎧=0

if y = 0

^ ı ı ⋅ v (x, y) = y   ⎨ > 0 ⎩ 0

⇀

if y < 0

and the vertical component. ⇀

^ ȷ ȷ ⋅ v (x, y) = −

g

⎧=0

sin x   ⎨ > 0 ⎩ ℓ 0 leftward motion when y < 0 downward motion when 0 < x < π, 2π < x < 3π, ⋯ and upward motion when −π < x < 0, π < x < 2π, ⋯ . This gives us the collection of arrows in the figure

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Our full sketch will be less cluttered if we make all arrows the same length. This gives

which is a sketch of what is called the direction field of our vector field (see below). In the next section, we'll learn how to use vector field sketches to sketch solution trajectories.

 Definition 2.1.8 The direction field of a vector field

⇀

v (x, y, z)

is the vector field ⇀ v (x,y,z)

⎧ V(x, y, z) = ⎨ ⎩

⇀ | v (x,y,z)| ⇀

⇀

⇀

0

⇀

if  v (x, y, z) ≠ 0 ⇀

if  v (x, y, z) = 0

Exercises Stage 1  1 Below is a sketch of the vector field

⇀

v (x, y).

Find the regions where the x-coordinates and y -coordinates are positive, negative, and zero: ⎧>0 ⎪

 when 

v (x, y) ⋅ ^ ı ı ⎨=0

 when 

⇀

⎩ ⎪

0

 when 

v (x, y) ⋅ ^ ȷ ȷ ⎨=0

 when 

⇀

⎩ ⎪

0 for all u, or a(u) < 0 for all u. Let T (u) be an antiderivative of a(u). Then T (u) is strictly monotone (and continuous) and hence is invertible. That is, there is a continuous function U (t) that obeys U (T (u)) = u for all a < u < b and T (U (t)) = t for all t in the range of U . Differentiating T (U (t)) = t gives T (U (t)) U (t) = 1 and hence U (t) = . Set r(t) = R(U (t)). Then ′

′

1

′

T

′

(U(t))

′

′

1

′

r (t) = R (U (t))U (t) = a(U (t)) v(R(U (t))) ′

T (U (t)) 1 = a(U (t)) v(r(t)) a(U (t)) = v(r(t))

So r(t) is a field line and R(u) = r(T (u)) is a reparametrization of r(t). Here are a couple of examples that show that bad things can happen if we drop the requirement that v(R(u)) is nonzero.

 Example 2.2.9 Let the vector field v(x, y) be identically zero. Then any field line (x(t) , ′

y(t))

must obey

′

x (t) = 0

y (t) = 0

which forces both x(t) and y(t) to be constants. So each field line is just a single point. On the other hand every nonconstant R(u) obeys R (u) × v(R(u)) = 0 but is not contained in a field line. (As R(u) is not constant, it covers more than one point, while each field line is just a single point.) ⇀

′

Now here is a more interesting example.

 Example 2.2.10 ⇀

Consider the vector field v(x, y) = x ^ ı ı . This vector field takes the value 0 at each point on the y -axis, is a positive multiple of ^ ı ı at every point of the right half-plane and is a negative multiple of ^ ı ı at every point of the left half-plane. Let's find the field lines. Any field line must obey dx

dy (t) = x(t)

(t) = 0

dt

dt

So y(t) must be a constant. We can solve the linear ordinary differential equation hand side, and multiplying by the (integrating factor) e e

−t

−t

.

This gives

dx (t) − e

−t

dx (t) = x(t) dt

by moving the x(t) to the left

x(t) = 0

dt

By the product rule, this is the same as d (e

−t

x(t)) = 0

dt

which forces e

−t

x(t)

to be a constant. So our field lines are (C e

t

, D),

with C and D being arbitrary constants. Note that

if C = 0, the field line is just the single point (0, D) on the y -axis. It is illustrated by the black dot in the figure below. If C > 0, then as t runs from −∞ to +∞, the field line covers the horizontal half-line

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{(x, D)|x > 0}

in the right half-plane. It is illustrated by the red line in the figure below. If C < 0, then as t runs from −∞ to +∞, the field line covers the horizontal half-line {(x, D)|x < 0}

in the left half-plane. It is illustrated by the blue line in the figure below (with a different value of D than for the red line).

On the other hand, fix any constant D and set R(u) = u ^ ı ı + D^ ȷ ȷ . Then ⇀

′

R (u) × v(R(u)) = ^ ı ı × (u ^ ı ı) = 0

But as u runs from −∞ to +∞, R(u) runs over the full line {(x, D)| − ∞ < x < ∞} . It is not contained in any single field line and, in fact, completely covers three different field lines.

Exercises Stage 1  1 Suppose that the vector field quadrant of the xy-plane.

v(x, y)

sketched below represents the velocity of moving water at the point

(x, y)

in the first

Sketch the path followed by a rubber ducky dropped in at the point 1. (0, 2) 2. (1, 0) 3. (1, 2)

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 2 Find a vector field v(x, y) for which x(t) = e y(t) = e

−t

−t

cos t sin t

is a field line.

Stage 2  3✳ Consider the function f (x, y) = xy. ⇀

⇀

1. Explicitly determine the field lines (flow lines) of F(x, y) = ∇f . 2. Sketch the field lines of F and the level curves of f in the same diagram. ⇀

 4✳ ⇀

Find the field line of the vector field F = 2y ^ ı ı +

x y

2

y^ ^ ȷ ȷ +e k

that passes through (1, 1, e).

 5✳ ⇀

Find and sketch the field lines of the vector field F = x ^ ı ı + 3y ^ ȷ ȷ. 1. Think Poohsticks. 2. This is not such an unreasonable assumption. The flow often changes on a larger time scale. 3. This is also not an unreasonable approximation. 4. We'll have a more careful discussion of this in the optional §2.2.1. dx

5. If

dt

(t0 ) = 0,

but

dy dt

(t0 ) ≠ 0,

we should use y rather than x as the parameter. If

for all t and the streamline doesn't move. It is just a single point. 6. Of course

dy dx

dx dt

dy (t0 ) =

dt

(t0 ) = 0,

then r(t) = r(t

0)

is not the ratio of dy and dx. However pretending that it is provides a simple way to remember the technique that

is used to solve the equation. You may have used this mnemonic device before when you learned how to solve separable differential equations. Section 2.4 of the CLP-2 text contains a treatment of separable differential equations, including a justification for the mnemonic device. 7. Here is another nonrigorous, but intuitive way to come up with this equation. Suppose that our stick is at (x, y) and has velocity dx

(

dt

dy

(t) ,

dt

(t)).

In a tiny time interval dt the stick moves by (

(v1 (x, y) , v2 (x, y))

if

dx v1 (x,y)

=

dy v2 (x,y)

dx

dy (t) ,

dt

(t))dt = (dx, dy), dt

which is parallel to

.

8. In Example 2.1.7 we converted a second order ordinary differential equation into a system of first order ordinary differential equations. We are now just reversing the procedure we used there. 9. Even if you don't know how x(t) = A cos(Ωt − θ) was arrived at, you should be able to easily verify that it really does obey ′′

x

2

+ Ω x = 0.

This page titled 2.2: Optional — Field Lines is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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2.3: Conservative Vector Fields Not all vector fields are created equal. In particular, some vector fields are easier to work with than others. One important class of vector fields that are relatively easy to work with, at least sometimes, but that still arise in many applications are “conservative vector fields”.

 Definition 2.3.1: Conservative Fields ⇀

⇀

⇀

1. The vector field F is said to be conservative if there exists a function φ such that F = ∇φ. Then φ is called a potential for F. Note that if φ is a potential for F and if C is a constant, then φ + C is also a potential for F. 2. If F = ∇φ is a conservative field with potential φ and if C is a constant, then the set of points that obey φ(x, y, z) = C is called an equipotential surface. Similarly, in two dimensions, the set of points that obey φ(x, y) = C is called an equipotential curve. ⇀

⇀

⇀

⇀

⇀

 Warning 2.3.2 In physics, when a vector field is of the form ⇀

⇀

⇀

then

F = −∇φ,

is called a potential for

φ

⇀

F.

Note the minus 1 sign in

⇀

F = −∇φ. ↑

 Example 2.3.3. Potential energy The “conservative” in “conservative vector field” has nothing to do with politics. It comes from “conservation of energy”. Here is how. Suppose that you have a particle of mass m moving in a force field F that happens to be of the form F = ∇φ for some function φ. If the position of the particle a time t is (x(t), y(t), z(t)), then, by Newton's law of motion, ⇀

⇀

dv

⇀

⇀

ma = F

⟹ m

⇀

⇀

⇀

(t) = F(x(t), y(t), z(t)) dt ⇀

dv ⟹ m

⇀

(t) = ∇φ(x(t), y(t), z(t)) dt

Now dot both sides with

⇀

v (t). ⇀

⇀

dv

⟹ m v (t) ⋅

⇀

⇀

(t) = v (t) ⋅ ∇φ(x(t), y(t), z(t)) dt ∂φ

′

= x (t)

∂φ

′

(x(t), y(t), z(t)) + y (t) ∂x

(x(t), y(t), z(t)) ∂y

∂φ

′

+ z (t)

(x(t), y(t), z(t)) ∂z

Next use

d

⇀

⇀

⇀

⇀

dv

v ⋅ v = 2v ⋅

dt

dt

on the left hand side and the chain rule on the right hand side. d  ⟹   

1 (

dt

⇀

d 

⇀

m v (t) ⋅ v (t)) = 2 d 

⟹

1 (

dt

(φ(x(t), y(t), z(t))) dt

⇀

⇀

m v (t) ⋅ v (t) − φ(x(t), y(t), z(t))) = 0 2

1 ⟹

⇀

2

m| v (t)|

− φ(x(t), y(t), z(t)) = const

2

So m| v (t)| − φ(x(t), y(t), z(t)), which is called the energy 2 of the particle at time t, does not actually depend on time — it is conserved. Let's call the initial energy E. That is, E = m| v (0)| − φ(x(0), y(0), z(0)). Then m| v (t)| − φ(x(t), y(t), z(t)) = E for all t and, in particular 1

⇀

2

2

1

⇀

2

2

1

⇀

2

2

1 φ(x(t), y(t), z(t)) =

⇀

2

m| v (t)|

− E ≥ −E

2

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So even without having to find

we know that our particle can never escape the region

(x(t) , y(t) , z(t)),

{(x, y, z)|φ(x, y, z) ≥ −E} .

 Example 2.3.4. Gravity The gravitational force that a body of mass M at the origin exerts on a body of mass m at ⇀

GM m

⇀

F( r ) = −

⇀

r = (x, y, z)

is

⇀

r

3

r

where ⇀

− −−−−−−−− − ⇀ 2 2 2 r = | r | = √x + y + z

φ( r ) =

GMm r

.

and G is the gravitational constant. This force is conservative, with potential To verify that this is correct, observe that ∂  ∂x ∂ 

⇀

φ( r ) = ∂y ∂ 

GM m(2x)

∂ 

⇀

φ( r ) =

GM m 1 − −−−−−−−− − =− 2 2 2 ∂x √ x + y + z 2

[x

∂  GM m 1 − −−−−−−−− − =− ∂y √ x2 + y 2 + z 2 2

[x

∂ 

⇀

φ( r ) = ∂z

GM m

2

GM m

2 3/2

=−

+z ]

GM m(2y) +y

2

2 3/2

GM m(2z)

GM m =−

2

[x

+y

2

y

3

r

+z ]

=− 2

x

r3 GM m

=− 2

1

− −−−−−−−− − ∂z √ x2 + y 2 + z 2

+y

2

2 3/2

+z ]

z

3

r

We have already found conservation of energy very helpful a couple of times in Section 1.7 (Sliding on a Curve). So, at this point, there are probably several questions gnawing away at you. Is every vector field conservative? If not, is there an easy way to tell whether or not a vector field is conservative? If we know that a given vector field is conservative, how do you find a potential for it? Have no fear. We will consider those questions in some detail shortly. But first, a couple of more examples.

 Example 2.3.5 ⇀

In this example we will show that the vector field F(x, y) = x ^ ı ı −y ^ ȷ ȷ is conservative and find both its potential and its field lines. ⇀

1. The potential: Our vector field F(x, y) = x ^ ı ı −y ^ ȷ ȷ is conservative if we can find a φ obeying ∂φ (x, y) = x ∂x ∂φ (x, y) = −y ∂y

Recall that, when taking the partial derivative the coordinate y is viewed as a constant. So the first of these equations is satisfied if and only if there is a ψ(y), which does not depend on x, so that ∂ 

∂x

2

x φ(x, y) =

+ ψ(y) 2

For this to also satisfy the second equation, we need ∂φ −y =

2

∂  (x, y) =

∂y

x (

′

+ ψ(y)) = ψ (y)

∂y

2

which is the case if and only if there is a constant C with y

2

ψ(y) = −

+C 2

So, for any choice of the constant C ,

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2

x

y

2

− 2

is a potential. In particular, taking C

= 0,

+C 2

one possible potential is 2

y

x φ(x, y) =

2

− 2

2

Some equipotential curves for this potential are sketched in (c) below. They are the blue curves. ⇀

2. The field lines (Optional): Recalling (2.2.5), the field lines of the vector field F(x, y) = x ^ ı ı −y ^ ȷ ȷ are determined by dx

dy =

x

⟺

−ydx = xdy

−y ⟺

xdy + ydx = 0

⟺

d(xy) = 0

⟺

xy = C

by the product rule

for some constant C . If you are not comfortable with the use of the product rule above, here is another way to write the same computation. dy

y

dy

=− dx

⟺

x

x

= −y dx dy

⟺

x

+y = 0 dx

d ⟺

(xy) = 0

by the product rule

dx ⟺

xy = C

Some field lines are sketched in (c) below. They are the red curves. Note that they appear to cross the equipotential curves, the blue curves, at right angles. We shall see in Lemma 2.3.6, below, that this is not a coincidence. Also note that, while the above computation tells what the field lines are, it does not give us the direction of motion along the field lines. We determine the direction of motion next. 3. Direction of motion (Optional): The sign data ⎧>0

if x > 0 ⎫

^ ı ı ⋅ F(x, y) = x ⎨ = 0 ⎩

if x = 0 ⎬ ⎭

⇀

0

if y < 0 ⎫

^ ȷ ȷ ⋅ F(x, y) = −y ⎨ = 0 ⎩ 0

⇀

is visually displayed in the figure on the left below. The arrows in the figure on the left gives us the direction of motion along the field lines (in red) in the figure on the right below. Some equipotential curves are also sketched (in blue) in the figure on the right below.

We have just seen one example of a conservative vector field for which the field lines appear to cross the equipotential curves at right angles. Here is a result which says that that was no accident. The field lines of conservative fields always cross the equipotential surfaces at right angles.

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 Lemma 2.3.6. Optional ⇀

⇀

⇀

If F is a conservative vector field, then the field lines of F are perpendicular to the equipotential surfaces of F.

 Proof ⇀

⇀

⇀

Let F = ∇φ. Pick any point r and any nonzero vector T that is tangent to the equipotential surface at surface is φ(x, y, z) = φ( r ). Consider any curve r (t) = (x(t), y(t), z(t)) that ⇀

0

⇀

⇀

r 0.

That equipotential

⇀

0

⇀

lies in the equipotential surface of F through r (0) = r and ⇀

⇀

r 0,

so that φ( r (t)) = φ( r ⇀

for all t, and also obeys

⇀

0)

⇀ 0

⇀

dr

⇀

(0) = T. dt

Differentiating φ( r (t)) = φ( r ⇀

⇀

0)

with respect to t and applying the chain rule gives d  [φ(x(t), y(t), z(t))] = 0 dt

or ∂φ

∂φ

dx (x(t), y(t), z(t))

(t) +

∂x

dy (x(t), y(t), z(t))

dt

∂y ∂φ +

dz (x(t), y(t), z(t))

∂φ ∂x

∂φ

,

∂y

(t) = 0 dt

∂z

Notice that the left hand side is exactly the dot product of (

(t) dt

⇀

∂φ

,

∂z

) = ∇φ

with (

dx

dy ,

dt

dt

⇀

dz ,

dr ) =

dt

. dt

So

⇀

⇀

dr

⇀

∇φ( r (t)) ⋅

(t) = 0 dt

⇀

⇀

dr

⇀

F( r (t)) ⋅

(t) = 0 dt

Then set t = 0 to get ⇀

⇀

⇀

F( r 0 ) ⋅ T = 0 ⇀

This says that the vector T, which is tangent to the equipotential surface at which is a tangent vector to the field line of F through r . ⇀

⇀ r 0,

is perpendicular to the vector field at

⇀ r 0,

⇀ 0

Here is another example in which we try to find a potential for a vector field.

 Example 2.3.7 Let's try to find a potential for the vortex vector field obey

⇀

v (x, y) = Ω( − y ^ ı ı + x^ ȷ ȷ)

of Example 2.1.4. The potential would have to

∂φ (x, y) = −Ωy ∂x ∂φ (x, y) = Ωx ∂y

We proceed just as we did in Example 2.3.5. The first of these equations is satisfied if and only if there is a ψ(y), which does not depend on x, so that φ(x, y) = −Ωxy + ψ(y)

For this to also satisfy the second equation, we need

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∂φ Ωx =

∂ 

′

(x, y) = ∂y

′

( − Ωxy + ψ(y)) = −Ωx + ψ (y) ⟺

ψ (y) = 2Ωx

∂y

If Ω ≠ 0, the right hand side of this equation depends on x while the left hand side in independent of x, no matter what ψ is. So no ψ can work, and v (x, y) = Ω( − y ^ ı ı + x^ ȷ ȷ ) is not conservative. ⇀

The previous example shows that not all vector fields are conservative. That answers the first of the questions that we posed just before Example 2.3.5. The next theorem provides a simple screening test for conservativeness, which partially answers the second question. The easy way to remember the screening test uses the curl, which we now define.

 Definition 2.3.8 ⇀

⇀

⇀

The curl of a vector field F(x, y, z) is denoted by ∇ × F(x, y, z) and is defined by ⇀

⇀

∇×F

=(

∂F3

−

∂y

)^ ı ı +(

∂F1

∂z

∂F3

−

∂z

)^ ȷ ȷ +(

∂x

∂F2

^ ȷ ȷ

^ k

∂ 

∂ 

∂ 

∂x

∂y

∂z

F1 (x, y, z)

F2 (x, y, z)

F3 (x, y, z)

= det ⎢ ⎢

−

∂F1

∂x

^ ı ı

⎡

⎣

∂F2

^ )k

∂y

⎤ ⎥ ⎥ ⎦

The determinant in the second row is really just a mnemonic device used to make it easy to remember the expression after the equals sign in the first row. One must be careful about the signs in this definition — the determinant helps with that.

 Theorem 2.3.9. Screening test for conservative vector fields. 1. Assume that F (x, y) and F (x, y) are continuously differentiable. If the vector field F conservative, then we must have 1

1 (x,

2

∂F1

=

1 (x,

y, z), F2 (x, y, z)

and F

∂F1

=

∂y

∂x

are continuously differentiable. If the vector field is conservative, then

3 (x,

^ F1 (x, y, z) ^ ı ı + F2 (x, y, z)^ ȷ ȷ + F3 (x, y, z)k

is

∂F2

∂y

2. Assume that F

y) ^ ı ı + F2 (x, y)^ ȷ ȷ

y, z)

∂F2

∂F1

∂x

∂z

=

∂F3

∂F2

∂x

∂z

=

∂F3 ∂y

Equivalently, ⇀

⇀

∇×F = (

∂F3 ∂y

−

∂F2

)^ ı ı +(

∂z

∂F1

−

∂z

∂F3

)^ ȷ ȷ +(

∂x

∂F2 ∂x

−

∂F1

⇀

^ )k = 0

∂y

⇀

That is, F is curl free.

 Proof (a) If the vector field F

1 (x,

y) ^ ı ı + F2 (x, y)^ ȷ ȷ

is conservative, then there is a potential φ(x, y) such that ∂φ ∂x

(x, y) = F1 (x, y)

∂φ ∂y

Applying

∂  ∂y

to the first equation and

∂  ∂x

(x, y) = F2 (x, y)

to the second equation gives

2.3.5

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∂

2

φ

∂F1

= ∂y∂x ∂

2

∂y

φ

∂F2

= ∂x∂y

Recall that, for any twice continuously differentiable function,

∂x ∂

2

φ

∂y∂x

∂

=

2

φ

∂x∂y

.

So the two left hand sides are equal, and the two

right hand sides must also be equal. (b) If the vector field F

1 (x,

^ y, z) ^ ı ı + F2 (x, y, z)^ ȷ ȷ + F3 (x, y, z)k

is conservative, then there is a potential φ(x, y, z) such that

∂φ ∂x

(x, y, z) = F1 (x, y, z)

∂φ ∂y

(x, y, z) = F2 (x, y, z)

∂φ ∂z

(x, y, z) = F3 (x, y, z)

We proceed just as in (a). Applying

∂  ∂y

to the first equation and

to the second equation gives

∂  ∂x

⎧ ⎪

∂

⎨ ⎩ ⎪

2

φ

∂y∂x ∂

2

∂  ∂z

to the first equation and

⎨ ⎩ ⎪

∂

2

φ

∂z∂x ∂

2

φ

∂x∂z ∂  ∂z

to the second equation and

∂y

⎫ ⎪

=

∂F2

⟹

∂F1

=

∂F2

∂y

⎭ ⎪

∂x

∂x

to the third equation gives

∂  ∂x

⎧ ⎪

Applying

∂F1

⎬ φ

∂x∂y

Applying

=

∂  ∂y

=

∂F1 ∂z ∂F3

⎫ ⎪ ⎬ ⎭ ⎪

⟹

∂F1

=

∂F3

∂z

∂x

∂x

to the third equation gives

⎧ ⎪ ⎨ ⎩ ⎪

∂

2

φ

∂z∂y ∂

2

⇀

=

φ

∂y∂z ⇀

=

=

∂F2 ∂z ∂F3

⎫ ⎪ ⎬

⟹

∂F2

=

∂F3

∂z

⎭ ⎪

∂y

∂y

⇀

Combining the three bullet points gives ∇ × F = 0 .

 Warning 2.3.10 As always, we have to be careful with the flow of logic 3. The screening test in Theorem 2.3.9 is a one-way test. If, for example, ⇀

F

∂F1 ∂y

≠

∂F2 ∂x

⇀

then the vector field F cannot be conservative. But if

∂F1 ∂y

is conservative. In fact there are fields that are not conservative but do obey

=

∂F2 ∂x

∂F1 ∂y

=

Theorem 2.3.9 does not guarantee that ∂F2 ∂x

.

We'll see one in Example 2.3.14,

below. We shall later find some additional regularity conditions which, when combined with

∂F1 ∂y

=

∂F2 ∂x

,

do imply

conservativeness. See Theorem 2.4.8, below.

 Example 2.3.11. Example 2.3.7 revisited In Example 2.3.7, we attempted to find a potential for the vector field ⇀

v (x, y) = Ω( − y ^ ı ı + x^ ȷ ȷ)

In the end we showed that, if Ω ≠ 0, no potential could exist, i.e. v (x, y) is not conservative. Had we known the screening test of Theorem 2.3.9.a, we could have concluded that v (x, y) is not conservative by simply observing that ⇀

⇀

2.3.6

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⇀

∂ v1

∂  =

∂y

( − Ωy) = −Ω ∂y

⇀

∂ v2

∂  =   

∂x

(Ωx)

= +Ω

∂x

are not equal, unless Ω = 0. But Ω = 0 makes a rather boring vector field.

 Example 2.3.12 Determine whether or not the vector field ⇀

^ F(x, y, z) = y ^ ı ı − z^ ȷ ȷ + xk

is conservative. If it is conservative, find a potential. Solution Let's start by applying the screening test Theorem 2.3.9.b. Since ⎡ ⇀

⇀

= det ⎢ ⎢

∇×F

⎣

is not

⇀

0,

^ k ⎤

^ ı ı

^ ȷ ȷ

∂ 

∂ 

∂ 

∂x

∂y

∂z

y

−z

x

^ ⎥ = ^ ı ı −^ ȷ ȷ −k ⎥ ⎦

⇀

the vector field F cannot be conservative.

 Example 2.3.13 Determine whether or not the vector field ⇀

F(x, y, z) = (y

2

+ 2x z

2

2 3 ^ − 1) ^ ı ı + (2x + 1)y ^ ȷ ȷ + (2 x z + z )k

is conservative. If it is conservative, find a potential. Solution Again start by applying the screening test Theorem 2.3.9.b. This time ⎡ ⇀

⇀

∇ × F = det ⎢ ⎢ ⎣

y

2

^ ı ı

^ ȷ ȷ

^ k

∂ 

∂ 

∂ 

∂x

∂y

∂z

+ 2x z

2

−1

(2x + 1)y

2

2x z + z

⎤ ⎥ ⎥ 3

⎦

^ = 0^ ı ı − (4xz − 4xz)^ ȷ ȷ + (2y − 2y)k ⇀

= 0 ⇀

So F passes the screening test. Let's look for a function φ(x, y, z) obeying ∂φ (x, y, z) = y

2

+ 2x z

2

−1

∂x ∂φ (x, y, z) = (2x + 1)y

(∗)

∂y ∂φ

2

(x, y, z) = 2 x z + z

3

∂z

The partial derivative ψ(y, z) with

∂  ∂x

treats y and z as constants. So

φ(x, y, z)

φ(x, y, z) = x y

2

2

+x z

obeys the first equation if and only if there is a function

2

− x + ψ(y, z)

This φ(x, y, z) will also obey the second equation if and only if

2.3.7

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∂  (x y

2

2

+x z

2

− x + ψ(y, z)) = (2x + 1)y

∂y ∂ψ ⟺

2xy +

(y, z) = (2x + 1)y ∂y

∂ψ ⟺

(y, z) = y ∂y y

⟺

2

ψ(y, z) =

+ ζ(z) 2

for some function ζ(z) which depends only on z. At this stage we know that φ(x, y, z) = x y

2

2

+x z

2

− x + ψ(y, z) = x y

2

2

+x z

2

y

2

−x +

+ ζ(z) 2

obeys the first two equations of (∗), for any function ζ(z). Finally to have the third equation of (∗) also satisfied, we also need to chose ζ(z) to obey ∂  (x y

2

2

+x z

2

2

y

2

−x +

∂z

+ ζ(z)) = 2 x z + z

3

2 2

′

2

⟺

2 x z + ζ (z) = 2 x z + z

⟺

ζ (z) = z

⟺

ζ(z) =

′

z

3

3

4

+C 4

for any constant C . So one possible potential, namely that with C φ(x, y, z) = x y

2

= 0,

2

+x z

2

is y

2

−x +

z

4

+ 2

4

4

Note, as a check , that ⇀

∇φ(x, y, z) = (y

2

+ 2x z

2

2 3 ^ − 1) ^ ı ı + (2xy + y)^ ȷ ȷ + (2 x z + z )k

as desired.

 Example 2.3.14. Optional: First look at −

y x2 +y 2

^ ı +

x

^ ȷ

x2 +y 2

Now is a good time to reread Warning 2.3.10. In this example we will show that the vector field y

⇀

F(x, y) = −

2

x

+y

2

^ ı ı +

x 2

x

+y

2

^ ȷ ȷ

2

defined for all (x, y) in R  except (x, y) = (0, 0)

passes the screening test of Theorem 2.3.9.a. We will also begin to see why it is not conservative on the domain R To verify the screening test, we compute

2

∂ 

2

y (−

∂y

2

x ∂ 

2

2

x

+y

2

2

(x 2

2

+ y ) − y(2y) (x

x (

∂x

+y

(x ) =−

2

+y )

2

y

2

(x

2

2

(x

2

+y )

2

2

−x

=

+ y ) − x(2x)

) =

2

y

2

2

+y )

2

2

2

−x

= (x

∖ {(0, 0)}.

2

+y )

2

and observe that the two right hand sides are identical. So the screening test is passed. In order for

⇀

F

to be conservative on the domain

partial derivatives

∂φ ∂x

(x, y)

and

∂φ ∂y

(x, y),

2

R

∖ {(0, 0},

there must exist a function

φ(x, y),

that, together with both

is defined for all (x, y) in R except (x, y) = (0, 0), and obeys 2

2.3.8

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∂φ

−

y (x, y) = −

∂x

2

x

+y

=

2

y

∂ 

x2

1 +(

y x

=

2

y ( arctan

∂x

)

) x

1

∂φ

x (x, y) =

2

∂y

x

+y

2

∂ 

x

=

1 +(

y x

=

2

y ( arctan

∂y

)

) x

by the chain rule, because ∂ 

y (

∂x

y ) =−

x

2

x

∂ 

y (

∂y

1 ) =

x

x

y

y

It looks like we have found a potential, namely arctan . But there is a problem. Recall that, by definition, arctan is an angle θ(x, y) that obeys tan θ(x, y) = arctan ; but for any (x, y) ∈ R ∖ {(0, 0} there are infinitely many angles having the tangent . To define φ(x, y) we have to select exactly one such angle. It is impossible to do so in such a way that φ(x, y) is continuous on all of R ∖ {(0, 0}. y

x

x

2

x

y

x

2

To see why, fix any r > 0, and imagine that you are walking on the circle x + y = r in the xy-plane. At time θ, you are at x = r cos θ, y = r sin θ and then = tan θ and you are allowed to define φ(x, y) = θ + kπ, for any integer k. 2

2

2

y

x

Suppose that at time θ = 0 you choose k = 0. That is, you choose φ(r, 0) = 0. Now start walking, choosing an allowed 5 φ(x, y), i.e. choosing a k, for each point (x, y) that you cross. Because φ(x, y) has to vary continuously with (x, y), you have to continue choosing k = 0. But you run off a cliff as θ approaches 2π, because then you are approaching (r, 0) from below, as in the figure below, and because you are choosing k = 0, φ(x, y) is just a little less than 2π, but you have already chosen φ(r, 0) = 0, not 2π. So φ(x, y) has a jump discontinuity 6 along the positive x-axis.

If you are having trouble following this argument, don't worry about it. We will return with a less hand-wavy argument later.

Exercises Stage 1  1 We've seen two calculations of the energy E of a system. Equation 1.7.1 told us says m| v (t)| − φ(x(t), y(t), z(t)) = E. 1

⇀

E =

1 2

⇀ 2

m| v |

+ mgy,

while Example 2.3.3

2

2

⇀

⇀

Consider a force given by F = ∇φ for some differentiable function φ : R and no other forces, and its position at time t is given by (x(t), y(t), 0).

3

⇀

→ R.

A particle of mass m is being acted on by F

True or false: mgy(t) = −φ(x(t), y(t), 0).

 2 For each of the following fields, decide which of the following holds: ⇀

1. The screening test for conservative vector fields tells us F is conservative. 2. The screening test for conservative vector fields tells us F is not conservative. ⇀

2.3.9

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⇀

3. The screening test for conservative vector fields does not tell us whether F is conservative or not. (The screening test is Theorem 2.3.9.) ⇀

^ 1. F = x ^ ı ı + z^ ȷ ȷ + yk ⇀

2. F = y

2

⇀

2

2

^ z^ ı ı + x z^ ȷ ȷ + x yk

3. F = (y e

xy

+ 1) ^ ı ı + (x e

xy

+ z)^ ȷ ȷ +(

1

^ + y) k

z

⇀

4. F = y cos(xy) ^ ı ı + x sin(xy)^ ȷ ȷ

 3 ⇀

Suppose F is conservative and let a, b, and c be constants. Find a potential for

⇀

F + (a, b, c),

OR give a conservative field

⇀

F

⇀

and constants a, b, and c for which F + (a, b, c) is not conservative.

 4 Prove, or find a counterexample to, each of the following statements. ⇀

⇀

1. If F is a conservative field and G is a non-conservative field, then F + G is non-conservative. ⇀

⇀

2. If F and G are both non-conservative fields, then F + G is non-conservative. 3. If F and G are both conservative fields, then F + G is conservative. ⇀

⇀

Stage 2  5✳ ⇀

Let D be the domain consisting of all (x, y) such that x > 1, and let F be the vector field y

⇀

F =−

2

x

+y

x

^ ı ı +

2

2

x

+y

2

^ ȷ ȷ

⇀

Is F conservative on D? Give reasons for your answer.

 6 ⇀

Find a potential φ for F(x, y) = (x + y) ^ ı ı + (x − y)^ ȷ ȷ , or prove none exists.

 7 ⇀

Find a potential φ for F(x, y) = (

1 x

1

−

y

) ^ ı ı +(

x y

2

)^ ȷ ȷ,

or prove none exists.

 8 ⇀

Find a potential φ for F(x, y, z) = (x

2

yz + xz) ^ ı ı +(

1 3

3

x z + y) ^ ȷ ȷ +(

1 3

3

x y+

1 2

2

x

^ + y) k,

or prove none exists.

 9 Find a potential φ for y

x

⇀

F(x, y) = (

2

x

+y

2

+z

2

) ^ ı ı +(

2

x

+y

2

+z

2

)^ ȷ ȷ +(

z 2

x

+y

2

+z

2

^ ) k,

or prove none exists.

2.3.10

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 10 Determine whether or not each of the following vector fields are conservative. Find the potential if it is. ⇀

^ 1. F(x, y, z) = x ^ ı ı − 2y ^ ȷ ȷ + 3zk ⇀

2. F(x, y) =

x^ ı ı − y^ ȷ ȷ 2

x

+y

2

 11 ⇀

Let F = e

(z

2

)

^ ı ı + 2By z

3

^ ȷ ȷ + (Axze

(z

2

)

2 2 ^ + 3By z ) k. ⇀

1. For what values of the constants A and B is the vector field F conservative on R 2. If A and B have values found in (a), find a potential function for F.

3

?

⇀

Stage 3  12 Find the velocity field for a two dimensional incompressible fluid when there is a point source of strength m at the origin. That is, fluid is emitted from the origin at area rate 2πm cm /sec. Show that this velocity field is conservative and find its potential. 2

 13 ⇀

⇀

A particle of mass 10 kg moves in the force field F = ∇φ, where φ(x, y, z) = −(x 0, the particle is at the origin, and it moves with a velocity 2 m/s.

2

+y

2

2

+ z ).

When its potential energy is

Following Example 2.3.3, give a region the particle can never escape.

 14 ⇀

^ A particle with constant mass m = 1/2 moves under a force field F = ^ȷȷ + 3√z k . At position is its speed at (1, 1, 1)? 3

(0, 0, 0),

its speed is

1.

What

(You may assume without proof that the particle does indeed reach the point (1, 1, 1).)

 15 For some differentiable, real-valued functions f , g, h : R → R, we define ⇀

′

′

′

F = 2f (x)f (x) ^ ı ı + g (y)h(z)^ ȷ ȷ + g(y)h (z). ⇀

Verify that F is conservative.

 16 Describe the region in R where the field 3

⇀

F = ⟨xy, xz, y

2

+ z⟩

has curl 0. 1. Physicists introduce this minus sign in order to eliminate the minus sign in the next footnote. 2. m| v (t)| is the kinetic energy and −φ is the potential energy. See Warning 2.3.2. 3. Use your favourite search engine to look up a list of common logical errors. One is “affirming the consequent”. An example would be concluding that because Shakespeare is dead, Elvis, who is also dead, must also be Shakespeare. 4. It is always worth doing this check. 1

⇀

2

2

2.3.11

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5. If φ(x, y) is not continuous, its gradient does not exist, and φ cannot be a potential. 6. Those who have taken some complex analysis may recognize this as the branch cut in ln z. This page titled 2.3: Conservative Vector Fields is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

2.3.12

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2.4: Line Integrals We have already seen, in §1.6, one type of integral along curves. We are now going to see a second, that turns out to have significant connections to conservative vector fields. It arose from the concept of “work” in classical mechanics. ⇀

Suppose that we wish to find the work done by a force

⇀

F( r )

moving a particle along a path

⇀

r (t).

During the “infinitesimal time

⇀

interval” 1 from t to t + dt the particle moves from

to

⇀

r (t)

⇀

⇀

r (t) + d r

with d r

⇀

dr =

By definition, the work done during

(t) dt. dt

that infinitesimal time interval is ⇀

⇀

⇀

⇀

⇀

dr

⇀

F( r (t)) ⋅ d r = F( r (t)) ⋅

(t) dt dt

The total work done during the time interval from t to t is then 0

1

t1

Work = ∫

⇀

⇀

dr

⇀

F( r (t)) ⋅

(t) dt dt

t0

There are some useful shorthand notations for this work.

 Definition 2.4.1 Denote by C the parametrized path

⇀

r (t)

⇀

∫

with t

0

≤ t ≤ t1 .

⇀

⇀

F ⋅ dr = ∫

C

Then

⇀

t1

⇀

(F1 dx + F2 dy + F3 dz) = ∫

C ⇀

If C is a closed path, the notation ∮

C

⇀

F ⋅ dr

⇀

⇀

dr

⇀

F( r (t)) ⋅

(t) dt dt

t0

is also used.

⇀

In the event that F is conservative, and we know the potential φ, the following theorem provides a really easy way to compute “work integrals”. The theorem is a generalization of the fundamental theorem of calculus, and indeed some people call it the fundamental theorem of line integrals.

 Theorem 2.4.2 ⇀

Let F = ∇φ be a conservative vector field. Then if C is any curve that starts at P and ends at P

1,

0

⇀

∫

we have 2

⇀

F ⋅ d r = φ(P1 ) − φ(P0 )

C

Proof Let r (t) = (x(t), y(t), z(t)), definition, ⇀

⇀

∫

t0 ≤ t ≤ t1 ,

t1 ⇀

F ⋅ dr

⇀

=∫

C

be any parametrization of ⇀

⇀

t1

dr

F( r (t)) ⋅

(t) dt = ∫ dt

t0 t1

=∫

∂x

t0

r (t0 ) = P0

and

r (t1 ) = P1 .

Then, by

⇀

(t) dt

∂φ (t) +

dt

dy (x(t), y(t), z(t))

∂y

(t) dt

∂φ +

dz (x(t), y(t), z(t))

∂z t1

⇀

dt

dx (x(t), y(t), z(t))

⇀

dr

⇀

∇φ( r (t)) ⋅

t0

∂φ [

⇀

with

C

(t)]dt dt

d 

=∫

[φ(x(t), y(t), z(t))]dt

by the chain rule in reverse

dt

t0 ⇀

⇀

= φ( r (t1 )) − φ( r (t0 )) = φ(P1 ) − φ(P0 )

by the fundamental theorem of calculus.

2.4.1

https://math.libretexts.org/@go/page/91901

⇀

Observe that, in Theorem 2.4.2, the value, φ(P ) − φ(P ), of the integral ∫ F ⋅ d r depended only on the endpoints P and P of the curve, not on the path that the curve followed to get to P from P . We shall see, in Theorem 2.4.7, below, that this happens only for conservative vector fields. Here are several examples of line integrals of vector fields that are not conservative. 1

0

⇀

0

C

0

1

1

 Example 2.4.3 Set P

0

and 3

= (0, 0), P1 = (1, 1)

⇀

F(x, y) = xy ^ ı ı + (y

2

+ 1) ^ ȷ ȷ

We shall consider three curves, all starting at P and ending at P

1.

0

1. Let C be the straight line from P to P . 2. Let C be the path, made from two straight lines, which follows the x-axis from P to (1, 0) and then follows the line x = 1 from (1, 0) to P . 3. Let C be the part of the parabola x = y from P to P . 1

0

1

2

0

1

2

3

0

We shall calculate the work ∫

⇀

Ci

1. We parametrize C by 1

⇀

F ⋅ dr

for each of the curves. with t running from 0 to 1. Then x(t) = t and y(t) = t so that

⇀

r (t) = t ^ ı ı +t ^ ȷ ȷ

⇀

1

⇀

⇀

2

dr

2

F( r (t)) = t

^ ı ı + (t

+ 1) ^ ȷ ȷ

and

(t) = ^ ı ı +^ ȷ ȷ dt

so that 1

⇀

∫

⇀

F ⋅ dr

=∫

C1

⇀

⇀

1

dr

⇀

F( r (t)) ⋅ dt

0

2

(t) dt = ∫

[t

2

^ ı ı + (t

+ 1) ^ ȷ ȷ] ⋅ [^ ı ı +^ ȷ ȷ ] dt

0

1 2

=∫

[2 t

+ 1] dt

0

5 = 3

2. We split C into two parts, C running from P to (1, 0) along the x-axis and then C running from (1, 0) to P along the line x = 1. We parametrize C by r (x) = x ^ ı ı with x running from 0 to 1 and C by r (y) = ^ ı ı +y ^ ȷ ȷ with y running 4 from 0 to 1. Then 2

2,x

0

2,y

⇀

1

⇀

2,x

2,y

⇀

∫

⇀

⇀

F ⋅ dr = ∫

C2

⇀

⇀

F ⋅ dr + ∫

C2,x

⇀

F ⋅ dr

C2,y ^ ı ı

 1 2

=∫

[(x)(0) ^ ı ı + (0

d  + 1) ^ ȷ ȷ] ⋅

(x ^ ı ı ) dx dx

0

^ ȷ ȷ

 1

+∫

[(1)(y) ^ ı ı + (y

1

0

+ 1) ^ ȷ ȷ] ⋅

d 

(^ ı ı +y ^ ȷ ȷ ) dy

dy

0

=∫

2

1

0 dx + ∫

(y

2

+ 1) dy

0

4 = 3

2.4.2

https://math.libretexts.org/@go/page/91901

3. We parametrize C by 3

⇀

2

^ ı ı +t ^ ȷ ȷ

r (t) = t

with t running from 0 to 1. Then x(t) = t and y(t) = t so that 2

⇀

⇀

⇀

3

dr

2

F( r (t)) = t

^ ı ı + (t

+ 1) ^ ȷ ȷ

and

(t) = 2t ^ ı ı +^ ȷ ȷ dt

so that ⇀

∫

1 ⇀

F ⋅ dr

3

=∫

C3

2

^ ı ı + (t

[t

+ 1) ^ ȷ ȷ ] ⋅ [2t ^ ı ı +^ ȷ ȷ ] dt

0 1 4

=∫

[2 t

2

+t

+ 1] dt

0

2 =

1

Note that, despite the fact that ⇀

∫

C3

⇀

F ⋅ dr

C1 , C2

and

C3

26

+

+1 =

5

3

all start at

15

and all end at

P0

P1 ,

the three integrals

⇀

∫

C1

⇀

⇀

F ⋅ dr , ∫

C2

⇀

F ⋅ dr

and

all have different values.

 Example 2.4.4 Set 5 ⇀

F(x, y) = 2y ^ ı ı + 3x ^ ȷ ȷ

This time we consider two curves. 1. Let C be circle x + y = 1, traversed once counterclockwise, starting at (1, 0). 2. Let C be (trivial) curve which just consists of the single point (1, 0). 2

1

2

2

We shall calculate the work ∫

Ci

1. We parametrize C by 1

⇀

for each curve.

⇀

F ⋅ dr

with t running from 0 to 2π, just as we did in Example 1.0.1. Then

⇀

r (t) = cos t ^ ı ı + sin t ^ ȷ ȷ

2π

⇀

∮

⇀

F ⋅ dr

[2 sin t ^ ı ı + 3 cos t ^ ȷ ȷ ] ⋅ [− sin t ^ ı ı + cos t ^ ȷ ȷ ] dt

=∫

C1

0 2π 2

=∫

[ − 2 sin

2

t + 3 cos

t] dt

0

You could evaluate these integrals using double angle trig identities like you did in first year calculus. But there is a sneaky, much easier, way. Because sin t and cos t are translates of each other, and both are periodic of period π, the two integrals ∫ sin t dt and ∫ cos t dt represent the same area and so are equal. See the figure below. 2

2π

0

2

2π

2

2

0

Thus 2π

∫

2π 2

sin

2π 2

t dt = ∫

0

cos

0

0

∫ 2

2

2

[ sin

t + cos

t] dt

2

2π

1 =

1

t dt = ∫

dt = π

0

and ⇀

∮ C1

2π ⇀

F ⋅ d r = −2 ∫

2π 2

sin

t dt + 3 ∫

0

2

cos

t dt = π

0

2.4.3

https://math.libretexts.org/@go/page/91901

⇀

2. We parametrize C by 2

r (t) = ^ ı ı

Again, despite the fact that different.

and

C1

dr

for all t. Then

⇀

C2

⇀

both start at

⇀

and ∫

(t) = 0

and end at

(1, 0)

⇀

F ⋅ d r = 0.

C2

dt

(1, 0),

the two integrals

⇀

∫

C1

⇀

F ⋅ dr

and

⇀

∫

C2

are

⇀

F ⋅ dr

 Example 2.4.5. Example 2.3.14, again In Example 2.3.14, we saw that the vector field y

⇀

F(x, y) = −

2

x

x

+y

2

^ ı ı +

2

x

2

+y

^ ȷ ȷ

2

defined for all (x, y) in R  except (x, y) = (0, 0) ⇀

passed the screening test of Theorem 2.3.9.a, and yet was not conservative. In this example, we will see that this F violates the conclusion of Theorem 2.4.2, thereby providing a second proof that F(x, y) is not conservative on R with (0, 0) removed. For the curve C, of Theorem 2.4.2, we use the circle parametrized by x = a cos θ,  y = a sin θ, 0 ≤ θ ≤ 2π. Then dx = −a sin θ dθ and dy = a cos θ dθ so that ⇀

1

x dy − y dx ∫

2π

C

2

x

+y

2π

1 =

2

2

a

2

cos

2

2

θ dθ + a

2π

2

a

0

2

cos

2

sin

∫ 2

θ+a

θ dθ

2

sin

θ

2π

1 =

∫ 2π

dθ

0

=1

The curve C has initial point P0 = (a cos θ,  a sin θ)∣ ∣

θ=0

= (a, 0)

and final point P1

= (a cos θ,  a sin θ)∣ ∣

θ=2π

= (a, 0) = P0

⇀

So, if F were conservative with potential φ, Theorem 2.4.2 would give that 1

x dy − y dx ∫

2π

C

= φ(P1 ) − φ(P0 ) = 0

x2 + y 2

⇀

Consequently, F can't be conservative.

Path Independence ⇀

This brings us to the following question. Let F be any fixed vector field. When is it true that, given any two fixed points P and P , the integrals 0

1

⇀

∫

for all curves C, C that start at P and end at P independent of the path chosen” and we write ′

0

1?

⇀

⇀

F ⋅ dr = ∫

C

C

⇀

F ⋅ dr ′

When can we ignore the path taken? If this is the case we say that “∫

⇀

C

P1

∫

⇀

⇀

⇀

F ⋅ dr = ∫

P0

⇀

F ⋅ dr

is

⇀

F ⋅ dr

C

for any path C from P to P . The point of this section is that there is an intimate relation between path independence and conservativeness of vector fields, that we will get to in Theorem 2.4.7. 0

1

For simplicity, we will consider only vector fields that are defined and continuous on all of R (i.e. the xy-plane) or R (i.e. the usual three dimensional world). Some discussion about what happens for vector fields that are defined only on part of R or R is given in the optional §4.5. 2

3

2

First we show that if there is one pair of (not necessarily distinct) points P

0,

2.4.4

P1

3

such that

https://math.libretexts.org/@go/page/91901

⇀

∫

⇀

⇀

F ⋅ dr = ∫

C1

for all curves C

1,

C2

that start at P and end at P

C2

then it is also true that, for any other pair of points P

1,

0

⇀

⇀

⇀

F ⋅ dr = ∫ ′

for all curves C

1

, C2

that start at P and end at P ′

′

0

1

, P

′

1

⇀

F ⋅ dr ′

C1 ′

′

0

∫

′

⇀

F ⋅ dr

C2

This might seem unlikely at first, but the idea of the proof is really intuitive.

.

 Theorem 2.4.6 ⇀

Let F be a vector field that is defined and continuous on all of R ). Assume that

(or R ). Let P

2

3

R

0,

′

′

0

1

P1 , P , P

be any four points in

2

R

(or

3

⇀

∫

⇀

⇀

F ⋅ dr = ∫

C1

for all curves C

1,

C2

that start at P and end at P

1.

0

C2

Then ⇀

∫

⇀

⇀

F ⋅ dr = ∫

C

′

for all curves C

1

′

, C2

that start at P and end at P ′

′

0

1

⇀

F ⋅ dr

C

1

′

⇀

F ⋅ dr

′

2

.

Proof Let C and C be any two curves that start at P and end at P ′

′

′

′

1

2

0

1

.

We start by choosing any two (auxiliary) curves Cℓ Cr

that starts at P and ends at P and that starts at P and ends at P . ′

0

0

′

1

1

and then we define the curves C1 C2

to be C , followed by C , followed by C and to be C , followed by C , followed by C . ′

ℓ

r

1 ′

ℓ

r

2

Then both C and C start at P and end at P , so that, by hypothesis, 1

2

0

1

⇀

∫

⇀

⇀

F ⋅ dr = ∫

C1

⇀

F ⋅ dr

C2

and, from the construction of C and C

2,

1

⇀

∫

⇀

⇀

F ⋅ dr + ∫

Cℓ ⇀

⟹

∫

C1 ⇀

⇀

F ⋅ dr = ∫ ′

C1

⇀

⇀

F ⋅ dr + ∫ ′

⇀

⇀

F ⋅ dr = ∫

Cr

Cℓ

⇀

⇀

F ⋅ dr + ∫

⇀

⇀

F ⋅ dr + ∫ ′

C2

⇀

F ⋅ dr

Cr

⇀

F ⋅ dr ′

C2

as desired. We are now ready for our main theorem on conservative fields.

2.4.5

https://math.libretexts.org/@go/page/91901

 Theorem 2.4.7 ⇀

Let F be a vector field that is defined and continuous on all of R (or R ). Then the following three statements are equivalent. 2

⇀

3

⇀

⇀

1. F is conservative. That is, there exists a function φ such that F = ∇φ. ⇀

2. The integral ∮ F ⋅ d r = 0 for any closed curve C. 3. The integral ∫ F ⋅ d r is path independent. That is, for any points P C , C that start at P and end at P . ⇀

C ⇀

1

⇀

2

0,

0

P1

⇀

we have ∫

C1

⇀

⇀

F ⋅ dr = ∫

C2

⇀

F ⋅ dr

for all curves

1

That is, if any one of the three statements are true, then all three are true. Proof It suffices for us to prove 6 that the truth of (a) implies the truth of (b) and the truth of (b) implies the truth of (c) and the truth of (c) implies the truth of (a). That's exactly what we will do. (a) ⟹ (b): Let C be a closed curve that starts at P and then ends back at P . Then, by Theorem 2.4.2 with P 0

0

⇀

∮

1

= P0 ,

⇀

F ⋅ d r = φ(P0 ) − φ(P0 ) = 0

C

(b) ⟹ (c): Pick any point

P0

and set

⇀

⇀

Then we are assuming that

P1 = P0 .

∮

C

end at P . In particular ∫ F ⋅ d r takes the same value for all curves that start at immediately yields property (c). 1

⇀

C

(c) ⟹ (a): We are to show that ⇀

⇀

F

is conservative. We'll start by guessing

for all curves that start at

⇀

F ⋅ dr = 0 P0

and end at

P1 .

P0

So Theorem 2.4.6

and then we'll verify that, for our chosen

φ

⇀

and

φ,

we

⇀

really do have F = ∇φ. Our guess for φ is motivated by Theorem 2.4.2. If our F really is conservative, its potential is going ⇀

⇀

to have to obey ∫ F ⋅ d r = φ(P ) − φ(P ) for any curve C that starts at P and ends at P . Let's choose P = 0 . Remembering, from Definition 2.3.1.a, that adding a constant to a potential always yields another potential, we can always ⇀

1

C

choose

⇀

φ( 0 ) = 0. ⇀

⇀

φ(x) = ∫

F ⋅ dr

integral ∫

F ⋅ dr

C

⇀

C

⇀

Then

0

⇀

φ(P1 ) = ∫

C

for any curve

C

⇀

F ⋅ dr

0

for any curve

that starts at

⇀

0

C

and ends at

0

and ends at

P1 .

0

So define, for each point

x,

Note that, since we we are assuming that (c) is true, the

x. ⇀

takes the same value for all curves C that start at ⇀

⇀

that starts at

1

and end at x.

0

⇀

We now verify that, for this chosen φ, we really do have F = ∇φ. Fix any point and ends at x. For any vector u, let D be the curve with parametrization

x

and any curve

Cx

that starts at the origin

u

⇀

r u (t) = x + tu

This curve is a line segment that starts at assumption, φ(x + su) = ∫

C

⇀

⇀

F ⋅ dr

x

at

0 ≤t ≤1

and ends at

t =0

for any curve C that starts at

x+u

⇀

at

t = 1.

Observe that

⇀ ′

r

u

(t) = u.

Recall that, by

and ends at x + su. So

0

⇀

φ(x + su) = ∫

⇀

F⋅ dr

Cx +Dsu

where C

x

+ Dsu

is the curve which first follows C from the origin to x and then follows D x

⇀

∫

⇀

F⋅ dr

su

⇀

=∫

Cx +Dsu

Cx

=∫

⇀

F⋅ dr

Dsu ⇀

1 ⇀

F⋅ dr +∫

Cx

In the second integral, make the change of variables τ

⇀

⇀

F⋅ dr +∫

from x to x + su. We have

⇀

F(x + tsu) ⋅ (su) dt

0

= ts, dτ = sdt.

2.4.6

This gives

https://math.libretexts.org/@go/page/91901

s

⇀

φ(x + su) = ∫

⇀

F⋅ dr +∫

Cx

⇀

F(x + τ u) ⋅ u dτ

0

By the fundamental theorem of calculus, applied to the second integral, d  ds

⇀

∣ φ(x + su) ∣

= F(x + su) ⋅ u

s=0

⇀

∣ ∣

= F(x) ⋅ u

s=0

^ Applying this with u = ^ ıı ,  ^ ȷ ȷ ,  k gives us ∂φ (

∂φ (x) ,

∂x

∂φ (x) ,

∂y

⇀

⇀

⇀

^ (x)) = (F(x) ⋅ ^ ıı , F(x) ⋅ ^ ȷ ȷ , F(x) ⋅ k)

∂z

which is ⇀

∇φ(x) = F(x)

as desired. Using this result, we can completely characterize conservative fields on R and R 2

3

.

 Theorem 2.4.8 ⇀

Let F be a vector field that is defined and has continuous first order partial derivatives on all of conservative if and only if it passes the screening test ∇ × F = 0 , i.e. is curl free. ⇀

⇀

2

R

(or

3

R

). Then

⇀

F

is

⇀

 Warning 2.4.9 ⇀

Note that in Theorem 2.4.8 we are assuming that F passes the screening test on all of R or R . We have already seen, in Example 2.3.14, that if the screening test fails at even a single point, for example because the vector field is not defined at that point, then F need not be conservative. We'll explore what happens in such cases in the (optional) §4.5. We'll see that something can be salvaged. 2

3

⇀

Proof of Theorem 2.4.8. We'll give the proof for the R case. The proof for the R case is very similar. We have already seen, in Theorem 2.3.9, that if F is conservative, then it passes the screening test and there is nothing more to do. 2

3

⇀

⇀

So we now have to assume that F obeys

∂F1 ∂y

∂F2

(x, y) =

∂x

(x, y)

on all of R and prove that it is conservative. We'll do so using 2

the strategy of Example 2.3.13 to find a function φ(x, y), that obeys ∂φ (x, y) = F1 (x, y) ∂x ∂φ (x, y) = F2 (x, y) ∂y

The partial derivative

∂  ∂x

treats y as a constant. So φ(x, y) obeys the first equation if and only if there is a function ψ(y) with x

φ(x, y) = ∫

F1 (X, y) dX  +  ψ(y)

0

This φ(x, y) will also obey the second equation if and only if ∂φ F2 (x, y) =

(x, y) ∂y x

∂  =

(∫ ∂y x

=∫ 0

F1 (X, y) dX  +  ψ(y))

0

∂F1

′

(X, y) dX  +  ψ (y)

∂y

2.4.7

https://math.libretexts.org/@go/page/91901

So we have to find a ψ(y) that obeys x ′

ψ (y) = F2 (x, y) − ∫ 0

∂F1

(X, y) dX

∂y

This looks bad — no matter what ψ(y) is, the left hand side is independent of x, while it looks like the right hand side depends on x. Fortunately our screening test hypothesis now rides in to the rescue 7. (We haven't used it yet, and it has to come in somewhere.) x

F2 (x, y) − ∫ 0

∂F1

x

(X, y) dX = F2 (x, y) − ∫

∂y

0

∂F2

(X, y) dX

∂x X=x

∣ = F2 (x, y) − F2 (X, y) ∣X=0 = F2 (0, y)

In going from the first line to the second line we used the fundamental theorem of calculus. So choosing y

ψ(y) = ∫

F2 (0, Y ) dY + C

0

for any constant C , does the trick.

Exercises Stage 1  1 Evaluate ∫

C

2

x y

2

3

dx + x y dy

counterclockwise around the square with vertices (0, 0), (1, 0), (1, 1) and (0, 1).

 2 For each of the following fields, decide which of the following holds: ⇀

1. The characterization of conservative vector fields, Theorem 2.4.8 (with Theorem 2.3.9), tells us F is conservative. 2. The characterization of conservative vector fields, Theorem 2.4.8 (with Theorem 2.3.9), tells us F is not conservative. 3. The characterization of conservative vector fields, Theorem 2.4.8 (with Theorem 2.3.9), does not tell us whether F is conservative or not. ⇀

⇀

⇀

^ 1. F = x ^ ı ı + z^ ȷ ȷ + yk 2. F = y z ^ ı ı + x z^ ȷ ȷ +x ⇀

2

⇀

3. F = (y e

2

xy

2

^ yk

+ 1) ^ ı ı + (x e

xy

1 + z)^ ȷ ȷ +(

^ + y) k

z

⇀

4. F = y cos(xy) ^ ı ı + x sin(xy)^ ȷ ȷ

 3 ⇀

⇀

⇀

Let φ(x, y, z) = e + cos(z ), and define F = ∇φ. Evaluate ∫ F ⋅ d r over the closed curve C that is an ellipse – – traversed clockwise, centred at (1, 2, 3), passing through the points (√5 − 1, −2, √5 − 3), – – – – ((√5 − 2)/2, −1/2, (√5 − 6)/2), and (−2, √3 − 2, √3 − 3). 2

x +y

2

2

⇀

C

 4 Let P and P be points in R . Let A and B be paths from P to P 2

1

2

1

2.4.8

2,

as shown below.

https://math.libretexts.org/@go/page/91901

⇀

Suppose F is a conservative vector field in R with ∫ 2

A

⇀

⇀

F ⋅ d r = 5.

⇀

What is ∫

B

⇀

F ⋅ dr ?

 5✳ ⇀

Let F(x, y, z) = e closed paths C ?

x

sin y ^ ı ı + [ae

x

For which values of the constants

^ cos y + bz] ^ ȷ ȷ + cx k.

a, b, c

is

⇀

∫

C

⇀

F ⋅ dr = 0

for all

 6 Consider the four vector fields sketched below. Exactly one of those vector fields is conservative. Determine which three vector fields are not conservative and explain why.

(a)

(b)

(c)

(d)

 7✳ Consider the vector field x − 2y

⇀

F(x, y, z) =

2

x

+y

2

^ ı ı +

2x + y 2

x

+y

2

^ ^ ȷ ȷ +zk

⇀

1. Determine the domain of F. ⇀

⇀

2. Compute ∇ × F. Simplify the result. 3. Evaluate the line integral ⇀

∫

⇀

F ⋅ dr

C

where C is the circle of radius 2 in the plane z = 3, centered at (0, 0, 3) and traversed counter-clockwise if viewed from the positive z -axis, i.e. viewed “from above”. ⇀

4. Is F conservative?

 8 ⇀

⇀

^ Find the work, ∫ F ⋅ d r , done by the force field F = (x + y) ^ ı ı + (x − z)^ ȷ ȷ + (z − y)k in moving an object from to (0, −2, 3). Does the work done depend on the path used to get from (1, 0, −1) to (0, −2, 3)? ⇀

C

2.4.9

(1, 0, −1)

https://math.libretexts.org/@go/page/91901

Stage 2  9 Consider the vector field V(x, y) = (e

x

2

2

cos y + x , x y + 3)

Evaluate the line integral ∫ V ⋅ d r along the oriented curve C obtained by moving from (0, 0) to (1, 0) to (1, π) and finally to (0, π) along straight line segments. ⇀

C

 10 ⇀

Evaluate ∫

⇀

F ⋅ dr

C

for

⇀

1. F(x, y) = xy ^ ı ı −x ^ ȷ ȷ along y = x from (0, 0) to (1, 1). ^ 2. F(x, y, z) = (x − z) ^ ı ı + (y − z) ^ ȷ ȷ − (x + y) k along the polygonal path from (0, 0, 0) to (1, 0, 0) to (1, 1, 0) to (1, 1, 1). 2

2

⇀

 11 ✳ Let C be the part of the curve of intersection of Calculate

and

xyz = 8

⇀

∫

x = 2y

which lies between the points

(2, 1, 4)

and

(4, 2, 1).

⇀

F ⋅ dr

C

where ⇀

2

F =x

2 ^ ^ ı ı + (x − 2y) ^ ȷ ȷ +x y k

 12 ✳ ⇀

Let  F = e paths C ?

x

sin y ^ ı ı + [ae

x

^ cos y + bz] ^ ȷ ȷ + cx k.

For which values of the constants

is

a,  b,  c

⇀

 ∫

C

⇀

F ⋅ d r = 0 

for all closed

 13 ⇀

Let F = 6x

2

yz

2

3

^ ı ı + (2 x z

2

3 ^ + 2y − xz) ^ ȷ ȷ + 4 x yz k

^ and let G = yz ^ ı ı + xy k. ⇀

1. For what value of the constant λ is the vector field H = F + λG conservative on 3-space? 2. Find a scalar potential ϕ(x, y, z) for the conservative field H referred to in part (a). 3. Find ∫ F ⋅ d r if C is the curve of intersection of the two surfaces z = x and y = e from the point (0, 1, 0) to the point ⇀

⇀

xz

C

(1, e, 1).

 14 ✳ ⇀

Find the work done by the force field F(x, y, z) = (x − y from (0, 0, 1) to (2, 1, 0).

2

, y −z

2

2

, z−x )

on a particle that moves along the line segment

 15 ✳ ⇀

Let  F =

x x2 +y 2

^ ı ı +

y x2 +y 2

3

^ ȷ ȷ +x

^ k.

Let P be the path which starts at (1, 0, 0), ends at ( 2

x

+y

2

=1

xe

z

1 √2

,

1 √2

,

1 2

ln 2)

and follows

=1

⇀

Find the work done in moving a particle along P in the field F.

2.4.10

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 16 ✳ Let

⇀

F = (yz cos x , z sin x + 2yz , y sin x + y

Evaluate ∫

⇀

2

and let

− sin z)

C

be the line segment

⇀

r (t) = (t, t, t),

for

0 ≤ t ≤ π/2.

⇀

F ⋅ dr .

C

 17 ✳ Let C be the upper half of the unit circle centred on clockwise. Compute the line integral ∫ xy dy.

(1, 0)

(i.e. that part of the circle which lies above the x-axis), oriented

C

 18 ✳ Show that the following line integral is independent of path and evaluate the integral. ∫

(y e

x

+ sin y) dx + (e

x

+ sin y + x cos y) dy

C

where C is any path from (1, 0) to (0, π/2).

 19 ✳ Evaluate the integral ∫

xy dx + yz dy + zx dz

C

around the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1), oriented clockwise as seen from the point (1, 1, 1).

 20 ✳ Evaluate the line integral ∫

C

⇀

⇀

F ⋅ dr ,

⇀

where F is the conservative vector field ⇀

x

F(x, y, z) = (y + ze , x + e

y

sin z, z + e

x

+e

y

cos z)

and C is the curve given by the parametrization ⇀

t

r (t) = (t, e , sin t),

t from 0 to π.

 21 ✳ 1. For which values of the constants α, β and γ is the vector field ⇀

F(x, y, z) = α e

y

^ ı ı + (x e

conservative? 2. For those values of α, β and γ found in part (a), calculate ∫

C

y

⇀

^ + β cos z) ^ ȷ ȷ − γy sin z k

⇀

F ⋅ dr ,

where C is the curve parametrized by x = t

2

t

, y =e ,

z = πt, 0 ≤ t ≤ 1.

 22 ✳ ⇀

Consider the vector field F(x, y, z) = (cos x, 2 + sin y, e

z

).

⇀

1. Compute the curl of F. 2. Is there a function f such that F = ∇f ? Justify your answer. 3. Compute the integral ∫ F ⋅ d r along the curve C parametrized by ⇀

⇀

⇀

⇀

C

2.4.11

⇀

r (t) = (t, cos t, sin t)

with 0 ≤ t ≤ 3π.

https://math.libretexts.org/@go/page/91901

 23 ✳ 1. Consider the vector field ⇀

y

F(x, y, z) = (z + e , x e ⇀

y

−e

z

sin y, 1 + x + e

z

cos y)

⇀

Find the curl of F. Is F conservative? 2. Find the integral ∫ F ⋅ d r of the field F from (a) where C is the curve with parametrization ⇀

⇀

⇀

C

⇀

2

2

r (t) = (t , sin t, cos

t)

where t ranges from 0 to π.

 24 ✳ ⇀

⇀

A physicist studies a vector field F. From experiments, it is known that F is of the form ⇀

F = (x − a)y e

x

^ ı ı + (x e

x

3

+z )^ ȷ ȷ + by z

2

^ k ⇀

where a and b are some real numbers. From theoretical considerations, it is known that F is conservative. 1. Determine a and b. ⇀

⇀

2. Find a potential f (x, y, z) such that ∇f = F. 3. Evaluate the line intgeral ∫ F ⋅ d r where C is the curve defined by ⇀

⇀

C

⇀

r (t) = (t , cos 2t , cos t),

0 ≤t ≤π

4. Evaluate the line integral I =∫

(x + 1)y e

x

dx + (x e

x

3

+ z ) dy + 4y z

2

dz,

C

where C is the same curve as in part (c). [Note: the “4” in the last term is not a misprint!]. Questions 2.4.2.25 and 2.4.2.26 ask you to evaluate line integrals of vector fields that are not conservative, but that can be expressed as a sum of a conservative vector field and another vector field that can be written concisely.

 25 ✳ Let ⇀

2

F = (y e

3z

3

+ Ax y ) ^ ı ı + (2xy e

3z

2

2

2

+ 3x y ) ^ ȷ ȷ + Bx y e

3z

^ k

⇀

1. Find all values of A and B for which the vector field F is conservative. ⇀

2. If A and B have values found in (a), find a potential function for F. ^ ^ 3. Let C be the curve with parametrization r (t) = e ^ ı ı +e ȷ ȷ + ln(1 + t) k from (1, 1, 0) to (e ⇀

∫

2

(y e

3z

2t

−t

3

+ x y ) dx + (2xy e

3z

2

2

2

+ 3 x y ) dy + 3x y e

3z

2

,

1 e

, ln 2).

Evaluate

dz.

C

 26 ✳ 1. For which value(s) of the constants a, b is the vector field ⇀

z

2

F = (2x sin(πy) − e ) ^ ı ı + (ax

z z^ cos(πy) − 3 e )^ ȷ ȷ − (x + by)e k

conservative? 2. Let F be a conservative field from part (a). Find all functions φ for which F = ∇φ. ⇀

⇀

2.4.12

⇀

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⇀

⇀

3. Let F be a conservative field from part (a). Evaluate ∫ from (0, 0, 0) to (1, 1, ln 2). 4. Evaluate ∫ G ⋅ d r where

⇀

F ⋅ dr

C

where C is the intersection of y = x and z = ln(1 + x)

⇀

C

z

2

G = (2x sin(πy) − e ) ^ ı ı + (π x

z

cos(πy) − 3 e ) ^ ȷ ȷ − xe

z

^ k

and C is the intersection of y = x and z = ln(1 + x) from (0, 0, 0) to (1, 1, ln 2).

 27 ✳ Consider the vector field ⇀

2

F(x, y, z) = −2y cos x sin x ^ ı ı + (cos ⇀

x + (1 + yz)e

yz

2

)^ ȷ ȷ +y e

yz

^ k

⇀

1. Find a real valued function f (x, y, z) such that F = ∇f . 2. Evaluate the line integral ⇀

∫

⇀

F ⋅ dr

C

where C is the arc of the curve

⇀

t

2

r (t) = (t, e , t

, 0 ≤ t ≤ π, traversed from (0, 1, −π

2

−π )

2

)

to (π, e

π

, 0).

 28 ✳ ⇀

^ Consider the vector field F(x, y, z) = 2x ^ ı ı + 2y ^ ȷ ȷ + 2z k. ⇀

⇀

1. Compute ∇ × F. 2. If C is any path from (0, 0, 0) to (a

1,

a2 , a3 )

and a = a

1

^ ^ ı ı + a2 ^ ȷ ȷ + a3 k,

show that ∫

C

⇀

⇀

F ⋅ d r = a ⋅ a.

 29 ✳ Let C be the parameterized curve given by π

⇀

r (t) = ( cos t, sin t, t),

0 ≤t ≤ 2

and let ⇀

F = (e ⇀

yz

, xze

yz

+ ze

y

, xy e

yz

y

+e )

⇀

1. Compute and simplify ∇ × F. 2. Compute the work integral ∫ F ⋅ d r . ⇀

⇀

C

 30 ✳ 1. Show that the planar vector field ⇀

2

2

F(x, y) = (2xy cos(x ) , sin(x ) − sin(y))

is conservative. 2. Find a potential function for F. ⇀

⇀

3. For the vector field F from above compute ∫

C

⇀

⇀

F ⋅ dr ,

where C is the part of the graph x = sin(y) from y = π/2 to y = π.

 31 ✳ Consider the following force field, in which m, n, p, q are constants: ⇀

F = (mxyz + z

2

2 2 2 3 ^ − ny ) ^ ı ı + (x z − 4xy) ^ ȷ ȷ + (x y + pxz + q z ) k

2.4.13

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⇀

1. Find all values of m, n, p, q such that ∮

C

⇀

F ⋅ dr = 0

for all piecewise smooth closed curves C in R

3

.

⇀

2. For every possible choice of m, n, p, q in (a), find the work done by F in moving a particle from the bottom to the top of ^ the sphere x + y + z = 2z. (The direction of k defines “up”.) 2

2

2

Stage 3  32 Let C be the curve from (0, 0, 0) to (1, 1, 1) along the intersection of the surfaces y = x and z = x 2

1. Find ∫ 2. Find ∫

C

.

if s is arc length along C and ρ = 8x + 36z. ^ F ⋅ d r if F = sin y ^ ı ı + (x cos y + z) ^ ȷ ȷ + (y + z) k. ρ ds ⇀

C

3

⇀

⇀

 33 ✳ The curve C is the helix that winds around the cylinder x + y = 1 (counterclockwise, as viewed from the positive z -axis, looking down on the xy-plane). It starts at the point (1, 0, 0), winds around the cylinder once, and ends at the point (1, 0, 1). Compute the line integral of the vector field 2

2

⇀

2

F(x, y, z) = (−y, x, z )

along C .

 34 ✳ Evaluate the line integrals below. (Use any method you like.) 1. ∫

C

2

(x

+ y) dx + x dy,

⇀

where C is the arc of the parabola y = 9 − x from (−3, 0) to (3, 0). 2

⇀

^ ds, where F(x, y) = 2 x ^ 2. ∫ F ⋅ n ı ı + ye ^ ȷ ȷ , C is the boundary of the square 0 ≤ x ≤ 1, normal vector pointing outward from the square, and s is arc length. 2

x

C

0 ≤ y ≤ 1.

^ is the unit Here n

 35 ✳ A particle of mass ⇀

^ F(t) = ^ ȷ ȷ − sin t k,

has position r where t denotes time.

⇀

m =1

0

=^ ȷ ȷ

and velocity

⇀

^ v0 = ^ ı ı +k

at time

t = 0.

The particle moves under a force

1. Find the position r (t) of the particle as a function of t. 2. Find the position r of the particle when it crosses the plane x = π/2 for the first time after time t = 0. 3. Determine the work done by F in moving the particle from r to r . ⇀

⇀ 1

⇀

⇀ 0

⇀ 1

Questions 2.4.2.36 and 2.4.2.37 ask you to find a path that leads to a particular value of a line integral. Many such paths are possible — you only need to find one.

 36 ✳ ⇀

1. Consider the vector field F(x, y) = (3y, x − 1) in R . Compute the line integral 2

⇀

∫

⇀

F ⋅ dr

L

where L is the line segment from (2, 2) to (1, 1). 2. Find an oriented path C from (2, 2) to (1, 1) such that ⇀

∫

⇀

F ⋅ dr = 4

C ⇀

where F is the vector field from (a).

2.4.14

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 37 ✳ ⇀

⇀

Let F = (2y + 2) ^ ı ı be a vector field on R . Find an oriented curve C from (0, 0) to (2, 0) such that ∫ 2

C

⇀

F ⋅ d r = 8.

 38 ✳ Let ⇀

F(x, y) = (1, yg(y))

and suppose that g(y) is a function defined everywhere with everywhere continuous partials. Show that for any curve C whose endpoints P and Q lie on the x-axis, ∣ ⇀ ∣ ⇀ distance between P  and Q = ∣∫ F ⋅ dr ∣ ∣ C ∣

 39 ✳ Let S be the surface z = 2 + x − 3y and P = (0, 0, 2) on the surface S. 2

and let

2

⇀

2 2^ F(x, y, z) = (xz + ax y ) ^ ı ı + yz^ ȷ ȷ + z k.

Consider the points

P1 = (1, 1, 0)

2

⇀

Find a value of the constant a so that ∫

⇀

⇀

F ⋅ dr = ∫

C1

C2

⇀

F ⋅ dr

for any two curves C and C on the surface S from P to P 1

2

1

2.

 40 ✳ ⇀

Consider the vector field F defined as ⇀

2

2

F(x, y, z) = ((1 + ax )y e

3x

2

2

− bxz cos(x z) , x e

3x

2

, x

2

cos(x z))

where a and b are real valued constants. ⇀

⇀

1. Compute ∇ × F. ⇀

2. Determine for which values a and b the vector field F is conservative. 3. For the values of a and b obtained in part (b), find a potential function f such that ∇f 4. Evaluate the line integral ⇀

2

∫

(y e

3x

2

2

+ 2xz cos(x z))dx + x e

3x

2

dy + x

⇀

= F.

2

cos(x z)dz

C

where C is the arc of the curve (t, t, t

3

)

starting at the point (0, 0, 0) and ending at the point (1, 1, 1).

 41 ✳ Let C be the curve from (0, 0, 0) to (1, 1, 1) along the intersection of the surfaces y = x and z = x 2

1. Find ∫ 2. Find ∫ 3. Find ∫

C C

C

⇀

3

.

⇀

^ if F = (xz − y) ^ ı ı + (z + x) ^ ȷ ȷ + y k. ρ ds if s is arc length along C and ρ(x, y, z) = 8x + 36z. ^ F ⋅ d r if F = sin y ^ ı ı + (x cos y + z) ^ ȷ ȷ + (y + z) k. ⇀

F ⋅ dr

⇀

⇀

⇀

 42 ✳ ⇀

The vector field F(x, y, z) = Ax

3

2

y z ^ ı ı + (z

3

4

+ Bx yz) ^ ȷ ȷ + (3y z

2

1. Find the values of the constants A and B. 2. Find a potential φ such that F = ∇φ on R . 3. If C is the curve y = −x,  z = x from (0, 0, 0) to (1, −1, 1), evaluate I ⇀

⇀

is conservative on R

4 2 ^ −x y )k

3

.

3

2

2.4.15

⇀

=∫

C

⇀

F ⋅ dr .

https://math.libretexts.org/@go/page/91901

4. Evaluate J = ∫ (z − 4x y z)dx + (z − x yz)dy + (3y z − x y )dz, where C is the curve of part (c). 5. Let T be the closed triangular path with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1), oriented counterclockwise as seen from the point (1, 1, 1). Evaluate ∫ (z ^ ı ı + F) ⋅ d r . 3

2

3

4

2

4

2

C

⇀

⇀

T

 43 ✳ A particle of mass m =2

is acted on by a force ⇀

2

F = (4t , 6 t

, −4t)

At t = 0, the particle has velocity zero and is located at the point (1, 2, 3). 1. Find the velocity vector v (t) for t ≥ 0. 2. Find the position vector r (t) for t ≥ 0. 3. Find κ(t) the curvature of the path traversed by the particle for t ≥ 0. 4. Find the work done by the force on the particle from t = 0 to t = T . ⇀

⇀

 44 ✳ The position of an airplane at time t is given by x = y =

4 √2 3

3/2

t

,  z = t(2 − t)

from take-off at t = 0 to landing at t = 2.

1. What is the total distance the plane travels on this flight? 2. Find the radius of curvature κ at the apex of the flight, which occurs at t = 1. ^ 3. Two external forces are applied to the plane during the flight: the force of gravity G = −M g k , where M is the mass of the plane and g is a constant; and a friction force F = −| v | v , where v is the velocity of the plane. Find the work done by each of these forces during the flight. 4. One half-hour later, a bird follows the exact same flight — path as the plane, travelling at a constant speed v = 3. One can ⇀

⇀ 2⇀

⇀

4 √2

4 √2

3

3

^ show that at the apex of the path, i.e. when the bird is at ( , , 1), the principal unit normal N to the path points in ^ the −k direction. Find the bird's (vector) acceleration at that moment.

1. Yes, yes. We should first consider short time intervals Δt > 0 and then take the limit Δt → 0 at the end. But you have undoubtedly used this type of argument so many times before that you would be thoroughly bored by it. 2. So φ acts a bit like the antiderivative of first year calculus. 3. The reader should check that this vector field is not conservative. 4. You might like to think about why we can split up the integral like this. 5. Again, the reader should verify that this vector field is not conservative. 6. This is a pretty efficient, and standard, way to structure the proof of the equivalence of three statements. 7. or bails us out, or saves our bacon, or … This page titled 2.4: Line Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

2.4.16

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2.5: Optional — The Pendulum Model a pendulum by a mass m that is connected to a hinge by an idealized rod that is massless and of fixed length ℓ. Denote by θ the angle

between the rod and vertical. The forces acting on the mass are gravity, which has magnitude mg and direction (0, −1), tension in the rod, whose magnitude, τ , automatically adjusts itself so that the distance between the mass and the hinge is fixed at ℓ and whose direction, (− sin θ, cos θ), is always parallel to the rod and possibly some frictional forces, like friction in the hinge and air resistance. We shall assume that the total frictional force has magnitude proportional to the speed 1 of the mass and has direction opposite to the direction of motion of the mass. If we use a coordinate system centered on the hinge, the (x, y) coordinates of the mass at time t are ℓ( sin θ(t), − cos θ(t)). Hence its velocity vector is d 

⇀

v (t) =

dθ [ℓ( sin θ(t), − cos θ(t))] = ℓ( cos θ(t), sin θ(t))

dt

(t) dt

and the total frictional force is −βℓ(cos θ, sin θ)

dθ ,

for some constant β. The acceleration vector of the mass is

dt d  a(t) =

2

d θ

⇀

v (t) = ℓ(cos θ, sin θ)

dt

2

dθ + ℓ(− sin θ, cos θ)(

2

) dt

dt

⇀

so that Newton's law of motion, F = ma, now tells us 2

d θ ma(t) = mℓ(cos θ, sin θ)

dθ + mℓ(− sin θ, cos θ)(

2

dt

2

) dt dθ

⇀

= F = mg(0, −1) + τ (− sin θ, cos θ) − βℓ(cos θ, sin θ) dt

To eliminate the (unknown) coefficient

τ

we dot this equation with

direction of motion of the mass. Dotting with (cos θ, sin θ) gives  mℓ 2

d θ

β

dθ

+

2

which extracts the component parallel to the dθ

2

2

dt

= −mg sin θ − βℓ

  dt

or

g +

m dt

dt

(cos θ, sin θ), d θ

sin θ = 0 ℓ

which is the equation of motion of the (nonlinear) pendulum. In general, it can be hard to analyse nonlinear differential equations. But if the amplitude of oscillation is small enough that we may approximate sin θ by θ we get the equation of motion of the linear pendulum 2 which is 2

d θ 2

dt

β

g

dθ

+

+ m dt

θ =0 ℓ

These equations may be reformulated as systems of first order ordinary differential equation, that is as equations for the flow lines of a vector field, by the simple expedient of defining (as we did in Example 2.1.7) x(t) = θ(t)

′

y(t) = θ (t)

2.5.1

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Then, for the full, nonlinear, equation

2

d θ 2

dt

+

β m

dθ + dt

g ℓ

sin θ = 0

′

′

′

′′

x (t) = θ (t) = y(t) g

y (t) =  θ (t) = −

β sin x(t) −

ℓ

y(t) m

The solutions of this first order system of ordinary differential equations are flow lines for the vector field g V((x, y)) = (y , −

β sin x −

ℓ

y) m

When β = 0, this is exactly the vector field of Example 2.1.7.

1. The dependence of air resistance (drag) on the speed v is relatively complex. At low speed drag tends to be approximately proportional to v, while at high speed it tends to be approximately proportional to v . 2

2. When β = 0, this equation reduces to the equation solutions exhibit simple harmonic motion.

2

d θ 2

dt

+

g ℓ

θ = 0,

which occurs in many different applications, and whose

This page titled 2.5: Optional — The Pendulum is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

2.5.2

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CHAPTER OVERVIEW 3: Surface Integrals 3.1: Parametrized Surfaces 3.2: Tangent Planes 3.3: Surface Integrals 3.4: Interpretation of Flux Integrals 3.5: Orientation of Surfaces Thumbnail: The total flux through the surface is found by adding up for each patch. In the limit as the patches become infinitesimally small, this is the surface integral. (CC0; Chetvorno via Wikipedia) This page titled 3: Surface Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1

3.1: Parametrized Surfaces For many applications we will need to use integrals over surfaces. One obvious one is just computing surface areas. Another is computing the rate at which fluid traverses a surface. The first step is to simply specify surfaces carefully. There are three common ways to specify a surface in three dimensions. 1. Graph of a function: Probably the most common way to specify a surface is to give its equation in the form 2

z = f (x, y)

(x, y) ∈ D ⊂ R

Here “(x, y) ∈ D ⊂ R ” just means that (x, y) runs over the subset D of R . For example, if the surface is the top half of the sphere of radius one centred on the origin 2

2

− −−−−−−− − 2

z = √1 −x

−y

2

2

with x

+y

2

≤1

2. Implicitly: We can also specify that the surface is the set of points (x, y, z) that satisfy the equation G(x, y, z) = 0, or, more generally 1, satisfy the equation G(x, y, z) = K, with K a constant. For example, the sphere of radius one centred on the origin is the set of points that obey 2

x

+y

2

+z

2

=1

We shall explore this surface a little more in Example 3.1.2 below. 3. Range of a function: Probably the most useful way to specify a surface, when one needs to integrate over the surface, is as the range of a function ⇀

2

r : D ⊂R

3

→ R

⇀

(u, v) ∈ D ↦ r (u, v) = (x(u, v) , y(u, v) , z(u, v))

The upper line means that r is a function which is defined on the subset D of R and which assigns to each point on D a point in R . The second line means that the function r assigns to the element (u, v) of D the element r (u, v) = (x(u, v) , y(u, v) , z(u, v)) in R . Such a surface is called a parametrized surface — each point of the surface is labelled by the values of the two parameters u and v. Parametrized surfaces are of course the two parameter analog of parametrized curves. Examples of parametrized surfaces come next. ⇀

2

3

⇀

⇀

3

 Example 3.1.1 One simple, even trivial, way to parametrize the surface which is the graph 2

z = f (x, y)

(x, y) ∈ D ⊂ R

is to choose x and y as the parameters. That is, to choose ⇀

r (u, v) = (u, v, f (u, v)),

or

⇀

r (x, y) = (x, y, f (x, y)),

(u, v) ∈ D (x, y) ∈ D

Let's do something a bit more substantial.

 Example 3.1.2. Sphere The sphere of radius 1 centred on the origin is the set of points (x, y, z) that obey 2

G(x, y, z) = x

+y

2

+z

2

=1

We cannot express this surface as the graph of a function because, for each (x, y) with x x + y + z = 1, namely

2

2

2

+y

2

< 1,

there are two z 's that obey

2

− −−−−−−− − 2

z = ±√ 1 − x

3.1.1

−y

2

https://math.libretexts.org/@go/page/91904

On the other hand, locally, this surface is the graph of a function. This means that, for any point (x , y , z ) on the sphere, all points of the surface that are sufficiently near (x , y , z ) can be expressed in one of the forms z = f (x, y) or x = g(y, z), or – y = h(x, z). For example, the part of the sphere that is within a distance √2 of the point (0, 0, 1) is 0

0

2

{(x, y, z)| x

0

+y 2

= {(x, y, z)| x

= {(x, y, z)| x = {(x, y, z)| x

+z 2

+y

2

2

2

+y

+y

0

0

0

– = 1,  |(x, y, z) − (0, 0, 1)| < √2} 2

+z

2

2

2

+z

+z

2

= 1,  x

2

2

+y

2

= 1,  x

2

+y

2

2

+ (z − 1 ) +z

2

< 2}

− 2z + 1 < 2}

= 1,  z > 0}

− −−−−−−− − 2

= {(x, y, z)|z = √ 1 − x

−y

2

2

,  x

+y

2

< 1}

This is illustrated in the figure below which shows the y = 0 section of the sphere x + y + z = 1 and also the – section of the set of points that are within a distance √2 of (0, 0, 1). (They are the points inside the dashed circle.) 2

2

2

Similarly, as illustrated schematically in the figure below, the part of the sphere that is within a distance (1, 0, 0) is 2

{(x, y, z)| x

+y 2

= {(x, y, z)| x

2

= {(x, y, z)| x

+z

+y

2

= {(x, y, z)| x

2

2

+y

+y

of the point

– = 1,  |(x, y, z) − (1, 0, 0)| < √2}

+z

2

2

2

– √2

y =0

2

+z

+z

2

= 1,  (x − 1 )

2

2

2

= 1,  x

+y

2

+z

− 2x + 1 + y

2

2

< 2}

+z

2

< 2}

= 1,  x > 0}

− −−−−−−− − = {(x, y, z)|x = √ 1 − y

2

−z

2

,  y

2

+z

2

< 1}

The figure below shows the y = 0 section of the sphere x + y + z = 1 and also the y = 0 section of the set of points that – are within a distance √2 of (1, 0, 0). (Again, they are the points inside the dashed circle.) 2

2

2

We can parametrize the unit sphere by using spherical coordinates, which you should have seen before. As a reminder, here is a figure showing the definitions of the three spherical coordinates 2 ρ =  distance from (x, y, z) to (0, 0, 0) ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯

φ =  angle between the line (0, 0, 0) (x, y, z) and the z axis ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯

θ =  angle between the line (0, 0, 0) (x, y, 0) and the x axis

3.1.2

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and here are two more figures giving the side and top views of the previous figure.

From the figure, we see that Cartesian and spherical coordinates are related by x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ

The points on the sphere x

2

+y

2

+z

2

=1

are precisely the set of points with ρ = 1. So we can use the parametrization

⇀

r (θ, φ) = ( sin φ cos θ , sin φ sin θ , cos φ)

Here is how to see that as φ runs over (0, π) and θ runs over [0, 2π), r (θ, φ) covers the whole sphere x + y + z = 1 except for the north pole (φ = 0 gives the north pole for all values of θ ) and the south pole (φ = π gives the south pole for all values of θ ). ⇀

2

2

2

Fix θ and have φ run over the interval 0 < φ ≤ . Then r (θ, φ) traces out one quarter of a circle starting at the north pole r (θ, 0) = (0, 0, 1) (but excluding the north pole itself) and ending at the point r (θ, ) = (cos θ, sin θ, 0) in the xy-plane. π

⇀

2

⇀

⇀

π 2

Keep θ fixed at the same value and extend the interval over which φ runs to 0 < φ < π. Now r (θ, φ) traces out a semicircle starting at the north pole r (θ, 0) = (0, 0, 1), ending at the south pole r (θ, π) = (0, 0, −1) (but excluding both the north and south poles themselves) and passing through the point r (θ, ) = (cos θ, sin θ, 0) in the xy-plane. ⇀

⇀

⇀

⇀

π 2

3.1.3

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Finally have θ run over 0 ≤ θ < 2π. Then the semicircle rotates about the z -axis, sweeping out the full sphere, except for the north and south poles. Recall that φ is the angle between the radius vector and the z -axis. If you hold φ fixed and increase θ by a small amount dθ, r (θ, φ) sweeps out the red circular arc in the figure on the left below. If you hold φ fixed and vary θ from 0 to 2π, r (θ, φ) sweeps out a line of latitude. The figure on the right below gives the lines of latitude (or at least the parts of those lines in the first octant) for φ = , , , and = . ⇀

⇀

π

2π

3π

4π

5π

π

10

10

10

10

10

2

On the other hand, if you hold θ fixed and increase φ by a small amount dφ, r (θ, φ) sweeps out the red circular arc in the figure on the left below. If you hold θ fixed and vary φ from 0 to π, r (θ, φ) sweeps out a line of longitude. The figure on the right below gives the lines of longitude (or at least the parts of those lines in the first octant) for θ = 0, , , , and ⇀

⇀

5π 10

=

π 2

π

2π

3π

4π

10

10

10

10

.

3.1.4

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 Example 3.1.3. Cylinder The surface x

2

+z

2

=1

is an infinite cylinder. Part of this cylinder in the first octant is sketched below.

Note that the section of this cylinder that lies in the xz-plane, and in fact in any plane y = c, is the circle x + z = 1. We can of course parametrize this circle by x = cos θ, z = sin θ. So we can parametrize the whole cylinder by using θ and y as parameters. 2

⇀

r (θ, y) = ( cos θ , y , sin θ)

2

0 ≤ θ < 2π,    − ∞ < y < ∞

 Example 3.1.4. Surface of Revolution In this example, we are going to parametrize a surface of revolution. In your first integral calculus course, you undoubtedly encountered many surfaces created by rotating a curve y = f (x) about the x-axis or the y -axis. In this course, we are used to having the z -axis, rather than the y -axis, run vertically. So in this example, we'll parametrize the surface constructed by rotating the curve z = g(y) = e

y

0 ≤y ≤1

about the z -axis. Exactly the same procedure can be used to parametrize surfaces created by rotating about the x-axis or the y axis too. We start by just sketching the curve, considering the yz-plane as the plane x = 0 in R . The specified curve is the red curve in the figure below. Concentrate on any one point on that curve. It is the blue dot at (0, Y , e ) 3

Y

in the figure. When our curve is rotated about the z -axis, the blue dot sweeps out a circle. The circle that the blue dot sweeps out lies in the horizontal plane z = e is centred on the z -axis and has radius Y .

Y

and

We can parametrize the circle swept out in the usual way. Here is a top view of the circle, with the parameter, named indicated.

3.1.5

θ,

https://math.libretexts.org/@go/page/91904

The coordinates of the red dot are (Y

sin θ , Y cos θ , e

Y

This also gives a parametrization of the surface of revolution

).

x(Y , θ) = Y sin θ y(Y , θ) = Y cos θ z(Y , θ) = e

Y

0 ≤ Y ≤ 1,

0 ≤ θ < 2π

Notice, by way of checks, that when θ = 0, (x(Y , 0) , y(Y , 0) , z(Y , 0)) = (0, Y , e

Y

)

runs over the entire desired curve (namely z = g(y), 0 ≤ y ≤ 1 ), when Y runs over 0 ≤ Y ≤ 1 and for any fixed 0 ≤ Y ≤ 1, (x(Y , θ) , y(Y , θ) , z(Y , θ)) runs over the circle x + y = Y , in the plane z = e runs over 0 ≤ θ < 2π. 2

2

2

Y

,

when θ

Also notice that 2

x(Y , θ)

2

+ y(Y , θ)

=Y

2

so that − −−−−−−−−−−−−− − 2

Y = √ x(Y , θ)

2

+ y(Y , θ)

and 2

z(Y , θ) = e

Y

=e

2

√x(Y ,θ) +y(Y ,θ)

That is, the surface of revolution is contained in the (infinite) surface z =e

Remembering that 0 ≤ Y

≤ 1,

we have that 1 ≤ z = e

Y

≤ e. 2

z =e

√x +y

√x2 +y 2

Thus the surface of revolution is

2

1 ≤z ≤e

Finally here is a sketch of the part of the surface in the first octant, x, y, z ≥ 0.

3.1.6

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 Example 3.1.5. Torus In this example, we are going to parametrize a donut (well, its surface), or an inner tube.

The formal mathematical name for the surface of a donut is a torus. Our strategy will be to first parametrize the section of the torus in the right half of the yz-plane, and then built up the full torus by rotating the circle about the z -axis. The section is a circle, sketched below.

We'll assume that the centre of the circle is a distance R from the z -axis, and that the circle has radius the circle is at

r.

Then the red dot on

x =0 y = R + r cos θ z = r sin θ

In particular the red dot is a distance r sin θ above the xy-plane and is a distance R + r cos θ from the z -axis. So when we rotate the section about the z -axis, the red dot sweeps out a circle which is sketched below.

The circle that the red dot sweeps out lies in the plane z = r sin θ and is centred on the z -axis and has radius ρ = R + r cos θ. We can parametrize the circle swept out in the usual way. Here is a top view of the circle, with the parameter, named indicated.

3.1.7

ψ,

https://math.libretexts.org/@go/page/91904

So the parametrization of the circle swept out by the red dot, and also the parametrization of the torus, is x = ρ cos ψ = (R + r cos θ) cos ψ y = ρ sin ψ = (R + r cos θ) sin ψ z = r sin θ

or ⇀

^ r (θ, ψ) = (R + r cos θ) cos ψ  ^ ı ı + (R + r cos θ) sin ψ ^ ȷ ȷ + r sin θ k

0 ≤ θ, ψ < 2π

Exercises Stage 1  1 Parametrize the surface given by z = e

x+1

+ xy

in terms of x and y.

 2✳ Let S be the surface given by ⇀

2

r (u, v) = (u + v , u

2

+v

, u − v),

−2 ≤ u ≤ 2,   − 2 ≤ v ≤ 2

This is a surface you are familiar with. What surface is it (it may be just a portion of one of the following)? sphere / helicoid / ellipsoid / saddle / parabolic bowl / cylinder / cone / plane

Stage 2  3✳ Suppose S is the part of the hyperboloid z = 1 (i.e. for which z ≥ 1 ).

2

x

+y

2

− 2z

2

=1

that lies inside the cylinder

2

x

+y

2

=9

and above the plane

Which of the following are parameterizations of S? 1. The vector function − −−−−−−− − √ u2 + v2 − 1

⇀

r (u, v) = u ^ ı ı +v^ ȷ ȷ +

with domain D = {(u, v)|2 ≤ u 2. The vector function

2

2

+v

– √2

^ k

≤ 9} .

− − − − − − − 2 u 1 ^ − k 2 2

⇀

r (u, v) = u sin v ^ ı ı − u cos v ^ ȷ ȷ +√

–

with domain D = {(u, v)|√3 ≤ u ≤ 3,  0 ≤ v ≤ 2π} .

3.1.8

https://math.libretexts.org/@go/page/91904

3. The vector function − − − − − − − − − − − − 2 2 ^ r (u, v) = √ 1 + 2v cos u ^ ı ı + √ 1 + 2v sin u ^ ȷ ȷ +vk

⇀

with domain D = {(u, v)|0 ≤ u ≤ 2π,  1 ≤ v ≤ 2} . 4. The vector function − − u − − − − − − − − − − ^ r (u, v) = √ 1 + u sin v ^ ı ı + √ 1 + u cos v ^ ȷ ȷ +√ k 2

⇀

with domain D = {(u, v)|2 ≤ u ≤ 8,  0 ≤ v ≤ 2π} . 5. The vector function − − − − − √u + 1 − − ^ r (u, v) = √u cos v ^ ı ı − √u sin v ^ ȷ ȷ + k – √2

⇀

with domain D = {(u, v)|3 ≤ u ≤ 9,  0 ≤ v ≤ 2π} .

 4✳ Suppose the surface S is the part of the sphere x + y + z z ≥ 0. Which of the following are parameterizations of S? 2

1.

2

2

=2

that lies inside the cylinder

2

x

+y

2

=1

and for which

⇀

^ r (ϕ, θ) = 2 sin ϕ cos θ ^ ı ı + 2 cos ϕ ^ ȷ ȷ + 2 sin ϕ sin θ k π 0 ≤ϕ ≤

,  0 ≤ θ ≤ 2π 4 − −−−−−−− −

2.

⇀

2

r (x, y) = x ^ ı ı −y ^ ȷ ȷ +√2 −x

2

x

+y

2

−y

2

^ k

≤1

− −−− − 2 ^ r (u, θ) = u sin θ ^ ı ı + u cos θ ^ ȷ ȷ + √2 − u k

3.

⇀

0 ≤ u ≤ 2,  0 ≤ θ ≤ 2π – – – ^ r (ϕ, θ) = √2 sin ϕ cos θ ^ ı ı + √2 sin ϕ sin θ ^ ȷ ȷ + √2 cos ϕ k

4.

⇀

π 0 ≤ϕ ≤

,  0 ≤ θ ≤ 2π 4

− − − − − − − − − − 2 2 ^ r (ϕ, z) = −√ 2 − z sin ϕ ^ ı ı + √ 2 − z cos ϕ ^ ȷ ȷ +zk

5.

⇀

– 0 ≤ ϕ ≤ 2π,  1 ≤ z ≤ √2

 5✳ Let S be the part of the paraboloid z + x + y = 4 lying between the planes indicate whether or not it correctly parameterizes the surface S. 2

1. 2. 3.

⇀

2

r (u, v) = u ^ ı ı +v^ ȷ ȷ + (4 − u

2 ^ − v ) k,

2

2

0 ≤u

2

+v

^ r (u, v) = (u cos v) ^ ı ı + (u sin v) ^ ȷ ȷ + (4 − u ) k, 2

and

z = 1.

For each of the following,

≤1

− − − − − − − − − − ^ r (u, v) = (√4 − u cos v) ^ ı ı + (√4 − u sin v) ^ ȷ ȷ + u k,

⇀

⇀

z =0

0 ≤ u ≤ 1, 0 ≤ v ≤ 2π

– √3 ≤ u ≤ 2, 0 ≤ v ≤ 2π

Stage 3  6✳ Consider the following surfaces S1

is the hemisphere given by the equation x

2

+y

2

+z

2

=4

3.1.9

with z ≥ 0.

https://math.libretexts.org/@go/page/91904

S2 S3

is the cylinder given by the equation x + y = 1. is the cone given by the equation z = x + y with z ≥ 0. 2

2

2

2

2

Consider the following parameterizations: – – – r (θ, ϕ) = (√4 cos θ sin ϕ , √4 sin θ sin ϕ , √4 cos ϕ)

1.

⇀

0 ≤ θ ≤ 2π,

0 ≤ ϕ ≤ π/6

– – – r (θ, ϕ) = (√4 cos θ sin ϕ , √4 sin θ sin ϕ , √4 cos ϕ)

2.

⇀

3.

⇀

0 ≤ θ ≤ 2π,

0 ≤ ϕ ≤ π/4

– – – r (θ, ϕ) = (√4 cos θ sin ϕ , √4 sin θ sin ϕ , √4 cos ϕ)

0 ≤ θ ≤ 2π,

0 ≤ ϕ ≤ π/3

− − − − − − − − − − 2 2 r (θ, z) = (√ 4 − z cos θ , √ 4 − z sin θ , z)

4.

⇀

5.

⇀

0 ≤ θ ≤ 2π,

1 ≤z ≤2

− − − − − − − − − − 2 2 r (θ, z) = (√ 4 − z cos θ , √ 4 − z sin θ , z)

0 ≤ θ ≤ 2π,

– √2 ≤ z ≤ 2

− − − − − − − − − − 2 2 r (θ, z) = (√ 4 − z cos θ , √ 4 − z sin θ , z)

6.

⇀

0 ≤ θ ≤ 2π,

7.

⇀

8.

⇀

– √3 ≤ z ≤ 2

r (θ, z) = (z cos θ , z sin θ , z)

0 ≤ θ ≤ 2π,

0 ≤z ≤1

r (θ, z) = (z cos θ , z sin θ , z)

0 ≤ θ ≤ 2π,

9.

– 0 ≤ z ≤ √2

⇀

r (θ, z) = (z cos θ , z sin θ , z)

0 ≤ θ ≤ 2π,

– 0 ≤ z ≤ √3 − −− −− −

10.

⇀

2

r (x, y) = (x , y , √ x 2

x

+y

2

+y

2

)

≤1 − −− −− −

11.

⇀

2

r (x, y) = (x , y , √ x 2

x

+y

2

+y

2

)

– ≤ √2 − −− −− −

12.

⇀

2

r (x, y) = (x , y , √ x 2

x

+y

2

+y

2

)

≤2

For each of the following, choose from above all of the valid parameterization of each of the given surfaces. Note that there may be one or more valid parameterization for each surface, and not necessarily all of the above parameterizations will be used. 1. The part of S 2. The part of S 3. The part of S 4. The part of S

1 1 3 3

contained inside S contained inside S contained inside S contained inside S

2: 3: 2: 1:

 7 Parametrize a solid of rotation about a line not parallel to an axis. Maybe first show that the plane you're rotating is normal to that axis. 1. Give a parametric equation for the circle of radius 1, centred at (2, 2, 4), lying in the plane x = y. 2. Give a parametrized equation for the surface formed by rotating the circle from part (a) about the line ^ r (t) = 4 ^ ı ı + 4^ ȷ ȷ + tk.

⇀

3.1.10

https://math.libretexts.org/@go/page/91904

1. Of courses we can always convert the equation G(x, y, z) = K into H (x, y, z) = 0 with H (x, y, z) = G(x, y, z) − K. But it is often more convenient to use G(x, y, z) = K. 2. The symbols ρ, φ, θ are the standard mathematics symbols for the spherical coordinates. Appendix A.7 gives another set of symbols that is commonly used in the physical sciences and engineering. This page titled 3.1: Parametrized Surfaces is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

3.1.11

https://math.libretexts.org/@go/page/91904

3.2: Tangent Planes If you are confronted with a complicated surface and want to get some idea of what it looks like near a specific point, probably the first thing that you will do is find the plane that best approximates the surface near the point. That is, find the tangent plane to the surface at the point. In general, a good way to specify a plane is to supply a nonzero vector n (called a normal vector) perpendicular to the plane 1 (to determine the orientation of the plane) and one point (x , y , z ) on the plane. ⇀

0

0

0

If (x, y, z) is any other point on the plane, then the vector (x, y, z) − (x0 , y0 , z0 ) = (x − x0 , y − y0 , z − z0 )

lies entirely in the plane and so is perpendicular to n . This gives the following very neat the equation for the plane. ⇀

⇀

n ⋅ (x − x0 , y − y0 , z − z0 ) = 0

The following theorem provides formulae for normal vectors n to general surfaces, assuming first that the surface is parametrized, second that the surface is a graph and finally the surface is given by an implicit equation. The formulae are developed in the proof of the theorem. ⇀

 Theorem 3.2.1. Normal vectors to surfaces 1. Let ⇀

2

3

r : D ⊂R

→ R

⇀

(u, v) ∈ D ↦ r (u, v) = (x(u, v) , y(u, v) , z(u, v))

be a parametrized surface and let (x

0,

∂ 

⇀

Tu =

∂ 

⇀

∂x

∣ r (u, v0 ) ∣

⇀

∂u

=( ∂u

u=u0

∂v

∂y (u0 , v0 ) ,

∂x

∣ r (u0 , v) ∣

⇀

Tv =

be a point on the surface. Set

⇀

y0 , z0 ) = r (u0 , v0 )

=(

v=v0

∂u

∂z (u0 , v0 ) ,

∂u

∂y (u0 , v0 ) ,

∂v

(u0 , v0 ))

∂z (u0 , v0 ) ,

∂v

(u0 , v0 )) ∂v

Then ∣

^ ı ı

∣ ⇀

⇀

⇀

n = Tu × Tv = det

∣

∂x

∣

∂u

∣

∂x

∣

∂v

^ k

^ ȷ ȷ ∂y

(u0 , v0 )

∂u

(u0 , v0 )

∂y

(u0 , v0 )

∂v

(u0 , v0 )

∂z ∂u ∂z ∂v

(u0 , v0 )

∣ ∣ ∣ ∣

∣ (u0 , v0 ) ∣

is normal (i.e. perpendicular) to the surface at (x , y , z ). 2. Let (x , y , z ) = f (x , y ) be a point on the the surface z = f (x, y). Then, 0

0

0

0

0

0

0

0

^ n = −fx (x0 , y0 ) ^ ı ı − fy (x0 , y0 ) ^ ȷ ȷ +k

⇀

is normal to the surface at (x , y , z ). 3. Consider the surface given implicitly by the equation G(x, y, z) = K, where K is a constant. Let (x the surface and assume that the gradient ∇G(x , y , z ) ≠ 0 . Then 0

0

0

0,

⇀

y0 , z0 )

be a point on

⇀

0

0

⇀

0

⇀

n = ∇G(x0 , y0 , z0 )

is normal to the surface at (x

0,

y0 , z0 ).

Note that none of the normal vectors n above need be of unit length. ⇀

3.2.1

https://math.libretexts.org/@go/page/91905

Note

that

if

we

apply

part

(c)

to

we get the normal vector which is the same as the normal vector provided by part (b). Of course

G(x, y, z) = z − f (x, y)

⇀

^ n = ∇G(x0 , y0 , z0 ) = −fx (x0 , y0 ) ^ ı ı − fy (x0 , y0 ) ^ ȷ ȷ + k,

⇀

they had to be at least parallel. Proof (a) First fix v = v and let u vary. Then 0

⇀

u ↦ r (u, v0 ) = (x(u, v0 ) , y(u, v0 ) , z(u, v0 ))

is a curve on the surface (the red curve in the figure on the right below) that passes through figure) when u = u .

(x0 , y0 , z0 )

(the black dot in the

0

The tangent vector to this curve at (x

0,

∂ 

⇀

y0 , z0 ),

which is also a tangent vector to the surface at (x

0,

∂x

⇀

Tu = ∂u

∣ r (u, v0 ) ∣u=u

=(

∂y

is

∂z

(u0 , v0 ) , ∂u

0

y0 , z0 ),

(u0 , v0 ) , ∂u

(u0 , v0 )) ∂u

It is the red arrow in the figure on the right above. Next fix u = u and let v vary. Then 0

⇀

v ↦ r (u0 , v) = (x(u0 , v) , y(u0 , v) , z(u0 , v))

is a curve on the surface (the blue curve in the figure on the right above) that passes through tangent vector to this curve at (x , y , z ), which is also a tangent vector to the surface at (x , y 0

0

0

∂ 

⇀

Tv = ∂v

∂x

⇀

∣ r (u0 , v) ∣

=(

v=v0

∂y (u0 , v0 ) ,

∂v

(x0 , y0 , z0 )

0,

0

z0 ),

when

v = v0 .

The

is

∂z (u0 , v0 ) ,

(u0 , v0 ))

∂v

∂v

It is the blue arrow in the figure on the right above. ⇀

⇀

We now have two vectors, namely T and T , that are tangent to the surface at (x u

0,

v

∣

^ ıı

∣ ⇀

⇀

⇀

n = Tu × Tv = det

∣

∂x

∣

∂u

∣

∂x

∣

is normal (i.e. perpendicular) to the surface at need not be of unit length.

∂v

∂u

(u0 , v0 )

∂y

(u0 , v0 )

(x0 , y0 , z0 ).

∂v

(u0 , v0 )

So their cross product

^ k

^ ȷ ȷ ∂y

(u0 , v0 )

y0 , z0 ).

∂z ∂u ∂z ∂v

(u0 , v0 )

∣ ∣ ∣ ∣

∣ (u0 , v0 ) ∣

Note however that this vector need not be normalized. That is, it

(b) Next assume that the surface is given by the equation z = f (x, y). Then, renaming u to x and v to y, we may reuse part (a): ⇀

r (x, y) = (x, y, f (x, y))

parametrizes the surface and, at (x

0,

y0 , z0 ) = f (x0 , y0 )),

⇀

⇀

∂r

Tx =

(x0 , y0 ) = (1 , 0 , fx (x0 , y0 )) ∂x

⇀

Ty =

⇀

∂r

∂y

(x0 , y0 ) = (0 , 1 , fy (x0 , y0 ))

3.2.2

https://math.libretexts.org/@go/page/91905

and

⇀

n

^ k

∣ ^ ıı ∣

^ ȷ ȷ

= Tx × Ty = det ∣ 1

0

∣ ^ ıı − fy (x0 , y0 ) ^ ȷ ȷ +k fx (x0 , y0 ) ∣ = −fx (x0 , y0 ) ^

1

∣ fy (x0 , y0 ) ∣

⇀

⇀

∣ ∣0

∣

(c) Finally assume that the surface is given implicitly by the equation G(x, y, z) = 0 or, more generally by the equation, G(x, y, z) = K, where K is a constant. If r (t) = (x(t) y(t) , z(t)) is any curve with r (0) = (x , y , z ) that lies on the surface, then ⇀

⇀

0

⇀

G( r (t)) = K

0

0

for all t

d  ⟹

G(x(t), y(t), z(t)) = 0

for all t

dt

Applying the chain rule gives ∂G

dx (x(t), y(t), z(t))

∂x

dy

∂G (t) +

(x(t), y(t), z(t))

dt

∂y ∂G

dz

+

(x(t), y(t), z(t)) ∂z

The left hand side is exactly the dot product of (

∂G

,

∂x

⇀

∂G ∂y

,

⇀

∂G ∂z

⟹ ⇀

This tell us that ∇G(x

0,

y0 , z0 )

⇀

) = ∇G

with (

dy ,

⇀′

⇀

dz ,

dt

dr ) =

dt

,

so that

dt

for all t

⇀′

∇G(x0 , y0 , z0 ) ⋅ r (0) = 0

is perpendicular to

⇀′

r (0),

which is a tangent vector to G = K at (x

0,

all curves r (t) on G = K and so is true for all tangent vectors to vector to G(x, y, z) = K at (x , y , z ). 0

dx dt

⇀

0

(t) = 0 dt

∇G( r (t)) ⋅ r (t) = 0 ⇀

(t) dt

G=K

at

(x0 , y0 , z0 ).

So

y0 , z0 ).

This is true for

⇀

∇G(x0 , y0 , z0 )

is a normal

0

 Example 3.2.2 Consider the surface x = x(u, v) = u cos v y = y(u, v) = u sin v z = z(u, v) = u

Observe that 2

x(u, v)

2

+ y(u, v)

2

=u

2

= z(u, v)

So our surface is also 2

G(x, y, z) = x

+y

2

−z

2

=0

We shall sketch it shortly. But first, let's find it's tangent plane at (x , y , z ) = r (u , v the parametrization and once using its implicit equation. First, using the parametrization we have ⇀

0

= ∂u

0

0 ).

In fact, let's do it twice. Once using

^ r (u, v) = u cos v ^ ı ı + u sin v ^ ȷ ȷ + u k,

⇀

^ (u0 , v0 ) = cos v0 ^ ı ı + sin v0 ^ ȷ ȷ +k

⇀

∂r

⇀

Tv

0

⇀

∂r

⇀

Tu

0

= ∂v

(u0 , v0 ) = −u0 sin v0 ^ ı ı + u0 cos v0 ^ ȷ ȷ

so that ⇀

n

^ = ( cos v0 ^ ı ı + sin v0 ^ ȷ ȷ + k) × ( − u0 sin v0 ^ ı ı + u0 cos v0 ^ ȷ ȷ) = ( − u0 cos v0 , −u0 sin v0 , u0 ) = (−x0 , −y0 , z0 )

3.2.3

https://math.libretexts.org/@go/page/91905

Next using the implicit equation G(x, y, z) = x

2

+y

2

−z

2

we have the normal vector

= 0,

⇀

∇G(x0 , y0 , z0 ) = (2 x0 , 2 y0 , −2 z0 ) = −2(−x0 , −y0 , z0 )

Of course the two vectors (−x , −y , z ) and −2(−x , −y , z ) are parallel to each other. Either can be used as a normal vector and the tangent plane to x + y − z = 0 at (x , y , z ) is 0

0

2

0

0

2

0

0

2

0

0

0

⇀

0 = n ⋅ (x − x0 , y − y0 , z − z0 ) = −x0 (x − x0 ) − y0 (y − y0 ) + z0 (z − z0 ) ⇀

provided (x , y , z clearly a problem. 0

0)

0

⇀

≠ 0.

⇀

More generally, if T

u

In the event that ⇀

× Tv = 0

⇀

(x0 , y0 , z0 ) = 0

⇀

⇀

(or ∇G(x

0,

y0 , z0 ) = 0

the “tangent plane equation” reduces to

and there is

0 =0

), then either 2

the surface fails to have a tangent plane at (x , y , z ), or our parametrization is screwy 3 there. For example, we can parametrize the xy-plane, z = 0, by 0

0

0

⇀

⇀

(This is just polar coordinates.) Then T ^ ^ ȷ ȷ , so that T × T = u k is 0 when u

r (u, v) = u cos v ^ ı ı + u sin v ^ ȷ ȷ.

⇀

u

⇀

Tv = −u0 sin v0

^ ı ı + u0 cos v0

⇀

u

v

0

0

and But the plane z = 0 is its own tangent plane

= cos v0 ^ ı ı + sin v0 ^ ȷ ȷ

⇀

= 0.

everywhere. The surface of current interest is x + y = z . The intersection of this surface with the horizontal plane z = z is x + y = z , which is the circle of radius | z | centred on x = y = 0. So our surface is a stack of circles. The radius of the circle in the xy-plane is zero. The radius increases linearly as we move away from the xy-plane. Our surface is a cone. It does not have a tangent plane at (0, 0, 0). 2

2

2

0

2

2

2

0

0

 Example 3.2.3 This time we shall find the tangent planes to the surface 2

x

+y

2

−z

2

=1

As for the cone of the last example, the intersection of this surface with the horizontal plane z = z is a circle — the circle of 0

− − − − −

radius √1 + z centred on x = y = 0. Our surface is again a stack of circles. The radius of the circle in the xy-plane is 1. The 2

0

radius increases as we move away from the xy-plane. Here is a sketch of the surface.

It is called a hyperboloid 4 of one sheet. Using the implicit equation G(x, y, z) = x

2

+y

2

−z

2

= 1,

we have

⇀

∇G(x0 , y0 , z0 ) = (2 x0 , 2 y0 , −2 z0 ) = 2(x0 , y0 , −z0 )

and we may take (x

0,

y0 , −z0 )

as a normal vector at (x

0,

y0 , z0 ).

So the tangent plane to x

2

+y

2

−z

2

=1

at (x

0,

y0 , z0 )

is

⇀

0 = n ⋅ (x − x0 , y − y0 , z − z0 ) = x0 (x − x0 ) + y0 (y − y0 ) − z0 (z − z0 )

3.2.4

https://math.libretexts.org/@go/page/91905

This time n = (x ⇀

⇀

0,

n = (x0 , y0 , −z0 )

⇀

y0 , −z0 ) ≠ 0 ,

at (x

0,

so that we have a tangent plane, at every point of the surface. In particular, the vanishing of is not a problem because (0, 0, 0) is not on the surface.

y0 , z0 ) = (0, 0, 0)

 Example 3.2.4. Optional — Parametrizing the Hyperboloid of One Sheet The hyperboloid of one sheet, x + y − z = 1, has a symmetry. It is invariant under rotation about the z -axis. So it is natural to parametrize the surface using cylindrical coordinates. 2

2

2

x = r cos θ y = r sin θ z =z

In cylindrical coordinates the surface

2

x

+y

2

is r − z = 1, and we could parametrize it by Alternatively, we can eliminate the square roots in the parametrization

−z

− − − − − − − − − − 2 2 ^ r (θ, z) = √1 + z cos θ ^ ı ı + √1 + z sin θ ^ ȷ ȷ + z k.

⇀

2

2

=1

2

by exploiting the hyperbolic trig functions 1 sinh u =

(e

u

−e

−u

1 )

cosh u =

2

(e

u

+e

−u

)

2

The functions have properties 5 that are very similar to those of sin θ and cos θ. d

d cosh u = sinh u

We can set r = cosh u,

2

sinh u = cosh u

du

cosh

u − sinh

2

u =1

du

z = sinh u

to yield the parametrization ⇀

^ r (θ, u) = cosh u cos θ ^ ı ı + cosh u sin θ ^ ȷ ȷ + sinh u k

^. As an exercise in working with hyperbolic trig functions, we'll use this parametrization to find n x = cosh u cos θ

xu = sinh u cos θ

xθ = − cosh u sin θ

y = cosh u sin θ

yu = sinh u sin θ

yθ =

z = sinh u

zu = cosh u

zθ = 0

cosh u cos θ

So ∣ ⇀

⇀

n = Tu × Tθ = det ∣ ∣

^ k

^ ı ı

^ ȷ ȷ

sinh u cos θ

sinh u sin θ

∣

⇀

∣ − cosh u sin θ 2

= ( − cosh

cosh u cos θ 2

u cos θ , − cosh

∣ ∣

cosh u ∣ ∣ 0

∣

u sin θ , sinh u cosh u)

Exercises Stage 1  1 Is it reasonable to say that the surfaces

2

x

+y

2

2

+ (z − 1 )

=1

and

2

x

+y

2

2

+ (z + 1 )

=1

are tangent to each other at

(0, 0, 0)?

3.2.5

https://math.libretexts.org/@go/page/91905

 2 Let the point r = (x , y , z ) lie on the surface G(x, y, z) = 0. Assume that parametrized curve r (t) = (x(t), y(t), z(t)) is contained in the surface and that r (t curve at r lies in the tangent plane to G = 0 at r . ⇀

0

0

0

0

⇀

⇀

⇀

⇀

⇀

Suppose that the Show that the tangent line to the

∇G(x0 , y0 , z0 ) ≠ 0 .

0)

⇀

= r 0.

⇀

0

0

 3 Find the parametric equations of the normal line to the surface z = f (x, y) at the point (x , y , z =f (x , y )). By definition, the normal line in question is the line through (x , y , z ) whose direction vector is perpendicular to the surface at 0

0

0

0

0

0

0

0

(x0 , y0 , z0 ).

 4 ⇀

⇀

Let F (x , y , z ) = G(x , y , z ) = 0 and let the vectors ∇F (x , y , z ) and ∇G(x , y , z ) be nonzero and not be parallel to each other. Find the equation of the normal plane to the curve of intersection of the surfaces F (x, y, z) = 0 and G(x, y, z) = 0 at (x , y , z ). By definition, that normal plane is the plane through (x , y , z ) whose normal vector is the tangent vector to the curve of intersection at (x , y , z ). 0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

 5 Let f (x , y ) = g(x , y ) and let (f (x , y ), f (x , y )) ≠ (g (x , y ), g (x the curve of intersection of the surfaces z = f (x, y) and z = g(x, y) at (x , y 0

0

0

0

x

0

0

y

0

0

x

0

0

y

0

0,

0

y0 )) .

Find the equation of the tangent line to

, z0 = f (x0 , y0 )).

Stage 2  6✳ 2

Let f (x, y) =

x y 4

x

+ 2y

2

.

Find the tangent plane to the surface z = f (x, y) at the point (−1 ,

1,

1 3

).

 7✳ Find the tangent plane to 27 − −−−−−−−−−−− − =9 2 √ x + y2 + z2 + 3

at the point (2, 1, 1).

 8✳ Consider the surface z = f (x, y) defined implicitly by the equation xy z + y vector to find the equation of the tangent plane to this surface at the point z = ax + by + c, where a, b and c are constants. 2

2

z

3

Use a 3--dimensional gradient Write your answer in the form

2

= 3 +x .

(−1, 1, 2).

 9✳ A surface is given by 2

z =x

2

− 2xy + y .

1. Find the equation of the tangent plane to the surface at x = a, y = 2a. 2. For what value of a is the tangent plane parallel to the plane x − y + z = 1?

3.2.6

https://math.libretexts.org/@go/page/91905

 10 ✳ A surface S is given by the parametric equations 2

x = 2u 2

y =v

2

z =u

3

+v

Find an equation for the tangent plane to S at the point (8, 1, 5).

 11 ✳ Let S be the surface given by ⇀

2

r (u, v) = (u + v , u

2

+v

, u − v),

−2 ≤ u ≤ 2,   − 2 ≤ v ≤ 2

Find the tangent plane to the surface at the point (2, 2, 0).

 12 ✳ Find the tangent plane and normal line to the surface z = f (x, y) =

2y x2 +y 2

at (x, y) = (−1, 2).

 13 ✳ Find all the points on the surface x

2

+ 9y

2

+ 4z

2

where the tangent plane is parallel to the plane x − 8z = 0.

= 17

 14 ✳ Let S be the surface z = x + 2y contains the origin (0, 0, 0). 2

2

+ 2y − 1.

Find all points

P (x0 , y0 , z0 )

on

S

with

x0 ≠ 0

such that the normal line at

P

 15 ✳ Find all points on the hyperboloid z

2

2

= 4x

+y

2

−1

where the tangent plane is parallel to the plane 2x − y + z = 0.

Stage 3  16 ✳ 1. Find a vector perpendicular at the point (1, 1, 3) to the surface with equation x + z = 10. 2. Find a vector tangent at the same point to the curve of intersection of the surface in part (a) with surface y 3. Find parametric equations for the line tangent to that curve at that point. 2

2

2

+z

2

= 10.

 17 ✳ Let P be the point where the curve ⇀

3

r (t) = t

2

^ ı ı +t ^ ȷ ȷ +t

^ k,

(0 ≤ t < ∞)

intersects the surface z

3

+ xyz − 2 = 0

Find the (acute) angle between the curve and the surface at P .

3.2.7

https://math.libretexts.org/@go/page/91905

 18 Find all horizontal planes that are tangent to the surface with equation 2

z = xye

−( x +y

2

)/2

What are the largest and smallest values of z on this surface? 1. Alternatively, you could find two vectors that are in the plane (and not parallel to each other), and then construct a normal vector by taking their cross product. 2. We saw the same dichotomy when considering what happened for a curve when r (t) = 0. See Example 1.1.10. 3. Of course “screwy” is not a mathematically precise word. One way a parametrization r (u, v) could be “screwy” is if it failed to give a one-to-one correspondence between parameter values (u, v) and points on (part of) the surface. For example, polar coordinates r (u, v) = u cos v ^ ı ı + u sin v ^ ȷ ȷ give r (0, v) = (0, 0) for all values of v. 4. There are also hyperboloids of two sheets. See Appendix A.8. 5. This is no accident: cosh u = cos(iu) and sinh u = −i sin(iu), where i is the usual complex number that obeys i = −1. You can verify these formulae by just checking that cosh u and cos(iu) have the same Taylor expansions and that sinh u and −i sin(iu) have the same Taylor expansions. ⇀′

⇀

⇀

⇀

2

This page titled 3.2: Tangent Planes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

3.2.8

https://math.libretexts.org/@go/page/91905

3.3: Surface Integrals We are now going to define two types of integrals over surfaces. Integrals that look like ∬ surface S.

S

ρ dS

are used to compute the area and, when ρ is, for example, a mass density, the mass of the

⇀

^ dS, with n ^ (x, y, z) being a unit vector that is perpendicular to S at (x, y, z), are called flux Integrals that look like ∬ F ⋅ n integrals. We shall see in §3.4, that when v is the velocity field of a moving fluid and ρ is the density of the fluid, then ^ dS is the rate at which fluid is crossing the surface S. ∬ ρv ⋅ n S

⇀

⇀

S

Parametrized Surfaces Suppose that we wish to integrate over part, S, of a surface that is parametrized by r (u, v). We start by cutting S up into small pieces by drawing a bunch of curves of constant u (the blue curves in the figure below) and a bunch of curves of constant v (the red curves in the figure below). ⇀

Concentrate on any one the small pieces. Here is a greatly magnified sketch.

For example, the lower red curve was constructed by holding v fixed at the value v , varying u and sketching r (u, v ), and the upper red curve was constructed by holding v fixed at the slightly larger value v + dv, varying u and sketching r (u, v + dv). So the four intersection points in the figure are ⇀

0

0

⇀

0

⇀

P2 = r (u0 , v0 + dv) ⇀

0

⇀

P3

= r (u0 + du, v0 + dv) ⇀

P0 = r (u0 , v0 )

P1 = r (u0 + du, v0 )

Now if ⇀

R(t) = r (u0 + tdU , v0 + tdV )

(where dU and dV are any small constants) then, by Taylor expansion, ⇀

r (u0 + dU , v0 + dV ) = R(1) ′

≈ [R(0) + R (0) (t − 0)]

t=1

⇀

⇀

= r (u0 , v0 ) +

Applying this three times, once with dU

= du, dV = 0,

⇀

∂r

∂u

once with dU

3.3.1

∂r (u0 , v0 ) dU +

= 0 dV = dv,

∂v

(u0 , v0 ) dV

and once with dU

= du, dV = dv,

https://math.libretexts.org/@go/page/91906

⇀

P0 = r (u0 , v0 ) ⇀

⇀

∂r

⇀

P1 = r (u0 + du, v0 )

≈ r (u0 , v0 ) +

∂u

(u0 , v0 ) du

⇀

⇀

∂r

⇀

P2 = r (u0 , v0 + dv)

≈ r (u0 , v0 ) +

∂v

(u0 , v0 ) dv

⇀

⇀

P3

= r (u0 + du, v0 + dv)

⇀

∂r

⇀

≈ r (u0 , v0 ) +

∂u

∂r (u0 , v0 ) du +

(u0 , v0 ) dv

∂v

We have dropped all Taylor expansion terms that are of degree two or higher in du, dv. The reason is that, in defining the integral, we take the limit du, dv → 0. Because of that limit, all of the dropped terms contribute exactly 0 to the integral. We shall not prove this. But we shall show, in the optional §3.3.5, why this is the case. The small piece of our surface with corners P

0,

is approximately a parallelogram with sides

P1 , P2 , P3

− − − →

⇀

− − − →

∂r

P0 P1 ≈ P2 P3 ≈

(u0 , v0 ) du ∂u

− − − →

⇀

− − − →

∂r

P0 P2 ≈ P1 P3 ≈

(u0 , v0 ) dv ∂v

− − − →

Denote by θ the angle between the vectors P

0 P1

− − − →

of the parallelogram is ∣∣P

∣ sin θ.

0 P2 ∣

− − − →

and P

− − − →

0 P2 .

The base of the parallelogram, P

0 P1 ,

− − − →

has length ∣∣P

∣

0 P1 ∣,

and the height

So the area of the parallelogram is 1

− − − →

− − − →

− − − →

− − − →

| P0 P1 | | P0 P2 |  sin θ = ∣ ∣P0 P1 × P0 P2 ∣ ∣ ⇀

⇀

∂r ∣∂r ∣ ≈∣ (u0 , v0 ) × (u0 , v0 )∣dudv ∣ ∂u ∣ ∂v

Furthermore,

⇀ ∂ r ∂u

curves lie in ⇀ ∂ r ∂u

(u0 , v0 ) ×

to the surface.

(u0 , v0 )

S. ⇀ ∂ r ∂v

So

and

⇀ ∂ r ∂u

(u0 , v0 )

⇀ ∂ r ∂v

(u0 , v0 )

(u0 , v0 )

and

are tangent vectors to the curves ⇀ ∂ r ∂v

0,

r (t , v0 )

are tangent vectors to

(u0 , v0 )

is perpendicular to S at (u

⇀

v0 ).

S

and at

⇀

r (u0 , t) (u0 , v0 )

respectively. Both of these and the cross product

We have found both dS and n ^ dS, where n ^ is a unit normal vector

 Equation 3.3.1 For the parametrized surface

⇀

r (u, v), ⇀

^ dS = ± n

⇀

∂r

∂r (u , v) ×

(u , v) dudv

∂u

∂v

⇀

⇀

∂r ∣∂r ∣ dS = ∣ (u , v) × (u , v)∣ dudv ∣ ∂u ∣ ∂v

The ± sign in 3.3.1 is there because there are two unit normal vectors at each point of a surface, one on each side of the surface. Typically, the application itself tells you which of the two normal vectors should be used. We shall see many examples shortly.

Graphs The surface which is the graph z = f (x, y) can be parametrized by ^ r (x, y) = x ^ ı ı +y ^ ȷ ȷ + f (x, y) k

⇀

As

3.3.2

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⇀

∂r

⇀

∂f

∂x

∂r

^ k

= ^ ı ı +

∂f

and

=^ ȷ ȷ +

∂x

∂y

^ k

∂y

we have ı ı ⎡ ^ ⇀

^ k ⎤

^ ȷ ȷ

⇀

∂r

∂r ×

∂x

∂y

∂f

⎢ = det ⎢ 1 ⎢

0

⎣0

1

∂x ∂f

⎥ ^ ı ı − fy (x, y) ^ ȷ ȷ +k ⎥ = −fx (x, y) ^ ⎥ ⎦

∂y

So, 3.3.1 gives the following.

 Equation 3.3.2 For the surface z = f (x, y), ^ dS n

^ = ±[ − fx (x, y) ^ ı ı − fy (x, y) ^ ȷ ȷ + k] dxdy − −−−−−−−−−−−−−−−−− − 2

dS = √ 1 + fx (x, y )

2

+ fy (x, y )  dxdy

Similarly, for the surface x = g(y, z), ^ ^ dS = ±[ ^ n ı ı − gy (y, z) ^ ȷ ȷ − gz (y, z) k] dydz − −−−−−−−−−−−−−−−− − 2

dS = √ 1 + gy (y, z)

2

+ gz (y, z)

 dydz

and for the surface y = h(x, z), ^ dS n

^ = ±[ − hx (x, z) ^ ı ı +^ ȷ ȷ − hz (x, z) k] dxdz − −−−−−−−−−−−−−−−−− − 2

dS = √ 1 + hx (x, z)

2

+ hz (x, z)

 dxdz

Again, in any given application, some care must be taken in choosing the sign in 3.3.2, so as to get the appropriate normal vector. The formulae like

− −−−−−−−−−−−−−−−−− − 2

dS = √1 + fx (x, y )

2

+ fy (x, y )

 dxdy

in 3.3.2 have geometric interpretations. The red parallelogram in the

sketch

represents a little piece of our surface. It has area

− −−−−−−−−−−−−−−−−− − 2

dS = √1 + fx (x, y )

2

+ fy (x, y )

 dxdy.

The blue parallelogram in the same

sketch represents the projection of the red parallelogram onto the xy-plane. It has area dxdy. The vector normal for the red parallelogram. We have seen that it is parallel to ⇀

in the sketch is a unit

⇀

∂r

∂r ×

∂x

^ n

∂y

^ = −fx (x, y) ^ ı ı − fy (x, y) ^ ȷ ȷ +k

^ so that the angle θ between n ^ and k obeys

3.3.3

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^ ^ (−fx (x, y) ^ ı ı − fy (x, y) ^ ȷ ȷ + k) ⋅ k cos θ = ^ ^ ∣ ı ı − fy (x, y) ^ ȷ ȷ + k∣ ∣ − fx (x, y) ^ ∣ | k| 1 =

− −−−−−−−−−−−−−−−−− − 2

√ 1 + fx (x, y )

2

+ fy (x, y )

− −−−−−−−−−−−−−−−−− −

The geometric interpretation of dS = √1 + f

x (x,

2

y)

2

+ fy (x, y )

 dxdy

is that the area dS of a little piece of surface is the area

^ ^ (which is perpendicular to the surface) and k of its projection on the xy-plane times the factor where θ is the angle between n (which is perpendicular to the xy-plane). Notice that 1

cos θ

when θ is close to zero, which corresponds the f being almost constant and our surface being almost parallel to the xy-plane, dS reduces to almost dxdy. On the other hand, in the limit θ → , which corresponds to f and/or f becoming infinite and our surface becoming perpendicular to the xy-plane, dS becomes “infinity times” dxdy. In this case, we should represent our surface either in the form x = g(y, z) or in the form y = h(x, z), rather than in the form z = f (x, y). π

x

2

y

Surfaces Given by Implicit Equations Finally suppose that the surface is given by the equation G(x, y, z) = K, with K a constant. Suppose further that at some point on 2 the surface ≠ 0. Then near that point we may solve the equation G(x, y, z) = K for z as a function of x and y. That is, the surface also obeys z = f (x, z) for a function f (x, y) that satisfies ∂G ∂z

G(x, y, f (x, y)) = K

near the point. Differentiating this with respect to x and y gives, by the chain rule, ∂ 0 = ∂x

[G(x, y, f (x, y))]  =  Gx (x, y, f (x, y)) + Gz (x, y, f (x, y)) fx (x, y)

∂ 0 = ∂y

[G(x, y, f (x, y))]  =  Gy (x, y, f (x, y)) + Gz (x, y, f (x, y)) fy (x, y)

which implies Gx (x, y, f (x, y)) fx (x, y) = −

Gy (x, y, f (x, y)) fy (x, y) = −

Gz (x, y, f (x, y))

Gz (x, y, f (x, y))

and ^ −fx (x, y) ^ ı ı − fy (x, y) ^ ȷ ȷ +k

Gx (x, y, f (x, y))

Gy (x, y, f (x, y)) ^ ı ı +

= Gz (x, y, f (x, y))

^ ^ ȷ ȷ +k

Gz (x, y, f (x, y))

⇀

∇G(x, y, f (x, y)) = Gz (x, y, f (x, y))

So, by (3.3.1),

 Equation 3.3.3 For the surface G(x, y, z) = K, when G

z (x,

y, z) ≠ 0, ⇀

∇G(x, y, z) n ^ dS = ±

 dxdy

⇀

^ ∇G(x, y, z) ⋅ k ⇀

∣ dS = ∣ ∣

Similarly, for the surface G(x, y, z) = K, when G

x (x,

∇G(x, y, z)

∣ ∣ dxdy ⇀ ^ ∣ ∇G(x, y, z) ⋅ k

y, z) ≠ 0,

3.3.4

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⇀

∇G(x, y, z) n ^ dS = ±

 dydz

⇀

∇G(x, y, z) ⋅ ^ ı ı ⇀

∣ dS = ∣ ∣

and for the surface G(x, y, z) = K, when G

y (x,

∇G(x, y, z)

∣ ∣ dydz ∣

⇀

∇G(x, y, z) ⋅ ^ ı ı

y, z) ≠ 0, ⇀

∇G(x, y, z) ^ dS = ± n

 dxdz

⇀

∇G(x, y, z) ⋅ ^ ȷ ȷ ⇀

∣ dS = ∣ ∣

∇G(x, y, z)

∣ ∣ dxdz ∣

⇀

∇G(x, y, z) ⋅ ^ ȷ ȷ

If, for some point (x , y , z ) we have G (x , y , z ) = G (x , y , z ) = G (x , y , z ) = 0, we also have problem! Often this is a sign that our surface is not smooth at (x , y , z ) and in fact does not have a normal vector there. For an example of this, see Example 3.2.2. 0

0

0

x

0

0

Examples of ∬

S

0

0

0

y

0

0

0

z

0

0

0

0

ρ dS

We'll start by computing, in several different ways, the surface area of the hemisphere 2

x

+y

2

+z

2

2

=a

z ≥0

(with a > 0 ). You probably know, from high school, that the answer is × 4π a = 2π a . But you have probably not seen a derivation of this answer. Note that, since x + y = a − z on the hemisphere, the set of (x, y)'s for which there is a z with (x, y, z) on the hemisphere is exactly {(x, y) ∈ R | x + y ≤ a } . 1

2

2

2

2

2

2

2

2

2

2

2

 Example 3.3.4. Area of a hemisphere — using cylindrical coordinates Let's parametrize the hemisphere x coordinates 3

2

+y

2

+z

2

2

=a ,

using as parameters the polar coordinates

z ≥ 0,

r, θ

of the cylindrical

x = r cos θ y = r sin θ z =z

and then apply 3.3.1. In cylindrical coordinates the equation x +y ≤ a is 0 ≤ r ≤ a, 0 ≤ θ < 2π. 2

2

2

x

+y

2

+z

2

2

=a

becomes

2

r

+z

2

2

=a ,

and the condition

2

So the hemisphere can be parametrized by − − − − − − 2 2 (x(r, θ) , y(r, θ) , z(r, θ)) = (r cos θ , r sin θ , √ a − r ) with  0 ≤ r ≤ a,  0 ≤ θ < 2π − − − − − − 2 2 −r

Note that we selected the positive solution z = √a

of r

2

+z

3.3.5

2

2

=a

in order to satisfy the condition that z ≥ 0. Since

https://math.libretexts.org/@go/page/91906

∂x

∂y

(

∂z

,

,

∂r ∂x

∂r

∂y

(

− − − − − − √ a2 − r2

)

∂z

,

,

∂θ

r ) = ( cos θ , sin θ , −

∂r

∂θ

) = (−r sin θ , r cos θ , 0) ∂θ

3.3.1 yields ^ dS = ±( n

∂y

∂x

∂z

,

,

∂r

⎣

∂θ

) drdθ ∂θ

^ k

cos θ

sin θ

−r sin θ

r cos θ

2

,

∂θ ^ ȷ ȷ

= ± det ⎢ ⎢

∂z

,

∂r

^ ı ı

⎡

∂y

∂x )×(

∂r

−

⎤

r

⎥  drdθ ⎥

√a2 −r2

⎦

0

2

r

cos θ r sin θ − − − − − − , − − − − − − , r) drdθ 2 2 √a − r √ a2 − r2 − −−−−−−−− − − − − − − − −

= ±(

4

2

r dS = √

2

2

2

a

+r

2

a r  drdθ = √

−r

2

a

2

 drdθ =

−r

ar − − − − − −  drdθ 2 √ a − r2

So the area of the hemisphere is a

∫

2π

dr ∫

0

a

ar dθ 

0

− − − − − − √ a2 − r2

= 2πa ∫

r dr 

0 0

− − − − − − √ a2 − r2

−du/2

= 2πa ∫ 2

a

− √u 2

with u = a

2

− r ,  du = −2r dr

0

− = 2πa[ − √u ]

2

a 2

= 2πa

as it should be.

 Example 3.3.5. Area of a hemisphere — using an implicit equation This time we'll compute the area of the hemisphere by using that, if (x, y, z) is on the hemisphere, then G(x, y, z) = a with G(x, y, z) = x + y + z . Since 2

2

2

2

⇀

∇G(x, y, z) = (2x , 2y , 2z)

3.3.3 yields ⇀

∣ dS = ∣ ∣

∣ =∣ ∣

∇G(x, y, z)

∣ ∣ dxdy ⇀ ^ ∣ ∇G(x, y, z) ⋅ k (2x , 2y , 2z)

∣ ∣ dxdy ∣ 2z − −−−−−−−− − 2 2 2 √x +y +z

=

 dxdy |z|

=

So the area is ∬

a 2

x +y

y = r sin θ.

2

2

≤a

√a2 −x2 −y 2

a − −−−−−−−− −  dxdy 2 2 2 √a −x −y

 dxdy.

2

on x

+y

2

+z

2

2

=a

To evaluate this integral, we switch to polar coordinates, substituting x = r cos θ,

This gives

3.3.6

https://math.libretexts.org/@go/page/91906

a

a area = ∬ x2 +y 2 ≤a2

− −−−−−−−− − 2 2 2 √a −x −y

a

= 2πa ∫

 dxdy = ∫

2π

dr r ∫

0

a dθ 

0

− − − − − − √ a2 − r2

r dr 

0

− − − − − − √ a2 − r2

We already showed, in Example 3.3.4, that the value of this integral is 2π a

2

.

 Example 3.3.6. Area of a hemisphere — using spherical coordinates Of course “integrating over a sphere” cries out for spherical coordinates. So this time we parametrize the hemisphere x + y + z = a , z ≥ 0, using as parameters the angular coordinates θ, φ of the spherical coordinates 2

2

2

2

x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ

and then apply 3.3.1. In spherical coordinates the equation x + y + z = a is 0 ≤ φ ≤ , 0 ≤ θ < 2π. So the hemisphere can be parametrized 4 by 2

2

2

2

becomes just ρ

2

2

=a ,

and the condition z ≥ 0

π 2

(x(θ, φ) , y(θ, φ) , z(θ, φ)) = (a sin φ cos θ , a sin φ sin θ , a cos φ ) π 0 ≤φ ≤

,  0 ≤ θ < 2π 2

Since ∂x

∂y

(

,

∂z ,

∂θ

∂θ

∂x

∂y

(

, ∂φ

) = ( − a sin φ sin θ , a sin φ cos θ , 0) ∂θ ∂z

, ∂φ

) = (a cos φ cos θ , a cos φ sin θ , −a sin φ) ∂φ

3.3.1 yields ^ dS = ±( n

∂x

∂y ,

∂θ

∂z ,

∂θ

∂x )×(

∂θ

∂y ,

∂φ

∂z ,

∂φ

) dθdφ ∂φ

= ±(− a sin φ sin θ, a sin φ cos θ, 0)×(a cos φ cos θ, a cos φ sin θ, −a sin φ) dθdφ 2

= ±( − a 2

= ∓a

2

sin

2

φ cos θ , −a

2

sin

2

φ sin θ , −a

sin φ cos φ) dθdφ

sin φ( sin φ cos θ , sin φ sin θ , cos φ) dθdφ

and −−−−−−−−−−−−−−−−−−−−−−−−−− − 2

dS = a

2

=a

2

sin φ √ sin

2

φ cos

2

θ + sin

2

φ sin

2

θ + cos

φ  dθdφ

sin φ dθdφ

So the area of the hemisphere is

3.3.7

https://math.libretexts.org/@go/page/91906

π

2

a

π

2π

2

∫

dφ ∫

0

dθ  sin φ

2

= 2π a

0

2

∫

π/2

2

dφ  sin φ = 2π a [ − cos φ ] 0

0 2

= 2πa

There is an easier way to do this, using a little geometry.

 Example 3.3.7. Area of a hemisphere — using spherical coordinates again We are now going to again compute the surface area of the hemisphere using spherical coordinates. But this time instead of determining dS using the canned formula 3.3.1, we are going to read it off of a sketch. Sketch the part of the hemisphere that is in the first octant, x ≥ 0, y ≥ 0, z ≥ 0. Slice it up into small pieces by drawing in curves of constant θ (the blue lines in the figure below) and curves of constant φ (the red lines in the figure below).

Each piece is approximately a little rectangle. Concentrate on one of them, like the piece with the thick sides in the figure above. The area, dS, of that piece is (essentially) the product of its height and its width. Each of the two sides of the piece is a segment of a circle of radius a (a fat blue line in both the figure above and in the figure on the left below) that subtends an angle dφ and hence is the fraction

dφ 2π

of a full circle of radius a and hence is of length

dφ 2π

2πa = adφ.

The top of the piece is a segment of a circle of radius a sin φ (a fat red line in both the figure above and in the figure on the right below) that subtends an angle dθ and hence is the fraction of a full circle of radius a sin φ and hence is of length 2πa sin φ = a sin φdθ. dθ

dθ

2π

2π

These are drawn in the figure below.

So the area of our piece is 2

dS = (adφ)(a sin φdθ) = a

This is exactly the same formula that we found for hemisphere of radius a is 2π a . (Phew!)

dS

sin φ dθdφ

in Example 3.3.6 so that we will, yet again, get that the area of a

2

3.3.8

https://math.libretexts.org/@go/page/91906

But wait! We can do it again, by yet another method!

 Example 3.3.8. Area of a hemisphere — using z = f (x, y) We'll compute the area of the hemisphere one last time 5. This time we'll use that the equation of the hemisphere is − −−−−−−−− − 2

2

z = f (x, y) = √ a

−x

−y

2

2

with (x, y) running over x

+y

2

2

≤a

So 3.3.2 yields − −−−−−−−−−−−−−−−−− − 2

2

dS = √ 1 + fx (x, y )

+ fy (x, y )  dxdy

− −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− − 2 2 −x −y = √1 +( − −−−−−−−− −) +( − −−−−−−−− − )  dxdy √ a2 − x2 − y 2

√ a2 − x2 − y 2

− −−−−−−−−−−−− − 2

x = √1 +

2

+y 2

a

−x

2

−y

2

 dxdy

− − − − − − − − − − − a2 =√

2

a

So the area is ∬

a 2

x +y

2

2

≤a

2

2

√a −x −y

2

 dxdy.

2

−x

−y

 dxdy

2

We already found, in Example 3.3.5, that the value of this integral in 2π a

2

.

Let's do some more substantial examples, where the integrand is not 1.

 Example 3.3.9 Evaluate   ∬

S

2

2

2

x y z  dS 

where S is the part of the cone x

2

+y

2

=z

2

with 0 ≤ z ≤ 1.

Solution 1 We can express S as

− −− −− − 2

z = f (x, y) = √ x

+y

2

2

x

+y

2

≤1

Now since x fx (x, y) =

y fy (x, y) =

− −− −− − 2 2 √x + y

− −− −− − 2 2 √x + y

3.3.2 gives 6 2

y

x dS = [1 +

2

x

+y

2

+

2

x

2

1/2

+y

2

]

–  dxdy = √2 dxdy

Our integral is then ∬ S

– 2 2 2 x y z  dS = √2 ∬

2

2

2

x y (x 2

x +y

2

2

+ y ) dxdy

≤1

Since we are integrating over a circular domain, let's convert to polar coordinates.

3.3.9

https://math.libretexts.org/@go/page/91906

2π

1

– 2 2 2 x y z  dS = √2 ∫

∬ S

2

dθ ∫

0

2

2π

– = √2 [∫

1 2

dθ  cos

2

θ sin

– √2

dr r ]

0

– √2

2π 2

∫ 8

7

θ] [∫

0

=

2

dr r(r cos θ) (r sin θ) r

0

2

dθ  cos

θ sin

θ =

2π

32

0

– √2

2

∫

dθ  sin (2θ)

0

2π

=

∫ 64

dθ [1 − cos(4θ)]

0

Remembering 7 that the integral of cos(θ), or cos(4θ), over a full period is 0, we end up with ∬

2

2

– √2

2

x y z  dS =

– π √2 (2π) =

64

S

32

Solution 2 We may parametrize 8 the cone by ⇀

r (z, θ)

^ = z cos θ ^ ıı + z sin θ ^ ȷ ȷ +zk

0 ≤ z ≤ 1,  0 ≤ θ ≤ 2π

Then because ⇀

∂r

⇀

∂r

^ = cos θ ^ ıı + sin θ ^ ȷ ȷ +k

= −z sin θ ^ ıı + z cos θ ^ ȷ ȷ

and

∂z

∂θ

3.3.1 yields 9 ^ ȷ ȷ

cos θ

sin θ

−z sin θ

z cos θ

^ dS = ± det ⎢ n ⎣

^ k⎤

^ ıı

⎡

1 ⎥ dzdθ 0

⎦

^ = ±[ − z cos θ ^ ıı − z sin θ ^ ȷ ȷ + z k] dzdθ – dS = √2z dzdθ

So our integral becomes 2π

∬

– 2 2 2 x y z  dS = √2 ∫

S

1

0

2

1

dθ ∫

0

dz z

7

2

cos

2

θ sin

θ

0

– √2

2π 2

∫ 8

2

dz z(z cos θ) (z sin θ) z

0 2π

– = √2 ∫

=

2

dθ ∫

dθ  cos

2

θ sin

θ

0

We evaluated this integral in Solution 1. So again ∬

2

2

2

– π √2

x y z  dS = 32

S

Let's do something more celestial.

 Example 3.3.10 Consider a spherical shell of radius a with mass density μ per unit area. Think of it as a hollow planet 10. We are going to determine the gravitational force that it exerts on a particle of mass m a distance b away from its centre. This particle can be either outside the shell (b > a ) or inside the shell (b < a ). We can choose the coordinate system so that the centre of the shell is at the origin and the particle is at (0, 0, b). By Newton's law of gravitation, the force exerted on the particle by a tiny piece of the shell of surface area dS located at r is ⇀

3.3.10

https://math.libretexts.org/@go/page/91906

G (μdS) m

⇀

3

⇀

( r − (0, 0, b))

| r − (0, 0, b)|

Here G is the gravitational constant, μdS is the mass of the tiny piece of shell, m is the mass of the particle

and r − (0, 0, b) is the vector from the particle to the piece of shell. If we work in spherical coordinates, as we did in Example 3.3.6, ⇀

2

dS = a

sin φ dφdθ

and ⇀

^ = a sin φ cos θ ^ ı ı + a sin φ sin θ ^ ȷ ȷ + a cos φ k

r

^ = a sin φ cos θ ^ ı ı + a sin φ sin θ ^ ȷ ȷ + (a cos φ − b) k

⇀

r − (0, 0, b) 2

⇀

2

| r − (0, 0, b)|

2

=a

+b

− 2ab cos φ

The total force is then π

⇀

2

F = Gμm a

∫

^ a sin φ cos θ ^ ı ı +a sin φ sin θ ^ ȷ ȷ +(a cos φ−b) k

2π

dφ ∫

0

dθ  sin φ  2

0

2

[a

+b

3/2

− 2ab cos φ ]

Note for future reference that the square root in [a + b − 2ab cos φ] [ b + a − 2ab cos φ ] is the length of r − (0, 0, b), which is positive. 2

2

2

2

3/2

is the positive square root because

⇀

1/2

This integral is a little different than other integrals that we have encountered so far in that the integrand is a vector. By definition 11, ∬

^ [ G1 ^ ı ı + G2 ^ ȷ ȷ + G3 k] dS = ^ ı ı ∬

S

G1  dS + ^ ȷ ȷ ∬

S

^ G2  dS + k ∬

S

G3  dS

S

so we just have to compute the three components separately. In our case, the ^ ı ı and ^ ȷ ȷ components π

⇀

2

F⋅ ^ ı ı = Gμm a

∫

⎡ dφ

⎣

0

π

⇀

2

F⋅^ ȷ ȷ = Gμm a

∫

2π

0

cos θ dθ = ∫

2π

0

2

+b

2

π

2

+b

0

2π

3/2

⎦

0

− 2ab cos φ ]

a cos φ − b dθ  sin φ  3/2

[ a2 + b2 − 2ab cos φ ]

π

0

⎤ dθ  sin θ

so that

0

2^ = 2πGμm a k ∫

⎦

0

− 2ab cos φ ]

2π

dφ ∫

⎤ dθ  cos θ

∫ [a

sin θ dθ = 0

2^ F = Gμm a k ∫

∫

a sin φ

⎣

⇀

3/2

sin φ 

0

are both zero 12 because ∫

2

[a

⎡ dφ

2π

a sin φ

sin φ 

a cos φ − b dφ  sin φ  2

[a

2

+b

3/2

− 2ab cos φ ]

To evaluate this integral we substitute

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2

2

u =a

2

+b

2

a − 2ab cos φ

du = 2ab sin φ dφ

+b

−u

cos φ = 2ab

When φ = 0,

2

u = (a − b)

and when φ = π, u = (a + b)

2

so

,

2

F =

(a+b)

^ k∫

b

3/2

2

u

(a−b)

πGμma =

2

3/2

u

2

πGμma

2

^ k[(

2

a

−1/2

−b

u )

b 1/2

2b

du 

(a−b)

=

2

a −b −u

(a+b)

^ k∫

b

−b

2b

du 

2

Recalling that u

2

a +b −u

2

πGμma

⇀

2b

(a+b)

u )

2b

−1/2

2

1/2

1 −(

] 1/2

2

(a−b)

is the positive square root, πGμma

⇀

F =

2

^ k[(

b

2

−a

1 )

b

b

2

a+b −

b

2

−a

−(

a+b

|a − b|

1 )

b

b

+

] b

|a − b|

If b > a, so that |a − b| = b − a πGμma

⇀

F

=

^ k[

b −a

a+b −

b

b

a+b −

b

2

G(4π a μ)m

b −a +

b

] =− b

2

^ k

b

If b < a, so that |a − b| = a − b πGμma

⇀

F

=

^ k[

b

b −a

a+b −

b

a+b +

b

a−b +

b

] =0 b

The moral 13 is if the particle is inside the shell, it feels no gravitational force at all, and if the particle is outside the shell, it feels the same gravitational force as it would if the entire mass of the shell (4π a were concentrated at the centre of the shell.

2

)

μ

 Example 3.3.11. Optional — Gravity Train The “Gravity Train” 14 refers to the following curious, though admittedly not very practical, thought experiment. Pretend that the Earth is a perfect sphere of radius R and that it has a constant mass density ρ. Pick any two distinct points on the surface of the Earth. Call them V and M . Bore a tunnel straight through the Earth from V to M . Place a train in the tunnel at V . Assume that the only forces acting on the train are gravity, G, and a normal force, N, that the tunnel imposes on the train to keep it in the tunnel. In particular, there are no frictional forces, like air resistance, and the train does not have an engine. Release the train and assume that it does not melt as it passes through the centre of the Earth.

What happens? We'll simplify our analysis of the motion of the train by picking a convenient coordinate system. First translate our coordinate system so that the centre of the Earth, call it O, is at the origin, (0, 0, 0). Then rotate our coordinate system about the origin so that the origin, V and M all lie in the xz-plane.

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Then rotate our coordinate system about the y -axis so that V and M have the same z -coordinate Z ≥ 0. So the coordinates − − − − − − − − − − − − − − of V and M are ( ± √R − Z , 0 , Z). Let's suppose that V is at (√R − Z , 0 , Z) and M is at − − − − − − − ( − √R − Z , 0 , Z). It really doesn't matter which is which, but we can always arrange that it is V at − − − − − − − √ (+ R −Z , 0 , Z) by rotating around the z -axis by 180 if necessary. 2

2

2

2

2

2

2

2

∘

The y - and z -coordinates of the train are always fixed at and look at the x-component of Newton's law of motion.

and

0

respectively. So let's call the x-coordinate at time

Z,

t x(t),

ma = G + N

It is ′′

m x (t) = G ⋅ ^ ı ı

because the normal force N has no ^ ı ı component. Recall that Newton's law of gravity says that GM m G =−

⇀ 3

⇀

r

|r|

where G is the gravitational constant, r is the vector from O to the train, and m is the mass of the train. In this case, because of our computation in Example 3.3.10, the train only feels gravity from shells of the Earth that are inside the train, so that M is the mass of the ⇀

part of the Earth whose distance to the centre of the Earth is no more than | r |. So ⇀

4 M =

⇀ 3

π| r | ρ 3

and Gm

′′

m x (t) = −

⇀ 3

|r|

4  

⇀ 3

⇀

π| r | ρ  r ⋅ ^ ı ı 3

so that ′′

4πGρ

x (t) +

x(t) = 0 3

This is exactly the differential equation of simple harmonic motion. We have seen it before in Example 2.2.7. Except for the 4πGρ

constant , it is identical to the equation solved in Example A.9.4 of the Appendix A.9, entitled “Review of Linear Ordinary Differential Equations”. The general solution is 3

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− −−− − − −−− − 4πGρ 4πGρ  t) + C2 sin( √  t) 3 3

x(t) = C1 cos( √

with C and C being arbitrary constants. If we release the train, from rest, at − − − − − − − so that C = √R − Z , C = 0 and 1

2

2

1

2

t = 0,

− − − − − − − 2 2 −Z

then x(0) = √R

and

′

x (0) = 0

2

− −−− − − − − − − − − 4πGρ 2 2 √ √ x(t) = R −Z cos(  t) 3 − − − − − − − 2 2 −Z .

The train reaches M when x(t) = −√R

That is, when

− − − − cos( √

4πGρ 3

So the transit time, T , from V to

 t) = −1.

M

obeys − −−− − 4πGρ  T = π 3

√

− −−− −

− −− −

3 ⟹

T = π√

3π

4πGρ

Notice that this transit time depends only on the gravitational constant completely independent of

G

=√

4Gρ

and the density of the Earth

ρ.

In particular it is

where V and M are and, in particular, how close together V and M are, and also of the radius of the Earth. In the case of the Earth, the transit time is about 42 minutes.

Optional — Dropping Higher Order Terms in du, dv ^ dS and dS formulae for In the course of deriving 3.3.1, that is, n

we approximated, for example, the vectors − − − → P0 P1 − − − → P0 P2

⇀

⇀

⇀

∂r

⇀

= r (u0 + du, v0 ) − r (u0 , v0 ) =

∂u

∂r (u0 , v0 ) du + E1 ≈

⇀

⇀

= ∂v

∂r (u0 , v0 ) dv + E2 ≈

where E is bounded 15 by a constant times du and E is bounded by a constant times dv drop E and E . 2

2

1

1

(u0 , v0 ) du

⇀

∂r

⇀

= r (u0 , v0 + dv) − r (u0 , v0 )

∂u

2

∂v .

(u0 , v0 ) dv

That is, we assumed that we could just

2

So we approximated − − − → − − − → ∣P P × P P ∣ ∣ 0 1 0 2∣

⇀

∣ = [ ∣

⇀

∂r

∂u

∂r (u0 , v0 ) du + E1 ] × [

⇀

=

∣ ∣

∂r (u0 , v0 ) du ×

⇀

≈

∣ ∣

∂r (u0 , v0 ) du ×

where the length of the vector E is bounded by a constant times does not change the value of the integral at all 16. 3

∂v

(u0 , v0 ) dv + E3

∣ ∣

⇀

∂r

∂u

∣ (u0 , v0 ) dv + E2 ] ∣

⇀

∂r

∂u

∂v

∂v 2

du

∣ (u0 , v0 ) dv ∣ 2

dv + du dv .

We'll now see why dropping terms like

E3

Suppose that our domain of integration consists of all (u, v)'s in a rectangle of width A and height B, as in the figure below.

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Subdivide the rectangle into a grid of n × n small subrectangles by drawing lines of constant v (the red lines in the figure) and lines of constant v (the blue lines in the figure). Each subrectangle has width du = and height dv = . Now suppose that in setting up the integral we make, for each subrectangle, an error that is bounded by some constant times 2

du

2

dv + du dv

2

A =(

B

A

) n

+ n

B (

n

2

)

A

B

n

n

AB(A + B) =

3

n

n

Because there are a total of n subrectangles, the total error that we have introduced, for all of these subrectangles, is no larger than a constant times 2

2

n

AB(A + B) ×

AB(A + B) =

3

n

n

When we define our integral by taking the limit n → 0 of the Riemann sums, this error converges to exactly 0.

Exercises Stage 1  1 Let 0 < θ
0. Denote by S the part of the surface z = y

tan θ

with 0 ≤ x ≤ a,

0 ≤ y ≤ b.

1. Find the surface area of S without using any calculus. 2. Find the surface area of S by using (3.3.2).

 2 Let a, b, c > 0. Denote by S the triangle with vertices (a, 0, 0), (0, b, 0) and (0, 0, c). 1. Find the surface area of S in three different ways, each using (3.3.2). 2. Denote by T the projection of S onto the xy-plane. (It is the triangle with vertices (0, 0, 0) (a, 0, 0) and (0, b, 0).) Similarly use T to denote the projection of S onto the xz-plane and T to denote the projection of S onto the yz-plane. Show that xy

xz

yz

− −−−−−−−−−−−−−−−−−−−−−−−−−−−− − 2

Area(S) = √ Area(Txy )

2

+ Area(Txz )

2

+ Area(Tyz )

 3 Let a, h > 0. Denote by S the part of the cylinder x

2

+z

2

2

=a

with x ≥ 0, 0 ≤ y ≤ h and z ≥ 0.

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1. Find the surface area of S without using any calculus. 2. Parametrize S by π

^ r (θ, y) = a cos θ  ^ ı ı +y ^ ȷ ȷ + a sin θ k

⇀

0 ≤θ ≤

,  0 ≤ y ≤ h 2

Find the surface area of S by using (3.3.1).

Stage 2  4 Let S be the part of the surface axis, that is,

z = xy

lying inside the cylinder

2

I =∬

(x

2

x

+y

2

= 3.

Find the moment of inertia of

S

about the z -

2

+ y ) dS

S

 5✳ Find the surface area of the part of the paraboloid z = a

2

2

−x

−y

2

which lies above the xy--plane.

 6✳ Find the area of the portion of the cone z

2

2

=x

+y

2

lying between the planes z = 2 and z = 3.

 7✳ Determine the surface area of the surface given by z =

2 3

3/2

(x

+y

3/2

),

over the square 0 ≤ x ≤ 1,

0 ≤ y ≤ 1.

 8✳ 1. To find the surface area of the surface z = f (x, y) above the region D, we integrate ∬ F (x, y) dA. What is F (x, y)? – 2. Consider a “Death Star”, a ball of radius 2 centred at the origin with another ball of radius 2 centred at (0, 0, 2√3) cut out of it. The diagram below shows the slice where y = 0. D

1. The Rebels want to paint part of the surface of Death Star hot pink; specifically, the concave part (indicated with a thick line in the diagram). To help them determine how much paint is needed, carefully fill in the missing parts of this integral: surface area = ∫

∫

 dr dθ

2. What is the total surface area of the Death Star?

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 9✳ Find the area of the cone z

2

2

=x

+y

2

between z = 1 and z = 16.

 10 ✳ − −−−−−−−− − 2 2 2 z = √a − x − y

Find the surface area of that part of the hemisphere (x −

a 2

2

)

+y

2

=(

a 2

which lies within the cylinder

2

) .

 11 The cylinder  x

2

+y

2

= 2x 

cuts out a portion S of the upper half of the cone  x

2

4

∬

(x

−y

4

2

+y z

2

2

2

−z x

+y

2

2

=z .

Compute

+ 1) dS

S

 12 Find the surface area of the torus obtained by rotating the circle (x − R) about the z -axis.

2

+z

2

2

=r

(the circle is contained in the

-plane)

xz

 13 A spherical shell of radius a is centred at the origin. Find the centroid (i.e. the centre of mass with constant density) of the part of the sphere that lies in the first octant.

 14 Find the area of that part of the cylinder x

2

+y

2

= 2ay

lying outside z

2

2

=x

2

+y .

 15 ✳ Let a and b be positive constants, and let S be the part of the conical surface 2

a z

2

2

2

= b (x

2

+y )

where 0 ≤ z ≤ b. Consider the surface integral 2

I =∬

(x

2

+ y ) dS.

S

1. Express I as a double integral over a disk in the xy-plane. 2. Use the parametrization x = t cos θ, y = t sin θ, etc., to express I as a double integral over a suitable region in the tθ plane. 3. Evaluate I using the method of your choice.

 16 Evaluate, for each of the following, the flux   ∬

S

⇀

⇀

^ dS  F⋅n

^ is the outward normal to the surface S. where n

n

^ 1.  F = (x + y + z ) (x ^ ı ı +y ^ ȷ ȷ + z k)  and the surface S is the sphere x + y + z = a . ^ 2.  F = x ^ ı ı +y ^ ȷ ȷ + z k  and S is the surface of the rectangular box 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c. 2

2

2

2

2

2

2

⇀

− −− −− − 2 2 +y .

⇀

^ 3.  F = y ^ ı ı + z k  and S is the surface of the solid cone 0 ≤ z ≤ 1 − √x

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 17 ✳ Let S be the part of the surface x

2

2

1. Find ∬ S

+y

2

that lies above the square −1 ≤ x ≤ 1,

+ 2z = 2

−1 ≤ y ≤ 1.

2

x +y − −−−−−−− −  dS. 2 2 √1 + x + y ⇀

^ 2. Find the flux of F = x ^ ı ı + y^ ȷ ȷ + zk upward through S.

 18 ✳ Let S be the part of the surface z = xy that lies above the square 0 ≤ x ≤ 1,

0 ≤y ≤1

in the xy-plane.

2

1. Find ∬ S

x y − −−−−−−− −  dS. 2 2 √1 + x + y ⇀

^ 2. Find the flux of F = x ^ ı ı + y^ ȷ ȷ + k upward through S.

 19 ✳ Find the area of the part of the surface z = y

that lies above 0 ≤ x, y ≤ 1.

3/2

 20 ✳ Let S be spherical cap which consists of the part of the sphere x f (x, y, z) = (2 − z)(x + y ). Calculate

2

2

+y

2

2

+ (z − 2 )

which lies under the plane z = 1. Let

=4

2

∬

f (x, y, z) dS

S

 21 ✳ 1. Find a parametrization of the surface S of the cone whose vertex is at the point (0, 0, 3), and whose base is the circle x + y = 4 in the xy-plane. Only the cone surface belongs to S, not the base. Be careful to include the domain for the parameters. 2. Find the z -coordinate of the centre of mass of the surface S from (a). 2

2

 22 ✳ Let S be the surface of a cone of height a and base radius interiour of the cone. Find the centre of mass of S.

a.

The surface

S

does not include the base of the cone or the

Locate the cone in a coordinate system so that its base is in the xy-plane, and its vertex on the z -axis. So the vertex will be the point (0, 0, a). The base is a circle of radius a in the xy-plane with centre at the origin. The cone surface is characterized by the fact that for every point of S, the distance from the z -axis and the distance from the xy-plane add up to a.

 23 ✳ Let S be the portion of the elliptical cylinder outward normal to S. Let parameterization of S.

2

x

⇀

F =x ^ ı ı + xyz ^ ȷ ȷ + zy

+ 4

1 4

^ k.

y

2

=1

lying between the planes

Calculate the flux integral

z =0

and

⇀

∬

S

^ dS F⋅n

z =1

and let

^ n

denote the

directly, using an appropriate

 24 ✳ Evaluate the flux integral

3.3.18

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⇀

^ dS F⋅n

∬ S ⇀

^ where F(x, y, z) = (x + 1) ^ ı ı + (y + 1) ^ ȷ ȷ + 2z k, and S is the part of the paraboloid z = 4 − x − y triangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x. S is oriented so that its unit normal has a negative z -component. 2

2

that lies above the

 25 ✳ Evaluate the surface integral ∬

2

x y  dS

S

where S is the part of the sphere x

2

+y

2

+z

2

=2

− − − − − − 2 2 +z .

for which x ≥ √y

 26 ✳ Let S be the surface given by the equation 2

x

+z

2

2

= sin

y

lying between the planes y = 0 and y = π. Evaluate the integral − − − − − − − − 2

√ 1 + cos

∬

y dS

S

 27 ✳ Let S be the part of the paraboloid z = 1 − x

2

−y

2

lying above the xy-plane. At (x, y, z) S has density z ρ(x, y, z) =

− −−− − √ 5 − 4z

Find the centre of mass of S.

 28 ✳ Let S be the part of the plane x +y +z = 2 ^ ^ has a positive k that lies in the first octant oriented so that n component. Let ⇀

^ F =x ^ ı ı +y ^ ȷ ȷ +zk

Evaluate the flux integral ⇀

^ dS F⋅n

∬ S

 29 ✳ ⇀

⇀

^ dS of the vector field F(x, y, z) = (x, y, z) upwards (with respect to the z -axis) through the surface Find the net flux ∬ F ⋅ n S parametrized r = (u v , u v , uv) for 0 ≤ u ≤ 1, 0 ≤ v ≤ 3. S

⇀

2

2

3.3.19

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 30 ✳ Let S be the surface obtained by revolving the curve z = e , 0 ≤ y ≤ 1, around the y -axis, with the orientation of S having n ^ pointing toward the y -axis. y

1. Draw a picture of S and find a parameterization of S. 2. Compute the integral ∬ e dS. 3. Compute the flux integral ∬ F ⋅ n ^ dS where F = (x, 0, z). y

S

⇀

⇀

S

 31 ✳ Compute the net outward flux of the vector field ⇀

^ x ^ ı ı +y ^ ȷ ȷ +zk

r

⇀

F =

=

⇀

|r|

− −−−−−−−− − √ x2 + y 2 + z 2

across the boundary of the region between the spheres of radius 1 and radius 2 centred at the origin.

 32 ✳ Evaluate the surface integral ∬

S

z

2

dS

where S is the part of the cone x

2

+y

2

= 4z

2

where 0 ≤ x ≤ y and 0 ≤ z ≤ 1.

 33 ✳ Compute the flux integral ∬

S

⇀

^ dS, F⋅n

where 1

⇀

F =(−

3

x

− xy

2

1 , −

2

and S is the part of the paraboloid downwards.

2

z = 5 −x

−y

y

3

2

, z )

2

lying inside the cylinder

2

2

x

+y

2

≤ 4,

with orientation pointing

 34 ✳ Let the thin shell S consist of the part of the surface z surface density given by ρ(x, y, z) = 3z kg per unit area.

2

= 2xy

with

x ≥ 1, y ≥ 1

and

z ≤ 2.

Find the mass of

S

if it has

 35 ✳ Let S be the portion of the paraboloid x = y + z ^ Find the flux of F = 2 ^ ı ı +z^ ȷ ȷ + y k out of S. 2

2

^ is so chosen that n ^ ⋅ ^ that satisfies x ≤ 2y. Its unit normal vector n ı ı > 0.

⇀

 36 ✳ Let S denote the portion of the paraboloid z = 1 − ^ k

1 4

2

x

−y

2

for which z ≥ 0. Orient S so that its unit normal has a positive

component. Let ⇀

F(x, y, z) = (3 y

Evaluate the surface integral ∬

S

2

2 ^ + z) ^ ı ı + (x − x ) ^ ȷ ȷ +k

⇀

^ dS. ∇×F⋅ n

 37 Let

S

be the boundary of the apple core bounded by the sphere

Find the flux integral   ∬

S

⇀

^ dS  F⋅n

where

⇀

^ F =x ^ ı ı +y ^ ȷ ȷ +zk

2

x

+y

2

+z

2

= 16

and the hyperboloid

2

x

+y

2

−z

2

= 8.

^ is the outward normal to the surface S. and n

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Stage 3  38 ✳ 1. Consider the surface S given by the equation 2

x

+z

Find an equation for the tangent plane to S at the point ( 2. Compute the integral

1

2

= cos

π

,

2

2

∬

1

,

4

2

y

).

sin y dS

S

where S is the part of the surface from (a) lying between the planes y = 0 and y =

1 2

π.

 39 ✳ Let

f

be a function on

3

R

such that all its first order partial derivatives are continuous. Let ⇀

{(x, y, z)|f (x, y, z) = c}

for some c ∈ R. Assume that ∇f

⇀

≠ 0

⇀

⇀

be the surface

S

⇀

on S. Let F be the gradient field F = ∇f .

1. Let C be a piecewise smooth curve contained in S (not necessarily closed). Must it be true that ∫ why. 2. Prove that for any vector field G,

⇀

C

⇀

F ⋅ d r = 0?

Explain

⇀

∬

(F × G) ⋅ n ^ dS = 0.

S

 40 ✳ 1. Give parametric descriptions of the form the domains of your parametrizations.

⇀

r (u, v) = (x(u, v) , y(u, v) , z(u, v))

for the following surfaces. Be sure to state

1. The part of the plane 2x + 4y + 3z = 16 in the first octant {(x, y, z)|x ≥ 0,  y ≥ 0,  z ≥ 0}

2. The cap of the sphere x + y 3. The hyperboloid z = 1 + x 2

2

2

2

+z +y

2

= 16

2

–

for 4/√2 ≤ z ≤ 4.

for 1 ≤ z ≤ 10.

2. Use your parametrization from part (a) to compute the surface area of the cap of the sphere x

2

+y

2

+z

2

= 16

for

– 4/ √2 ≤ z ≤ 4.

 41 ✳ Let S be the part of the sphere x

2

+y

2

+z

2

=2

where y ≥ 1, oriented away from the origin.

1. Compute ∬

y

3

dS

S

2. Compute ∬

^ ^ dS (xy ^ ı ı + xz ^ ȷ ȷ + zy k) ⋅ n

S

3.3.21

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 42 ✳ Let S be the part of the surface (x + y + 1) where

2

+z

2

⇀

=4

which lies in the first octant. Find the flux of F downwards through S

⇀

F = xy ^ ı ı + (z − xy) ^ ȷ ȷ

1. As we mentioned above, the approximation below becomes exact when the limit du, dv → 0 is taken in the definition of the integral. See the optional §3.3.5. 2. This is called the implicit function theorem. We will not prove it. But it is not so hard to understand why it is true, if one thinks in terms of the Taylor expansion of G about the point. For simplicity, let's suppose that the point is (0, 0, 0) and G happens to be exactly equal to its first order Taylor expansion about (0, 0, 0). That is, G(x, y, z) = A + Bx + C y + Dz, for some constants A, B, C , D. Since (0, 0, 0) is on the surface, A = K. As = D ≠ 0 we can easily solve G(x, y, z) = K for z as a function of x and y. Namely z = (−Bx − C y). The general proof is based on the fact that, under reasonable hypotheses, the first order Taylor expansion is a good approximation to G near (0, 0, 0). 3. The symbols r, θ, z are the standard mathematics symbols for the cylindrical coordinates. Appendix A.7 gives another set of symbols that is commonly used in the physical sciences and engineering. 4. As we have noted before, the spherical coordinate system really breaks down at φ = 0, because ρ = 1, φ = 0 gives the same point, namely the north pole (0, 0, 1), for all values of θ. We should really treat our integral like an improper integral, first integrating over ε < φ ≤ and then taking the limit ε → 0 . However the breakdown of the spherical coordinate system at φ = 0, just like the breakdown of polar coordinates at r = 0, rarely causes problem and it is routine to skip the “improper integral” step. 5. We promise! 6. This answer for dS is a very clean. Think about why. Hint: review the discussion following 3.3.2. 7. If you have forgotten why, sketch the graph. 8. We did so previously, with different variable names, in Example 3.2.2. 9. Again the formula for dS is very neat. Think about why. 10. A favourite of science fiction and fantasy writers. Plug “subterranean fiction” into your favourite search engine. While you're at it, also try “gravity train”. We'll look at it in the optional Example 3.3.11. 11. Under this definition we still have ∬ (A + B) dS = ∬ A dS + ∬ B dS. 12. Think about why the ^ ı ı and ^ ȷ ȷ components should both be zero. Think symmetry. 13. These two results appeared in Isaac Newton's Principia Mathematica (1687). They are known as Newton's “superb theorems”. 14. The British physicist and architect (he was Surveyor to the City of London and chief assistant to Christopher Wren) Robert Hooke (1635--1703) wrote about the gravity train idea in a letter to Isaac Newton. A gravity train was used in the 2012 movie Total Recall. 15. Remember the error in the Taylor polynomial approximations. 16. See the optional §1.1.6 of the CLP-2 text for an analogous argument concerning Riemann sums. ∂G ∂z

1

D

π

+

2

This page titled 3.3: Surface Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

3.3.22

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3.4: Interpretation of Flux Integrals We defined, in §3.3, two types of integrals over surfaces. We have seen, in §3.3.4, some applications that lead to integrals of the ^ dS. Recall that integrals of this type are type ∬ ρ dS. We now look at one application that leads to integrals of the type ∬ F ⋅ n called flux integrals. Imagine a fluid with ⇀

S

S

the density of the fluid (say in kilograms per cubic meter) at position (x, y, z) and time t being ρ(x, y, z, t) and with the velocity of the fluid (say in meters per second) at position (x, y, z) and time t being v (x, y, z, t). ⇀

We are going to determine the rate (say in kilograms per second) at which the fluid is flowing through a tiny piece dS of surface at (x, y, z). During a tiny time interval of length dt about time t, fluid near dS moves v (x, y, z, t)dt. The green line in the figure ^ =n ^ (x, y, z) is a unit normal vector to dS. below is a side view of dS and n ⇀

So during that tiny time interval the red line moves to the green line and the green line moves to the blue line so that the fluid filling the dark grey region below the green line crosses through dS and moves to light grey region above the green line. ^ and If we denote by θ the angle between n

⇀

v dt,

the volume of fluid that crosses through dS during the time interval dt is the volume whose side view is the dark grey region below the green line. This region has base dS and height | v dt| cos θ and so has volume ⇀

⇀

⇀

^ (x, y, z) dt dS | v (x, y, z, t)dt| cos θ dS = v (x, y, z, t) ⋅ n ^ (x, y, z) has length one. because n The mass of fluid that crosses dS during the time interval dt is then ⇀

^ (x, y, z) dt dS ρ(x, y, z, t) v (x, y, z, t) ⋅ n

and the rate at which fluid is crossing through dS is ⇀

^ (x, y, z) dS ρ(x, y, z, t) v (x, y, z, t) ⋅ n

Integrating dS over a surface S, we conclude that

 Lemma 3.4.1 The rate at which fluid mass is crossing through a surface S is the flux integral ∬

⇀

ρ(x, y, z, t) v (x, y, z, t) ⋅ n ^ (x, y, z) dS

S

^ (x, y, z) is a unit normal to S at (x, y, z). If the flux Here ρ is the density of the fluid, v is the velocity field of the fluid, and n ^ ^. integral is positive the fluid is crossing in the direction n. If it is negative the fluid is crossing opposite to the direction of n The rate at which volume of fluid is crossing through a surface S is the flux integral ⇀

∬

⇀

^ (x, y, z) dS v (x, y, z, t) ⋅ n

S

3.4.1

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Examples of Flux Integrals  Example 3.4.2. Point Source In Example 2.1.2, we found that the vector field of a point source 1 (in three dimensions) that creates 4πm liters per second is m

⇀

v (x, y, z) =

2

^ r(x, y, z)

r(x, y, z)

where ^ x^ ı ı + y^ ȷ ȷ + zk

− −−−−−−−− − 2

r(x, y, z) = √ x

+y

2

+z

2

^ r(x, y, z) =

− −−−−−−−− − √ x2 + y 2 + z 2

We sketched it in Figure 2.1.3. We'll now compute the flux of this vector field across a sphere centred on the origin. Suppose that the sphere has radius R.

Then the outward 2 pointing normal at a point (x, y, z) on the sphere is ^ x^ ı ı + y^ ȷ ȷ + zk

^ (x, y, z) = ^ n r(x, y, z) =

^ x^ ı ı + y^ ȷ ȷ + zk

− −−−−−−−− − = √ x2 + y 2 + z 2

R

Note that ^ r(x, y, z) ⋅ ^ r(x, y, z) = 1 and that, on the sphere, r(x, y, z) = R. So the flux of ∬

m

⇀

^  dS = ∬ v ⋅n

2

S

r(x, y, z)

S

R

S

m =∬

2

v

outward through the sphere is

^ r(x, y, z) ⋅ ^ r(x, y, z) dS

m  dS =

⇀

2

2

4π R

R

= 4πm

This is the rate at which volume of fluid is exiting the sphere. In our derivation of the vector field we assumed that the fluid is incompressible, so it is also the rate at which the point source is creating fluid.

 Example 3.4.3. Vortex In Figure 2.1.6, we sketched the vector field (in two dimensions) ⇀

v (x, y) = Ω( − y ^ ı ı + x^ ȷ ȷ)

We'll now compute the flux of this vector field across a circle C centred on the origin. Suppose that the circle has radius R.

By definition, in two dimensions, the flux of a vector field across a curve C is ∫

C

⇀

^  ds. v ⋅n

This is the natural analog of the flux in three dimensions — the surface S has been replaced by the curve area dS of a tiny piece of S has been replaced by the arc length ds of a tiny piece of C .

3.4.2

C,

and the surface

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The outward pointing unit normal at a point (x, y) on our circle C is x^ ı ı + y^ ȷ ȷ

^ (x, y) = n

x^ ı ı + y^ ȷ ȷ

− −− −− − = 2 2 √x + y

R

So Ω

⇀

^ (x, y) = v (x, y) ⋅ n

( − y^ ı ı + x^ ȷ ȷ ) ⋅ (x ^ ı ı + y^ ȷ ȷ) = 0 R

and the flux across C is ⇀

^  ds = 0 v ⋅n

∫ C

This should not be a surprise — no fluid is crossing C at all. This is exactly what we would expect from looking at the arrows in Figure 2.1.6 or at the stream lines in Example 2.2.6.

 Example 3.4.4 Evaluate   ∬

S

⇀

^  dS  F⋅n

where ⇀

^ F(x, y, z) = (x + y) ^ ı ı + (y + z) ^ ȷ ȷ + (x + z) k

and S is the boundary of V

2

= {(x, y, z)|0 ≤ x

+y

2

≤ 9,  0 ≤ z ≤ 5} ,

3 ^ is the outward normal to S. and n

Solution The volume V looks like a tin can of radius 3 and height 5.

It is natural to decompose its surface S into three parts St Sb Ss

2

= {(x, y, z)|0 ≤ x

2

= {(x, y, z)|0 ≤ x 2

= {(x, y, z)| x

+y

+y +y

2

2

2

≤ 9,  z = 5} = the top ≤ 9,  z = 0} = the bottom

= 9,  0 ≤ z ≤ 5} = the side

We'll compute the flux through each of the three parts separately and then add them together. ^ The Top: On the top, the outward pointing normal to S is n ^ = k and dS = dxdy. This is probably intuitively obvious. But if it ^ isn't, you can always derive it by parametrizing the top by r (x, y) = x ^ ı ı +y ^ ȷ ȷ + 5 k with x + y ≤ 9. So the flux through the top is ⇀

2

2

⇀

∬

^  dS F⋅n

=∬

(x + z) dxdy = ∬ 2

St

x +y

2

(x + 5) dxdy 2

≤9

x +y

2

≤9

z=5

The integral ∬

2

x +y

2

≤9

x dxdy = 0

since x is odd and the domain of integration is symmetric about x = 0. So ⇀

∬ St

2

F⋅n ^  dS = ∬

5 dxdy = 5π(3 ) 2

x +y

2

= 45π

≤9

^ ^ = −k The Bottom: On the bottom, the outward pointing normal to S is n and dS = dxdy. So the flux through the bottom is

3.4.3

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⇀

^  dS F⋅n

∬

= −∬

(x + z) dxdy = − ∬ 2

Sb

x +y

2

x dxdy = 0 2

≤9

2

x +y

≤9

z=0

again since x is odd and the domain of integration is symmetric about x = 0. The Side: We can parametrize the side by using cylindrical coordinates. ⇀

r (θ, z) = (3 cos θ , 3 sin θ , z)

0 ≤ θ < 2π,  0 ≤ z ≤ 5

Then, using 3.3.1,

⇀

∂r

= (−3 sin θ , 3 cos θ , 0) ∂θ ⇀

∂r

= (0 , 0 , 1) ∂z ⇀

^ dS = n

⇀

∂r

∂r ×

 dθ dz

∂θ

∂z

= (3 cos θ , 3 sin θ , 0) dθ dz ^ = (cos θ , Note that n

sin θ , 0)

is outward pointing 4 , as desired. Continuing,

⇀

^ = 3(cos θ+sin θ) ^ ı ı + (3 sin θ+z) ^ ȷ ȷ + (3 cos θ+z) k

F(x(θ, z), y(θ, z), z(θ, z)) ⇀

2

F⋅n ^ dS

2

= {9 cos

θ+3 sin θ cos θ+9 sin

θ+3z sin θ} dθ dz

3 = {9 +

sin(2θ) + 3z sin θ} dθ dz 2

So the flux through the side is 2π

⇀

∬

^  dS F⋅n

Ss

=∫

5

dθ ∫

0

3 dz {9 +

2π

=9∫

sin(2θ) + 3z sin θ} 2

0 5

dθ ∫

0

2π

dz

since  ∫

0

2π

sin θ dθ = ∫

0

sin(2θ) dθ = 0

0

= 9 × 2π × 5 = 90π

and the total flux is ⇀

∬

^  dS F⋅n

=∬

S

⇀

⇀

⇀

^  dS + ∬ F⋅n

^  dS + ∬ F⋅n

^  dS F⋅n

St

Sb

Ss

= 45π + 0 + 90π = 135π

 Example 3.4.5 ⇀

⇀

Evaluate   ∬ F ⋅ n ^  dS  where  F(x, y, z) = x ^ ı ı + 2y z ≥ 0, and n ^ is the upward pointing unit normal. S

4

2

^ ^ ȷ ȷ + zk,  

3.4.4

S

is the half of the surface

 

1 4

2

x

+

1 9

y

2

+z

2

= 1 

with

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Solution 1 We start by parametrizing the surface, which is half of an ellipsoid. By way of motivation for the parametrization, recall that spherical coordinates, with ρ = 1, provide a natural way to parametrize the sphere x + y + z = 1. Namely x = cos θ sin φ, y = sin θ sin φ, z = cos φ. The reason that these spherical coordinates work is that the trig identity cos α + sin α = 1 implies 2

2

2

2

2

2

x

+y

2

2

= cos

2

θ sin

2

φ + sin

2

2

θ sin

φ = sin

φ

and then 2

(x

2

+y )+z

2

2

2

= sin

φ + cos

φ =1

The equation of our ellipsoid is 2

x (

)

2

y +(

2

)

+z

2

=1

3

so we can parametrize the ellipsoid by replacing x with choose the parametrization

y

and y with

x 2

in our parametrization of the sphere. That is, we

3

x(θ, φ) = 2 cos θ sin φ y(θ, φ) = 3 sin θ sin φ z(θ, φ) = cos φ

with (θ, φ) running over 0 ≤ θ ≤ 2π,  0 ≤ φ ≤ π/2. Note that 1

2

x(θ, φ )

1 +

4

2

2

y(θ, φ )

+ z(θ, φ )

=1

9

as desired. Then, using 3.3.1, ∂y

∂x (

∂z

, ∂θ

∂x (

, ∂θ ∂y

, ∂φ

) = (−2 sin θ sin φ , 3 cos θ sin φ , 0) ∂θ ∂z

, ∂φ

) = (2 cos θ cos φ , 3 sin θ cos φ , − sin φ) ∂φ ^ dS = −( n

∂y

∂x ,

∂z

∂θ

∂y

∂x

,

)×(

∂θ

∂θ 2

= −(−3 cos θ sin

, ∂φ

∂z ,

∂φ 2

φ, −2 sin θ sin

) dθdφ ∂φ

φ, −6 sin φ cos φ)dθdφ

^ dS was put there to make the z component of n ^ positive. (The problem specified that n ^ is to be The extra minus sign in n upward unit normal.) As ⇀

F(x(θ, φ) , y(θ, φ) , z(θ, φ)) 4

=2

4

cos

4

θ sin

2

φ  ^ ıı + 2 × 3

2

sin

2

θ sin

^ φ ^ ȷ ȷ + cos φ k

we have ⇀

^ dS F⋅n

4

= [3 × 2

5

cos

6

θ sin

2

3

φ+2 × 2 × 3

sin

4

θ sin

2

φ+6 sin φ cos

φ] dθdφ

and the desired integral π

⇀

∬

2π

2

^  dS F⋅n

S

=∫

dφ ∫

0

4

dθ [3 × 2

5

cos

6

θ sin

2

φ +2 ×2 ×3

3

sin

4

θ sin

2

+ 6 sin φ cos

Since   ∫

2π

0

m

cos

θ dθ = ∫

2π

0

m

sin

θ dθ = 0 

φ

0

φ]

for all odd 5 natural numbers m,

3.4.5

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π/2

⇀

^ dS F⋅n

∬

=∫

S

2π

0

π/2 2

dφ ∫

dθ 6 sin φ cos

0

dφ  sin φ cos

φ

0

1

π/2

3

= 12π[ −

2

φ = 12π ∫

cos

φ]

3

= 4π

0

The integral was evaluated by guessing (and checking) that done by substituting u = cos φ, du = − sin φ dφ.

−

1 3

3

cos

is an antiderivative of

φ

2

sin φ cos

φ.

It can also be

Solution 2 This time we'll parametrize the half-ellipsoid using a variant of cylindrical coordinates. x(r, θ) = 2r cos θ y(r, θ) = 3r sin θ − − − − − 2 z(r, θ) = √ 1 − r

with (r, θ) running over have

Because we built the factors of

0 ≤ θ ≤ 2π,  0 ≤ r ≤ 1.

2

and

2

3

into

x(r, θ)

and

y(r, θ),

we

2

x(r, θ)

y(r, θ)

2

+

=r

4

2

cos

2

2

θ+r

2

sin

θ =r

9 2

2

x(r, θ) ⟹

y(r, θ)

2

+ 4

2

− − − − −

2

+ z(r, θ)

2

+ (√ 1 − r

=r

)

=1

9

as desired. Further z(r, θ) ≥ 0 by our choice of square root in the definition of z(r, θ). So, using 3.3.1, ∂x (

∂y , ∂θ

∂x (

∂z ,

∂θ

∂y ,

) = (−2r sin θ, 3r cos θ, 0) ∂θ ∂z

,

∂r

∂r

r ) = (2 cos θ, 3 sin θ, −

∂r ∂x

^ dS = −( n

∂y ,

∂θ

∂θ 2

3r = −( −

∂z ,

∂x )×(

∂y ,

∂θ

2

2r ,−

∂z ,

∂r

cos θ

− − − − − √ 1 − r2

)

− − − − − √ 1 − r2

∂r

)dr dθ ∂r

sin θ

− − − − − √ 1 − r2

, −6r)dr dθ

^ dS was put there to make the z component of n ^ positive. Continuing, Once again, the extra minus sign in n ⇀

F(x(r, θ) , y(r, θ) , z(r, θ))

4

4

=2 r

4

cos

2

2

θ^ ıı + 2 × 3 r

2

sin

− − − − − 2 ^ θ^ ȷ ȷ + √1 − r k

6

⇀

^ dS F⋅n

4

= [3 × 2

4

r

5

− − − − − √ 1 − r2

cos

2

2

θ+2 3

r

− − − − − √ 1 − r2

3

sin

θ

− − − − − 2

+ 6r√ 1 − r

Again using that   ∫

2π

0

m

cos

θ dθ = ∫

2π

0

m

sin

θ dθ = 0  1

⇀

^ dS = ∫ F⋅n

∫ S

] dr dθ

for all odd natural numbers m,

2π

− − − − − 2 dθ 6r√ 1 − r

dr ∫

0

0 1

= 12π ∫

1 − − − − − 1 3/2 2 2 dr r√ 1 − r = 12π[ − (1 − r ) ]

3

0

0

= 4π

The integral was evaluated by guessing (and checking) that − done by substituting u = 1 − r , du = −2r dr.

1 3

2

(1 − r )

3/2

− − − − − 2 .

is an antiderivative of r√1 − r

It can also be

2

Solution 3

3.4.6

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The surface is of the form G(x, y, z) = 0 with G(x, y, z) =

1 4

2

x

1

+

9

⇀

x

∇G

^ dS = n

y

⇀

2y 9

^ ∇G ⋅ k x

y

4z

^ ^ ȷ ȷ + 2zk dx dy

^ ^ ȷ ȷ + k)dx dy

9z

5

x

⇀

^ dS F⋅n

⟹

Hence, using 3.3.3,

− 1.

2z

^ ıı +

=(

2

+z

^ ıı +

2

dx dy =

2

2y

=(

3

+

+ z)dx dy

4z

9z

⇀

^ dS, and consequently F ⋅ n ^ dS become infinite 6 It is true that n

as z → 0. So we should really treat the integral as an improper integral, first integrating over z ≥ ε and then taking the limit ε → 0 . But, as we shall see, the singularity is harmless. So it is standard to gloss over this point. On S, +

− − − − − − − − − − 2

z = z(x, y) = √1 −

y

2

x

4

−

9

2

and

x

y

+

4

2

so

≤ 1,

9

5

x

⇀

^ dS F⋅n

∫

=∬ x

S

Both

5

x

4z(x,y)

and

2y

3

9z(x,y)

2

4

y

+

9

2y

(

2

+ z(x, y)) dx dy

4z(x, y)

≤1

3

+ 9z(x, y)

are odd under x → −x,  y → −y and the domain of integration is even under x → −x,  y → −y,

so their integrals are zero and ⇀

^ dS = ∬ F⋅n

∫

x

S

2

4

y

+

z(x, y) dx dy

2

9

≤1

−−−−−−−−− − 2 y2 x

=∬ x

2

4

y

+

√1−

2

9

− 4

≤1

To evaluate this integral, first make the change of variables 7 x = 2X,

S

^ dS = ∫ F⋅n

S

+Y

2

to give

2π

− − − − − 2 dθ 6r√ 1 − r = 12π ∫

dr ∫

0

to give

≤1

Y = r sin θ, dXdY = r drdθ

1

⇀

∫

2

X

Then switch to polar coordinates, X = r cos θ,

dx = 2dX, y = 3Y , dy = 3dY

−−−−−−−−− − 2 2 √1 −X −Y  6 dX dY

⇀

^ dS = ∬ F⋅n

∫

 dx dy 9

0

− − − − − 2 dr r√ 1 − r

0

1 = 12π[ −

1

1

3/2

2

(1 − r )

]

3

= 4π

0

Solution 4 − − − − − − − − − − 2

The surface is of the form z = f (x, y) with f (x, y) = √1 −

^ dS n

∂f

∂f ^ ıı −

= [− ∂x

⎡ ⇀

^ dS = ⎢ ⟹ F⋅n

4

4

2y

+

^ ıı +

y 9

⎤

^ ȷ ȷ

^ + k⎥ dx dy − − − − − − − − − − 2

⎣√1−

y

2

x

4

−

⎦

9

−−−−−−−−− − 2 ⎤ 2 y x 1− − ⎥ dx dy 4 9 ⎦

√

y

−

Hence, using 3.3.2,

4

3

2

4

.

x

9

x

9

⎡

+ − − − − − − − − − − 2

⎣√1−

−

^ ^ ȷ ȷ +k] dx dy = ⎢

∂y

5

x

y

2

x

9

Note that our unit normal is upward pointing, as required. As in Solution 3, by the oddness of the integrand,

3.4.7

5

x

and y terms in the 3

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5

∫

x

⎡

⇀

^ dS = ∬ F⋅n x

S

2

4

y

+

⎢

2

9

4

≤1

2y

+

3

9

− − − − √  ⋯  

⎣

−−−−−−−−− − 2 2 ⎤ x y − ⎥ dx dy 4 9 ⎦

+√1 −

−−−−−−−−− − 2

√

=∬ x

2

4

y

+

2

9

y

x 1−

4

≤1

2

−

 dx dy 9

Now continue as in Solution 3. 1. You can imagine that a very small pipe pumps water to the origin. 2. It doesn't really matter which unit normal we pick here. We just have to be clear which one we're using. With the outward normal, the flux gives the rate at which fluid crosses the sphere in the outward direction. If we were to use the inward pointing normal, the flux would give the rate at which fluid crosses the sphere in the inward direction. 3. It is necessary that the problem specify, one way or another, whether n ^ is the inward pointing normal or the outward pointing ⇀

^  dS  is ambiguous. Think about where the orientation of the normal vector gets normal. Without this, the meaning of   ∬ F ⋅ n used in your solution. 4. To check, draw, in your head, a sketch of the top view of the can. “Top view” just means “ignore the z -coordinate”. The top view of the can is a circle of radius 3. Then, at a generic point, r = (cos θ, sin θ), on the can, draw the unit normal ^ = (cos θ , sin θ) with its tail at r . It is pointing away from the origin, just like r is. That is, n ^ is pointing outward. n 5. Look at the graphs of cos φ and sin φ. 6. That's because the ellipsoid is becoming vertical as z → 0, so that x and y are not really good parameters there. 7. The reader interested in general changes of variables in multidimensional integrals should look up “Jacobian determinant”. S

⇀

⇀

m

⇀

m

This page titled 3.4: Interpretation of Flux Integrals is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

3.4.8

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3.5: Orientation of Surfaces One thing that made the flux integrals of the last section possible is that we could choose sensible unit normal vectors section, we explain this more carefully.

^. n

In this

Consider the sphere x + y + z = 1. We can think of this surface as having two sides — an inside (the side you see when you are living inside the sphere) and an outside (the side you see when you are living outside the sphere). Concentrate on one point (x , y , z ) on the sphere. The surface x + y + z = 1 has precisely two unit normal vectors at (x , y , z ), namely 2

2

2

2

0

0

2

2

0

0

^ + = +(x0 , y0 , z0 ) n

0

0

^ − = −(x0 , y0 , z0 ) n

and

^ ^ We can view n as being associated to (or attached to) the outside of the sphere and n as being associated to (or attached to) the ^ ^ inside of the sphere. Note that, as we move over the sphere, both n and n change continuously. +

−

+

−

 Definition 3.5.1 An oriented surface is a surface together with a continuous function 3 ^ N : S → R

^ such that, for each point p of S, N (p) is a unit normal to S at p.

 Example 3.5.2. Sphere One orientation of the sphere S = {(x, y, z)|x

2

+y

2

+z

2

= 1}

is

^ N(x, y, z) = (x, y, z)

It associates to each point p of S the outward pointing unit normal to S at p. We can think of S with this orientation as being the outer side of S. The other orientation of the sphere S = {(x, y, z)|x

2

+y

2

+z

2

= 1}

is

^ N(x, y, z) = −(x, y, z)

It associates to each point p of S the inward pointing unit normal to S at p. We can think of S with this orientation as being the inner side of S. While this discussion might seem inordinately picky, it turns out that not all surfaces can be oriented. Our next example exhibits one.

 Example 3.5.3. Optional — The Möbius Strip ^ There are some surfaces S for which it is not possible to choose a continuous orientation map N : S → R . Such surfaces are said to be non-orientable. The most famous non-orientable surface is the Möbius 1 strip 2, which you can construct as follows. Take a rectangular strip of paper. 3

Lay it flat and then introduce a half twist so that the arrow on the right hand end points upwards, rather than downwards. Then glue the two ends of the strip together, with the two arrows coinciding. That's the Möbius strip.

3.5.1

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Let's parametrize it. Think of the strip of paper that we used to construct it as consisting of a backbone (the horizontal black line in the figure below) with a bunch of ribs (like the thick blue line in the figure) emanating from it.

When we glue the two ends of the strip together, the black line forms a circle. If the strip has length circumference ℓ and hence radius . We'll parametrize it as the circle

ℓ,

the circle will have

ℓ

2π

ℓ

^ r(θ)

where ^ r(θ) = cos(θ) ^ ı ı + sin(θ) ^ ȷ ȷ

2π

This circle is in the xy-plane. It is the black circle in the figure below. (The figure only shows the part of the circle in the first octant, i.e. with x, y, z ≥ 0.)

Now we'll add in the blue ribs. We'll put the blue rib, that is attached to the backbone at vectors ^ r(θ) and

ℓ 2π

^ r(θ),

in the plane that contains the

^ k.

^ A side view of the plane that contains the vectors ^ r(θ) and k is sketched in the figure below.

To put the half twist into the strip of paper, we want the blue rib to rotate about the backbone by 180 , i.e. π radians, as θ runs from 0 to 2π. That will be the case if we pick the angle φ in the figure to be . The vector that is running along the blue rib in the figure is ∘

θ

2

^ u(v, θ, φ) = v cos(φ) ^ r(θ) + v sin(φ) k

where the length, v, of the vector is a parameter. If the width of our original strip of paper is w, then as the parameter v runs from − to + , the tip of the vector u(v, θ, φ) runs over the entire blue rib. So, choosing φ = , our parametrization of the Möbius strip is w

w

θ

2

2

2

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ℓ

⇀

θ

^ r(θ) + u (v, θ,

r (θ, v) =

)

2π ℓ

2

^ r(θ) + v cos(

= 2π

θ

)^ r(θ) + v sin(

2

θ

^ )k

2

w

w

0 ≤ θ < 2π,   −

≤v≤ 2

2

where ^ r(θ) = cos(θ) ^ ı ı + sin(θ) ^ ȷ ȷ. Now that we have parametrized the Möbius strip, let's return to the question of orientability. Recall, from Definition 3.5.1, that, ^ if the Möbius strip were orientable, there would exist a continuous function N which assigns to each point r of the strip a unit ^ normal vector N( r ) at r . First, we'll find the normal vectors to the surface using 3.3.1. The partial derivatives ⇀

⇀

⇀

⇀

∂r

ℓ (θ, v)

=

∂θ

′

^ r (θ) + v cos(

2π

′

)^ r (θ) −

2

⇀

∂r

θ (θ, v) = cos(

∂v

θ

v

θ sin(

2

)^ r(θ) + sin(

2

θ

)^ r(θ) +

2

v

θ cos(

2

^ )k

2

^ )k

2

are relatively messy, so let's just consider the case v = 0 (i.e. find the normal vectors on the backbone). Then ⇀

∂r

ℓ

′

^ r (θ)

(θ, 0) = ∂θ

2π

⇀

∂r

θ (θ, 0) = cos(

∂v

)^ r(θ) + sin(

2

θ

^ )k

2

Since ′

^ r (θ) × ^ r(θ) ′

^ ^ r (θ) × k

^ = ( − sin(θ) ^ ı ı + cos(θ) ^ ȷ ȷ ) × ( cos(θ) ^ ı ı + sin(θ) ^ ȷ ȷ ) = −k ^ = ( − sin(θ) ^ ı ı + cos(θ) ^ ȷ ȷ) × k = ^ r(θ)

we have ⇀

⇀

∂r

∂r (θ, 0) ×

∂θ

ℓ (θ, 0) = −

∂v

θ [ cos(

2π

^ As k and ^ r(θ) are mutually perpendicular unit vectors, vectors to the Möbius strip at r (θ, 0) are

cos (

^ ) k − sin(

2 θ 2

θ

)^ r(θ)]

2

^ ) k − sin (

θ 2

)^ r(θ)

has length one, and the two unit normal

⇀

θ ±[ cos(

^ ) k − sin(

2

So, for each θ,

^ ⇀ N( r (θ, 0))

θ

)^ r(θ)]

2

must be either θ

cos(

^ ) k − sin (

2

θ

θ

)^ r(θ)

or

− [ cos(

2

^ ) k − sin(

2

θ

)^ r(θ)]

2

^ Imagine walking along the Möbius strip. The normal vector N ( r (θ, v)) is our body when we are at ⇀

the tail of the vector ^( N

ℓ 2π

^ ⇀ N( r (θ, v)).

⇀

+( cos (

θ 2

θ 2

ℓ

⇀

θ.

Because

^ ⇀ N( r (θ, 0))

has to be +( cos ( )^ r(θ)). We keep increasing θ. By continuity, )^ r(θ)) for bigger and bigger θ. Eventually we get to θ = 2π, i.e. to

^ ⇀ N( r (θ, 0))

^ ) k − sin (

— our feet are at 2π

upright.) Now we start walking along the backbone of the Möbius strip, increasing continuous,

r (θ, v)

and our head is at the arrow of We start walking at r (0, 0) = ^ ı ı . Our body, ^ ^ ^ ^ has to be one of ±( cos(0) k − sin(0) ^ r(0)) = ±k. Let's suppose that N( r (0, 0)) = +k. (We start

^ ⇀ N( r (θ, v))

^ (⇀ ^ ı ı) = N r (0, 0))

⇀

θ

2

⇀

^ ) k − sin (

ℓ

r (2π, 0) =

θ

2

^ r(2π) =

2π

ℓ

ℓ ^ ı ı =

2π

^ ⇀ N( r (θ, 0))

has to be has to be

⇀

^ r(0) = r (0, 0)

2π

We are back to our starting point. Continuity has forced

3.5.3

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^ ⇀ ^ ⇀ ∣ N( r (2π, 0)) = N( r (θ, 0)) ∣

θ = +[ cos( 2

θ=2π

So we have arrived back upside down. That's a problem — not Strip II (Red Ants). ^ N(

ℓ

2π

^ ^ ı ı ) = +k,

^ −k.

^ ) k − sin(

θ 2

∣ )^ r(θ)] ∣

^ ⇀ ^ N( r (2π, 0)) = N(

^ = −k

θ=2π

ℓ 2π

^ ı ı)

and we have already defined

So the Möbius strip is not orientable. The interested reader should look up M. C. Escher's Möbius

1. August Ferdinand Möbius (1790--1868) was a German mathematician and astronomer. He was a descendant of Martin Luther and a student of Gauss. 2. Another famous non-orientable surface is the Klein bottle. You can easily find discussions of it using your favourite search engine. This page titled 3.5: Orientation of Surfaces is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

3.5.4

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CHAPTER OVERVIEW 4: Integral Theorems 4.1: Gradient, Divergence and Curl 4.2: The Divergence Theorem 4.3: Green's Theorem 4.4: Stokes' Theorem 4.5: Optional — Which Vector Fields Obey ∇ × F = 0 4.6: Really Optional — More Interpretation of Div and Curl 4.7: Optional — A Generalized Stokes' Theorem Thumbnail: Diagram of an arbitrary volume partitioned into several parts, illustrating that the flux out of the original volume is equal to the sum of the flux out of the component volumes. (CC0; Chetvorno via Wikipedia) This page titled 4: Integral Theorems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1

4.1: Gradient, Divergence and Curl “Gradient, divergence and curl”, commonly called “grad, div and curl”, refer to a very widely used family of differential operators and related notations that we'll get to shortly. We will later see that each has a “physical” significance. But even if they were only shorthand 1, they would be worth using. For example, one of Maxwell's equations (relating the electric field notation is (

∂E3

−

∂y

∂E2

)^ ı ı −(

∂E3

∂z

−

∂E1

∂x

and the magnetic field

E

∂E2

)^ ȷ ȷ +(

∂z

∂E1

−

∂x 1

=− c

) written without the use of this

^ )k

∂y

∂B1

(

B

∂B2

^ ı ı +

∂t

^ ȷ ȷ +

∂t

∂B3

^ k)

∂t

The same equation written using this notation is 1 ∂B

⇀

∇×E = − c

∂t ⇀

The shortest way to write (and easiest way to remember) gradient, divergence and curl uses the symbol “∇” which is a differential operator like . It is defined by ∂

∂x

∂

⇀

∂

∇ = ^ ı ı

^ +k

+^ ȷ ȷ ∂x

∂y

∂ ∂z

and is called “del” or “nabla”. Here are the definitions.

 Definition 4.1.1 1. The gradient of a scalar-valued function f (x, y, z) is the vector field ∂f

⇀

grad f = ∇f =

∂f ^ ı ı +

∂f ^ ȷ ȷ +

∂x

∂y

^ k

∂z ⇀

Note that the input, f , for the gradient is a scalar-valued function, while the output,∇f , is a vector-valued function. 2. The divergence of a vector field F(x, y, z) is the scalar-valued function ⇀

⇀

⇀

⇀

div F = ∇ ⋅ F =

∂F1

+

∂F2

∂x

∂F3

+

∂y

∂z

⇀

⇀

⇀

Note that the input, F, for the divergence is a vector-valued function, while the output, ∇ ⋅ F, is a scalar-valued function. ⇀

3. The curl of a vector field F(x, y, z) is the vector field ⇀

⇀

⇀

curl F = ∇ × F = (

∂F3

−

∂y

∂F2

)^ ı ı −(

∂z

∂F3

−

∂x

∂F1

)^ ȷ ȷ +(

∂F2

∂z

⇀

∂x ⇀

−

∂F1

^ )k

∂y

⇀

Note that the input, F, for the curl is a vector-valued function, and the output, ∇ × F, is a again a vector-valued function. 4. The Laplacian 2 of a scalar-valued function f (x, y, z) is the scalar-valued function ⇀2

⇀

⇀

∂

2

f

Δf = ∇ f = ∇ ⋅ ∇f =

2

∂

2

f

+

∂x

∂y

2

∂ +

2

f

∂z 2

⇀

The Laplacian of a vector field F(x, y, z) is the vector field ⇀

⇀2 ⇀

⇀

⇀⇀

ΔF = ∇ F = ∇ ⋅ ∇F =

∂

2

⇀

F 2

∂x

∂

2

⇀

F

+ ∂y

2

∂

2

⇀

F

+ ∂z

2

Note that the Laplacian maps either a scalar-valued function to a scalar-valued function, or a vector-valued function to a vector-valued function.

4.1.1

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The gradient, divergence and Laplacian all have obvious generalizations to dimensions other than three. That is not the case for the curl. It does have a, far from obvious, generalization, which uses differential forms. Differential forms are well beyond our scope, but are introduced in the optional §4.7.

 Example 4.1.2 As an example of an application in which both the divergence and curl appear, we have Maxwell's equations 3 4 5, which form the foundation of classical electromagnetism. ⇀

∇ ⋅ E = 4πρ ⇀

∇⋅B = 0 1 ∂B

⇀

∇×E+

=0 c

∂t

1 ∂B

⇀

4π

∇×B−

= c

J

∂t

c

Here E is the electric field, B is the magnetic field, ρ is the charge density, J is the current density and c is the speed of light.

Vector Identities Two computationally extremely important properties of the derivative d 

d 

are linearity and the product rule.

dx

df (af (x) + bg(x)) = a

dx

dg (x) + b

dx d 

(x) dx

df (f (x) g(x)) = g(x)

dx

dg (x) + f (x)

dx

(x) dx

Gradient, divergence and curl also have properties like these, which indeed stem (often easily) from them. First, here are the statements of a bunch of them. (A memory aid and proofs will come later.) In fact, here are a very large number of them. Many are included just for completeness. Only a relatively small number are used a lot. They are in red.

 Theorem 4.1.3. Gradient Identities ⇀

⇀

⇀

1. ∇(f + g) = ∇f + ∇g 2. ∇(cf ) = c ∇f , for any constant c 3. ∇(f g) = (∇f )g + f (∇g) ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

4. ∇(f /g) = (g ∇f − f ∇g)/g at points x where g(x) ≠ 0. 5. ∇(F ⋅ G) = F × (∇ × G) − (∇ × F) × G + (G ⋅ ∇)F + (F ⋅ ∇)G ⇀

⇀

⇀

2

⇀

⇀

⇀

⇀

⇀

⇀

⇀

Here 6 ⇀ ⇀

⇀

∂F

⇀

(G ⋅ ∇)F = G1

∂x

⇀

∂F + G2

∂y

∂F + G3

∂z

 Theorem 4.1.4. Divergence Identities ⇀

⇀

⇀

⇀

⇀

1. ∇ ⋅ (F + G) = ∇ ⋅ F + ∇ ⋅ G 2. ∇ ⋅ (cF) = c ∇ ⋅ F, for any constant c 3. ∇ ⋅ (f F) = (∇f ) ⋅ F + f ∇ ⋅ F ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

4. ∇ ⋅ (F × G) = (∇ × F) ⋅ G − F ⋅ (∇ × G)

 Theorem 4.1.5. Curl Identities ⇀

⇀

⇀

⇀

⇀

1. ∇ × (F + G) = ∇ × F + ∇ × G ⇀

⇀

⇀

⇀

2. ∇ × (cF) = c ∇ × F, for any constant c

4.1.2

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⇀

⇀

⇀

⇀

⇀

⇀

3. ∇ × (f F) = (∇f ) × F + f ∇ × F ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

4. ∇ × (F × G) = F(∇ ⋅ G) − (∇ ⋅ F)G + (G ⋅ ∇)F − (F ⋅ ∇)G Here ⇀

⇀

(G ⋅ ∇)F = G1

⇀

⇀

⇀

∂F

∂F

∂F

∂x

+ G2

∂y

+ G3

∂z

 Theorem 4.1.6. Laplacian Identities ⇀2

1. ∇

⇀2

⇀2

2. ∇

⇀2

(cf ) = c ∇ f ,

⇀2

3. ∇

⇀2

(f + g) = ∇ f + ∇ g

⇀2

for any constant c ⇀

⇀2

⇀

(f g) = f ∇ g + 2 ∇f ⋅ ∇g + g ∇ f

 Theorem 4.1.7. Degree Two Identities ⇀

⇀

⇀

1. ∇ ⋅ (∇ × F) = 0 ⇀

(divergence of curl)

⇀

2. ∇ × (∇f ) = 0 (curl of gradient) 3. ∇ ⋅ (f {∇g × ∇h}) = ∇f ⋅ (∇g × ∇h) ⇀

⇀

⇀

⇀

⇀

⇀

⇀2

⇀

4. ∇ ⋅ (f ∇g − g∇f ) = f ∇ ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀2

g−g∇ f ⇀2 ⇀

5. ∇ × (∇ × F) = ∇(∇ ⋅ F) − ∇

(curl of curl)

F

Memory Aid. Most of the vector identities (in fact all of them except Theorem 4.1.3.e, Theorem 4.1.5.d and Theorem 4.1.7) are really easy to guess. Just combine the conventional linearity and product rules with the facts that if the left hand side is a vector (scalar), then the right hand side must also be a vector (scalar) and the only valid products of two vectors are the dot and cross products and the product of a scalar with either a scalar or a vector cannot be either a dot or cross product and A × B = −B × A. (The cross product is antisymmetric.) ⇀

⇀

⇀

⇀

⇀

⇀

For example, consider Theorem 4.1.4.c, which says ∇ ⋅ (f F) = (∇f ) ⋅ F + f ∇ ⋅ F. ⇀

⇀

⇀

⇀

The left hand side, ∇ ⋅ (f F), is a scalar, so the right hand side must also be a scalar. ⇀

The left hand side, ∇ ⋅ (f F), is a derivative of the product of f and F, so, mimicking the product rule, the right hand side will be a sum of two terms, one with F multiplying a derivative of f , and one with f multiplying a derivative of F. The derivative acting on f must be ∇f , because ∇ ⋅ f and ∇ × f are not well-defined. To end up with a scalar, rather than a vector, we must take the dot product of ∇f and F. So that term is (∇f ) ⋅ F. ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

The derivative acting on F must be either ∇ ⋅ F or ∇ × F. We also need to multiply by the scalar f and end up with a scalar. So the derivative must be a scalar, i.e. ∇ ⋅ F and that term is f {∇ ⋅ F}. Our final guess is ∇ ⋅ (f F) = (∇f ) ⋅ F + f ∇ ⋅ F, which, thankfully, is correct. ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

Proof of Theorems 4.1.3, 4.1.4, 4.1.5, 4.1.6 and 4.1.7. All of the proofs (except for those of Theorem 4.1.7.c,d, which we will return to later) consist of writing out the definition of the left hand side and writing out the definition of the right hand side and observing (possibly after a little manipulation) that they are the same. For Theorem 4.1.3.a,b, Theorem 4.1.4.a,b, Theorem 4.1.5.a,b and Theorem 4.1.6.a,b, the computation is trivial — one line per identity, if one uses some efficient notation. Rename the coordinates x, y, z to x , x , x and the standard unit basis vectors ^ ıı , 1

^ ^ ȷ ȷ, k

to

⇀

^ ıı 1 , ^ ıı 2 , ^ ıı 3 .

Then ∇ = ∑

3 n=1

^ ıı n

∂  ∂xn

2

3

and the proof of, for example, Theorem 4.1.4.a is

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3 ⇀

∂ 

⇀

∇ ⋅ (F + G)

⇀

=∑ ∂xn

n=1 3

^ ıı n ⋅ (F + G)

∂  ∂xn

n=1

3

⇀

∂ 

^ ıı n ⋅ F + ∑

=∑

⇀

n=1

⇀

⇀

^ ıı n ⋅ G = ∇ ⋅ F + ∇ ⋅ G

∂xn

For Theorem 4.1.3.c,d, Theorem 4.1.4.c, Theorem 4.1.5.c and Theorem 4.1.6.c, the computation is easy — a few lines per identity. For example, the proof of Theorem 4.1.5.c is ⇀

3

⇀

∇ × (f F)

∂

3

⇀

∂xn

n=1 3

∂

^ ıı n × (f F) = ∑

=∑

∂xn

n=1

∂f

3

⇀

∂

^ ıı n × F + f ∑

=∑ ∂xn

n=1 ⇀

⇀

⇀

^ ıı n × F

∂xn

n=1

⇀

⇀

(f { ^ ıı n × F})

⇀

= (∇f ) × F + f ∇ × F

In the second line we used Theorem 1.1.3.b. The similar verification of Theorems 4.1.3.c,d, 4.1.4.c, and 4.1.6.c are left as exercises. The latter two are parts (a) and (c) of Question 3 in Section 4.1 of the CLP-4 problembook. For Theorem 4.1.4.d, the computation is also easy if one uses the fact that a ⋅ (b × c) = (a × b) ⋅ c

which is Lemma 4.1.8.a below. The verification of Theorem 4.1.4.d is part (b) of Question 3 in Section 4.1 of the CLP-4 problembook. That leaves the proofs of Theorem 4.1.3.e, Theorem 4.1.5.d, Theorem 4.1.7.a,b,c,d,e, which we write out explicitly. Theorem 4.1.3.e: First write out the left hand side as 3 ⇀

⇀

∇(F ⋅ G) = ∑ ^ ıı n n=1

⇀

3

∂ ∂xn

Then rewrite a × (b × c) = (c ⋅ a)b − (b ⋅ a)c,

n=1

3

∂F

⇀

(F ⋅ G) = ∑ ^ ıı n (

⇀

∂xn

⋅ G) + ∑ ^ ıı n (F ⋅ n=1

∂G ) ∂xn

which is Lemma 4.1.8.b below, as

(c ⋅ a)b = a × (b × c) + (b ⋅ a)c

Applying it once with b = ^ ıı

n,

⇀

c =

⇀ ∂F ∂xn

and once with b = ^ ıı

, a=G

3

⇀

∇(F ⋅ G)

= ∑ [G × ( ^ ıı n × n=1

n,

∂xn

⇀

⇀

∂F

∂F

) + (G ⋅ ^ ıı n )

∂xn 3

⇀

+ ∑ [F × ( ^ ıı n ×

⇀

gives

, a=F

] ∂xn

∂G

⇀

) + (F ⋅ ^ ıı n )

∂xn

n=1 ⇀

∂G

c =

⇀

⇀

⇀

⇀

∂G ] ∂xn

⇀

⇀

⇀

= G × (∇ × F) + (G ⋅ ∇)F + F × (∇ × G) + (F ⋅ ∇)G

Theorem 4.1.5.d: We use the same trick. Write out the left hand side as ⇀

3

⇀

∂

=∑^ ıı n ×

∇ × (F × G)

∂xn

n=1

⇀

3

=∑^ ıı n × (

⇀

⇀

∇ × (F × G)

3

∂F

⇀

× G) + ∑ ^ ıı n × (F ×

∂xn

n=1

Applying a × (b × c) = (c ⋅ a)b − (b ⋅ a)c,

⇀

(F × G)

∂G ) ∂xn

n=1

which is Lemma 4.1.8.b below, ⇀

3

∂F

= ∑ [Gn n=1 ⇀

∂xn ⇀

3

∂Fn − ∂xn ⇀

∂Gn

G] + ∑ [

⇀

n=1 ⇀

∂xn ⇀

∂G

⇀

F − Fn ⇀

] ∂xn

⇀

= (G ⋅ ∇)F − (∇ ⋅ F)G + (∇ ⋅ G)F − (F ⋅ ∇)G

4.1.4

https://math.libretexts.org/@go/page/91909

Theorem 4.1.7.a: Substituting in ⇀

⇀

∇×F

=(

∂F3

∂F2

−

∂y

∂F3

)^ ıı − (

∂z

−

∂F1

∂x

∂F2

)^ ȷ ȷ +(

∂z

−

∂x

∂F1

^ )k

∂y

gives ⇀

∂

⇀

∇ ⋅ (∇ × F )

=

(

∂F3

∂x ∂

2

=

∂F2

−

∂y

F3

∂z ∂

2

−

∂x∂y

∂ )−

F2

∂F3

( ∂y

∂

2

−

∂x∂z

F3

−

∂F1

∂x ∂

+

∂y∂x

2

∂ )+

(

∂z

F1

∂

∂z

2

+

∂y∂z

∂F2

F2

∂ −

∂z∂x

−

∂x 2

∂F1

)

∂y

F1

∂z∂y

=0

because the two red terms have cancelled, the two blue terms have cancelled and the two black terms have cancelled. Theorem 4.1.7.b: Substituting in ∂f

⇀

∇f

∂f

^ ıı +

= ∂x

^ ȷ ȷ+

∂y

∂f

^ k

∂z

gives ⇀

∂

⇀

∇ × (∇f )

∂f

=(

∂

∂f

∂

)^ ıı − (

− ∂y ∂z

∂z ∂y

∂f

∂

∂f

∂ +(

∂f

− ∂x ∂z ∂f

∂ −

∂x ∂y

)^ ȷ ȷ

∂z ∂x ^ )k

∂y ∂x

=0

Theorem 4.1.7.c: By Theorem 4.1.4.c, followed by Theorem 4.1.4.d, ⇀

⇀

⇀

⇀

∇ ⋅ [f (∇g × ∇h)]

⇀

⇀

⇀

⇀

⇀

= ∇f ⋅ (∇g × ∇h) + f ∇ ⋅ (∇g × ∇h) ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

= ∇f ⋅ (∇g × ∇h) + f [(∇ × ∇g) ⋅ ∇h − ∇g ⋅ (∇ × ∇h)] ⇀

⇀

⇀

⇀

By Theorem 4.1.7.b, ∇ × ∇g = ∇ × ∇h = 0, so ⇀

⇀

⇀

⇀

⇀

⇀

∇ ⋅ [f (∇g × ∇f )] = ∇f ⋅ (∇g × ∇h)

Theorem 4.1.7.d: By Theorem 4.1.4.c, ⇀

⇀

⇀

∇ ⋅ (f ∇g − g∇f )

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

= (∇f ) ⋅ (∇g) + f ∇ ⋅ (∇g) − (∇g) ⋅ (∇f ) + g ∇ ⋅ (∇f ) ⇀2

⇀2

= f ∇ g−g∇ f

Theorem 4.1.7.e: ⇀

⇀

3

⇀

∇ × (∇ × F)

=∑^ ıı ℓ ℓ=1

∂

3

∂xℓ

m=1

∑

^ ıı ℓ × ( ^ ıı m × ^ ıı n ) 

ℓ,m,n=1

Using a × (b × c) = (c ⋅ a)b − (b ⋅ a)c,

×∑ ^ ıı n Fn )

∂xm

3

=

3

∂

×( ∑ ^ ıı m

∂

2

n=1

Fn

∂ xℓ ∂ xm

we have

^ ıı ℓ × ( ^ ıı m × ^ ıı n ) = ( ^ ıı ℓ ⋅ ^ ıı n ) ^ ıı m − ( ^ ıı ℓ ⋅ ^ ıı m ) ^ ıı n = δℓ,n ^ ıı m − δℓ,m ^ ıı n

where 7 δm,n = {

1

if m = n

0

if m ≠ n

Hence

4.1.5

https://math.libretexts.org/@go/page/91909

3 ⇀

⇀

⇀

∇ × (∇ × F) =

δℓ,n ^ ıı m  

∑ ℓ,m,n=1 3

=

^ ıı m  

∑ m,n=1 ⇀

⇀

∂

∂

2

3

Fn

−

∂ xℓ ∂ xm

∂Fn

ℓ,m,n=1 3

− ∑

∂xm ∂xn

δℓ,m ^ ıı n  

∑

^ ıı n  

m,n=1

∂

2

∂

2

Fn

∂ xℓ ∂ xm

Fn 2

∂xm

⇀2 ⇀

⇀

= ∇(∇ ⋅ F) − ∇ F

 Lemma 4.1.8 1. a ⋅ (b × c) = (a × b) ⋅ c 2. a × (b × c) = (c ⋅ a)b − (b ⋅ a)c Proof (a) Here are two proofs. For the first, just write out both sides a ⋅ (b × c)

= (a1 , a2 , a3 ) ⋅ (b2 c3 − b3 c2 , b3 c1 − b1 c3 , b1 c2 − b2 c1 )

(a × b) ⋅ c

= (a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ) ⋅ (c1 , c2 , c3 )

= a1 b2 c3 − a1 b3 c2 + a2 b3 c1 − a2 b1 c3 + a3 b1 c2 − a3 b2 c1

= a2 b3 c1 − a3 b2 c1 + a3 b1 c2 − a1 b3 c2 + a1 b2 c3 − a2 b1 c3

and observe that they are the same. For the second proof, we again write out both sides, but this time we express them in terms of determinants.

a⋅ b ×c

ıı ⎡ ^

^ ȷ ȷ

^ k ⎤

= (a1 , a2 , a3 ) ⋅ det ⎢ b 1

b2

b3 ⎥

c2

c3

⎣

= a1 det [

b3

c2

c3

] − a2 det [

a1

a2

a3

= det ⎢ b1

b2

b3 ⎥

c1

c2

c3

ıı ⎡ ^

^ ȷ ȷ

⎡

⎣

a×b ⋅ c

b2

c1

= det ⎢ a 1 ⎣

b1

= c1 det [

c1

c3

] + a3 det [

b1

b2

c1

c2

]

^ k ⎤

a2

a3 ⎥ ⋅ (c1 , c2 , c3 )

b2

b3

a2

a3

b2

b3

⎦

] − c2 det [

c2

c3

a2

a3 ⎥

b2

b3

b1

b3

⎦

c1

⎣

b1

⎤

= det ⎢ a1

⎡

⎦

a1

a3

b1

b3

] + c3 det [

a1

a2

b1

b2

]

⎤

⎦

Exchanging two rows in a determinant changes the sign of the determinant. Moving the top row of a bottom row requires two exchanges of rows. So the two 3 × 3 determinants are equal.

3 ×3

determinant to the

(b) The proof is not exceptionally difficult — just write out both sides and grind. Substituting in ^ b × c  =  (b2 c3 − b3 c2 ) ^ ıı − (b1 c3 − b3 c1 )^ ȷ ȷ + (b1 c2 − b2 c1 )k

gives, for the left hand side,

4.1.6

https://math.libretexts.org/@go/page/91909

a × (b × c) =

⎡

^ ıı

^ ȷ ȷ

^ k

⎤

det ⎢

a1

a2

a3

⎥

b2 c3 − b3 c2

−b1 c3 + b3 c1

b1 c2 − b2 c1

⎣

⎦

^ ıı [ a2 (b1 c2 − b2 c1 ) − a3 (−b1 c3 + b3 c1 )]

=

−^ ȷ ȷ [ a1 (b1 c2 − b2 c1 ) − a3 (b2 c3 − b3 c2 )] ^ + k[ a1 (−b1 c3 + b3 c1 ) − a2 (b2 c3 − b3 c2 )]

On the other hand, the right hand side (a ⋅ c)b − (a ⋅ b)c  =  

^ (a1 c1 + a2 c2 + a3 c3 )(b1 ^ ıı + b2 ^ ȷ ȷ + b3 k) ^ − (a1 b1 + a2 b2 + a3 b3 )(c1 ^ ıı + c2 ^ ȷ ȷ + c3 k)

=  ^ ıı  [ a1 b1 c1 + a2 b1 c2 + a3 b1 c3 − a1 b1 c1 − a2 b2 c1 − a3 b3 c1 ] + ^ ȷ ȷ  [ a1 b2 c1 + a2 b2 c2 + a3 b2 c3 − a1 b1 c2 − a2 b2 c2 − a3 b3 c2 ] ^ + k [ a1 b3 c1 + a2 b3 c2 + a3 b3 c3 − a1 b1 c3 − a2 b2 c3 − a3 b3 c3 ] =  ^ ıı  [ a2 b1 c2 + a3 b1 c3 − a2 b2 c1 − a3 b3 c1 ] +^ ȷ ȷ  [ a1 b2 c1 + a3 b2 c3 − a1 b1 c2 − a3 b3 c2 ] ^ + k [ a1 b3 c1 + a2 b3 c2 − a1 b1 c3 − a2 b2 c3 ]

The last formula that we had for the left hand side is the same as the last formula we had for the right hand side.

 Example 4.1.9. Screening tests ⇀

⇀

⇀

We have seen the vector identity Theorem 4.1.7.b before. It says that if a vector field F is of the form F = ∇φ for some some function φ (that is, if F is conservative), then ⇀

⇀

⇀

⇀

⇀

∇ × F = ∇ × (∇φ) = 0 ⇀

Conversely, we have also seen, in Theorem 2.4.8, that, if F is defined and has continuous first order partial derivatives on all of 3

R ,

⇀

⇀

⇀

and if ∇ × F = 0, then F is conservative. The vector identity Theorem 4.1.7.b is our screening test for conservativeness.

Because its right hand side is zero, the vector identity Theorem 4.1.7.a is suggestive. It says that if a vector field ⇀

⇀

F

is of the

⇀

form F = ∇ × A for some some vector field A, then ⇀

⇀

⇀

⇀

∇ ⋅ F = ∇ ⋅ (∇ × A) = 0 ⇀

⇀

⇀

⇀

When F = ∇ × A, A is called a vector potential for F. We shall see in Theorem 4.1.16, below, that, conversely, if F(x) is defined and has continuous first order partial derivatives on all of R , and if ∇ ⋅ F = 0, then F has a vector potential 8. The vector identity Theorem 4.1.7.a is indeed another screening test. ⇀

3

⇀

⇀

As an example, consider the Maxwell's equations ⇀

∇⋅B = 0 1 ∂B

⇀

∇×E+

=0 c

∂t

that we saw in Example 4.1.2. The first equation implies that (assuming

B

is sufficiently smooth) there is a vector field

A,

⇀

called the magnetic potential, with B = ∇ × A. Substituting this into the second equation gives ⇀

1 ∂ 

0 = ∇×E+

⇀

⇀

c ∂t

So E +

1

∂A

c

∂t

1 ∂A

∇ × A = ∇ × (E +

) c

∂t

passes the screening test of Theorem 4.1.7.b and there is a function φ (called the electric potential) with 1 ∂A E+

= −∇φ c

∂t

We have put in the minus sign just to provide compatibility with the usual physics terminology.

4.1.7

https://math.libretexts.org/@go/page/91909

 Example 4.1.10 Let

⇀

^ r (x, y, z) = x ^ ı ı +y ^ ȷ ȷ +zk

and let ψ(x, y, z) be an arbitrary function. Verify that ⇀

⇀

⇀

∇ ⋅ ( r × ∇ψ) = 0

Solution By the vector identity Theorem 4.1.4.d, ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

∇ ⋅ ( r × ∇ψ) = (∇ × r ) ⋅ ∇ψ − r ⋅ (∇ × (∇ψ))

By the vector identity Theorem 4.1.7.b, the second term is zero. Now since ⇀

∂z

⇀

∇× r = (

⇀

∂y −

∂y

)^ ı ı −(

∂z

∂z

∂x −

∂x

)^ ȷ ȷ +(

∂z

⇀

the first term is also zero. Indeed ∇ ⋅ ( r × ∇ψ) = 0 holds for any curl free ⇀

∂y

∂x −

∂x

⇀

^ )k = 0

∂y

⇀

r (x, y, z).

Vector Potentials We'll now further explore the vector potentials that were introduced in Example 4.1.9. First, here is the formal definition.

 Definition 4.1.11 The vector field A is said to be a vector potential for the vector field B if ⇀

B = ∇×A

As we saw in Example 4.1.9, if a vector field B has a vector potential, then the vector identity Theorem 4.1.7.a implies that ∇ ⋅ B = 0. This fact deserves to be called a theorem. ⇀

 Theorem 4.1.12. Screening test for vector potentials If there exists a vector potential for the vector field B, then ⇀

∇⋅B = 0

Of course, we'll consider the converse soon. Also note that the vector potential, when it exists, is far from unique. Two vector fields ~ A and A are both vector potentials for the same vector field if and only if ⇀ ⇀ ~ ∇×A = ∇×A

⟺

⇀ ⇀ ~ ∇ × (A − A) = 0

~

That is, if and only if the difference A − A passes the conservative field screening test of Theorems 2.3.9 and 2.4.8. In particular, if A is one vector potential for a vector field B (i.e. if B = ∇ × A ), and if ψ is any function, then ⇀

⇀

⇀

⇀

⇀

⇀

∇ × (A + ∇ψ) = ∇ × A + ∇ × ∇ψ = B ⇀

by the vector identity Theorem 4.1.7.b. That is, A + ∇ψ is another vector potential for B. To simplify computations, we can always choose ⇀

^ (A + ∇ψ) ⋅ k = A3 +

∂ψ ∂z

,

ψ

so that, for example, the third component of

is zero — just choose ψ = − ∫ A

3

dz.

⇀

A + ∇ψ,

namely

We have just proven

 Lemma 4.1.13 If the vector field B has a vector potential, then, in particular, there is a vector potential A for B with 9 A

3

= 0.

Here is an example which exploits this choice to simplify the computations used to find a vector potential.

4.1.8

https://math.libretexts.org/@go/page/91909

 Example 4.1.14 Let ^ B = yz ^ ı ı + zx ^ ȷ ȷ + xy k

This vector field has been set up carefully to obey ∂

⇀

∂

∇⋅B =

∂

(yz) +

(zx) +

∂x

∂y

(xy) = 0 ∂z

and so passes the screening test of Theorem 4.1.12. Let's try and find a vector potential for ⇀

∇ × A = B,

B.

That is, let's try and find a vector field

^ A = A1 ^ ı ı + A2 ^ ȷ ȷ + A3 k

that obeys

or equivalently, ∂A3

∂A2

−

∂y −

∂A3

∂A1

+

∂x ∂A2

= B1 = yz

∂z

= B2 = zx

∂z ∂A1

−

∂x

= B3 = xy

∂y

This system is nasty to solve because every equation contains more than one of the three unknowns, A , A , A . Let us take advantage of our observation above that, if any vector potential exists, then, in particular, a vector potential A exists that also obeys A = 0. So let's also require that A = 0. Then the equations above simplify to 1

3

2

3

3

−

∂A2

= yz

∂z ∂A1

= zx

∂z ∂A2

−

∂A1

∂x

= xy

∂y

This system is much easier because, now that we have chosen A = 0, the first equation contains only a single unknown, namely A and we can find all A 's that obey the first equation simply by integrating with respect to z: 3

2

2

yz A2 = −

Note that, because

∂ ∂z

2

+ N (x, y) 2

treats x and y as constants, the constant of integration N is allowed to depend on x and y.

Similarly, the second equation contains only a single unknown, second equation is satisfied if and only if xz A1 =

A1 ,

and is easily solved by integrating with respect to

z.

The

2

+ M (x, y) 2

for some function M . Finally, the third equation is also satisfied if and only if M (x, y) and N (x, y) obey ∂

yz (−

∂x

2

∂ + N (x, y)) −

2

xz (

∂y

2

+ M (x, y)) = xy 2

which simplifies to ∂N

∂M (x, y) −

∂x

(x, y) = xy ∂y

This is one linear equation in two unknowns, M and N . Typically, we can easily solve one linear equation in one unknown. So we are free to eliminate one of the unknowns by setting, for example, M = 0, and then choose any N that obeys

4.1.9

https://math.libretexts.org/@go/page/91909

∂N (x, y) = xy ∂x 2

x y

Integrating with respect to x gives, as one possible choice, N (x, y) = xz

2

^ ı ı +(−

A =

yz

2

2

.

So we have found a vector potential. Namely

2

x y +

2

2

)^ ȷ ȷ

2

⇀

One can, and indeed should, quickly check that ∇ × A = B. Let's do another.

 Example 4.1.15 Let ^ B = (2x) ^ ı ı + (2z − 2x) ^ ȷ ȷ + (2x − 2z) k

This vector field obeys ∂

⇀

∂

∇⋅B =

∂

(2x) + ∂x

(2z − 2x) + ∂y

(2x − 2z) = 0 ∂z

and so passes the screening test of Theorem 4.1.12. We'll now find a vector potential A = A the last example, we'll simplify the computations by further requiring 10 that A = 0.

1

^ ^ ı ı + A2 ^ ȷ ȷ + A3 k

for B. As in

3

⇀

The requirements that ∇ × A = B and A

3

=0

come down to −

∂A2

= 2x

∂z ∂A1

= 2z − 2x

∂z ∂A2

−

∂A1

∂x

Because

∂ ∂z

= 2x − 2z

∂y

treats x and y as constants, the first equation is satisfied if and only if there is a function N (x, y) A2 = −2xz + N (x, y)

and second equation is satisfied if and only if there is a function M (x, y) A1 = z

2

− 2xz + M (x, y)

Finally, the third equation is also satisfied if and only if M (x, y) and N (x, y) obey ∂

∂ ( − 2xz + N (x, y)) −

∂x ∂N ⟺

(z

2

− 2xz + M (x, y))) = 2x − 2z

∂y −2z +

∂M (x, y) −

∂x ∂N ⟺

(x, y) = 2x − 2z ∂y ∂M

(x, y) − ∂x

(x, y) = 2x ∂y

All of the z 's in this equation have cancelled out 11 , and we can choose, for example, M (x, y) = 0 and N (x, y) = x have found a vector potential. Namely

2

A = (z

2

2

− 2xz) ^ ı ı + (x

.

So we

− 2xz)^ ȷ ȷ

⇀

Again it is a good idea to check that ∇ × A = B. We can use exactly the strategy of the last examples to prove

4.1.10

https://math.libretexts.org/@go/page/91909

 Theorem 4.1.16 Let B be a vector field that is defined and has all of its first order partial derivatives continuous on all of R . Then there exists a vector potential for B if and only if it passes the screening test ∇ ⋅ B = 0. 3

⇀

Proof ⇀

⇀

We already know that the existence of a vector potential implies that ∇ ⋅ B = 0. So we just have to assume that ∇ ⋅ B = 0 and prove that this implies the existence of a vector field A that obeys ∇ × A = B. Hence we need to solve ⇀

∂A3

∂A2

−

∂y

−

∂z

∂A3

∂A1

+

∂x

∂z

∂A2

∂A1

−

∂x

= B1 (x, y, z)

= B2 (x, y, z)

= B3 (x, y, z)

∂y

We'll explicitly find such an A using exactly the strategy of Example 4.1.14. In particular, we'll look for an A = 0. Then the equations simplify to

A

that also has

3

∂A2

−

∂z ∂A1

= B1 (x, y, z)

= B2 (x, y, z)

∂z ∂A2

−

∂A1

∂x

= B3 (x, y, z)

∂y

The first equation is satisfied if and only if z

~ ~ B1 (x, y, z ) dz + N (x, y)

A2 (x, y, z) = − ∫ 0

for some function N (x, y). And the second equation is satisfied if and only if z

~ ~ B2 (x, y, z ) dz + M (x, y)

A1 (x, y, z) = ∫ 0

So all three equations are satisfied if and only only if we can find M (x, y) and N (x, y) that obey A2 (x,y,z)

 z

∂ (−∫ ∂x

~ ~ B1 (x, y, z ) dz + N (x, y) )

0 A1 (x,y,z)

 z

∂ −

(∫ ∂y

~ ~ B2 (x, y, z ) dz + M (x, y) ) = B3 (x, y, z)

0

which is the case if and only if ∂N ∂x

z

∂M (x, y) −

(x, y) = B3 (x, y, z) + ∫ ∂y

(

∂B1

~ (x, y, z ) +

∂x

0

∂B2

~ ~ (x, y, z )) dz

∂y

Oof! At first sight, it looks like we have a very big problem here. No matter what N and M we pick the left hand side will depend on x and y only — not on z. But it appears like the right hand side depends on z too. Fortunately the screening test (which we have not used to this point in the proof) rides to the rescue and ensures that the right hand actually does does not depend on z. By the screening test, ⇀

∇⋅B =

∂B1 ∂x

+

∂B2 ∂y

4.1.11

+

∂B3

=0

∂z

https://math.libretexts.org/@go/page/91909

and we have ∂B1

∂B2

+

∂x

=−

∂B3

∂y

∂z

so that the right hand side is z

B3 (x, y, z) + ∫

∂B3

(−

~ ~ (x, y, z )) dz

~ = B3 (x, y, z) + [ − B3 (x, y, z ]

∂z

0

~ z =z ~ z =0

= B3 (x, y, 0)

by the fundamental theorem of calculus. So we just have to choose M and N to obey ∂N

∂M (x, y) −

∂x

For example, M = 0, N (x, y) = ∫ have found a formula for it.

x

0

∂y

~ ~ B3 (x, y, 0) dx

(x, y) = B3 (x, y, 0)

work. So not only have we proven that a vector potential exists, but we

 Warning 4.1.17 Note that in Theorem 4.1.16 we are assuming that B passes the screening test on all of R . If that is not the case, for example because the vector field is not defined on all of R , then B can fail to have a vector potential. An example (the point source) is provided in Example 4.4.8. 3

3

Interpretation of the Gradient ⇀

In this section we'll develop an interpretation of the gradient ∇f ( r before.

⇀

0 ).

This should just be a review of material that you have seen

Suppose that you are moving through space and that your position at time t is r (t) = (x(t), y(t), z(t)). As you move along, you measure, for example, the temperature. If the temperature at position (x, y, z) is f (x, y, z), then the temperature that you measure at time t is f (x(t), y(t), z(t)). So the rate of change of temperature that you feel is ⇀

d f (x(t), y(t), z(t)) dt ∂f =

dx (x(t), y(t), z(t))

∂x

∂f (t) +

dt

∂y

∂f

(t) dt

dz

+

(x(t), y(t), z(t)) ∂z

⇀

dy (x(t), y(t), z(t))

(t)

(by the chain rule)

dt ⇀′

⇀

= ∇f ( r (t)) ⋅ r (t) ⇀

⇀′

⇀

=∣ ∣∇f ( r (t))∣ ∣∣ ∣ r (t)∣ ∣ cos θ ⇀

where θ is the angle between the gradient vector ∇f ( r (t)) and the velocity vector r (t). This is the rate of change per unit time. We can get the rate of change per unit distance travelled by moving with speed one, so that ∣∣ r (t)∣∣ = 1 and then ⇀′

⇀

⇀′

d dt

If, at a given moment t = t

0,

you are at

⇀

⇀

r (t0 ) = r 0 , d dt

⇀

⇀

⇀

f ( r (t)) = ∣ ∣∇f ( r (t))∣ ∣ cos θ

then

⇀ ∣ f ( r (t)) ∣

⇀

t=t0

⇀

=∣ ∣∇f ( r 0 )∣ ∣ cos θ ⇀

Recall that θ is the angle between our direction of motion and the gradient vector ∇f ( r ). So to maximize the rate of change of temperature that we feel, as we pass through r , we should choose our direction of motion to be the direction of the the gradient vector ∇f ( r ). In conclusion ⇀ 0

⇀ 0

⇀

⇀ 0

4.1.12

https://math.libretexts.org/@go/page/91909

 Equation 4.1.18 ⇀

direction of maximum rate of change

⇀

∇f ( r 0 ) has direction 

{

⇀

of f  at  r 0 magnitude of maximum rate of change  has magnitude  {

⇀

 (per unit distance) of f  at  r 0

Interpretation of the Divergence ⇀

In this section we'll develop an interpretation of the divergence ∇ ⋅ v ( r in two steps. ⇀

⇀

⇀

0)

of the vector field

⇀

⇀

v( r )

at the point

First we'll express ∇ ⋅ v ( r ) in terms of flux integrals. Then we'll use the interpretation of flux integrals given in Lemma 3.4.1 to get an interpretation of ∇ ⋅ v ( r ⇀

We shall do so

⇀

r 0.

⇀

0

⇀

Think of

⇀

v (x, y, z)

as the velocity of a fluid at (x, y, z) and fix any point

⇀ r0

= (x0 , y0 , z0 ).

⇀

⇀

0 ).

Let, for any ε > 0,

Sε

be the sphere

centered at r of radius ε. ^ (x, y, z) the outward normal to S at (x, y, z). Denote by n ⇀

0

ε

⇀

We shall prove, in Lemma 4.1.20, below, that we can write ∇ ⋅ v ( r ⇀

⇀

⇀ 0)

as the limit

1

⇀

∇ ⋅ v (x0 , y0 , z0 ) = lim ε→0

4 3

⇀

∬ πε

3

v (x, y, z) ⋅ n ^ (x, y, z) dS

Sε

Once we have that lemma we can use that is the volume of the interior of the sphere S and 12 ^ (x, y, z) dS is the rate by Lemma 3.4.1, ∬ v (x, y, z) ⋅ n at which fluid is exiting S 4 3

πε

3

ε

⇀

ε

Sε

to conclude that

 Equation 4.1.19 ⇀

rate at which fluid is exiting an  ⎧ ⎪ ⇀

⇀

∇ ⋅ v ( r 0 ) = ⎨ infinitesimal sphere centred  ⎩ ⎪

⇀

at  r 0 , per unit time, per unit volume ⇀

= strength of the source at  r 0

Here is the critical computation.

 Lemma 4.1.20 ⇀

1

⇀

∇ ⋅ v (x0 , y0 , z0 ) = lim ε→0

4 3

∬ πε

3

⇀

^ (x, y, z) dS v (x, y, z) ⋅ n

Sε

Proof. (Optional). Here is one proof 13 of Lemma 4.1.20. By translating our coordinate system, it suffices to consider

⇀

r 0 = (x0 , y0 , z0 ) = (0, 0, 0).

Sε = {(x, y, z)||(x, y, z)| = ε}

We expand

⇀

v (x, y, z)

^ (x, y, z) = n

Then

1 (x, y, z) ε

in a Taylor expansion in powers of x, y, and z, to first order, with second order error term. ⇀

v (x, y, z) = A + B x + C y + D z + R(x, y, z)

4.1.13

https://math.libretexts.org/@go/page/91909

where ⇀

⇀

∂v

⇀

A = v (0, 0, 0)

⇀

∂v

B =

(0, 0, 0)

∂v

C =

(0, 0, 0)

∂x

D =

∂y

and the error term R(x, y, z) is bounded by a constant times 14

2

x

+y

2

In particular there is a constant K so that, on S

2

+z .

|R(x, y, z)| ≤ Kε

(0, 0, 0) ∂z ε,

2

So ⇀

^ (x, y, z) dS v (x, y, z) ⋅ n

∬ Sε

1 =

∬ ε

(A + B x + C y + D z + R(x, y, z)) ⋅ (x, y, z) dS

Sε

Multiply out the dot product so that the integrand becomes  A ⋅ ^ ıı x

^ +A ⋅ k z

+  A ⋅ ^ ȷ ȷ y

+B⋅ ^ ıı x

^ +  B ⋅ ^ ȷ ȷ xy + B ⋅ k xz

+C⋅ ^ ıı xy

+  C ⋅ ^ ȷ ȷ y

+D ⋅ ^ ıı xz 

^ 2 +  D ⋅ ^ ȷ ȷ yz + D ⋅ k z

2

2

^ + C ⋅ k yz

+ R(x, y, z) ⋅ (x, y, z)

That's a lot of terms. But most of them integrate to zero, simply because the integral of an odd function over an even domain is zero. Because S is invariant under x → −x and under y → −y and under z → −z we have ε

∬

x dS = ∬

Sε

y dS = ∬

Sε

z dS = ∬

Sε

xy dS = ∬

Sε

xz dS = ∬

Sε

yz dS = 0

Sε

which is a relief. We are now left with ∬

⇀

^ (x, y, z) dS v (x, y, z) ⋅ n

1 =

Sε

2

(B ⋅ ^ ıı x

∬ ε

+C⋅ ^ ȷ ȷ y

2

^ 2 + D ⋅ k z ) dS

Sε

1 +

∬ ε

R(x, y, z) ⋅ (x, y, z) dS

Sε

As well S is invariant 15 under the interchange of x and y and also under the interchange of x and z. Consequently ε

2

∬

x

dS

=∬

Sε

y

2

dS = ∬

Sε

∬ 3

dS =

ε

2

2

dS

∬ 3

since x

+y

2

[x

+y

2

2

+ z ] dS

Sε 2

+z

2

2

= ε  on Sε

Sε

4 =

1

2

Sε

1 =

z

πε

4

3

since the surface area of the sphere S is 4π ε . So far, we have 2

ε

∬

⇀

^ (x, y, z) dS = v (x, y, z) ⋅ n

4

3 ^ π ε (B ⋅ ^ ıı + C ⋅ ^ ȷ ȷ + D ⋅ k)

3

Sε

1 +

∬ ε

4 =

3

⇀

⇀

1

⇀

πε ∇ ⋅ v ( 0 ) + 3

R(x, y, z) ⋅ (x, y, z) dS

Sε

∬ ε

R(x, y, z) ⋅ (x, y, z) dS

Sε

(review the definitions of B, C, D)

which implies

4.1.14

https://math.libretexts.org/@go/page/91909

1 lim

4

ε→0

3

∬ πε

3

⇀

^ (x, y, z) dS v (x, y, z) ⋅ n

Sε

⇀

⇀

3

⇀

= ∇ ⋅ v ( 0 ) + lim ε→0

4πε

Finally, it suffices to recall that |R(x, y, z)| ≤ Kε and, on S 2

3 4πε

4

∣ ∣∬ ∣

Sε

ε,

∣ R(x, y, z) ⋅ (x, y, z) dS ∣ ≤ ∣

4

∬

R(x, y, z) ⋅ (x, y, z) dS

Sε

|(x, y, z)| = ε,

so that

3 ∬ 4πε

4

|R(x, y, z)| |(x, y, z)| dS

Sε

3 ≤ 4πε

4

∬

Kε

3

3 dS =

Sε

4πε

4

 K ε

3

2

(4π ε )

= 3Kε

converges to zero as ε → 0. So we are left with the desired result.

 Example 4.1.21 Here is a sketch of the vector field

⇀

^ v (x, y, z) = x ^ ı ı +y ^ ȷ ȷ +zk

and a sphere centered on the origin, like S

ε.

This velocity field has fluid being created and pushed out through the sphere. We have ⇀

⇀

⇀

∇ ⋅ v(0 ) = 3

consistent with our interpretation 4.1.19.

 Example 4.1.22 Here is a sketch of the vector field

⇀

v (x, y, z) = −y ^ ı ı +x ^ ȷ ȷ

and a sphere centered on the origin, like S

ε.

This velocity field just has fluid going around in circles. No fluid actually crosses the sphere. The divergence ⇀

⇀

⇀

∇ ⋅ v(0 ) = 0

consistent with our interpretation 4.1.19.

 Example 4.1.23 Here is a sketch of the vector field

⇀

v (x, y, z) = ^ ı ı

and a sphere centered on the origin, like S

ε.

4.1.15

https://math.libretexts.org/@go/page/91909

This velocity field just has fluid moving uniformly to the right. Fluid enters the sphere from the left and leaves through the right at precisely the same rate, so that the net rate at fluid crosses the sphere is zero. The divergence ⇀

⇀

⇀

∇ ⋅ v(0 ) = 0

again consistent with our interpretation 4.1.19.

Interpretation of the Curl We'll now develop the interpretation of the curl, or more precisely, of developing the interpretation of divergence, we'll ⇀

⇀

⇀

⇀

^ ∇ × v( r 0 ) ⋅ n

for any unit vector

^. n

As we did in

^ as a limit of integrals, and first express ∇ × v ( r ) ⋅ n then we'll interpret the integrals. ⇀

⇀ 0

To specify the integrals involved, let C be the circle which ε

is centered at r has radius ε ^ lies in the plane through r perpendicular to n ^ . Imagine standing on the circle with your feet on the plane through r is oriented in the standard way with respect to n ^ ^ and with your left arm point towards perpendicular to n, with the vector from your feet to your head in the same direction as n r . Then your are facing in the positive direction for C . ⇀

0

⇀

0

⇀

0

⇀

0

ε

We shall show in Lemma 4.1.25, below, that ⇀

⇀

^ = lim ∇ × v( r 0 ) ⋅ n ε→0

1 πε2

∮

⇀

⇀

⇀

v( r ) ⋅ d r

Cε

Now let's work on interpreting the right hand side, and in particular on interpreting the integral ∮ v ( r ) ⋅ d r , which is called the ^ and its paddles along C , as in the circulation of v around C . Place a tiny paddlewheel in the fluid with its axle running along n figure below, except that ⇀

⇀

⇀

Cε

⇀

ε

ε

the paddlewheel is really expensive and has a lot more than just four paddles. Pretend 16 that you are one of the paddles. If the paddlewheel is rotating at Ω radians per unit time, then in one unit of time you sweep out an arc of a circle of radius ε that subtends an angle Ω. That arc has length Ωε. So you are moving at speed Ωε. ⇀

If you are at

⇀

r,

the component of the fluid velocity in your direction of motion, i.e. tangential to C , is ε

⇀

^ t =

dr

, ds

⇀

⇀

dr

v( r ) ⋅

, ds

because

with s denoting arc length along the circle, is a unit vector tangential to C

ε.

4.1.16

https://math.libretexts.org/@go/page/91909

⇀

All paddles have to move at the same speed. So the speed of the paddles, Ωε, should be the average value of

⇀

⇀

dr

v( r ) ⋅ ds

the circle.

around

Thus the rate of rotation, Ω, of the paddlewheel should be determined by ⇀

∮

⇀

Cε

dr

⇀

v( r ) ⋅

Ωε = ∮ ⇀

⇀

0)

^ ⋅n

⇀

∮

Cε

=

⇀

⇀

v( r ) ⋅ d r 2πε

ds

Cε

Consequently, ∇ × v( r

ds ds

is the limit as ε (the radius of the paddlewheel) tends to zero of 1 πε2

∮

⇀

⇀

⇀

v ( r ) ⋅ d r = 2Ω

Cε

That's our interpretation.

 Equation 4.1.24 If a fluid has velocity field 1 2

⇀

⇀

⇀

^ ∇ × v( r 0 ) ⋅ n ⇀

⇀

⇀

v

and you place an infinitesimal paddlewheel at

⇀ r0

with its axle in direction

⇀

n,

then it rotates at

radians per unit time. In particular, to maximize the rate of rotation, orient the paddlewheel so that

⇀

^ ∥ ∇ × v ( r 0 ). n

There will be some examples at the end of this section. First, we show

 Lemma 4.1.25 ⇀

⇀

1

⇀

∇ × v( r 0 ) ⋅ n ^ = lim ε→0

πε

2

⇀

∮

⇀

⇀

v( r ) ⋅ d r

Cε

Proof. (Optional). Here is one proof 17 of Lemma 4.1.25. Just as we did in the proof of Lemma 4.1.20, we can always translate our coordinate system so that ⇀

^ ^ =k We can also rotate our coordinate system so that n . Because lies in the xy-plane, we can parametrize C by

r 0 = (x0 , y0 , z0 ) = (0, 0, 0).

that C

ε

⇀

r 0 = (0, 0, 0)

^ ^ =k and n , so

ε

⇀

r (t) = ε cos t ^ ıı + ε sin t ^ ȷ ȷ

Again as we did in the proof of Lemma 4.1.20, expand with second order error term.

⇀

v (x, y, z)

in a Taylor expansion in powers of x, y, and z, to first order,

⇀

v (x, y, z) = A + B x + C y + D z + R(x, y, z)

where ⇀

⇀

∂v

⇀

A = v (0, 0, 0)

B =

⇀

∂v (0, 0, 0)

C =

∂x

∂y

and the error term R(x, y, z) is bounded by a constant times x

2

+y

2

∂v (0, 0, 0)

2

+z .

|R(x, y, z)| ≤ Kε

D =

(0, 0, 0) ∂z

In particular there is a constant K so that, on C

ε,

2

So ∮

⇀

⇀

⇀

v( r ) ⋅ d r

Cε 2π

=∫

⇀

(A + B ε cos t + C ε sin t + R( r (t))) ⋅ ( − ε sin t ^ ıı + ε cos t ^ ȷ ȷ ) dt

0

4.1.17

https://math.libretexts.org/@go/page/91909

Again, multiply out the dot product so that the integrand becomes − εA ⋅ ^ ıı sin t

+  εA ⋅ ^ ȷ ȷ cos t

2

2

−ε B⋅ ^ ıı sin t cos t 2

2

−ε C⋅ ^ ıı sin

2

+ε B⋅ ^ ȷ ȷ cos

t

2

+  ε C ⋅ ^ ȷ ȷ sin t cos t

t

⇀

+ R( r (t)) ⋅ ( − ε sin t ^ ıı + ε cos t ^ ȷ ȷ)

Again most of these terms integrate to zero, because 2π

∫

2π

sin t dt =

∫

0 2π

∫

2

t

and cos

2

t

=0

2π

1 sin t cos t dt =

∫ 2

0

and the sin

cos t dt

0

sin(2t) dt = 0

0

terms are easily integrated using (see Example 2.4.4) 2π

2π 2

∫

sin

2

t dt = ∫

0

cos

2π

1 t dt =

2

∫ 2

0

[ sin

2

t + cos

t] dt = π

0

So we are left with 2π

∮

⇀

⇀

⇀

2

⇀

2

= πε B ⋅ ^ ȷ ȷ − πε C ⋅ ^ ıı + ∫

v( r ) ⋅ d r

Cε

R( r (t)) ⋅ ( − ε sin t ^ ıı + ε cos t ^ ȷ ȷ ) dt

0

which implies that ⇀

1 lim ε→0

πε

2

⇀

∮

⇀

⇀

v( r ) ⋅ d r

=

∂ v2

⇀

(0, 0, 0) −

∂x

Cε

∂ v1

(0, 0, 0)

∂y 2π

1 + lim ε→0

⇀

R( r (t)) ⋅ ( − ε sin t ^ ıı + ε cos t ^ ȷ ȷ ) dt

∫ πε2

0

⇀

⇀ ^ = (∇ × v (0, 0, 0)) ⋅ k 2π

1 + lim ε→0

Finally, it suffices to recall that |R(x, y, z)| ≤ Kε 1 πε2

∣ ∣∫ ∣

0

2π

2

,

πε

2

∫

⇀

R( r (t)) ⋅ ( − ε sin t ^ ıı + ε cos t ^ ȷ ȷ ) dt

0

so that

∣ ⇀ R( r (t)) ⋅ ( − ε sin t ^ ıı + ε cos t ^ ȷ ȷ ) dt∣ ∣

2π

1 ≤

⇀

∫ πε

0 2π

1 ≤

|R( r (t))| dt

∫ πε

2

dt

0

1 =

Kε

 K ε

2

(2π)

πε = 2Kε

converges to zero as ε → 0. Here are some examples. We will use the same vector fields as in Examples 4.1.21, 4.1.22 and 4.1.23. In all examples, we shall ^ orient the paddlewheel so that n ^ = k and sketch the top view, so that the paddlewheel looks like

4.1.18

https://math.libretexts.org/@go/page/91909

 Example 4.1.26 Here is a sketch of the vector field

^ v (x, y, z) = x ^ ı ı +y ^ ȷ ȷ +zk

⇀

and a circle centered on the origin, like C

ε.

This velocity field has fluid moving parallel to the paddles, so the paddlewheel should not rotate at all. The computation ^ k ⎤

⎡

^ ı ı

^ ȷ ȷ

∇ × v ( 0 ) = det ⎢ ⎢

∂

∂

∂

∂x

∂y

∂z

x

y

z

⇀

⇀

⇀

⎣

⇀

⎥ = 0 ⎥

⇀

⟹

⇀

⇀ ^ ∇ × v(0 ) ⋅ k = 0

⎦

is consistent with our interpretation 4.1.24.

 Example 4.1.27 Here is a sketch of the vector field

⇀

v (x, y, z) = −y ^ ı ı +x ^ ȷ ȷ

and a circle centered on the origin, like C

ε.

This velocity field has fluid going around in circles, counterclockwise. So the paddlewheel should rotate counterclockwise too. That is, it should have positive angular velocity. Our interpretation 4.1.24 predicts an angular velocity of half ⎡ ⇀

⇀

⇀

^ ∇ × v ( 0 ) ⋅ k = det ⎢ ⎢ ⎣

^ k ⎤

^ ı ı

^ ȷ ȷ

∂

∂

∂

∂x

∂y

∂z

−y

x

0

^ ^ ^ ⎥⋅k = 2k ⋅ k = 2 ⎥ ⎦

which is indeed positive 18.

 Example 4.1.28 Here is a sketch of the vector field

⇀

v (x, y, z) = ^ ı ı

and a circle centered on the origin, like C

ε.

The fluid pushing on the top paddle tries to make the paddlewheel rotate clockwise. The fluid pushing on the bottom paddle tries to make the paddlewheel rotate counterclockwise, at the same rate. So the paddlewheel should not rotate at all. Our interpretation 4.1.24 predicts an angular velocity of

4.1.19

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⎡ 1

⇀

⇀

⇀

^ ⋅ ∇ × v(0 ) ⋅ k =

2

1 2

det ⎢ ⎢ ⎣

^ k ⎤

^ ı ı

^ ȷ ȷ

∂

∂

∂

∂x

∂y

∂z

1

0

0

⇀

^ ^ ⎥⋅k = 0 ⋅k =0 ⎥ ⎦

as expected.

Exercises Stage 1  1✳ ⇀

Let F = P

^ ı ı +Q ^ ȷ ȷ

be the two dimensional vector field shown below.

1. Assuming that the vector field in the picture is a force field, the work done by the vector field on a particle moving from point A to B along the given path is: 1. 2. 3. 4.

Positive Negative Zero Not enough information to determine.

2. Which statement is the most true about the line integral ∫

C2

1.

⇀

∫

⇀

⇀

F ⋅ dr :

⇀

F ⋅ dr > 0

C2

2.

⇀

∫

⇀

F ⋅ dr = 0

C2

3.

⇀

∫

⇀

F ⋅ dr < 0

C2

4. Not enough information to determine.

⇀

⇀

∇⋅F

1. 2. 3. 4. 3. Q

x

at the point N (in the picture) is:

Positive Negative Zero Not enough information to determine. − Py

at the point Q is:

1. Positive 2. Negative

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3. Zero 4. Not enough information to determine. ⇀

4. Assuming that F = P 1.

∂P

2.

∂P

∂x

∂x

=0

at D.

>0

at D.

^ ı ı +Q ^ ȷ ȷ,

which of the following statements is correct about

∂P ∂x

at the point D?

3. < 0 at D. 4. The sign of at D can not be determined by the given information. ∂P

∂x

∂P

∂x

 2 ⇀

⇀

⇀

Does ∇ × F have to be perpendicular to F?

 3 Verify the vector identities ⇀

⇀

⇀

⇀

⇀

⇀

1. ∇ ⋅ (f F) = f ∇ ⋅ F + F ⋅ ∇f 2. ∇ ⋅ (F × G) = G ⋅ (∇ × F) − F ⋅ (∇ × G) ⇀

⇀

⇀

⇀2

3. ∇

⇀2

⇀

⇀

⇀

⇀

⇀2

⇀

(f g) = f ∇ g + 2 ∇f ⋅ ∇g + g ∇ f

Stage 2  4 ⇀

⇀

⇀

⇀

Evaluate ∇ ⋅ F and ∇ × F for each of the following vector fields. ⇀

^ 1. F = x ^ ı ı +y ^ ȷ ȷ +zk 2. F = x y ^ ı ı − yz ^ ȷ ȷ + zx ⇀

⇀

3. F = ⇀

4. F =

2

2

^ xı ı +y ^ ȷ ȷ 2

√x +y

2

^ −y ı ı +x^ ȷ ȷ 2

√x +y

2

2

^ k

(the polar basis vector ^ r in 2d) (the polar basis vector θ^θ in 2d)

 5✳ ⇀

1. Compute and simplify ∇ ⋅ ( ⇀

⇀ r r

)

for

2. Compute ∇ × (yz ^ ı ı + 2xz ^ ȷ ȷ +e

xy

⇀

r = (x, y, z)

and r = |(x, y, z)|. Express your answer in terms of r.

^ k).

 6✳ In the following, we use the notation

⇀

⇀

^ r =x ^ ı ı +y ^ ȷ ȷ + z k,

r = | r |,

and k is some number k = 0, 1, −1, 2, −2, … .

1. Find the value k for which ⇀

⇀

r

k

∇(r ) = −3

r5

2. Find the value k for which ⇀

k⇀

∇ ⋅ (r

2

r ) = 5r

3. Find the value k for which ⇀2

k

∇ (r ) =

2 4

r

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 7✳ Let

^ be the vector field r = x ^ ı ı +y ^ ȷ ȷ +zk and let r be the function r = | r |. Let a be the constant vector ^ a=a ^ ı ı +a ^ ȷ ȷ + a k. Compute and simplify the following quantities. Answers must be expressed in terms of a, r , and r. There should be no x's, y 's, or z 's in your answers. ⇀

⇀

r

⇀

⇀

1

⇀

2

3

1. ∇ ⋅ r 2. ∇(r ) ⇀

⇀

2

⇀

3. ∇ × ( r × a) 4. ∇ ⋅ (∇(r)) ⇀

⇀

⇀

 8✳ Let ⇀

⇀

^ r =x ^ ı ı +y ^ ȷ ȷ + z k,

⇀

1. Compute a where ∇( ⇀

1 r

r = |r|

a ⇀

) = −r

r.

2. Compute a where ∇ ⋅ (r r ) = ar. ⇀

⇀

⇀

3. Compute a where ∇ ⋅ (∇(r

3

)) = ar.

 9 ^ Find, if possible, a vector field A that has k component A

3

⇀

1. F = (1 + yz) ^ ı ı + (2y + zx)^ ȷ ȷ + (3 z ^ ^ ^ 2. G = yz ıı + zx ȷȷ + xy k

2

=0

and that is a vector potential for

^ + xy)k

Stage 3  10 ✳ Let −z

⇀

F =

2

x

+z

x 2

^ ı ı +y ^ ȷ ȷ +

2

x

+z

2

^ k

⇀

1. Determine the domain of F. ⇀

2. Determine the curl of F. Simplify if possible. 3. Determine the divergence of F. Simplify if possible. 4. Is F conservative? Give a reason for your answer. ⇀

⇀

 11 ✳ A physicist studies a vector field ⇀

F = ∇ × G,

⇀

F

in her lab. She knows from theoretical considerations that

⇀

F

must be of the form

⇀

for some smooth vector field G. Experiments also show that F must be of the form ⇀

^ F(x, y, z) = (xz + xy) ^ ı ı + α(yz − xy)^ ȷ ȷ + β(yz + xz)k

where α and β are constant. 1. Determine α and β. ^ 2. Further experiments show that G = xyz ^ ı ı − xyz^ ȷ ȷ + g(x, y, z)k. Find the unknown function g(x, y, z).

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 12 A rigid body rotates at an angular velocity of Ω rad/sec about an axis that passes through the origin and has direction a ^ . When you are standing at the head of a ^ looking towards the origin, the rotation is counterclockwise. Set Ω = Ω a ^. 1. Show that the velocity of the point r = (x, y, z) on the body is Ω × r . 2. Evaluate ∇ × (Ω × r ) and ∇ ⋅ (Ω × r ), treating Ω as a constant. 3. Find the speed of the students in a classroom located at latitude 49 N due to the rotation of the Earth. Ignore the motion of the Earth about the Sun, the Sun in the Galaxy and so on. The radius of the Earth is 6378 km. ⇀

⇀

⇀

⇀

⇀

⇀

∘

 13 ⇀

⇀

⇀

Suppose that the vector field F obeys ∇ ⋅ F = 0 in all of R . Let 3

^ r (t) = tx ^ ı ı + ty ^ ȷ ȷ + tz k,

⇀

0 ≤t ≤1

be a parametrization of the line segment from the origin to (x, y, z). Define 1

G(x, y, z) = ∫

⇀

⇀

dr

⇀

t F( r (t)) ×

⇀

⇀

Show that ∇ × G = F throughout R

3

(t) dt dt

0

.

1. Good shorthand is not only more brief, but also aids understanding “of the forest by hiding the trees”. 2. Pierre-Simon Laplace (1749–1827) was a French mathematician and astronomer. He is also the Laplace of Laplace's equation, the Laplace transform, and the Laplace-Bayes estimator. He was Napoleon's examiner when Napoleon attended the Ecole Militaire in Paris. 3. To be picky, these are Maxwell's equations in the absence of a material medium and in Gaussian units. 4. One important consequence of Maxwell's equations is that electromagnetic radiation, like light, propagate at the speed of light. 5. James Clerk Maxwell (1831–1879) was a Scottish mathematical physicist. In a poll of prominent physicists, Maxwell was voted the third greatest physicist of all time. Only Newton and Einstein beat him. ⇀⇀

6. This is really the only definition that makes sense. For example G ⋅ (∇F) does not make sense because you can't take the gradient of a vector-valued function. 7. δ is called the Kronecker delta function. It is named after the German number theorist and logician Leopold Kronecker (1823–1891). He is reputed to have said “God made the integers. All else is the work of man.” 8. Does this remind you of Theorem 2.4.8? It should. 9. There is nothing special about the subscript 3 here. By precisely the same argument, we could come up with another vector potential whose second component is zero, and with a third vector potential whose first component is zero. m,n

10. Of course, we could equally well pick A

1

=0

or A

2

= 0.

11. If the z 's had not cancelled out, no N (x, y) and M (x, y), which after all are independent of z, could satisfy the equation. That would have been a sure sign of a user error. 12. Lemma 3.4.1 is being applied with the density ρ set equal to one, so, more precisely, the rate is the number of units of volume of fluid exiting S per unit time. ε

13. There is another, easier to understand, proof of this result given in §4.4.1. We cannot give that proof here because it uses the divergence theorem, which we will get to later in the chapter. 14. Terms like xy, xz and yz are not needed because, for example, |xy| ≤

1 2

2

(x

2

+ y ).

This inequality is equivalent to

2

(|x| − |y| )

≥ 0.

15. Spheres have lots of symmetry! 16. Method acting might help you here. 17. There is another, easier to understand, proof of this result given in §4.4.1. We cannot give that proof here because it uses Stokes' theorem, which we will get to later in the chapter. 18. Even for small values of 2.

4.1.23

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This page titled 4.1: Gradient, Divergence and Curl is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

4.1.24

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4.2: The Divergence Theorem The rest of this chapter concerns three theorems: the divergence theorem, Green's theorem and Stokes' theorem. Superficially, they look quite different from each other. But, in fact, they are all very closely related and all three are generalizations of the fundamental theorem of calculus b

df

∫

(t) dt = f (b) − f (a)

a

dt

The left hand side of the fundamental theorem of calculus is the integral of the derivative of a function. The right hand side involves only values of the function on the boundary of the domain of integration. The divergence theorem, Green's theorem and Stokes' theorem also have this form, but the integrals are in more than one dimension. So the derivatives are multidimensional, like the curl and divergence, and the integrands can involve vector fields. For the divergence theorem, the integral on the left hand side is over a (three dimensional) volume and the right hand side is an integral over the boundary of the volume, which is a surface. For Green's and Stokes' theorems, the integral on the left hand side is over a (two dimensional) surface and the right hand side is an integral over the boundary of the surface, which is a curve. The divergence theorem is going to relate a volume integral over a solid V to a flux integral over the surface of V . First we need a couple of definitions concerning the allowed surfaces. In many applications solids, for example cubes, have corners and edges where the normal vector is not defined. On the other hand, to be able to compute a flux integral over a surface, we certainly need that the set of points where the normal vector is not well-defined is small enough that the existence of the flux integral is not jeopardized. This is the case for “piecewise smooth” surfaces, which we now define.

 Definition 4.2.1 ⇀ ∂ r

⇀ ∂ r

⇀ ∂ r

⇀ ∂ r

∂u

∂v

∂u

∂v

1. A surface is smooth if it has a parametrization r (u, v) with continuous partial derivatives and and with × nonzero. 2. A surface is piecewise smooth if it consists of a finite number of smooth pieces that meet along sharp curves and at sharp corners. ⇀

Here are sketches of a smooth surface (a sausage) and a piecewise smooth surface (an ice-cream cone), followed by the divergence theorem 1.

 Theorem 4.2.2. Divergence Theorem Let V ⇀

F

be a bounded solid with a piecewise smooth surface 2 ∂V be a vector field that has continuous first partial derivatives at every point of V .

Then ∬

⇀

⇀

^ dS = ∭ F⋅n

∇ ⋅ F dV

∂V

⇀

V

^ is the outward unit normal of ∂V . where n

4.2.1

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Like the fundamental theorem of calculus, the divergence theorem expresses the integral of a derivative of a function (in this case a vector-valued function) over a region in terms of the values of the function on the boundary of the region.

 Warning 4.2.3 ⇀

Note that in Theorem 4.2.2 we are assuming that the vector field F has continuous first partial derivatives at every point of V . If that is not the case, for example because F is not defined on all of V , then the conclusion of the divergence theorem can fail. ⇀

⇀

An example is F =

⇀ r ⇀ 3 | r |

2

, V = {(x, y, z)| x

+y

2

+z

2

See Example 4.2.7.

≤ 1} .

Proof We have to show that ⇀ ⇀

∬

⇀

=∭

∂V

(

+

∂x

V ⇀

Note that the left hand side is a sum of three terms — one involving

⇀

∂F1

⇀

^ ^ dS (F1 ^ ıı + F2 ^ ȷ ȷ + F3 k) ⋅ n

F1 ,

∂F2

⇀

+

∂y

one involving

⇀

∂F3

) dV

∂z ⇀

F2

and one involving

⇀

⇀

right hand side is a sum of three terms — one involving F , one involving F and one involving F 1

⇀

F3

3.

2

⇀

F3

— and the

We'll just show that the

terms on the left hand side and right hand side are equal, i.e. that ⇀ ⇀

∬

F3

^ ^ k⋅n dS = ∭

∂V ⇀

V

∂F3

 dV

∂z

⇀

Showing that the F terms match and the F terms match is done in the same way 3. 1

2

Special Geometry We'll first assume that the solid has the special form V = {(x, y, z)|B(x, y) ≤ z ≤ T (x, y),   (x, y) ∈ Rxy }

where R is some subset of the xy-plane. We can further assume that, for each After we're finished with this special case, we'll handle the general case. x,y

Let's work on ∬

⇀

∂V

^ ^ F3 k ⋅ n dS

(x, y) ∈ Rxy ,

we have

B(x, y) ≤ T (x, y).

first. As in the figure below,

the surface ∂V consists of three pieces — the top, the bottom and the side. We'll consider each in turn. The top is T

= {(x, y, z)|z = T (x, y),   (x, y) ∈ Rxy } .

By 3.3.2, on T

^ ^ dS = +[ − Tx (x, y) ^ n ıı − Ty (x, y) ^ ȷ ȷ + k] dxdy ^ is to be the outward normal, it must point upwards on T . That's why we have chosen, and emphasised, the “+ ” sign. As n ^ ^ dS = dxdy and So k ⋅n

4.2.2

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⇀

⇀

^ ^ F3 k ⋅ n dS = ∬

∬ T

F3 (x, y, T (x, y)) dxdy

Rxy

The bottom is B = {(x, y, z)|z = B(x, y),   (x, y) ∈ R

xy

By 3.3.2, on B

}.

^ ^ dS = −[ − Bx (x, y) ^ n ıı − By (x, y) ^ ȷ ȷ + k] dxdy ^ is to be the outward normal, it must point downwards on B. That's why we have chosen the “− ” sign. So As n ^ ^ k⋅n dS = −dxdy and

⇀

^ ^ F3 k ⋅ n dS = − ∬

∬ B

⇀

F3 (x, y, B(x, y)) dxdy

Rxy

The side is S = {(x, y, z)|(x, y) ∈ ∂ R ,  B(x, y) ≤ z ≤ T (x, y)} . It runs vertically. Hence on S the normal vector to ^ ^ ∂V is parallel to the xy-plane so that k ⋅ n = 0 and xy

⇀

∬

^ ^ F3 k ⋅ n dS = 0

S

So all together ⇀

^ ^ F3 k ⋅ n dS = ∬

∬ ∂V

⇀

T

^ ^ F3 k ⋅ n dS+∬

B

⇀

=∬

⇀

^ ^ F3 k ⋅ n dS+∬

⇀

^ ^ F3 k ⋅ n dS

S

⇀

[ F3 (x, y, T (x, y)) − F3 (x, y, B(x, y))] dxdy + 0

(∂V )

Rxy

Now let us examine ⇀

∂F3

∭

=∬

∂z

V

⇀

T (x,y)

 dV

dxdy ∫

Rxy

dz 

(x, y, z)

∂z

B(x,y) ⇀

=∬

∂F3

⇀

[ F3 (x, y, T (x, y)) − F3 (x, y, B(x, y))] dxdy

(V )

Rxy

by the fundamental theorem of calculus. That's exactly what we had to show. The integrals (∂V ) and (V ) are equal. General Geometry Now we'll drop the assumption on V that we imposed in the “Special Geometry” section above. The key idea that makes the proof work is that we can cut up any 4 V into pieces, each of which does obey the special assumption that we just considered. Consider, for example, the sausage shaped solid in the figure on the left below.

Call the sausage V . Cut it into two halves by running a cleaver horizontally through its centre. This splits the solid V into two halves, V and V as in the figure on the right above. It also splits the boundary ∂V of V into two halves S and S , also as in the figure on the right above. Note that 1

2

1

2

the boundary, ∂ V , of V is the union of S and the shaded disk S (the cut introduced by the cleaver). On the cut S , the ^ outward pointing normal to V is −k . The boundary, ∂ V , of V is the union of S and the shaded disk S . On the cut S , the outward pointing normal to V is 1

1

1

c

c

1

2

2

2

c

c

2

^ +k.

Now both V and V do satisfy the assumption of the “Special Geometry” section above. So 1

2

4.2.3

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⇀

⇀

∂F3

∭

∂z

V

⇀

∂F3

 dV = ∭ V1

∂z

⇀

∂V1

⇀

^ ^ F3 k ⋅ n dS

∂V2 ⇀

⇀

^ ^ dS + ∬ F3 k ⋅ n

=∬

 dV

∂z

V2

^ ^ F3 k ⋅ n dS + ∬

=∬

∂F3

 dV + ∭

S1

⇀

^ ^ dS + ∬ F3 k ⋅ n

Sc ↓

^ ^ dS F3 k ⋅ n

S2

⇀

^ ^ dS F3 k ⋅ n

+∬ Sc ↑

^ ^ Here S ↓ is the surface S with normal vector −k and S ↑ is the surface S with normal vector −k . So the second and ^ ^ ^ = −k in the second integral and n ^ = +k in the fourth integral. So they cancel fourth integrals are identical except that n exactly and c

c

c

c

⇀

∭

∂F3

⇀

 dV

V

⇀

^ ^ F3 k ⋅ n dS + ∬

=∬

∂z

S1

⇀

^ ^ F3 k ⋅ n dS = ∬

S2

^ ^ F3 k ⋅ n dS

∂V

as desired.

 Example 4.2.4 Evaluate the flux integral ∬

S

⇀

^ dS F⋅n

^ is the outward normal to S, which is the surface of the hemispherical region where n

2

V = {(x, y, z| x

+y

2

+z

2

2

≤ a ,  z ≥ 0}

and ⇀

F = xz

2

2

3

2

^ ı ı + (x y − z ) ^ ȷ ȷ + (2xy + y z + e

cos y

^ )k

Solution ⇀

The e in F suggests that a direct evaluation of the integral is difficult. So we'll use a little trickery to to evaluate it. Not surprisingly, considering that we have just proven the divergence theorem, the trick is to apply the divergence theorem 5. Since cos y

⇀

⇀

∇⋅F =

⇀

⇀

⇀

∂F1

∂F2

∂F3

+

∂x ∂ =

+

∂y

∂z ∂

2

(x z ) + ∂x

=z

2

2

∂

3

(x y − z ) + ∂y

2

+x

+y

2

(2xy + y z + e

cos y

)

∂z

2

The divergence theorem tell us that ⇀

∬ S

^ dS F⋅n

=∭

2

(x

+y

2

2

+ z ) dV

V

Spherical coordinates are perfect for this integral. (See Appendix A.6.3, if you need to refresh your memory.)

4.2.4

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π

2π 2

∭

(x

+y

2

2

+ z ) dV

=∫

V

a

2

dθ ∫

0

0

dρ ρ

2

sin φ ρ

0 π

2π

= [∫

2

dφ ∫

a

2

dθ][ ∫

0

sin φ dφ][ ∫

0

4

ρ

dρ]

0 π

= [2π][ − cos φ]

2

0

a

5

ρ [

] 5

0

5

2πa = 5

 Example 4.2.5 Evaluate the flux integral ∬ 1 ≤ z ≤ 2, and where

S

⇀

^ dS F⋅n

^ is the outward normal to S, which is the part of the surface z where n

⇀

F = 3x ^ ı ı + (5y + e

cos x

2

2

=x

+y

2

with

^ )^ ȷ ȷ +zk

Solution Again the e in F suggests that a direct evaluation is difficult 6 and again we'll apply the divergence theorem. But this time S is not the boundary of a solid V . It is the portion of the cone outlined in red in the figure on the left below and does not have a top or bottom “cap”. cos x

⇀

Fortunately, there is a solid V whose boundary, while not being equal to S, at least contains S. It is (unsurprisingly) 2

V = {(x, y, z)| x

+y

2

2

≤ z ,   1 ≤ z ≤ 2}

and is sketched in the figure on the right above. The boundary, ∂V , is the union of S and the two disks 2

D1

= {(x, y, z)| x

D2

= {(x, y, z)| x

2

+y +y

2

2

2

≤ z ,   z = 1} 2

≤ z ,   z = 2}

So the divergence theorem gives ⇀

∭

⇀

⇀

∇ ⋅ F dV

^ dS F⋅n

=∬

V

∂V

=∬

⇀

⇀

⇀

F⋅n ^ dS + ∬

F⋅n ^ dS + ∬

F⋅n ^ dS

S

D1

D2

which implies ⇀

∬

⇀

^ dS F⋅n

=∭

S

⇀

∇ ⋅ F dV − ∬

V

⇀

⇀

^ dS − ∬ F⋅n

^ dS F⋅n

D1

D2

The point of this exercise is that the left hand side, which is not easy to evaluate directly, is the integral we want, while the three integrals on the right hand side are all easy to evaluate. We do so now. The outward normal to (the horizontal disk) D is ^ +k. So 2

⇀

∬ D2

^ dS F⋅n

⇀

=∬

^ F ⋅ k dS = ∬

D2

4.2.5

z dS

D2

https://math.libretexts.org/@go/page/91910

As z = 2 on D , and D is a disk of radius 2, 2

2

⇀

2

^ dS = 2Area(D2 ) = 2π 2 F⋅n

∬

= 8π

D1

^ Similarly, the outward normal to (the horizontal disk) D is −k . So 1

⇀

⇀

^ dS F⋅n

∬

^ F ⋅ k dS = − ∬

= −∬

D1

D1

z dS

D1

As z = 1 on D , and D is a disk of radius 1, 1

1

⇀

∬

2

F⋅n ^ dS = Area(D1 ) = −π 1

= −π

D1 ⇀

⇀

Finally, as ∇ ⋅ F = 3 + 5 + 1 = 9 ⇀

∭

⇀

∇ ⋅ F dV = 9 Vol(V )

V

The volume of V can be easily computed using the first year technique 7 of slicing sketched in the figure below.

V

into thin horizontal pancakes like that

The pancake at height z has thickness dz, a circular cross-section of radius z (remember that the outer boundary of V has equation x cross-sectional area πz and volume π z dz.

2

+y

2

=z

2

), and hence has

2

2

So ⇀

∭

2

⇀

∇ ⋅ F dV = 9 Vol(V ) = 9 ∫

V

πz

2

πz

2

3

dz = 9 [

] 3

1

7 = 9 ×π

= 21π 3

1

and, all together ⇀

∬

⇀

^ dS F⋅n

=∭

S

⇀

⇀

⇀

^ dS − ∬ F⋅n

∇ ⋅ F dV − ∬

V

D1

^ dS F⋅n

D2

= 21π − (−π) − 8π = 14π

 Example 4.2.6 Evaluate the flux integral 0 ≤ z ≤ 1, and

⇀

∬

S

^ dS F⋅n

where

^ n

is the upward normal to

⇀

F = (x + e

y

2

S,

which is the part of

2

z = (x

2

+y )

2

with

^ ) ^ ı ı + (y + cos z) ^ ȷ ȷ +k

Solution This integral can be evaluated in much the same way as we evaluated the integral of Example 4.2.6. We first define a solid V whose boundary ∂V contains S. A good, and hopefully obvious, choice is 2

V = {(x, y, z)| (x

2

+y )

4.2.6

2

≤ z,   0 ≤ z ≤ 1}

https://math.libretexts.org/@go/page/91910

The boundary of V is the union of S, with outward pointing normal refers to the upward pointing normal) and the disk

⇀

−n

2

D = {(x, y, z)|z = 1,    (x

(recall that the problem specifies that the symbol

^ n

2

2

+y )

≤ 1}

^ with outward pointing normal k .

So the divergence theorem gives ⇀

∭

⇀

⇀

V

⇀

^ F ⋅ k dS

^ dS + ∬ F⋅n

∇ ⋅ F dV = − ∬ S

D

which implies ⇀

∬

⇀

^ dS F⋅n

= −∭

S

⇀

⇀

V

D

= −∭

2 dV + ∬

V

D

^ F ⋅ k dS

∇ ⋅ F dV + ∬

dS

D

is a circular disk of radius 1, and so has area π. To evaluate the volume integral we slice V into horizontal pancakes with the 1

pancake at height z having a circular cross-section of radius z

2

1

⇀

∬

(Recall that the boundary of V has (x

.

4

^ dS F⋅n

2

= −2 ∫

S

π √z dz + π = −2π ×

0

2

+y )

2

= z.

) So

π +π = −

3

3

Again, you can see that the actual integration is quite easy. All of the work (or at least all of the thinking) happens in the setup.

 Example 4.2.7 In Warning 4.2.3 we emphasised that the conclusion of the divergence Theorem 4.2.2 can fail if the vector field defined at even a single point of V . Here is an example. Set

⇀

F

is not

⇀

r

⇀

F =

⇀ ^ where  r = x ^ ı ı +y ^ ȷ ȷ +zk

⇀ 3

|r|

and V

2

= {(x, y, z)| x

+y

2

⇀

+z

2

⇀

≤ 1} .

Then, if (x, y, z) ≠ 0 , ∂

⇀

x

∂

∇ ⋅ F(x, y, z) =

y

+ 3/2

∂x

∂

3/2

∂y

[ x2 + y 2 + z 2 ]

[ x2 + y 2 + z 2 ]

z

+ ∂z 2

[x

+y

2

2

[x

2

+z ] −x

= 2

[x 2

[x

+y

+y

2

2

2

2

2

2

2

3 2

2

3/2

+z ] 2

(2x)

5/2

2

+z ] −z

+y

3

2

+z ]

+ [x

+y

[x

+y

2

2

+z ] −y

+ 2

[x

+y

2

2

3 2

(2y)

5/2

+z ]

(2z)

5/2

+z ]

=0

4.2.7

https://math.libretexts.org/@go/page/91910

On the other hand, the boundary of V is the unit sphere ∂V is n ^ =

⇀ r ⇀ | r |

2

= {(x, y, z)| x

+y

2

+z

2

= 1} .

The outward unit normal to ∂V

so that ⇀

∂V

⇀

r

⇀

^  dS F⋅n

∫

=∫ ⇀ | r |=1

⇀ 3

r

⋅

|r|

⇀

1  dS = ∫

|r|

⇀ | r |=1

⇀ 2

 dS = ∫

|r|

dS

⇀ | r |=1

= 4π ≠ 0

Optional — An Application of the Divergence Theorem — the Heat Equation Derivation of the Heat Equation Let T (x, y, z, t) be the temperature at time t at the point (x, y, z) in some object B. The heat equation 8 is the partial differential equation that describes the flow of heat energy and consequently the behaviour of T . We now use the divergence theorem to derive the heat equation from two physical “laws”, that we assume are valid: The amount of heat energy required to raise the temperature of an object by ΔT degrees is C M ΔT where, M is the mass of the object and C is a positive physical constant determined by the material contained in the object. It is called the specific heat, or specific heat capacity 9, of the object. Think of heat energy as a moving fluid. We will rig its velocity field so that heat flows in the direction opposite to the ⇀

temperature gradient. Precisely, we choose its velocity field to be −κ∇T (x, y, z, t). Here κ is another positive physical constant called the thermal conductivity of the object. So the rate at which heat is conducted across an element of surface area ^ is given by −κ n ^ ⋅ ∇T (x, y, z, t) dS at time t. (See Lemma 3.4.1.) For dS at (x, y, z) in the direction of its unit normal n example, in the figure ⇀

the temperature gradient, which points in the direction of increasing temperature, is opposite

^. n

Consequently the flow rate

⇀

^ . This is just what you would expect — heat flows is positive, indicating flow in the direction of n from hot regions to cold regions. Also the rate of flow increases as the magnitude of the temperature gradient increases. This also makes sense (and is reminiscent of Newton's law of cooling). ^ ⋅ ∇T (x, y, z, t) dS −κ n

^ the outward normal to ∂V . Let V ⊂ B be any three dimensional region in the object and denote by ∂V the surface of V and by n The amount of heat that enters V across an infinitesimal piece dS of ∂V in an infinitesimal time interval dt is ^ ⋅ ∇T (x, y, z, t) dS) dt. The amount of heat that enters V across all of ∂V in the time interval dt is given by the integral −( − κ n ⇀

⇀

∬

κn ^ ⋅ ∇T (x, y, z, t) dS dt

∂V

In this same time interval, the temperature at a point (x, y, z) in V changes by (x, y, z, t) dt. If the density of the object at (x, y, z) is ρ(x, y, z), the amount of heat energy required to increase the temperature of an infinitesimal volume dV of the object centred at (x, y, z) by (x, y, z, t) dt is C (ρdV ) (x, y, z, t) dt. The amount of heat energy required to increase the ∂T ∂t

∂T

∂T

∂t

temperature by

∂T ∂t

(x, y, z, t) dt

∂t

at all points (x, y, z) in V is then ∂T ∭ V

Cρ

(x, y, z, t) dV dt ∂t

4.2.8

https://math.libretexts.org/@go/page/91910

Assuming that the object is not generating or destroying 10 heat itself, this must be same as the amount of heat that entered V in the time interval dt. That is ∂T

⇀

^ ⋅ ∇T dS dt = ∭ κn

∬ ∂V

Cρ

dV dt ∂t

V

Now we cancel the common factor of dt. We can then rewrite the left hand side as an integral over V by applying the divergence theorem giving ⇀

∭

∂T

⇀

κ ∇ ⋅ ∇T dV = ∭

V

Cρ

dV ∂t

V

As both integrals are over the same volume V , we have ⇀

∭

∂T

⇀

κ ∇ ⋅ ∇T dV − ∭

V

Cρ

dV = 0 ∂t

V

∂T

⇀2

⟹

∭

[κ ∇ T − C ρ

⇀2

where ∇

⇀

⇀

= ∇⋅∇ =

∂

2

+

2

∂x

∂

2

∂y

2

+

∂

2

∂z

] dV = 0

(H)

∂t

V

is the Laplacian. This must be true for all volumes V in the object and for all times t. We

2

claim that this forces ∂T

⇀2

κ ∇ T (x, y, z, t) − C ρ

(x, y, z, t) = 0 ∂t

for all (x, y, z) in the object and all t. Suppose that to the contrary there was a point ⇀2

κ ∇ T (x0 , y0 , z0 , t0 ) − C ρ

∂T ∂t

must remain close to κ∇

By continuity, which we are assuming,

(x0 , y0 , z0 , t0 ) > 0.

⇀2

T (x0 , y0 , z0 , t0 ) − C ρ

∂T

(x0 , y0 , z0 , t0 )

∂t

0,

y0 , z0 ).

∂t

y0 , z0 ).

∂T ∂t

example, (x, y, z, t0 )

So we would have

(x, y, z, t0 ) > 0

Then, necessarily, ∂T

⇀

[κ ∇ ⋅ ∇T (x, y, z, t0 ) − C ρ

∂t

B

= B.

for

∂T

for all (x, y, z) in some small ball B centered on (x

which violates (H) for V

with,

κ ∇ T (x, y, z, t0 ) − C ρ 0,

⇀2

⇀

t0

⇀2

when (x, y, z) is close to (x

κ ∇ T (x, y, z, t0 ) − C ρ

∭

in the object and a time

(x0 , y0 , z0 )

(x, y, z, t0 )] dV > 0

This completes our derivation of the heat equation, which is

 Equation 4.2.8 ∂T

⇀2

(x, y, z, t) = α ∇ T (x, y, z, t) ∂t

where α =

κ Cρ

is called the thermal diffusivity.

An Application of the Heat Equation As an application, we look at the temperature a short distance below the surface of the Earth. For simplicity, we make the Earth flat 11 and we assume that the temperature, T , depends only on time, t, and the vertical coordinate, z. Then the heat equation simplifies to ∂T

∂

2

T

(z, t) = α ∂t

∂z

2

(z, t)

We choose a coordinate system having the surface of the Earth at z = 0 and having z increase downward. We also assume that the temperature T (0, t) at the surface of the Earth is primarily determined by solar heating and is given by T (0, t) = T0 + TA cos(σt) + TD cos(δt)

4.2.9

https://math.libretexts.org/@go/page/91910

Here T is the long term average of the temperature at the surface of the Earth, and T cos(δt) gives daily temperature variations. 0

TA cos(σt)

gives seasonal temperature variations

D

We measure time in days so that

and

δ = 2π

σ =

2π

2π

=

1 year

Then

.

365days

TA cos(σt)

has period one year and

TD cos(δt)

has

period one day. The solution to the initial value problem (HE)+(BC) can be found by separation of variables, a standard topic in courses on partial differential equations. The solution is T (z, t) = T0 + TA e

σ

−√

2α

− − − σ  z)

 z

cos (σt − √

2α −√

+ TD e

δ 2α

− − − δ  z) 2α

 z

cos (δt − √

(SLN)

Whether or not you can find this solution, you can, and should, check that (SLN) satisfies both (HE) and (BC). Now let's see what we can learn from the solution (SLN). For any fixed z, the time average of T (z, t) is T (just because the average value if cosine is zero), the same as the average temperature at the surface z = 0. That is, under the hypotheses that we have made, the long term average temperature at any depth z is is the same as the long term average temperature at the surface. 0

The term T

Ae

−√

σ 2α

−−

 z

cos (σt − √

σ

2α

 z)

oscillates in time with a period of one year, just like T

A

has an amplitude T −−

distance √

2α σ

Ae

−√

σ 2α

 z

cos(σt)

which is T at the surface and decreases exponentially as z increases. Increasing the depth z by a A

causes the amplitude of the oscillation to decrease by a factor of

1 e

.

Both of these first two bullet points are

probably very consistent with your intuition. But this term also has a third property that you may find less obvious. It has has a time lag of with respect to T cos(σt). The surface term T cos(σt) takes its maximum value when z

A

√2ασ

t = 0,  

2π σ

, 

4π σ

,  ⋯ .

At depth z, the corresponding term T

 σt − √

2α

 z = 0,  2π, 4π,   ⋯

Similarly, the term T

De

2α

Ae

−− σ

A

σ

−√

δ

−√

2α

 z

so that t =

z

, 

√2ασ

2π σ

z

+

, 

√2ασ

4π σ

−−

 z

cos (σt − √ z

+

δ 2α

,  ⋯ .

√2ασ

cos(δt)

has an amplitude which is T at the surface and decreases by a factor of D

√2αδ

takes its maximum value when

 z)

D

has a time lag of

 z)

−− cos (δt − √

oscillates in time with a period of one day, just like T z

σ

2α

with respect to T

1 e

−−

for each increase of √

2α δ

in depth.

cos(δt).

D

For water α is approximately 0.012 m /day. This α gives 2

− − − 2α

√

− − − 2α

≈ 1.2 m σ

√

≈ 0.062 m δ

z −− − ≈ 49 z days √2ασ

z − − − ≈ 2.6 z days √2αδ

for z measured in centimeters. So at a depth of a couple of meters, the temperature is pretty constant in time. What variation there is lags the surface variations by several months.

Variations of the Divergence Theorem Here are a couple useful variations of the divergence theorem.

 Theorem 4.2.9. Variations on the divergence theorem If V is a solid with surface ∂V , then

4.2.10

https://math.libretexts.org/@go/page/91910

∬

⇀

⇀

^ dS = ∭ F⋅n

∇ ⋅ F dV

∂V

⇀

V ⇀

^ dS = ∭ fn

∬ ∂V

∇f  dV

V ⇀

∬

⇀

n ^ × F dS = ∭

∂V

⇀

∇ × F dV

V

^ is the outward unit normal of ∂V . where n

Memory Aid. All three formulae can be combined into ~ ^ ∗ F dS = ∭ n

∬ ∂V

⇀ ~ ∇ ∗ F  dV

V ⇀ ~ F = F.

where ∗ can be either ⋅, × or nothing. When ∗ = ⋅ or ∗ = ×, then

~

When ∗ is nothing, F

= f.

Proof The first formula is exactly the divergence theorem and was proven in Theorem 4.2.2. ⇀

To prove the second formula, set F = f a, where a is any constant vector, and apply the divergence theorem. ⇀

^ dS = ∭ fa⋅ n

∬ ∂V

∇ ⋅ (f a) dV

V ⇀

=∭

⇀

[(∇f ) ⋅ a + f ∇ ⋅ a ] dV 

V

=0 ⇀

=∭

(∇f ) ⋅ a dV

V

To get the second line, we used the vector identity Theorem 4.1.4.c. To get the third line, we just used that that all of its derivatives are zero. Rewrite ⇀

∭

a

is a constant, so

⇀

(∇f ) ⋅ a dV = ∭

V

a ⋅ (∇f ) dV

V

Since a is a constant, we can factor it out of both integrals, so ⇀

^ dS = a ⋅ ∭ fn

a⋅ ∬ ∂V

∇f  dV

V ⇀

⟹

^ dS − ∭ fn

a⋅ {∬ ∂V

In particular, choosing a = ^ ıı ,

^ ȷ ȷ

∇f  dV } = 0

V

^ and k , we see that all three components of the vector ∬

∂V

⇀

^ dS − ∭ fn

V

∇f  dV

are zero. So

⇀

^ dS − ∭ fn

∬ ∂V

∇f  dV = 0

V

which is what we wanted show. ⇀

⇀

To prove the third formula, apply the divergence theorem, but with F replaced by a × F, where a is any constant vector. ⇀

∬

⇀

^  dS = ∭ (a × F) ⋅ n

∂V

⇀

∇ ⋅ (a × F) dV

V ⇀

=∭

⇀

⇀

⇀

[ F ⋅ (∇ × a) − a ⋅ (∇ × F)] dV 

V

⇀ =0 ⇀

= −∭

⇀

⇀

a ⋅ (∇ × F) dV = −a ⋅ ∭

V

⇀

∇ × F dV

V

4.2.11

https://math.libretexts.org/@go/page/91910

To get the second line, we used the vector identity Theorem 4.1.4.d. To get the third line, we again used that a is a constant, so that all of its derivatives are zero. For all vectors (a × b) ⋅ c = a ⋅ (b × c) (in case you don't remember this, it was Lemma 4.1.8.a) so that ⇀

⇀

^ = a ⋅ (F × n ^) (a × F) ⋅ n

and ⇀

a⋅ ∬

⇀

⇀

F × n  dS = −a ⋅ ∭

∂V

V ⇀

⟹

a⋅ {∬

⇀

⇀

F × n  dS + ∭

∂V

In particular, choosing zero. So

a= ^ ıı , ^ ȷ ȷ

⇀

∇ × F dV

⇀

∇ × F dV } = 0

V ⇀

^ and k , we see that all three components of the vector

⇀

∭

⇀

⇀

∇ × F dV = − ∬

V

∬

∂V

⇀

⇀

F × n  dS + ∭

V

⇀

are

∇ × F dV

⇀

⇀

F × n  dS = ∬

∂V

^ × F dS n

∂V

which is what we wanted show.

An Application of the Divergence Theorem — Buoyancy In this section, we use the divergence theorem to show that when you immerse an object in a fluid the net effect of fluid pressure acting on the surface of the object is a vertical force (called the buoyant force) whose magnitude equals the weight of fluid displaced by the object. This is known as Archimedes' principle 12. We shall also show that the buoyant force acts through the “centre of buoyancy” which is the centre of mass of the fluid displaced by the object. The design of self-righting 13 boats exploits the fact that the centre of buoyancy and the centre of gravity, where gravity acts, need not be the same. We start by computing the total force due to the pressure of the fluid pushing on the object. Recall that pressure is the force per unit surface area that the fluid exerts on the object acts perpendicularly to the surface pushes on the object Thus the force due to pressure that acts on an infinitesimal piece of the object's surface at r = (x, y, z) with surface area dS and ^ is −p( r ) n ^ dS. The minus sign is there because pressure is directed into the object. If the object fills the volume outward normal n V and has surface ∂V , then the total force on the object due to fluid pressure, called the buoyant force, is ⇀

⇀

⇀

^ dS p( r ) n

B = −∬ ∂V

⇀

We now wish to apply a variant of the divergence theorem to rewrite B = − ∭ ∇p dV . But there is a problem with this: p( r ) is the fluid pressure at r and is only defined where there is fluid. In particular, there is no fluid 14 inside the object, so p( r ) is not defined for any r in the interior of V . ⇀

V

⇀

⇀

⇀

So we pretend that we remove the object from the fluid and we call P ( r ) the fluid pressure at r when there is no object in the fluid. We also make the assumption that at any point r outside of the object, the pressure at r does not depend on whether the object is in the fluid or not. In other words, we assume that ⇀

⇀

⇀

⇀

⇀

⇀

p( r ) = {

⇀

P (r)

if  r  is not in V

not defined

if  r  is in the V

⇀

This assumption is only an approximation to reality, but, in practice, it is a very good approximation. So, by Theorem 4.2.9, B = −∬ ∂V

⇀

^ dS = − ∬ p( r ) n

⇀

⇀

^ dS = − ∭ P (r) n

∂V

⇀

∇P ( r ) dV

(4.2.1)

V

⇀

Our next job is to compute ∇P . Concentrate on an infinitesimal cube of fluid whose edges are parallel to the coordinate axes. Call the lengths of the edges dx, dy and dz and the position of the centre of the cube (x, y, z). The forces applied to the various faces of

4.2.12

https://math.libretexts.org/@go/page/91910

the cube by the pressure of fluid outside the cube are illustrated in the figure

The total force due to the pressure acting on the cube is the sum dx − P (x +

dx , y, z) dydz ^ ı ı + P (x −

, y, z) dydz ^ ı ı

2

2 dy

− P (x, y +

dy , z) dxdz ^ ȷ ȷ + P (x, y −

, z) dxdz ^ ȷ ȷ

2

2 dz

− P (x, y, z +

^ ) dxdy k + P (x, y, z −

2

dz

^ ) dxdy k

2

of the forces acting on the six faces. Consider the ^ ı ı component and rewrite it as dx − P (x +

dx , y, z) dydz ^ ı ı + P (x −

, y, z) dydz ^ ı ı

2

2 P (x +

=−

dx 2

, y, z) − P (x −

dx 2

, y, z) ^ ı ı  dxdydz

dx ∂P =−

(x, y, z) ^ ı ı  dxdydz ∂x

Doing this for the other components as well, we see that the total force due to the pressure acting on the cube is ∂P −{

(x, y, z) ^ ı ı+

∂x

∂P

(x, y, z) ^ ȷ ȷ+

∂y

∂P

⇀

^ (x, y, z) k}dxdydz = −∇P (x, y, z) dxdydz

∂z

We shall assume that the only other force acting on the cube is gravity and that the fluid is stationary (or at least not accelerating). Hence the total force acting on the cube is zero. If the fluid has density ρf , then the cube has mass ρf dxdydz so that the force of ^ gravity is −gρf dxdydz k . The vanishing of the total force now tells us that ⇀

⇀ ^ −∇P ( r ) dxdydz − gρf dxdydz k = 0

⇀

⟹

⇀ ^ ∇P ( r ) = −gρf k

Subbing this into (4.2.1) gives ^ B = gk∭

^ ρf dV = gMf k

V

where M = ∭ ρf dV is the mass of the fluid displaced by the object — not the mass of the object itself. Thus the buoyant force acts straight up and has magnitude equal to gM , which is also the magnitude of the force of gravity acting on the fluid displaced by the object. In other words, it is the weight of the displaced fluid. This is exactly Archimedes' principle. f

V

f

We next consider the rotational motion of our submerged object. The physical law that determines the rotational motion of a rigid body about a point r is analogous to the familiar Newton's law, m For the rotational law of motion, ⇀

0

⇀ d v dt

⇀

= F,

that determines the translational motion of the object.

the mass m is replaced by a physical quantity, characteristic of the object, called the moment of inertia, and the ordinary velocity v is replaced by the angular velocity, which is a vector whose length is the rate of rotation (i.e. angle rotated per unit time) and whose direction is parallel to the axis of rotation (with the sign determined by a right hand rule), and the force F is replaced by a vector called the torque about r . A force F applied at r = (x, y, z) produces the torque 15 ( r − r ) × F about r . ⇀

⇀

⇀

⇀

⇀

0

⇀

⇀

0

⇀

⇀

0

4.2.13

https://math.libretexts.org/@go/page/91910

This is derived in the optional §4.2.4, entitled “Torque”, and is all that we need to know about rotational motion of rigid bodies in this discussion. Fix any point

⇀ r 0.

The total torque about ⇀

produced by force of pressure acting on the surface of the submerged object is

⇀ r0

⇀

⇀

⇀

⇀

^ ) dS = ∬ ( r − r 0 ) × ( − p( r )n

T =∬ ∂V

⇀

⇀

^ × (P ( r ) ( r − r 0 )) dS n

∂V

Recall that in these integrals r = (x, y, z) is the position of the infinitesimal piece dS of the surface S. Applying the cross product variant of the divergence theorem in Theorem 4.2.9, followed by the vector identity Theorem 4.1.5.c, gives ⇀

⇀

⇀

T =∭

∇ × (P ( r ) ( r − r 0 )) dV

⇀

⇀

⇀

⇀

⇀

V ⇀

=∭

⇀

⇀

⇀

⇀

⇀

{ ∇P ( r ) × ( r − r 0 ) + P ( r ) ∇ × ( r − r 0 ) } dV 

V

⇀ =0 ⇀

=∭

⇀

⇀

⇀

∇P ( r ) × ( r − r 0 ) dV

V ⇀

since ∇ × r

⇀ 0

= 0,

because

⇀ r0

is a constant, and ^ ı ı

⎡ ⇀

⇀

∇ × r = det ⎢ ⎢ ⎣ ⇀

^ k ⎤

^ ȷ ȷ

∂

∂

∂

∂x

∂y

∂z

x

y

z

⎥ =0 ⎥ ⎦

^ We have already found that ∇P ( r ) = −gρf k . Substituting it in gives ⇀

⇀

⇀ ⇀ ^ gρf k × ( r − r 0 ) dV

T = −∭ V

⇀

^ = −gk × ∭

⇀

ρf ( r − r 0 ) dV

V

^ = −gk × { ∭

⇀

⇀

r ρf dV − r 0 ∭

V

^ ρf dV }k × {

= −g{ ∭

V

V

V

∭

V

r0

r ρf dV

⇀

− r 0} ρf dV

⇀

∭

So the torque generated at applied at the single point

⇀

− r 0} ρf dV

⇀

V

∭

⇀

r ρf dV

V

∭

∭

={

⇀

∭

V

= −B × {

ρf dV }

V

r ρf dV

⇀

− r 0} × B ρf dV

by pressure over the entire surface is the same the torque generated at $\vecs{r} _0$ by a force ⇀

∭

V

CB =

B

∭

V

r ρf dV ρf dV

This point is called the centre of buoyancy. It is the centre of mass of the displaced fluid. The moral of the above discussion is that the buoyant force, B, on a rigid body acts straight upward, has magnitude equal to the weight of the displaced fluid and acts at the centre of buoyancy, which is the centre of mass of the displaced fluid. As above, denoting by ρb the density of the object, the torque about ∭ V

⇀ ⇀ ^ ( r − r 0 ) × (−gρb k) dV = {

⇀ r0

due to gravity acting on the object is

⇀

∭

V

∭

V

r ρb dV

⇀

− r 0 } × (−g{ ∭ ρb dV

4.2.14

^ ρb dV } k)

V

https://math.libretexts.org/@go/page/91910

So the gravitational force, G, acts straight down, has magnitude equal to the weight gM

b

acts at the centre of mass, C

G

∭

V

=

= g∭

V

⇀ r ρb dV

,

∭

V

ρb dV

⇀

r ρb dV

(where ρb is the density of the object) of the object and

of the object.

Because the mass distribution of the object need not be the same as the mass distribution of the displaced fluid, buoyancy and gravity may act at two different points. This is exploited in the design of self-righting boats. These boats are constructed with a heavy, often lead (which is cheap and dense), keel. As a result, the centre of gravity is lower in the boat than the center of buoyancy, which, because the displaced fluid has constant density, is at the geometric centre of the boat. As the figure below illustrates, a right side up configuration of such a boat is stable, while an upside down configuration is unstable. The boat rotates so as to keep the centre of gravity straight below the centre of buoyancy. To see this pretend that you are holding on to the boat with one hand holding the centre of buoyancy and the other hand holding the centre of gravity. Use your hands to apply forces in the directions of the arrows and think about how the boat will respond.

Optional — Torque In this section, we derive the properties of torque that we used in the last section. Newton's law of motion says that the position ⇀

⇀′′

⇀

of a single particle moving under the influence of a force F obeys m r (t) = F. Similarly, the positions r (t), 1 ≤ i ≤ n, of a set of particles moving under the influence of forces F obey m r (t) = F , 1 ≤ i ≤ n. Very often systems of interest consist of some small number of rigid bodies. Suppose that we are interested in the motion of a single rigid body, say a piece of wood. The piece of wood is made up of a huge number 16 of atoms. So the system of equations determining the motion of all of the individual atoms in the piece of wood is huge. On the other hand, we shall see that because the piece of wood is rigid, its configuration is completely determined by the position of, for example, its centre of mass and its orientation (we won't get into what precisely is meant by “orientation”, but it is certainly determined by, for example, the positions of a few of the corners of the piece of wood). To be precise, we shall extract from the huge system of equations that determine the motion of all of the individual atoms, a small system of equations that determine the motion of the centre of mass and the orientation. We'll do so now. ⇀

r (t)

⇀

i

⇀

⇀′′

i

Imagine a piece of wood moving in R

3

⇀

i

i

.

Furthermore, imagine that the piece of wood consists of a huge number of particles joined by a huge number of weightless but very strong 17 steel rods. The steel rod joining particle number one to particle number two just represents a force acting between particles number one and two. Suppose that there are n particles, with particle number i having mass m , at time t, particle number i has position r (t), at time t, the external force (gravity and the like) acting on particle number i is F (t), and at time t, the force acting on particle number i, due to the steel rod joining particle number i to particle number j is F i

⇀

i

⇀

i

⇀

i,j (t).

⇀

there is no steel rod joining particles number i and j, just set F

i,j (t)

If

⇀

= 0.

In particular, F

i,i (t)

= 0.

The only assumptions that we shall make about the steel rod forces are

4.2.15

https://math.libretexts.org/@go/page/91910

(A1) ⇀

⇀

for each i ≠ j, F particles i and j.

i,j (t)

= −Fj,i (t).

In words, the steel rod joining particles

i

and

applies equal and opposite forces to

j

(A2) ⇀

for each i ≠ j, there is a function M (t) such that F (t) = M (t)[ r (t) − r (t)]. In words, the force due to the rod joining particles i and j acts parallel to the line joining particles i and j. For (A1) to be true, that is to have M (t)[ r (t) − r (t)] = −M (t)[ r (t) − r (t)], we need M (t) = M (t). i,j

⇀

i,j

⇀

i

⇀

j

j,i

⇀

i,j

i,j

⇀

i

j

⇀

j

i

i,j

j,i

Newton's law of motion, applied to particle number i, now tells us that n ⇀′′

mi r

i

⇀

⇀

(t) = Fi (t) + ∑ Fi,j (t) j=1

Adding up all of the equations (N ), for i = 1,  2,  3,   ⋯ ,  n gives i

n

n ⇀

⇀′′

⇀

∑ mi r i (t) = ∑ Fi (t) + i=1

⇀

The sum

∑

⇀

Fi,j (t)

i=1

∑

Fi,j (t)

1≤i,j≤n

⇀

contains F

1,2 (t)

exactly once and it also contains F

2,1 (t)

exactly once and these two terms cancel exactly,

1≤i,j≤n ⇀

by assumption (A1). In this way, all terms in

∑

Fi,j (t)

with i ≠ j exactly cancel. All terms with i = j are assumed to be zero.

1≤i,j≤n

So

⇀

∑

Fi,j (t) = 0

and the equation (ΣN ) simplifies to i

1≤i,j≤n

n

n ⇀′′

⇀

∑ mi r i (t) = ∑ Fi (t) i=1

Phew! Denote by

n

M = ∑ mi

i=1

the total mass of the body, by

n

R(t) =

i=1 ⇀

n

⇀

F(t) = ∑ Fi (t)

1 M

⇀

∑ mi r i (t)

the centre of mass 18 of the body and by

i=1

the total external force acting on the system. In this notation, equation (ΣN ) can be written as i

i=1

 Equation 4.2.10 ⇀

′′

M R (t) = F(t)

The upshot is that the centre of mass of the system moves just like a single particle of mass This is why we can often replace an extended object by a point mass at its centre of mass. Now take the cross product of

⇀ r i (t)

M

subject to the total external force.

and equation (N ) and sum over i. This gives i

n ⇀′′

⇀

∑ mi   r i (t) × r i (t) i=1 n ⇀

⇀

= ∑ r i (t) × Fi (t) + i=1

⇀

⇀

∑

⇀

r i (t) × Fi,j (t)

(Σ r i × Ni )

1≤i,j≤n

By the assumption (A2) ⇀

⇀

⇀

⇀

⇀

r 1 (t) × F1,2 (t) = M1,2 (t)  r 1 (t) × [ r 1 (t) − r 2 (t)] ⇀

⇀

⇀

⇀

⇀

r 2 (t) × F2,1 (t) = M2,1 (t)  r 2 (t) × [ r 2 (t) − r 1 (t)] ⇀

⇀

⇀

= −M1,2 (t)  r 2 (t) × [ r 1 (t) − r 2 (t)]

so that ⇀

⇀

⇀

⇀

r 1 (t) × F1,2 (t) + r 2 (t) × F2,1 (t)

⇀

⇀

⇀

⇀

= M1,2 (t) [ r 1 (t) − r 2 (t)] × [ r 1 (t) − r 2 (t)] = 0

4.2.16

https://math.libretexts.org/@go/page/91910

because the cross product of any two parallel vectors is zero. The last equation says that the i = 1, j = 2 term in

⇀

⇀

∑

exactly cancels the i = 2, j = 1 term. In this way all of

r i (t) × Fi,j (t)

1≤i,j≤n

the terms in

∑

⇀

⇀

with

r i (t) × Fi,j (t)

cancel. Each term with

i ≠j

i =j

is exactly zero because

⇀

Fii = 0.

So

1≤i,j≤n

∑

⇀

⇀

r i (t) × Fi,j (t) = 0

and (Σ r

⇀ i

× Ni

) simplifies to

1≤i,j≤n

n

n ⇀

⇀′′

⇀

⇀

∑ mi   r i (t) × r i (t) = ∑ r i (t) × Fi (t) i=1

i=1

At this point it makes sense to define vectors n ⇀′

⇀

L(t) = ∑ mi   r i (t) × r i (t) i=1 n ⇀

⇀

⇀

T(t) = ∑ r i (t) × Fi (t) i=1

because, in this notation, (

⇀ Σri

× Ni

) becomes

 Equation 4.2.11 d

⇀

L(t) = T(t) dt

⇀

Equation 4.2.11 plays the role of Newton's law of motion for rotational motion. T(t) is called the torque and plays the role of “rotational force”. L(t) is called the angular momentum (about the origin) and is a measure of the rate at which the piece of wood is rotating. For example, if a particle of mass m is travelling in a circle of radius ρ in the xy-plane at ω radians per unit time, then r (t) = ρ cos(ωt) ^ ı ı + ρ sin(ωt)^ ȷ ȷ and ⇀

⇀

⇀′

m r (t) × r (t)

= m[ρ cos(ωt) ^ ı ı + ρ sin(ωt)^ ȷ ȷ ] × [ − ωρ sin(ωt) ^ ı ı + ωρ cos(ωt)^ ȷ ȷ] 2 ^ = m ρ  ω k

^ is proportional to ω, which is the rate of rotation about the origin and is in the direction k , which is normal to the plane containing the circle. ⇀

⇀

In any event, in order for the piece of wood to remain stationary, equations 4.2.10 and 4.2.11 force F(t) = T(t) = 0. Now suppose that the piece of wood is a seesaw 19 that is supported on a fulcrum at acting on particle number i, for each 1 ≤ i ≤ n, and the

force ⇀

Φ

p.

imposed by the fulcrum that is pushing up on the particle at n

^ ^ F = Φ − ∑ mi gk = Φ − M gk.

The forces consist of gravity,

p.

The

total

external

^ −mi gk,

force

is

^ If the seesaw is to remain stationary, this must be zero so that Φ = M gk .

i=1

The total torque (about the origin) is n

n

⇀

⇀ ⇀ ^ ^ T = p × Φ − ∑ mi g r i × k = g(M p − ∑ mi r i ) × k i=1

i=1

4.2.17

https://math.libretexts.org/@go/page/91910

If the seesaw is to remain stationary, this must also be zero. This will be the case if the fulcrum is placed at n

1

⇀

p =

∑ mi r i M

i=1

which is just the centre of mass of the piece of wood. ⇀

More generally, suppose that the external forces acting on the piece of wood consist of

acting on particle number

Fi ,

n

⇀

1 ≤ i ≤ n,

⇀

and a “fulcrum force” Φ acting on a particle at p. The total external force is F = Φ + ∑ F

i.

i,

for each

If the piece of wood is to

i=1 n

⇀

remain stationary, this must be zero so that Φ = − ∑ F

The total torque (about the origin) is

i.

i=1

n

⇀

n

⇀

⇀

⇀

⇀

T = p × Φ + ∑ r i × Fi = ∑( r i − p) × Fi i=1

i=1

If the piece of wood is to remain stationary, this must also be zero. That is, the torque about point 1 ≤ i ≤ n, must be zero.

p

due to all of the forces

⇀

Fi ,

Optional — Solving Poisson's Equation In this section we shall use the divergence theorem to find a formula for the solution of Poisson's equation ⇀2

∇ φ = 4πρ

Here ρ = ρ( r ) is a given (continuous) function and φ is the unknown function that we wish to find. This equation arises, for example, in electrostatics, where ρ is the charge density and φ is the electric potential. ⇀

The main step in finding this solution formula will be to consider an arbitrary (smooth) function φ and an arbitrary (smooth) region V in R and an arbitrary point r in the interior of V 3

⇀ 0

and to find an auxiliary formula which expresses φ( r

⇀

⇀2

0)

in terms of

with r running over V and ∇φ( r ) and φ( r ), with r running only over ∂V . ⇀

∇ φ( r ), ⇀

⇀

⇀

⇀

⇀

This auxiliary formula, which we shall derive below, is ⇀2 ⇀ φ( r 0 )

1 =−

⇀

⇀

∇ φ( r ) {∭

4π

⇀

⇀

 d

3⇀

| r − r 0|

V

⇀

⇀

r − r0

⇀

r −∬

φ( r )

⇀

⇀

3

⋅n ^  dS

| r − r 0|

∂V ⇀

∇φ( r ) −∬

⇀

^  dS} ⋅n

⇀

(V )

| r − r 0|

∂V

When we take the limit as V expands to fill all of R then, assuming that φ and ∇φ go to zero sufficiently quickly 20 at two integrals over ∂V will converge to zero and we will end up with the formula ⇀

3

⇀2 ⇀ φ( r 0 )

1 =− 4π

This expresses φ evaluated at an arbitrary point, ⇀2

want, since ∇

φ = 4πρ

⇀ r 0,

R

3

⇀

⇀

 d

3⇀

r

| r − r 0| ⇀2

of R in terms of ∇ 3

the

⇀

∇ φ( r ) ∭

∞,

⇀

φ( r ),

with

⇀

r

running over R , which is exactly what we 3

for any solution of Poisson's equation. So once we have proven (V) we will have proven 21

4.2.18

https://math.libretexts.org/@go/page/91910

 Theorem 4.2.12 Assume that ρ( r ) is continuous and decays sufficiently quickly as decay sufficiently quickly as r → ∞, then ⇀

⇀

r → ∞.

If

φ

obeys

⇀2

∇ φ = 4πρ

on

3

R ,

and

φ

and

⇀

∇φ

⇀

⇀

ρ( r )

⇀

φ( r 0 ) = − ∭ R

for all

⇀

r0

in R

3

⇀

⇀

 d

3⇀

r

| r − r 0|

3

.

Let ^ r (x, y, z) = x ^ ı ı +y ^ ȷ ȷ +zk

⇀

⇀

^ r 0 = x0 ^ ı ı + y0 ^ ȷ ȷ + z0 k

We shall exploit three properties of the function

1 ⇀ ⇀ | r − r 0|

The first two properties are ⇀

1

⇀

∇

.

⇀

⇀

⇀

| r − r 0|

⇀

⇀

⇀

r − r0

⇀ ⇀

(P1)

3

⇀

| r − r 0|

1

⇀2

∇

⇀

r − r0

=−

= −∇ ⋅

⇀

| r − r 0|

⇀

3

=0

(P2)

| r − r 0|

and are valid for all r ≠ r . Verification of the first property is a simple one line computation. Verification of the second property is a simple three line computation. (See the solution to Question 6 in Section 4.1.) ⇀

⇀

0

The other property of

1 ⇀ ⇀ | r − r 0|

that we shall use is the following. Let

Sε

be the sphere of radius

ε

centered on

⇀

r 0.

Then, for any

continuous function ψ( r ), ⇀

⇀

ψ( r ) lim ∬ ε→0+

⇀

Sε

⇀

⇀

1 p

 dS = lim ε→0+

| r − r 0|

ε

p

ψ( r 0 )

⇀

∬

ψ( r ) dS = lim ε→0+

Sε

ε

p

∬

 dS

Sε

⇀

ψ( r 0 ) = lim ε

ε→0+

4π ε

p

⇀

⎧ 4πψ( r 0 ) =⎨0 ⎩ undefined

2

if p = 2 (P3)

if p < 2 if p > 2

Derivation of (V): Here is the derivation of (V ). Let V be the part of V outside of S

ε.

ε

Note that the boundary ∂V of V consists of two parts — the boundary ε

normal to ∂V on S is − ε

ε

ε

⇀ ⇀ r − r 0 ⇀ ⇀ | r − r 0|

,

because it points towards

⇀

r0

∂V

of V and the sphere

Sε

— and that the unit outward

and hence outside of V

ε.

Recall the vector identity Theorem 4.1.7.d, which says ⇀

⇀

⇀

⇀2

⇀2

∇ ⋅ (f ∇g − g∇f ) = f ∇ g − g ∇ f

Applying this identity with f

=

1 ⇀ ⇀ | r − r 0|

and g = φ gives

4.2.19

https://math.libretexts.org/@go/page/91910

=0 by (P2)



⇀2

1

⇀

∇⋅(

⇀

⇀

1

⇀

∇φ − φ ∇

⇀

⇀

| r − r 0|

∇ φ ) =

⇀

⇀

| r − r 0|

1

⇀2

−φ ∇

⇀

⇀

| r − r 0|

⇀

| r − r 0|

⇀2

∇ φ =

⇀

⇀

| r − r 0|

which is the integrand of the first integral on the right hand side of (V). So, by the divergence theorem ⇀2

∇ φ ∭

⇀

Vε

|r −

dV = ∭

(

∇⋅(

⇀

⇀

| r − r 0|

∂V

⇀

(

⇀

)dV

⇀

| r − r 0|

^  dS )⋅n

⇀

⇀

⇀

⇀

1

⇀

∇φ − φ ∇

| r − r 0|

Sε

1

⇀

∇φ − φ ∇

| r − r 0| 1

+∬

⇀ ⇀ r 0|

1

⇀

∇φ − φ ∇

⇀

⇀

|r −

Vε

1 =∬

1

⇀

⇀ r 0|

⇀

| r − r 0|

⇀

⇀

⇀

⇀

r − r0

)⋅(−

⇀

)dS

(M)

| r − r 0|

To see the connection between (M) and the rest of (V), note that, by (P1), the first term on the right hand side of (M) is 1 ∬

(

∂V

⇀

⇀

1

⇀

∇φ − φ ∇

⇀

⇀

| r − r 0|

^  dS )⋅n

⇀

| r − r 0| ⇀

⇀

⇀

∇φ( r ) =∬

⇀

⇀

| r − r 0|

∂V

φ( r )

⇀

r − r0

⇀

^  dS + ∬ ⋅n

⇀

⇀

^  dS ⋅n

3

(R1)

| r − r 0|

∂V

which is 4π times the second and third terms on the right hand side of (V), ⇀

and substituting in ∇

1 ⇀ ⇀ | r − r 0|

=−

⇀ ⇀ r − r 0 ⇀ ⇀ 3 | r − r 0|

,

from (P1), and applying (P3) with p = 2, the limit of the second term on the right

hand side of (M) is 1 lim ∬ ε→0+

(

⇀

⇀ ⇀

| r − r 0|

Sε

1

⇀

∇φ − φ ∇

⇀

)⋅(−

⇀

| r − r 0| ⇀

= − lim ∬ ε→0+

⇀

⇀

⇀

)dS

1

⇀

[ ∇φ ⋅ ( r − r 0 ) + φ]

⇀

⇀

r − r0

| r − r 0|

⇀

⇀

2

dS

| r − r 0|

Bε

⇀

⇀

⇀

= −4π[∇φ ⋅ ( r − r 0 ) + φ ]

⇀ ⇀ r = r 0

=

So applying 22 lim

⇀ −4πφ( r 0 )

(R2)

to (M) and substituting in (R1) and (R2) gives

ε→0+

∭ V

⇀2

⇀

∇ φ

∇φ( r )

⇀

⇀

| r − r 0|

 dV = ∬

⇀

∂V

⇀

⇀

⇀

⋅n ^  dS + ∬

| r − r 0|

∂V

⇀

φ( r )

⇀

r − r0

⇀

⇀

⇀

3

⋅n ^  dS − 4πφ( r 0 )

| r − r 0|

which is exactly equation (V).

Exercises Stage 1  1 Let V be the cube V = {(x, y, z)|0 ≤ x ≤ 1,  0 ≤ y ≤ 1,  0 ≤ z ≤ 1}

and R be the square R = {(x, y)|0 ≤ x ≤ 1,  0 ≤ y ≤ 1}

4.2.20

https://math.libretexts.org/@go/page/91910

and let f (x, y, z) have continuous first partial derivatives. 1. Use the fundamental theorem of calculus to show that ∂f ∭

(x, y, z) dx dy dz = ∬

V

∂z

f (x, y, 1) dx dy −∬

R

f (x, y, 0) dx dy

R

2. Use the divergence theorem to show that ∂f ∭

(x, y, z) dx dy dz = ∬

V

∂z

f (x, y, 1) dx dy −∬

R

f (x, y, 0) dx dy

R

 2 ⇀

1. By applying the divergence theorem to F = ϕ a, where a is an arbitrary constant vector, show that ⇀

∭

^ dS ϕn

∇ϕ dV = ∬

V

∂V

2. Show that the centroid (x ¯, y ¯, z ¯) of a solid V with volume |V | is given by 1 (x ¯, y ¯, z ¯) =

∬ 2|V |

2

(x

+y

2

2

+z )n ^ dS

∂V

Stage 2  3 Let S be the unit sphere centered at the origin and oriented by the outward pointing normal. If ⇀

2

F(x, y, z) = (x, y, z ) ⇀

evaluate the flux of F through S 1. directly and 2. by applying the divergence theorem.

 4 Evaluate, by two methods, the integral ∬ pointing unit normal to S.

S

⇀

⇀

F⋅n ^ dS,

^ where F = z k ,

S

is the surface x

2

+y

2

+z

2

2

=a

and n ^ is the outward

1. First, by direct computation of the surface integral. 2. Second, by using the divergence theorem.

 5 Let ⇀

and be the solid in 3-space defined by

 F = zy V

3

^ ^ ı ı + yx ^ ȷ ȷ + (2z + y )k  2

2

9 −x 0 ≤z ≤

2

9 +x

−y +y

2

2

and D

be the bottom surface of V . Because

surface is z = 0,

2

x

+y

2

2

2

2

2

9−x −y 9+x +y

is positive for x

2

+y

2

9,

the bottom

≤ 9.

4.2.21

https://math.libretexts.org/@go/page/91910

Let S be the curved portion of the boundary of V . It is z =

2

2

2

2

9−x −y 9+x +y

2

, x

+y

2

≤ 9.

Here is a sketch of the first octant part

of S and D.

Denote by |V | the volume of V and compute, in terms of |V |, ⇀

1. ∬

^ dS F⋅n

^ pointing downward with n

D ⇀

2. ∭

∇ ⋅ F dV

V ⇀

3. ∬

F⋅n ^ dS

with n ^ pointing outward

S

Use the divergence theorem to answer at least one of parts (a), (b) and (c).

 6 Evaluate the integral methods.

⇀

∬

S

⇀

where

^ dS, F⋅n

F = (x, y, 1)

and

S

is the surface

2

z = 1 −x

for

2

−y ,

2

x

+y

2

≤ 1,

by two

1. First, by direct computation of the surface integral. 2. Second, by using the divergence theorem.

 7✳ ⇀

1. Find the divergence of the vector field F = (z + sin y, zy, sin x cos y). ⇀

2. Find the flux of the vector field F of (a) through the sphere of radius 3 centred at the origin in R . 3

 8 The sides of a grain silo are described by the portion of the cylinder x + y = 1 with 0 ≤ z ≤ 1. The top of the silo is given by the portion of the sphere x + y + z = 2 lying within the cylinder and above the xy-plane. Find the flux of the vector field 2

2

2

2

2

2

x

2

V(x, y, z) = (x yz , yz + e z , x

+ y)

out of the silo.

 9 Let B be the ball of volume V centered at the point (x , y , z ), and let ^ flux of F = x ^ ı ı + xy ^ ȷ ȷ + (3z − yz)k outward (from B ) through S. 0

⇀

0

0

S

be the sphere that is the boundary of

B.

Find the

2

 10 ✳ Let ⇀

F(x, y, z) = (1 + z

1+z

1+z

4.2.22

, 1 +z

1+z

1+z

, 1)

https://math.libretexts.org/@go/page/91910

Let S be the portion of the surface 2

x

+y

2

= 1 −z

4

⇀

which is above the xy-plane. What is the flux of F downward through S?

 11 ✳ ^ Use the divergence theorem to find the flux of x ^ ı ı + y^ ȷ ȷ + 2zk through the part of the ellipsoid 2

x

with z ≥ 0. [Note: the ellipsoid

2

x

2

a

+

2

y

2

+

b

2

z

2

c

+y

2

+ 2z

has volume

=1

2

=2

]

4

πabc.

3

 12 ✳ ⇀

Let F(x, y, z) = ⇀

⇀

3

r /r

where

^ r =x ^ ı ı +y ^ ȷ ȷ +zk

⇀

and r = | r |. ⇀

⇀

1. Find ∇ ⋅ F. 2. Find the flux of F outwards through the spherical surface x + y + z = a . 3. Do the results of (a) and (b) contradict the divergence theorem? Explain your answer. 4. Let E be the solid region bounded by the surfaces z − x − y + 1 = 0, z = 1 and z = −1. Let σ be the bounding ⇀

2

2

2

2

2

2

2

⇀

surface of E. Determine the flux of F outwards through σ. 5. Let R be the solid region bounded by the surfaces z − x − y surface of R. Determine the flux of F outwards through Σ. 2

2

2

+ 4y − 3 = 0,

z =1

and z = −1. Let Σ be the bounding

⇀

 13 ✳ Consider the ellipsoid S given by 2

x

y

2

+

z

2

+ 4

=1 4

with the unit normal pointing outward. 1. Parameterize S. 2. Compute the flux ∬

S

⇀

^ dS F⋅n

of the vector field ⇀

F(x, y, z) = (x, y, z)

3. Verify your answer in (b) using the divergence theorem.

 14 ✳ Evaluate the flux integral ∬

S

⇀

^ dS, F⋅n ⇀

where 3

F(x, y, z) = (x

2

+ cos(y ) , y

3

+ ze

and S is the surface of the solid region bounded by the cylinder surface is positively oriented (its unit normal points outward).

2

x

x

, z

+y

2

2

+ arctan(xy))

=2

and the planes

z =0

and

z = 2x + 3.

The

 15 ✳ Find the flux of the vector field (x + y, x + z, y + z) through the cylindrical surface whose equation is x + z = 4, and which extends from y = 0 to y = 3. (Only the curved part of the cylinder is included, not the two disks bounding it on the left and right.) The orientation of the surface is outward, i.e., pointing away from the y -axis. 2

4.2.23

2

https://math.libretexts.org/@go/page/91910

 16 ✳ The surface S is the part above the xy-plane of the surface obtained by revolving the graph of z = 1 − x around the z -axis. The surface S is oriented such that the normal vector has positive z -component. The circle with radius 1 and centre at the origin in the xy-plane is the boundary of S. 4

⇀

Find the flux of the divergenceless vector field F(x, y, z) = (yz, x + z, x

2

2

+y )

through S.

 17 ✳ Let S be the part of the paraboloid direction. Let

2

z = 2 −x

−y

2

contained in the cone

⇀

2

3

F = (tan √z + sin(y )) ^ ı ı +e

−x

− −− −− − 2 2 z = √x + y

and oriented in the upward

^ ^ ȷ ȷ + zk

⇀

Evaluate the flux integral ∬

S

^ dS. F⋅n

 18 ✳ Evaluate the surface integral ⇀

∬

F⋅n ^ dS

S ⇀

where F(x, y, z) = ( cos z + x y , x e , sin y + x z) and S is the boundary of the solid region enclosed by the paraboloid z = x +y and the plane z = 4, with outward pointing normal. 2

2

−z

2

2

 19 ✳ Let S be the part of the sphere x

2

+y

2

+z

2

=4

⇀

F = (e

y

between the planes z = 1 and z = 0 oriented away from the origin. Let

+ xz) ^ ı ı + (zy + sin(x)) ^ ȷ ȷ + (z

2

^ − 1) k

Compute the flux integral ⇀

^ dS. F⋅n

∬ S

 20 ✳ Let B be the solid region lying between the planes x = −1, x = 1, y = 0, y = 2 and bounded below by the plane above by the plane z + y = 3. Let S be the surface of B. Find the flux of the vector field

z =0

and

⇀

2 2 ^ F(x, y, z) = (x z + cos πy) ^ ı ı + (yz + sin πz) ^ ȷ ȷ + (x − y ) k

 21 ✳ Let S be the hemisphere x

2

+y

2

+z

2

= 1,

z ≥ 0,

oriented with n ^ pointing away from the origin. Evaluate the flux integral ⇀

^ dS F⋅n

∬ S

where ⇀

− −− −− − 2

2

F = (x + cos(z )) ^ ı ı + (y + ln(x

4.2.24

5

2

+ z )) ^ ȷ ȷ + √x

+y

2

^ k

https://math.libretexts.org/@go/page/91910

 22 ✳ Let E be the solid region between the plane z = 4 and the paraboloid z = x

2

1

⇀

F =(−

3

x

+e

z

1

2

)^ ı ı +(−

y

3

3

Let

2

+y .

^ + x tan z)^ ȷ ȷ + 4zk

3

⇀

1. Compute the flux of F outward through the boundary of E. ^ ^ has a positive k 2. Let S be the part of the paraboloid z = x + y lying below the z = 4 plane oriented so that n component. Compute the flux of F through S. 2

2

⇀

 23 ✳ Consider the vector field ^ x ^ ı ı +y ^ ȷ ȷ +zk

⇀

F(x, y, z) = 2

[x ⇀

+y

2

2

3/2

+z ]

⇀

1. Compute ∇ ⋅ F. 2. Let S be the sphere given by 1

2

2

x

oriented outwards. Compute ∬

+ (y − 2 )

+z

2

=9

⇀

^ dS. F⋅n

S1

3. Let S be the sphere given by 2

2

2

x

oriented outwards. Compute ∬

+ (y − 2 )

+z

2

=1

⇀

^ dS. F⋅n

S2

4. Are your answers to (b) and (c) the same or different? Give a mathematical explanation of your answer.

 24 ✳ ⇀

Let F be the vector field defined by ⇀

3

F(x, y, z) = (y z + 2x) ^ ı ı + (3y − e

Calculate the flux integral ∬

S

⇀

^ dS F⋅n

sin z

2

)^ ȷ ȷ + (e

x +y

2

^ + z) k

where S is the boundary surface of the solid region E  :  0 ≤ x ≤ 2,

0 ≤ y ≤ 2,

0 ≤ z ≤ 2 +y

with outer normal.

 25 ✳ Consider the vector field ⇀

2

F(x, y, z) = (z arctan(y ) , z

Let the surface S be the part of the sphere x

2

+y

2

+z

2

=4

3

2

ln(x

+ 1) , 3z)

that lies above the plane z = 1 and be oriented downwards.

⇀

1. Find the divergence of F. 2. Compute the flux integral ∬

S

⇀

^ dS. F⋅n

4.2.25

https://math.libretexts.org/@go/page/91910

 26 ✳ Let S be the sphere x

2

+y

2

+z

2

=3

oriented inward. Compute the flux integral ⇀

^ dS F⋅n

∬ S

where ⇀

F = (x y

2

4

+y z

6

, yz

2

4

2

+ x z , zx

4

+ xy )

 27 ✳ ⇀

Consider the vector field F(x, y, z) = −2xy ^ ı ı + (y

2

2

+ sin(xz)) ^ ȷ ȷ + (x

2 ^ + y ) k.

⇀

1. Calculate ∇ ⋅ F. ⇀

2. Find the flux of F through the surface S defined by 2

x

+y

2

2

+ (z − 12 )

2

= 13 ,  z ≥ 0

using the outward normal to S.

 28 ✳ Let

be the portion of the hyperboloid

S

⇀

F = (x + e

yz

2

x

+y

2

2 ^ )^ ı ı + (2yz + sin(xz)) ^ ȷ ȷ + (xy − z − z ) k

=1 between z = −1 and out of S (away from the origin).

−z

2

Find the flux of

z = 1.

 29 ✳ ⇀

⇀

Let F be the vector field F(x, y, z) = (x − y − 1) ^ ı ı + (e is the part of the ellipsoid x + y + 2z = 1 with z ≥ 0. 2

2

2

3 5 ^ +z )^ ȷ ȷ + (2xz + z ) k.

cos y

Evaluate ∬

S

⇀

⇀

^ dS ∇×F⋅ n

where

S

2

 30 ✳ Let

be the portion of the sphere

S

⇀

2

F = (x

+e

y

2

2

)^ ı ı + (e

2

x

+y

^ +y )^ ȷ ȷ + (4 + 5x) k

x

2

2

2

+ (z − 1 )

that lies above the

=4

-plane. Find the flux of

xy

outward across S.

 31 ✳ ⇀

Find the flux of F = x y

2

^ ^ ı ı + x y^ ȷ ȷ +k 2

outward through the hemispherical surface 2

x

+y

2

+z

2

= 4,

z ≥0

 32 ✳ Let D be the cylinder x

2

+y

2

≤ 1, 0 ≤ z ≤ 5.

Calculate the flux of the vector field

⇀

z

F = (x + xy e ) ^ ı ı +

1

2

y ze

z

z ^ ^ ȷ ȷ + (3z − yze ) k

2

outward through the curved part of the surface of D.

 33 Find the flux of 2

x

+y

2

+z

2

⇀

2 ^ F = (y + xz) ^ ı ı + (y + yz)^ ȷ ȷ − (2x + z )k

upward through the first octant part of the sphere

2

=a .

4.2.26

https://math.libretexts.org/@go/page/91910

 34 ⇀

^ Let  F = (x − yz) ^ ı ı + (y + xz)^ ȷ ȷ + (z + 2xy)k  S1 S2 V

and let

be the portion of the cylinder  x + y = 2  that lies inside the sphere  x + y + z = 4 be the portion of the sphere  x + y + z = 4  that lies outside the cylinder  x + y = 2  be the solid bounded by S and S 2

2

2

2

1

2

2

2

2

2

2

2

Compute ⇀

1. ∬

^ dS F⋅n

^ pointing inward with n

S1 ⇀

2. ∭

∇ ⋅ F dV

V ⇀

3. ∬

F⋅n ^ dS

with n ^ pointing outward

S2

Use the divergence theorem to answer at least one of parts (a), (b) and (c).

Stage 3  35 Let E( r ) be the electric field due to a charge configuration that has density ρ( r ). Gauss' law states that, if V is any solid in R with surface ∂V , then the electric flux ⇀

⇀

3

∬

E⋅n ^ dS = 4πQ

where

Q =∭

∂V

ρ dV

V

^ is the outward pointing unit normal to ∂V . Show that is the total charge in V . Here, as usual, n ⇀

⇀

⇀

∇ ⋅ E( r ) = 4πρ( r )

for all r in everywhere. ⇀

3

R .

This is one of Maxwell's equations. Assume that

⇀

⇀

∇ ⋅ E( r )

and

⇀

ρ( r )

are well--defined and continuous

 36 Let V be a solid in R with surface ∂V . Show that 3

∬

⇀

^ dS = 3 Volume(V ) r ⋅n

∂V

where

⇀

^ r =x ^ ı ı +y ^ ȷ ȷ +zk

^ is the outer normal to ∂V . See if you can explain this result geometrically. and, as usual, n

 37 ✳ Let

S

be the sphere of radius

3,

centered at the origin and with outward orientation. Given the vector field

⇀

F(x, y, z) = (0, 0, x + z): ⇀

1. Calculate (using the definition) the flux of F through S ⇀

∬

F⋅n ^ dS

S

That is, compute the flux by evaluating the surface integral directly. 2. Calculate the same flux using the divergence theorem.

4.2.27

https://math.libretexts.org/@go/page/91910

 38 ✳ Consider the cube of side length 1 that lies entirely in the first octant (x ≥ 0, y ≥ 0, z ≥ 0 ) with one corner at the origin and another corner at point (1, 1, 1). As such, one face lies in the plane x = 0, one lies in the plane y = 0, and another lies in the plane z = 0. The other three faces lie in the planes x = 1, y = 1, and z = 1. Denote S as the open surface that consists of the union of the 5 faces of the cube that do not lie in the plane z = 0. The surface S is oriented in such a way that the unit normal vectors point outwards (that is, the orientation of S is such that the unit normal vectors on the top face point towards positive z -directions). Determine the value of ⇀

^ dS F⋅n

I =∬ S ⇀

where F is the vector field given by ⇀

z

2

F = (y cos(y ) + z − 1 ,

+ 1 , xy e

z

2

)

x +1

 39 ✳ 1. Find an upward pointing unit normal vector to the surface z = xy at the point (1, 1, 1). 2. Now consider the part of the surface z = xy, which lies within the cylinder x + y = 9 and call it S. Compute the upward flux of F = (y, x, 3) through S. 3. Find the flux of F = (y, x, 3) through the cylindrical surface x + y = 9 in between z = xy and z = 10. The orientation is outward, away from the z-axis. 2

2

⇀

⇀

2

2

 40 ✳ ⇀

1. Find the divergence of the vector field F = (x + sin y, z + y, z ). 2. Find the flux of F through the upper hemisphere x + y + z = 25, z ≥ 0, oriented in the positive z -direction. 3. Specify an oriented closed surface S, such that the flux ∬ F ⋅ n ^ dS is equal to −9. 2

⇀

2

2

2

⇀

S

 41 ✳ Evaluate the surface integrals. (Use any method you like.) 1. ∬

S

z

2

dS,

if S is the part of the cone x

2

⇀

+y

2

= 4z

where 0 ≤ x ≤ y and 0 ≤ z ≤ 1.

2

⇀

^ dS, if 2. ∬ F ⋅ n and S is the rectangle with vertices (0, 2, 0), (0, 0, 4), (5, 2, 0), (5, 0, 4), oriented so that the normal vector points upward. ^ F = zk

S

⇀

3. ∬

S

⇀

^ where F = (y − z ) ^ ı ı + (z − x )^ ȷ ȷ +z k with the normal vector pointing outward.

^ dS, F⋅n

0 ≤ z ≤ 3,

2

2

2

and S is the boundary surface of the box 0 ≤ x ≤ 1,

0 ≤ y ≤ 2,

 42 ✳ Let σ be the open surface given by z = 1 − x Let σ be the planar surface given by z = 0, where a, b, c, and d are constants.

2

1

3

2

−y , 2

x

+y

z ≥ 0. 2

Let σ be the open surface given by Let F = [a(y + z ) + bxz] ^ ı ı + [c(x 2

⇀

≤ 1.

2

2

2

z =x

2

+y

2

− 1,

z ≤ 0.

2

2

+ z ) + dyz] ^ ȷ ȷ +x

^ k

⇀

1. Find the flux of F upwards across σ . 2. Find all values of the constants a, b, c, and d so that the flux of F outwards across the closed surface σ 3. Find all values of the constants a, b, c, and d so that the flux of F outwards across the closed surface σ 1

⇀

1

∪ σ3

1

∪ σ2

⇀

4.2.28

is zero. is zero.

https://math.libretexts.org/@go/page/91910

 43 ✳ Let S be the ellipsoid x

2

⇀

1. Find ∬

2

+ 3z

2

= 16

and n ^ its outward unit normal. (x, y, z) − (2, 1, 1)

⇀

^ dS F⋅n

S

+ 2y

if F(x, y, z) =

. 2

2

[(x − 2 )

2. Find ∬

G⋅n ^ dS

S

3/2

2

+ (y − 1 )

+ (z − 1 ) ]

(x, y, z) − (3, 2, 2)

if G(x, y, z) =

. 2

2

[(x − 3 )

+ (y − 2 )

2

3/2

+ (z − 2 ) ]

 44 ✳ Let Ω ⊂ R be a smoothly bounded domain, with boundary which is continuously differentiable in Ω ∪ ∂Ω, 3

⇀

∭

∂Ω

and outer unit normal

⇀

n ^.

Prove that for any vector field

⇀

F

⇀

^ dS F×n

∇ × F dV = − ∬

Ω

∂Ω

 45 ✳ ^ , then for any smooth function p(x, y, z) on R , we have Recall that if S is a smooth closed surface with outer normal field n 3

^  ds = ∭ pn

∬ S

∇p dV

E

where E is the solid bounded by S. Show that as a consequence, the total force exerted on the surface of a solid body contained in a gas of constant pressure is zero. (Recall that the pressure acts in the direction normal to the surface.)

 46 ✳ ⇀

Let

be a smooth 3-dimensional vector field such that the flux of + 2 a ) for every a > 0. Calculate ∇ ⋅ F(0, 0, 0).

F 3

π(a

⇀

4

⇀

F

out of the sphere

2

x

+y

2

+z

2

2

=a

is equal to

⇀

 47. ✳ ⇀

Let  F = (x

2

+y

2

2

2

+z ) ^ ı ı + (e

x

and let S be the part of the surface  x

2 ^ +y )^ ȷ ȷ + (3 + x + z) k 

2

+y

2

+z

2

2

= 2az + 3 a   ⇀

having  z ≥ 0, oriented with normal pointing away from the origin. Here a > 0 is a constant. Compute the flux of F through S.

 48 ✳ Let u = u(x, y, z) be a solution of Laplace's Equation, ∂

2

u 2

∂

∂x

in R . Let R be a smooth solid in R 3

3

2

u

+ ∂y

2

∂

2

u

+ ∂z

2

= 0,

.

⇀

1. Prove that the total flux of F = ∇u out through the boundary of R is zero. 2. Prove that the total flux of G = u∇u out through the boundary of R equals ∂u ∭ R

[(

2

) ∂x

∂u +(

2

) ∂y

4.2.29

∂u +(

2

) ] dV ∂z

https://math.libretexts.org/@go/page/91910

 49 ✳ Let ⇀

R 2

F =x

be the part of the solid cylinder

2

x

2

+ (y − 1 )

≤1

satisfying

2

0 ≤z ≤y ;

let

S

be the boundary of

R.

Given

^ ^ ı ı + 2y ^ ȷ ȷ − 2z k, ⇀

1. Find the total flux of F outward through S. ⇀

2. Find the total flux of F outward through the (vertical) cylindrical sides of S. π

Hint: ∫ 0

n

sin

π

n−1 θ dθ =

n−2

∫ n

sin

θ dθ

for n = 2, 3, 4, … .

0

 50. ✳ A smooth surface S lies above the plane z = 0 and has as its boundary the circle x + y = 4y in the plane z = 0. This circle also bounds a disk D in that plane. The volume of the 3-dimensional region R bounded by S and D is 10 cubic units. Find the flux of 2

2

⇀

2 2 ^ F(x, y, z) = (x + x y) ^ ı ı + (y − x y )^ ȷ ȷ + (z + 2x + 3y)k

through S in the direction outward from R. 1. It is also known as Gauss's theorem. Johann Carl Friedrich Gauss (1777–1855) was a German mathematician. Throughout the 1990's Gauss's portrait appeared on the German ten-mark banknote. In addition to Gauss's theorem, the Gaussian distribution (the bell curve), degaussing and the CGS unit for the magnetic field, and the crater Gauss on the Moon are named in his honour. 2. We are going to consistently use the notation ∂ (thing) to denote the boundary of (thing). 3. Mutatis mutandis. 4. We are assuming that V is “reasonable”. 5. It's almost as though someone rigged the example with this in mind. 6. In fact, it is possible to evaluate this integral directly, if one recognizes that the ugly part of the integrand is odd under y → −y and integrates to exactly zero. 7. You can review in §1.6 of the CLP-2 text. 8. The heat equation was formulated by the French mathematician and physicist Jean-Baptiste Joseph Fourier in 1807. He lived from 1768 to 1830, a period which included both the French revolution and the reign of Napoleon. Indeed Fourier served on his local Revolutionary Committee, was imprisoned briefly during the Terror, and was Napoleon Bonaparte's scientific advisor on his Egyptian expedition of 1798. Fourier series and the Fourier transform are named after him. Fourier is also credited with discovering the greenhouse effect. 9. Heat is now understood to arise from the internal energy of the object. In an earlier theory, heat was viewed as measuring an invisible fluid, called the caloric. The amount of caloric that an object could hold was called its “heat capacity” by the Scottish physician and chemist Joseph Black (1728–1799). 10. The caloric theory of heat was itself destroyed by the cannon boring experiment of 1798. In this experiment the American/British physicist Benjamin Thompson (1753–1814) boiled water just using the heat generated by friction during the boring of a cannon. 11. Insert sarcastic footnote here. 12. The interested reader should do a net search for the story of Archimedes and the golden crown. 13. The first design of a self-righting boat was entered by William Wouldhave in a lifeboat design competition organised by South Shield's Law House committee in 1789. 14. A cup of tea in the galley doesn't count. 15. This is what Archimedes was referring to when he said “Give me a lever and a place to stand and I will move the earth.” 16. Just 12 grams of carbon contains about 6 × 10 atoms. 17. Mathematicians and their idealizations! Really the rods just represent the atomic/chemical forces that hold the wood together. 18. Note that this is just the weighted average (no pun intended) of the positions of the particles. 19. Or teeter-totter for those who speak a different English dialect. 23

4.2.30

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20. Suppose, for example, that, for large | r − r ⇀

constant times

1 ⇀ ⇀ 2 | r − r 0|

.

⇀ 0 |,

⇀

|φ( r )|

is bounded by a constant times

Then, if ∂V is the sphere of radius R centred on

over ∂V are bounded by a constant times

1 R

⇀ r 0,

∂V

⇀

has surface area 4πR and the two integrals 2

. ⇀2

21. Note that the theorem does not claim that the φ defined in the theorem obeys ∇ scope. 22. You might worry about the singularity in

⇀

and |∇φ( r )| is bounded by a

1 ⇀ ⇀ | r − r 0|

⇀2 ∇ φ ⇀ ⇀ | r − r 0|

when applying lim

may be seen using spherical coordinates centred on completely eliminates the singularity.

ε→0+

⇀

r 0.

φ = 4πρ.

to ∭

Vε

It does, but the proof is beyond our

⇀2 ∇ φ ⇀ ⇀ | r − r 0|

That this singularity is harmless

dV .

Then dV contains a factor of | r − r ⇀

⇀

2

0|

(see §A.6.3), which

This page titled 4.2: The Divergence Theorem is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

4.2.31

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4.3: Green's Theorem Our next variant of the fundamental theorem of calculus is Green's 1 theorem, which relates an integral, of a derivative of a (vectorvalued) function, over a region in the xy-plane, with an integral of the function over the curve bounding the region. First we need to define some properties of curves.

 Definition 4.3.1 1. A curve C with parametrization r (t), a ≤ t ≤ b, is said to be closed if r (a) = r (b). 2. A curve C is said to be simple if it does not cross itself. More precisely, if r (t), a ≤ t ≤ b, is a parametrization of the curve and if a ≤ t , t ≤ b obey t ≠ t and {t , t } ≠ {a, b}, then r (t ) ≠ r (t ). That is, if r (t ) = r (t ), then either t =t or t = a, t = b, or t = b, t = a. 3. A curve C is piecewise smooth if it has a parametrization r (t) which ⇀

⇀

⇀

⇀

⇀

1

1

2

1

2

1

2

2

1

1

2

⇀

1

⇀

2

⇀

1

2

2

⇀

is continuous and which is differentiable except possibly at finitely many points with the derivative being continuous and nonzero except possibly at finitely many points. Here are sketches of some examples.

And here is Green's theorem.

 Theorem 4.3.2. Green's Theorem Let be a finite region in the xy-plane, the boundary, C , of R consist of a finite number of piecewise smooth, simple closed curves R

that are oriented (i.e. arrows are put on C ) consistently with R in the sense that if you walk along C in the direction of the arrows, then R is on your left

F1 (x, y)

and F

2 (x,

y)

have continuous first partial derivatives at every point of R.

Then ∮

[ F1 (x, y) dx + F2 (x, y) dy] = ∬

C

R

(

∂F2 ∂x

−

∂F1

)  dxdy

∂y

 Warning 4.3.3 Note that in Theorem 4.3.2 we are assuming that F and F have continuous first partial derivatives at every point of R. If that is not the case, for example because F or F is not defined on all of R, then the conclusion of Green's theorem can fail. An example is F = − , F = , R = {(x, y)| x + y ≤ 1} . See Examples 4.3.7 and 4.3.8. 1

1

y

1

2

x +y

x

2

2

2

x +y

2

2

2

2

2

4.3.1

https://math.libretexts.org/@go/page/91911

Here are three notational remarks before we start the proof. ⇀

⇀

^ One way to remember the integrand on the right hand side is to write it as (∇ × F) ⋅ k . Many people use M instead of F and N instead of F . Then Green's theorem becomes 1

∮

C

2

[M (x, y) dx + N (x, y) dy] = ∬

R

(

∂N ∂x

∂M

−

∂y

) dxdy

The symbol ∮ is just an alternate notation for ∫ that is sometimes used when C is a closed curve. See Notation 2.4.1. C

C

Proof We prove the result by reformulating it as a divergence theorem statement. To that end, we define V

= {(x, y, z)|(x, y) ∈ R,   0 ≤ z ≤ 1}

G(x, y, z) = F2 (x, y) ^ ıı − F1 (x, y) ^ ȷ ȷ

Notice that V is exactly the volume obtained by expanding R vertically upward by one unit.

The definition of G does not contain a typo — the x -component of G really is (More or less the reverse of what you would normally write down.)

F2

and the y-component of G really is

−F1 .

These definitions have been rigged so that the divergence theorem applied to G and V , namely ⇀

^ dS = ∭ G⋅n

∬ ∂V

∇ ⋅ G dV

V

gives us exactly Green's theorem, as we shall now see. ⇀

Since ∇ ⋅ G =

∂F2 ∂x

−

∂F1 ∂y

the right hand side is just

,

1

⇀

∭

∇ ⋅ G dV

=∬

V

dxdy ∫

R

⇀

dz ∇ ⋅ G

0 1

=∬

dxdy ∫

R

=∬

dz  (

R

(x, y) −

∂x

0

dxdy  (

∂F2

∂F2

(x, y) −

∂x

∂F1

(x, y))

∂y ∂F1

(x, y))

∂y

because the integrand is independent of z. This is exactly the right hand side of Green's theorem. Now for the left hand side. The boundary, ∂V , of V is the union of the (flat) bottom, the (flat) top and the (curved) side. The ^ ^ outward unit normal on the (horizontal, flat) top is +k and the outward unit normal on the (horizontal, flat) bottom is −k so that ^ dS G⋅n

∬

=∬

∂V

^ G ⋅ k dS + ∬

top

=∬

^ G ⋅ (−k) dS + ∬

bottom

^ dS G⋅n

side

^ dS G⋅n

side

^ We have used the fact that the k component of G is exactly zero to discard the integrals over the top and bottom of ∂V . To evaluate the integral over the side, we'll parametrize the side. Suppose that r (t) = x(t) ^ ıı + y(t) ^ ȷ ȷ, a ≤ t ≤ b, is a parametrization of C , with the arrow in the figure above giving the direction of increasing t. Then we can use ⇀

⇀ ^ ^ R(t, z) = r (t) + z k = x(t) ^ ıı + y(t) ^ ȷ ȷ +zk

4.3.2

a ≤ t ≤ b,  0 ≤ z ≤ 1

https://math.libretexts.org/@go/page/91911

^ dS for the side. Since as a parametrization of the side. We'll use (3.3.1) to determine n ∂R

′

′

(t, z) = x (t) ^ ıı + y (t) ^ ȷ ȷ

∂t ∂R

^ (t, z) = k

∂z

(3.3.1) gives ^ dS = n

∂R

∂R (t, z) ×

∂t

(t, z) dtdz ∂z

′ ′ ^ = (x (t) ^ ıı + y (t) ^ ȷ ȷ ) × k dtdz ′

′

= ( − x (t) ^ ȷ ȷ + y (t) ^ ıı ) dtdz

Note that with this choice of ± sign (that is, pointing normal, as we see from the sketch

∂R ∂t

×

∂R ∂z

 dtdz

rather than −

∂R ∂t

×

∂R ∂z

 dtdz

^ really is the outward ), the vector n

We can now compute the surface integral directly. ^ dS = ∬ G⋅n

∬ ∂V

^ dS G⋅n

side b

1

=∫

′

a

′

dz G(R(t, z)) ⋅ ( − x (t) ^ ȷ ȷ + y (t) ^ ıı )

dt ∫ 0 b

=∫

1

dt ∫

a

′

′

dz (F2 (x(t), y(t)) ^ ıı − F1 (x(t), y(t)) ^ ȷ ȷ ) ⋅ ( − x (t) ^ ȷ ȷ + y (t) ^ ıı )

0

b

=∫

′

′

dt [ F2 (x(t), y(t)) y (t) + F1 (x(t), y(t)) x (t)]

a

since the integrand is independent of z =∮

[ F1 (x, y) dx + F2 (x, y) dy]

C

This is exactly the left hand side of Green's theorem.

 Example 4.3.4 Evaluate ∮

[(x − xy) dx + (y

3

+ 1) dy

C

where C is the curve given in the figure

Solution

4.3.3

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Let R = {(x, y)|1 ≤ x ≤ 2,  0 ≤ y ≤ 1} . By Green's theorem ∮

[(x − xy) dx + (y

3

∂ + 1) dy

=∬

C

[

(y

3

∂ + 1) −

∂x

R 2

1

=∫

dx ∫

1

(x − xy)]dxdy ∂y 2

x dy x =

0

2

3 ∣ ∣ = 2 ∣1 2

Here is a simple corollary of Green's theorem that tells how to compute the area enclosed by a curve in the xy-plane.

 Corollary 4.3.5 Let be a finite region in the xy-plane whose boundary consists of a finite number of piecewise smooth, simple closed curves. Orient C (i.e. put arrows on C ) so that if you walk along C in the direction of the arrows, then R is on your left. R C

Then 1 Area(R) = ∮

xdy = − ∮

C

ydx =

∮ 2

C

[xdy − ydx]

C

Proof This is just Green's theorem applied first with

⇀

F =x^ ȷ ȷ,

then with

⇀

F = −y ^ ıı

and finally with

⇀

F =

⇀

1 2

[ −y ^ ıı + x ^ ȷ ȷ ].

For all

three of these F 's, ∂F2

−

∂F1

∂x

=1

∂y

so that Green's theorem gives ∮

[ F1 (x, y) dx + F2 (x, y) dy]

=∬

C

R

(

∂F2

−

∂x

∂F1

) dxdy = ∬

∂y

 dxdy

R

= Area(R)

 Example 4.3.6 In this example we will use Green's theorem to compute the area enclosed by the astroid x

2/3

+y

2/3

2/3

=a

.

In Example 1.1.9 we found the parametrization ⇀

3

r (t) = x(t) ^ ı ı + y(t) ^ ȷ ȷ = a cos

3

t ^ ı ı + a sin

t^ ȷ ȷ

0 ≤ t ≤ 2π

for the astroid. So, by Corollary 4.3.5,

4.3.4

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1 Area

= 2

[xdy − ydx] =

′

[x(t)y (t) − y(t)x (t)]dt

0

2π

2

3a

3

∫ 2

[ cos

2

t sin

3

t cos t + sin

2

t cos

t sin t]dt

0 2

2π

3a =

2

∫ 2

cos

2

t sin

2

t[ cos

2

t + sin

t]dt

0

2

2π

3a =

2

∫ 2

cos

2

t sin

t dt

0 2

2π

3a =

∫ 8 3

′

∫ 2

C

=

=

2π

1 ∮

2

2π

3a

2

sin (2t) dt =

∫ 16

0

[1 − cos(4t)] dt

0

2

a π 8

 Example 4.3.7. Trick Question Evaluate ⇀

∮

B ⋅ dr

C

where −y ^ ı ı +x ^ ȷ ȷ B =

2

x

+y

2

and C is the curve x(t) = sin(cos t) y(t) = sin(sin t) z(t) = 0

with 0 ≤ t ≤ 2π. Solution First let's think about the curve C . If the curve were just X(t) = cos t, Y (t) = sin t, Z(t) = 0, it would be the unit circle centred on the origin in the xy-plane, traversed counterclockwise. For − ≤ u ≤ , the function sin u increases monotonically with u and is of the same sign as u so that, since | sin t|, | cos t| ≤ 1 < , π

π

2

2

π 2

x(t) = sin ( cos t) y(t) = sin ( sin t)

has the same sign as X(t) = cos t and is increasing at precisely the same t 's as is X(t) has the same sign as Y (t) = sin t and is increasing at precisely the same t 's as is Y (t)

So the extra sine in our parametrization of C just distorts the circle, straightening the sides a little as depicted here.

It looks like our problem is a straightforward Green's theorem problem like Example 4.3.4. Let's just try using the strategy of Example 4.3.4. Because

4.3.5

https://math.libretexts.org/@go/page/91911

∂B2

−

∂B1

∂x

∂ =

∂y

x

∂ −

∂x x2 + y 2

2

1 =

2

x

+y 2

(x

−y

∂y x2 + y 2 2x

1

−

2

2

(x2 + y 2 ) 2

2

+ y ) − 2x

2

2

2

+y )

2

x

+y

2

+ (x

= (x

+

2y 2

+ y ) − 2y

2

− (x2 + y 2 )

2

2

2

=0

it looks like Green's theorem gives us, trivially, ∮

⇀

B ⋅ dr = ∮

C

[ B1 dx + B2 dy] = ∬

C

R

(

∂B2

−

∂x

∂B1

)  dxdy = 0

∂y

where R is the region inside our curve C . That was easy — but it's also very wrong! Our next steps are to verify that ∮ B ⋅ d r ≠ 0, and explain why we got the wrong answer, and modify our computation so as to give the correct answer. We'll do this in Example 4.3.8. ⇀

C

Verification that ∮

C

⇀

B ⋅ d r ≠ 0:

}

Since ′

x (t) = − cos(cos t)  sin t ′

y (t) = cos(sin t)  cos t ′

z (t) = 0

our integral is ∮

⇀

B ⋅ dr

=∮

C

[ B1 dx + B2 dy]

C 2π ′

=∫

′

[ B1 (x(t), y(t)) x (t) + B2 (x(t), y(t)) y (t)]dt

0 2π

=∫ 0

sin(sin t) cos(cos t) sin t + sin(cos t) cos(sin t) cos t 2

2

dt

sin (cos t) + sin (sin t)

This is a very ugly looking integral 2. But even if we can't evaluate the integral, we can see that the integrand is strictly positive, and that forces ∮ B ⋅ r > 0. Because ⇀

C

π 0 ≤ | sin t|, | cos t| ≤ 1 < 2 cos(cos t) > 0,

and sin(sin t) has the same sign as sin t, and sin(sin t) is zero if and only if sin t = 0. So the first term in

the numerator, cos(cos t) sin(sin t) sin t ≥ 0

and is zero if and only if sin t = 0 cos(sin t) > 0, and sin(cos t) has the same sign as cos t, and sin(cos t) is zero if and only if cos t = 0. So the second term in the numerator, cos(sin t) sin(cos t) cos t ≥ 0

and is zero if and only if cos t = 0. There is no t for which both sin t and cos t are simultaneously zero. So the whole numerator sin(sin t) cos(cos t) sin t + sin(cos t) cos(sin t) cos t > 0

is strictly positive.

4.3.6

https://math.libretexts.org/@go/page/91911

Since the integrand is strictly positive, the integral is strictly positive. Why we got the wrong answer: In our initial and wrong calculation above, we assumed that

∂B2 ∂x

(x, y) −

∂B1 ∂y

(x, y) = 0

at all points

(x, y)

of the region

R

inside C . That's not true. While it is true for most points, it is not true for all points. The vector field B(x, y) is not defined at (x, y) = (0, 0).

So

∂B2 ∂x

(x, y) −

∂B1 ∂y

(x, y)

is also not defined at

(x, y) = (0, 0).

That's enough to invalidate Green's theorem.

Read the statement of Theorem 4.3.2 again carefully.

 Example 4.3.8. Example 4.3.7, again. Evaluate ⇀

∮

B ⋅ dr

C

where −y ^ ı ı +x ^ ȷ ȷ B =

2

x

+y

2

and C is the curve x(t) = sin(cos t) y(t) = sin(sin t) z(t) = 0

with 0 ≤ t ≤ 2π. Solution This is the same integral that we computed incorrectly in Example 4.3.7. We'll use two ingredients to compute correctly.

∮

C

⇀

B ⋅ dr

Let a > 0 and denote by C the clockwise oriented circle in the xy-plane that is of radius a and is centered on the origin. We can explicitly compute ∮ B ⋅ d r . To do so just parametrize C by x(t) = a cos t, y(t) = a sin t, z(t) = 0. Then x (t) = −a sin t, y(t) = a cos t and a

⇀

a

Ca

′

2π

∮

⇀

B ⋅ dr

=∫

Ca

−a sin t ^ ı ı + a cos t ^ ȷ ȷ [ 2

a

0

2

cos

2

t +a

2

sin

] ⋅ [ − a sin t ^ ı ı + a cos t ^ ȷ ȷ ]dt t

2π

=∫

dt = 2π

0

Pick an a that is small enough that C lies entirely inside C and apply Green's theorem with the region, R , that is between C and C . a

a

a

The curve bounding R has two components — C and C , but now C is oriented clockwise. (Recall that, in Green's theorem, when you walk along a boundary curve in the direction of the arrow, R has to be on your left.). Use −C to denote C oriented clockwise. (x, y) − (x, y) really is zero at all points (x, y) of the region R . So Green's theorem gives a

a

a

a

∂B2 ∂x

a

a

∂B1

a

∂y

4.3.7

https://math.libretexts.org/@go/page/91911

0

=∬

(

Ra

∂B2

−

∂B1

∂x ⇀

=∮

⇀

B ⋅ dr + ∮

C

B ⋅ dr

−Ca

⇀

B ⋅ dr − ∮

C

⇀

) dxdy = ∮

∂y

B ⋅ dr

Ca

and so ⇀

∮

⇀

B ⋅ dr = ∮

C

B ⋅ d r = 2π

Ca

Exercises Stage 1  1 Let R be the square R = {(x, y)|0 ≤ x ≤ 1,  0 ≤ y ≤ 1}

and let f (x, y) have continuous first partial derivatives. 1. Use the fundamental theorem of calculus to show that 1

∂f ∬ R

(x, y) dx dy = ∫ ∂y

1

f (x, 1) dx − ∫

0

f (x, 0) dx

0

2. Use Green's theorem to show that 1

∂f ∬ R

(x, y) dx dy = ∫ ∂y

1

f (x, 1) dx − ∫

0

f (x, 0) dx

0

 2 Let R be a finite region in the xy-plane, whose boundary, C , consists of a single, piecewise smooth, simple closed curve that is oriented couterclockwise. “Simple” means that the curve does not intersect itself. Use Green's theorem to show that ⇀

∬

⇀

⇀

∇ ⋅ F dx dy = ∮

R ⇀

where F = F

^ ı ı + F2 ^ ȷ ȷ,

1

^ n

^ ds F⋅n

C

is the outward unit normal to C and s is the arclength along C .

 3 Integrate

1

x dy − y dx ∮

2π

C

2

x

+y

2

counterclockwise around

1. the circle x + y = a 2. the boundary of the square with vertices (−1, −1), (−1, 1), (1, 1) and (1, −1) 3. the boundary of the region 1 ≤ x + y ≤ 2,  y ≥ 0 2

2

2

2

2

4.3.8

https://math.libretexts.org/@go/page/91911

 4 Show that ∂

x (

∂

−y

) =

x2 + y 2

∂x

( ∂y

x2 + y 2

)

for all (x, y) ≠ (0, 0). Discuss the connection between this result and the results of Q[4.3.1.3].

Stage 2  5 ⇀

⇀

Evaluate ∫ F ⋅ d r where F = x y ^ ı ı + 2xy ^ ȷ ȷ and C is the boundary of the square in the origin and diagonally opposite vertex at the point (3, 3), oriented counterclockwise. ⇀

2

2

C

-plane having one vertex at the

xy

 6 Evaluate

∮

(x sin y

2

2

2

− y ) dx + (x y cos y

2

where

+ 3x) dy

C

is the counterclockwise boundary of the trapezoid with

C

vertices (0, −2),  (1, −1),  (1, 1) and (0, 2).

 7. ✳ Evaluate

1 I =∮ C

(

− −−− − 2 0 ≤ y ≤ √4 − x .

2

x y

3

4

− x y) dx + (x y

4

3

counterclockwise around the boundary of the half-disk

2

+ x y ) dy

3

 8. ✳ Let C be the counterclockwise boundary of the rectangle with vertices (1, 0), (3, 0), (3, 1) and (1, 1). Evaluate ∮

(3 y

2

+ 2x e

y

2

2

) dx + (2y x e

y

2

) dy

C

 9. ✳ Consider the closed region enclosed by the curves y = x C is oriented counter-clockwise.

2

and y = 4 − x

2

+ 4x + 4

.

Let C be its boundary and suppose that

1. Draw the oriented curve C carefully in the xy-plane. 2. Determine the value of ∮

xy dx + (e

y

2

+ x )dy

C

 10. ✳ Let ⇀

F(x, y) = (y

2

−e

−y

2

+ sin x , 2xy e

−y

2

+ x)

Let C be the boundary of the triangle with vertices (0, 0), (1, 0) and (1, 2), oriented counter-clockwise. Compute ⇀

∫

⇀

F ⋅ dr

C

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 11. ✳ Suppose the curve C is the boundary of the region enclosed between the curves Determine the value of the line integral ∫

(2x e

2

y =x

− 4x + 3

and

2

y = 3 −x

+ 2x.

− −−− − 2 2 y + √ 2 + x ) dx + x (2 + e ) dy

y

C

where C is traversed counter-clockwise.

 12. ✳ Let ⇀

F(x, y) = (

Find ∫

C

⇀

⇀

F ⋅ dr ,

3 2

y

2

+e

−y

+ sin x) ^ ı ı +(

1 2

2

x

+ x − xe

−y

)^ ȷ ȷ

where C is the boundary of the triangle (0, 0), (1, −2), (1, 2), oriented anticlockwise.

 13. ✳ 1. Use Green's theorem to evaluate the line integral −y ∫ C

2

x

x

+y

2

dx +

2

x

+y

2

dy

where C is the arc of the parabola y = x + 1 from (−2, 2) to (2, 2). 2. Use Green's theorem to evaluate the line integral 1

2

4

−y ∫ C

where C is the arc of the parabola y = x 3. Is the vector field

2

−2

2

x

x

+y

2

dx +

2

x

+y

2

dy

from (−2, 2) to (2, 2). −y

⇀

F =

2

x

+y

2

^ ı ı +

x 2

x

+y

2

^ ȷ ȷ

conservative? Provide a reason for your answer based on your answers to the previous parts of this question.

 14. ✳ Suppose the curve C is the boundary of the region enclosed between the curves Determine the value of the line integral ∫

(2x e

y

2

y =x

− 4x + 3

and

2

y = 3 −x

+ 2x.

– 2 2 y + √2 + x )dx + x (2 + e )dy

C

where C is traversed counter-clockwise.

 15. ✳ Let

⇀

be a smooth plane vector field defined for (x, y) ≠ (0, 0), and suppose Q = P In the following I = ∫ F ⋅ d r for integer j, and all C are positively oriented circles. Suppose I

F(x, y) = P ^ ı ı +Q ^ ȷ ȷ

x

⇀

(x, y) ≠ (0, 0).

j

where C is the circle x

2

1

1. Find I for C 2. Find I for C 3. Find I for C 2

2

+y 2

: (x − 2 )

2

3

3

: (x − 2 )

4

4

: (x − 2 )

2

2

y

⇀

j

Cj

1

for =π

= 1.

+y +y

2 2

= 1. = 9. 2

+ (y − 2 )

Explain briefly. Explain briefly. = 9. Explain briefly.

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 16. ✳ ⇀

Consider the vector field F = P

^ ı ı +Q ^ ȷ ȷ,

where x +y P =

2

x

1. Compute and simplify Q

x

+y

y −x 2

,

Q =

2

x

+y

2

− Py .

⇀

2. Compute the integral ∫ F ⋅ d r directly using a parameterization, where C is the circle of radius R, centered at the origin, and oriented in the counterclockwise direction. ⇀

R

CR

⇀

3. Is F conservative? Carefully explain how your answer fits with the results you got in the first two parts. 4. Use Green's theorem to compute ∫ F ⋅ d r where C is the triangle with vertices (1, 1), (1, 0), (0, 1) oriented in the counterclockwise direction. ⇀

⇀

C

5. Use Green's theorem to compute ∫ counterclockwise direction.

⇀

C

where C is the triangle with vertices (−1, −1), (1, 0), (0, 1) oriented in the

⇀

F ⋅ dr

 17. ✳ 1. Evaluate − −−− − 3 2 2 √ 1 + x dx + (2x y + y ) dy

∫ C

where C is the unit circle x 2. Evaluate

2

+y

2

= 1,

oriented counterclockwise. − −−− − 3 2 2 √ 1 + x dx + (2x y + y ) dy

∫ C

where C is now the part of the unit circle x

2

+y

2

with x ≥ 0, still oriented counterclockwise.

= 1,

Stage 3  18. ✳ Evaluate the line integral 2

∫

(x

x

x

+ y e ) dx + (x cos y + e ) dy

C

where C is the arc of the curve x = cos y for −π/2 ≤ y ≤ π/2, traversed in the direction of increasing y.

 19. ✳ Use Green's theorem to establish that if C is a simple closed curve in the plane, then the area A enclosed by C is given by 1 A =

∮ 2

Use this to calculate the area inside the curve x

2/3

+y

2/3

x dy − y dx

C

= 1.

 20. ✳ ⇀

Let F(x, y) = (x + 3y) ^ ı ı + (x + y) ^ ȷ ȷ and for each circle C in the plane

G(x, y) = (x + y) ^ ı ı + (2x − 3y) ^ ȷ ȷ

⇀

∮

⇀

F ⋅ dr = A ∮

C

be vector fields. Find a number

A

such that

⇀

G ⋅ dr

C

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 21. ✳ ⇀

y

Let F(x, y) =

3

2

⇀

1. Compute ∮

C

⇀

F ⋅ dr

xy

^ ı ı −

( x2 +y 2 )

2

^ ȷ ȷ , (x, y) ≠ (0, 0).

2

( x2 +y 2 )

where C is the unit circle in the xy-plane, positively oriented. ⇀

2. Use (a) and Green's theorem to find ∮

C0

where C is the ellipse

⇀

F ⋅ dr

0

2

x

16

y

+

2

= 1,

25

positively oriented.

 22. ✳ Let

C1

be the circle

integrals ∮

C1

⇀

2

(x − 2 )

and ∮

⇀

F ⋅ dr

C2

+y

⇀

2

=1

and let

C2

be the circle

2

(x − 2 )

+y

2

= 9.

Let

⇀

F =−

y x2 +y 2

^ ı ı +

x x2 +y 2

^ ȷ ȷ.

Find the

2

Let C be

⇀

F ⋅ dr .

 23. ✳ Let R be the region in the first quadrant of the the boundary of R, oriented counterclockwise. 1. Evaluate ∫

x ds.

2. Evaluate ∫

F ⋅ dr ,

C

⇀

C

⇀

⇀

-plane bounded by the coordinate axes and the curve

xy

where F(x, y) = ( sin(x

2

2

) − xy) ^ ı ı + (x

y = 1 −x .

2

+ cos(y )) ^ ȷ ȷ.

 24. ✳ Let C be the curve defined by the intersection of the surfaces z = x + y and z = x

2

2

+y .

1. Show that C is a simple closed curve. 2. Evaluate ∮ F ⋅ d r where ⇀

⇀

C

1. 2.

⇀

2

F =x ⇀

F =y

2

^ ı ı +y

2

2

^ ı ı +x

^ ȷ ȷ + 3e ^ ȷ ȷ + 3e

z

^ k.

z

^ k.

 25 Find a smooth, simple, closed, counterclockwise oriented curve, C , in the xy-plane for the which the value of the line integral ∮ (y − y) dx − 2 x dy is a maximum among all smooth, simple, closed, counterclockwise oriented curves. 3

3

C

1. George Green (1793–1841) was a British mathematical physicist. He spent much of the early part of his life working in his father's bakery and grain mill. He was finally admitted as an undergraduate to Cambridge in 1832, aged nearly forty. 2. Indeed! This page titled 4.3: Green's Theorem is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

4.3.12

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4.4: Stokes' Theorem Our last variant of the fundamental theorem of calculus is Stokes' 1 theorem, which is like Green's theorem, but in three dimensions. It relates an integral over a finite surface in R with an integral over the curve bounding the surface. 3

 Theorem 4.4.1. Stokes' Theorem Let ^ has been chosen at each point of S and this choice depends be a piecewise smooth oriented surface (i.e. a unit normal n continuously on the point) the boundary, ∂S, of the surface S consist of a finite number of piecewise smooth, simple curves that are oriented ^ in the sense that consistently with n S

if you walk along ∂S in the direction of the arrow on ∂S, ^ with the vector from your feet to your head having direction n then S is on your left hand side.

⇀

F

be a vector field that has continuous first partial derivatives at every point of S.

Then ⇀

∮

⇀

⇀

∂S

⇀

^  dS ∇×F⋅ n

F ⋅ dr = ∬ S

Note that in Stokes' theorem, S must be an oriented surface. In particular, S may not be a Möbius strip. (See Example 3.5.3.) If S is part of the xy-plane, then Stokes' theorem reduces to Green's theorem. Our proof of Stokes' theorem will consist of rewriting the integrals so as to allow an application of Green's theorem. If ∂S is a simple closed curve and when you look at ∂S from high on the z -axis, it is oriented counterclockwise (look at the figure in Theorem 4.4.1), then ^ is upward pointing, i.e. has positive z -component, at least near ∂S. n Proof ⇀

Write F = F

1

^ ^ ıı + F2 ^ ȷ ȷ + F3 k.

Both integrals involve F terms and F terms and F terms. We shall show that the 1

2

3

F1

terms

⇀

in the two integrals agree. In other words, we shall assume that F = F ^ ıı . The proofs that the F and F terms also agree are similar. For simplicity, we'll assume 2 that the boundary of S consists of just a single curve, and that we can 1

2

3

pick a parametrization of S with ⇀

2

S = { r (u, v) = (x(u, v), y(u, v), z(u, v))|(u, v) in R ⊂ R } ⇀ ∂ r

^ dS = + and with r (u, v) orientation preserving in the sense that n pick a parametrization of the curve, ∂R, bounding R as (u(t), v(t)), ∂R in the direction of increasing t, then R is on your left. ⇀

∂u

Then the curve ∂S bounding S can be parametrized as R(t) =

×

⇀ ∂ r ∂v

Also in such a way that when you walk along

du dv.

a ≤ t ≤ b,

⇀

r (u(t), v(t)), a ≤ t ≤ b.

4.4.1

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The orientation of R(t): We'll now verify that the direction of increasing t for the parametrization R(t) of ∂S is the direction of the arrow on ∂S in the figure on the left above. By continuity, it suffices to check the orientation at a single point. Find a point (u , v ) on ∂R where the forward pointing tangent vector is a positive multiple of ^ ıı . The horizontal arrow on ∂R in the figure on the left below is at such a point. Suppose that t = t at this point — in other words, suppose that (u , v ) = (u(t ), v(t )). Because the forward pointing tangent vector to ∂R at (u , v ), namely (u (t ), v (t )), is a ^ positive multiple of ıı , we have u (t ) > 0 and v (t ) = 0. The tangent vector to ∂S at R(t ) = r (u , v ), pointing in the direction of increasing t, is 0

0

0

′

0

0

0

0

0

′

′

0

0

0

⇀

dt

0

0

⇀

⇀

d

′

R (t0 ) =

′

0

r (u(t), v(t))∣ ∣

t=t0

∂u

0

⇀

∂r

′

= u (t0 )

0

∂r

′

(u0 , v0 ) + v (t0 )

∂v

(u0 , v0 )

⇀

∂r

′

= u (t0 )

(u0 , v0 ) ∂u

and so is a positive multiple of

⇀ ∂ r ∂u

(u0 , v0 ).

See the figure on the right below.

If we now walk along a path in the uv-plane which starts at (u , v ), holds u fixed at u and increases v, we move into the interior of R starting at (u , v ). Correspondingly, if we walk along the path, r (u , v), in R with v starting at v and 0

0

0

⇀

0

3

0

0

increasing, we move into the interior of S. The forward tangent to this new path, interior of S. It's the blue arrow in the figure on the right below.

⇀ ∂ r ∂v

0

(u0 , v0 ),

points from

Now imagine that you are walking along ∂S in the direction of increasing t. At time t you are at R(t

0 ).

0

arm straight ahead of you. So it is pointing in the direction S.

It is pointing in the direction

orientation of

∂S,

⇀ ∂ r ∂v

(u0 , v0 ).

⇀ ∂ r ∂u

(u0 , v0 ).

If the direction of increasing

^ . And since n ^ dS = + same direction as n

∂u

×

⇀ ∂ r ∂v

du dv,

r (u0 , v0 )

into the

You point your right

You point your left arm out sideways into the interior of

then the vector from our feet to our head, which is ⇀ ∂ r

⇀

⇀ ∂ r ∂u

t

is the same as the forward direction of the

(u0 , v0 ) ×

⇀ ∂ r ∂v

(u0 , v0 ),

should be pointing in the

it is.

Now, with our parametrization and orientation sorted out, we can examine the integrals. The surface integral: Since F

= F1 ^ ıı ,

so that

4.4.2

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^ ȷ ȷ

∂

∂

∂

∂x

∂y

∂z

⇀

∇ × F = det ⎢ ⎢ ⎣

^ k ⎤

^ ıı

⎡ ⇀

F1

0

0

∂F1

⎥ = (0, ⎥

,−

∂z

∂F1

)

∂y

⎦

and ^ ıı

⎡ ⇀

^ dS = n

∂r

∂r ×

∂u

∂v

⎢  du dv = det ⎢ ⎢

∂x

∂y

∂z

∂u

∂u

∂u

⎣

∂x

∂y

∂z

∂v

∂v

∂v

∂y ∂z

∂z ∂y

=(

^ k ⎤

^ ȷ ȷ

⇀

− ∂u ∂v

) ^ ıı + (

∂u ∂v

∂z ∂x

∂x ∂y

∂y ∂x −

∂u ∂v

⎦

∂x ∂z −

∂u ∂v

+(

⎥ ⎥ ⎥

)^ ȷ ȷ

∂u ∂v ^ )k

∂u ∂v

and ⇀

⇀

^ dS = ∬ ∇×F⋅ n

∬ S

(0,

∂F1

{

∂F1

∂z ∂x (

∂z

R

∂F1

∂z

R

=∬

,−

∂y

∂x ∂z −

∂u ∂v

⇀

)− ∂u ∂v

⇀

∂r )⋅

∂r ×

∂u

∂F1

 du dv ∂v

∂x ∂y (

∂y

∂y ∂x −

∂u ∂v

)} du dv ∂u ∂v

Now we examine the line integral and show that it equals this one. The line integral: b

⇀

∮

⇀

F ⋅ dr = ∫

∂S

⇀

d

⇀

F( r (u(t), v(t))) ⋅

a b

=∫

⇀

⇀

r (u(t), v(t)) dt

dt ⇀

∂r

⇀

F( r (u(t), v(t))) ⋅ [ ∂u

a

⇀

du (u(t), v(t))

∂r (t) +

dt

dv (u(t), v(t))

∂v

(t)] dt dt

We can write this as the line integral ∮

M (u, v) du + N (u, v) dv

∂R b

=∫

du [M (u(t), v(t))  dt

a

dv (t) + N (u(t), v(t)) 

(t)] dt dt

around ∂R, if we choose ⇀

⇀

⇀

∂r

M (u, v) = F( r (u, v)) ⋅

∂x (u, v) = F1 (x(u, v), y(u, v), z(u, v))

∂u ⇀

⇀

∂r

N (u, v) = F( r (u, v)) ⋅

(u, v) ∂u

⇀

∂x (u, v) = F1 (x(u, v), y(u, v), z(u, v))

∂v

(u, v) ∂v

Finally, we show that the surface integral equals the line integral: By Green's Theorem, we have

4.4.3

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⇀

∮

⇀

F ⋅ dr

=∮

∂S

M (u, v) du + N (u, v) dv

∂R

∂N =∬

{

∂M −

} dudv

∂u

R

∂v

∂ =∬

∂x

{

R

[ F1 (x(u, v), y(u, v), z(u, v))] ∂u

∂v

∂

∂x

−

=∬

{(

∂x

R

−(

=∬

{ (

R ⇀

∂y

+

∂y

+

∂z

∂F1 ∂z ∂z

∂

+

∂z ∂x )

−(

∂u

∂v

∂ +F1

)

2

x

∂u∂v

∂x

∂

2

x

−F1

} dudv

∂u

∂v∂u

∂F1 ∂y

∂v

x } dudv

∂x

∂v

2

∂v∂u

)

∂u

∂F1 ∂z

∂v

∂F1 ∂z

∂u

+

∂u

∂F1 ∂y

∂v

∂F1 ∂y ∂y

∂F1 ∂y

∂u

∂F1 ∂x ∂x

+

x

− F1 ∂u

∂F1 ∂x

2

∂u∂v

[ F1 (x(u, v), y(u, v), z(u, v))] ∂v

∂ + F1

∂y

∂F1 ∂z

+

∂v

∂z

∂x )

∂v

} du dv ∂u

⇀

^ dS ∇×F⋅ n

=∬ S

which is the conclusion that we wanted. Before we move on to some examples, here are a couple of remarks. ⇀

⇀

⇀

^  dS Stokes' theorem says that ∮ F ⋅ d r = ∬ ∇ × F ⋅ n for any (suitably oriented) surface whose boundary is C . So if S and S are two different (suitably oriented) surfaces having the same boundary curve C , then ⇀

C

1

S

2

⇀

∬

⇀

⇀

^  dS = ∬ ∇×F⋅ n

S1

⇀

^  dS ∇×F⋅ n

S2

For example, if C is the unit circle 2

C = {(x, y, z)| x

+y

2

= 1,  z = 0}

oriented counterclockwise when viewed from above, then both 2

+y

2

S1

= {(x, y, z)| x

S2

= {(x, y, z)|z ≥ 0,  x

≤ 1,  z = 0} 2

+y

2

+z

2

= 1}

with upward pointing unit normal vectors, have boundary C . So Stokes' tells us that ∬

S1

It should not be a surprise that ∬

S1

⇀

⇀

⇀

^  dS = ∬ ∇×F⋅ n

S2

⇀

^  dS, ∇×F⋅ n 2

V = {(x, y, z)| x

+y

2

+z

2

⇀

⇀

⇀

^  dS = ∬ ∇×F⋅ n

S2

⇀

^  dS. ∇×F⋅ n

for the following reason. Let ≤ 1,  z ≥ 0}

be the solid between S and S . The boundary ∂V of V is the union of S and S 1

2

1

4.4.4

2.

https://math.libretexts.org/@go/page/91912

^ But beware that the outward pointing normal to ∂V (call it N ) is +n ^ on S and −n ^ on S . So the divergence theorem gives 2

⇀

∬

⇀

⇀

^  dS − ∬ ∇×F⋅ n

S2

1

⇀

^  dS ∇×F⋅ n

S1 ⇀

⇀

^  dS + ∬ ∇×F⋅ N

=∬ S2

⇀

⇀

^  dS ∇×F⋅ N

S1 ⇀

⇀

^ ∇ × F ⋅ N dS

=∬ ∂V

⇀

=∭

⇀

⇀

∇ ⋅ (∇ × F) dV

V

by the divergence theorem =0

by the vector identity Theorem 4.1.7.a. ⇀

⇀

⇀

⇀

As a second remark, suppose that the vector field F obeys ∇ × F = 0 everywhere. Then Stokes' theorem forces ⇀

⇀

are around all closed curves C , which implies that F is conservative, by Theorem 2.4.7. So Stokes' theorem provides another proof of Theorem 2.4.8. ∮

C

⇀

F ⋅ dr = 0

⇀

⇀

Here is an easy example which shows that Stokes' can be very useful when ∇ × F simplifies.

 Example 4.4.2 ⇀

⇀

Evaluate ∮ F ⋅ d r where F = [2z + sin (x counterclockwise when viewed from above. C

⇀

146

and the curve C is the circle x

^ )] ^ ı ı − 5z ^ ȷ ȷ − 5y k

2

+y

2

= 4, z = 1,

oriented

Solution ⇀

The x in F will probably make a direct evaluation of the integral difficult. So we'll use Stokes' theorem. To do so we need a surface S with ∂S = C . The simplest is just the flat disk 146

2

S = {(x, y, z)| x

+y

2

≤ 4,   z = 1}

Since

4.4.5

https://math.libretexts.org/@go/page/91912

⎡ ⇀

⇀

∇ × F = det ⎢ ⎢ ⎣

^ ı ı

^ ȷ ȷ

^ k

∂

∂

∂

∂x

∂y

∂z

−5z

−5y

146

2z + sin (x ∂

∂

∂y

∂z

−5z

−5y

= ^ ı ı det [

)

⎤ ⎥ ⎥ ⎦

] −^ ȷ ȷ det [

∂

∂

∂x

∂z 146

2z + sin (x

^ + k det [

∂

∂

∂x

∂y 146

2z + sin (x

)

)

]

−5y

]

−5z

= 2^ ȷ ȷ ^ and the normal to S is k , Stokes' theorem gives

⇀

∮

⇀

F ⋅ dr

⇀

=∬

C

⇀

^ dS = ∬ ∇×F⋅ n

S

^ (2 ^ ȷ ȷ ) ⋅ k dS = 0

S

Now we'll repeat the last example with a harder curve.

 Example 4.4.3 ⇀

⇀

^ Evaluate ∮ F ⋅ d r where F = [2z + sin (x )] ^ ı ı − 5z ^ ȷ ȷ − 5y k and the curve and z = y, oriented counterclockwise when viewed from above. ⇀

146

C

C

is the intersection of

2

x

+y

2

+z

2

=4

Solution The surface x + y + z = 4 is the sphere of radius 2 centred on the origin and z = y is a plane which contains the origin. So C , being the intersection of a sphere with a plane through the centre of the sphere, is a circle, with centre (0, 0, 0) and radius 2. The part of the circle in the first octant is sketched on the left below. 2

2

2

⇀

The x in F will probably make a direct evaluation of the integral difficult. So we'll use Stokes' theorem. To do so we need a surface S with ∂S = C . The simplest is the flat disk 146

2

S = {(x, y, z)| x

+y

2

+z

2

≤ 4,   z = y}

The first octant of S is shown in the figure on the right above. We saw in the last Example 4.4.2 that ⇀

⇀

∇ × F = 2^ ȷ ȷ

So Stokes' theorem gives ⇀

∮ C

⇀

F ⋅ dr

⇀

=∬

⇀

^ dS = 2 ∬ ∇×F⋅ n

S

^ ^ dS ȷ ȷ ⋅n

S

^ dS in two ways. The first way is more efficient, but also requires more insight. Since We'll evaluate the integral 2 ∬ ^ȷȷ ⋅ n ^ ^ ^ = ∇(z − y) = k − ^ ȷ ȷ, the upward unit normal to the plane z − y = 0, and hence to S, is n (k − ^ ȷ ȷ ). Consequently the ⇀

S

1

√2

integrand ^ −^ ȷ ȷ +k ^ ȷ ȷ ⋅n ^ =^ ȷ ȷ ⋅(

– √2

4.4.6

1 ) =−

– √2

https://math.libretexts.org/@go/page/91912

^ dS: is a constant and we do not need a formula for n

⇀

∮

⇀

F ⋅ dr

– ^ ^ dS = −√2 ∬ ȷ ȷ ⋅n

=2∬

C

S

– – 2 dS = −√2Area(S) = −√2π 2

S

– = −4 √2π

Alternatively, we can evaluate the integral ∬

S

using our normal protocol. As S is part of the plane z = f (x, y) = y,

^ dS ^ ȷ ȷ ⋅n

^ ^ ^ dS = ±( − fx ^ n ı ı − fy ^ ȷ ȷ + k) dxdy = ±(−^ ȷ ȷ + k) dxdy ^ ^ dS = (−^ To get the upward pointing normal pointing normal, we take the + sign so that n ȷ ȷ + k) dxdy. As (x, y, z) runs over

S

2

= {(x, y, z)| x

2

= {(x, y, z)| x

+y

(x, y)

runs over the elliptical disk R = {(x, y)|

2

x

4

+

y

x

4

y

+

2

2

+z

+ 2y

2

= {(x, y, z)|

2

≤ 1} .

2

2

≤ 4,   z = y}

≤ 4,   z = y}

2

≤ 1,   z = y}

2

The part of this ellipse in the first octant is the shaded region

in the figure below.

–

–

This ellipse has semiaxes a = 2 and b = √2 and hence area πab = 2√2π. So ⇀

∮

⇀

F ⋅ dr

=2∬

C

^ ȷ ȷ ⋅n ^ dS = 2 ∬

S

^ ^ ȷ ȷ ⋅ (−^ ȷ ȷ + k) dxdy = −2 ∬

R

dxdy

R

= −2Area(R) – = −4 √2π

 Example 4.4.4 Evaluate

⇀

⇀

^ where F = (x + y) ^ ı ı + 2(x − z) ^ ȷ ȷ + (y + z) k and C is the oriented curve obtained by going from (2, 0, 0) to (0, 3, 0) to (0, 0, 6) and back to (2, 0, 0) along straight line segments. ∮

C

⇀

F ⋅ dr

2

Solution 1 In this first solution, we'll evaluate the integral directly. The first line segment (C in the figure above) may be parametrized as 1

⇀

r (t) = (2, 0, 0) + t{(0, 3, 0) − (2, 0, 0)} = (2 − 2t , 3t , 0)

0 ≤t ≤1

So the integral along this segment is

4.4.7

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1

∫

⇀

⇀

1

dr

⇀

F( r (t)) ⋅

 dt dt

0

2

=∫

(2 + t , 2(2 − 2t) , (3t) ) ⋅ (−2 , 3 , 0) dt

0 1

=∫

(8 − 14t) dt

0 2

1

= [8t − 7 t ]

=1

0

The second line segment (C in the figure above) may be parametrized as 2

⇀

r (t) = (0, 3, 0) + t{(0, 0, 6) − (0, 3, 0)} = (0 , 3 − 3t , 6t)

0 ≤ t ≤ 1.

So the integral along this segment is 1

∫

⇀

⇀

⇀

1

dr

F( r (t)) ⋅

 dt dt

0

2

=∫

(3(1 − t) , −12t , 9(1 − t)

+ 6t) ⋅ (0, −3, 6) dt

0 1 2

=∫

[36t + 54(1 − t)

+ 36t] dt

0 2

= [18 t

3

− 18(1 − t)

2

1

+ 18 t ] 0

= 54

The final line segment (C in the figure above) may be parametrized as 3

⇀

r (t) = (0, 0, 6) + t{(2, 0, 0) − (0, 0, 6)} = (2t , 0 , 6 − 6t)

0 ≤t ≤1

So the line integral along this segment is 1

∫ 0

⇀

⇀

⇀

1

dr

F( r (t)) ⋅

 dt

=∫

dt

(2t , 4t − 12(1 − t) , 6(1 − t)) ⋅ (2, 0, −6) dt

0 1

=∫

2

2

[4t − 36(1 − t)] dt = [2 t

1

+ 18(1 − t) ]

= −16

0

0

The full line integral is ⇀

∮

⇀

F ⋅ d r = 1 + 54 − 16 = 39

C

Solution 2 ⇀

This time we shall apply Stokes' Theorem. The curl of F is ^ ıı

^ ȷ ȷ

^ k

∂

∂

∂

∂x

∂y

∂z

x +y

2(x − z)

⎡ ⇀

⇀

∇ × F = det ⎢ ⎢ ⎣

y

2

+z

⎤ ⎥ ⎥ ⎦

^ = (2y + 2) ^ ıı − (0 − 0)^ ȷ ȷ + (2 − 1)k ^ = 2(y + 1) ^ ıı + k

The curve C is a triangle and so is contained in a plane. Any plane has an equation of the form Ax + By + C z = D. Our plane does not pass through the origin (look at the figure above) so the D must be nonzero. Consequently we may divide Ax + By + C z = D through by D giving an equation of the form ax + by + cz = 1. Because (2, 0, 0) lies on the plane, a =

1

Because (0, 3, 0) lies on the plane, b = Because (0, 0, 6) lies on the plane, c =

1

2

3 1 6

. . .

x

y

2

3

So the triangle is contained in the plane + the plane + + = 1 that obeys x ≥ 0, this surface x

y

z

2

3

6

+

z 6

y ≥0

It is the boundary of the surface S that consists of the portion of and z ≥ 0. Rewrite the equation of the plane as z = 6 − 3x − 2y. For

= 1.

4.4.8

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^ ^  dS = (3 ^ n ıı + 2 ^ ȷ ȷ + k) dx dy

by 3.3.2, and we can write S

= {(x, y, z)|x ≥ 0,  y ≥ 0,  z ≥ 0,  z = 6 − 3x − 2y} = {(x, y, z)|x ≥ 0, y ≥ 0,  6 − 3x − 2y ≥ 0,  z = 6 − 3x − 2y}

As (x, y, z) runs over S, (x, y) runs over the triangle R

= {(x, y, z)|x ≥ 0,  y ≥ 0,  3x + 2y ≤ 6} = {(x, y, z)|x ≥ 0,  0 ≤ y ≤

3 2

(2 − x)}

Using horizontal strips as in the figure on the left below, ⇀

∮

⇀

F ⋅ dr

⇀

C

⇀

^ dS ∇×F⋅ n

=∬ S

^ ^ [2(y + 1) ^ ıı + k] ⋅ [3 ^ ıı + 2 ^ ȷ ȷ + k] dx dy

=∬ R

=∬

[6y + 7] dx dy

R 1

3

3

=∫

(6−2y)

dy ∫

0 3

1

=∫

dy  3

1 ∫ 3

[6y + 7][6 − 2y] 3

0

=

dy [−12 y

2

+ 22y + 42]

0

1 =

dx [6y + 7]

0

[ − 4y

3

+ 11 y

2

3

+ 42y ]

3

0

= [ − 4 × 9 + 11 × 3 + 42] = 39

Alternatively, using vertical strips as in the figure on the right above,

4.4.9

https://math.libretexts.org/@go/page/91912

⇀

∮

⇀

F ⋅ dr

=∬

C

[6y + 7] dx dy

R 3

2

=∫

2

(2−x)

dx ∫

0

dy [6y + 7]

0 2

=∫

2

3 dx [3

3 +7

(2 − x)] 2

2

0

27

1

= [−

3

(2 − x ) 4

9 =

2

(2 − x ) 2

21

1

−

2

2

(2 − x ) ]

3

2

2

0

21 8+

4

4 = 39 4

 Example 4.4.5 Evaluate ∮

⇀

C

⇀

F ⋅ dr

⇀

^ where F = (cos x + y + z) ^ ı ı + (x + z) ^ ȷ ȷ + (x + y) k

2

x

y

2

+

z

2

+ 2

and C is the intersection of the surfaces

=1

2

and

z =x

+ 2y

2

3

oriented counterclockwise when viewed from above. Solution y

2

2

First, let's sketch the curve. x + + = 1 is an ellipsoid centred on the origin and z = x + 2y paraboloid that passes through the origin. They are sketched in the figure below. The paraboloid is red. 2

z

2

2

2

3

is an upward opening

Their intersection, the curve C , is the blue curve in the figure. It looks like a deformed 3 circle. One could imagine parametrizing − z.

2

C.

For example, substituting

2

x

= z − 2y

into the equation of the ellipsoid gives

2

This can be solved to give y as a function of z and then x = z − 2y also gives x as a function of However this would clearly yield, at best, a really messy integral. So let's try Stokes' theorem.

3 2

y

2

+

1 3

(z +

3 2

)

=

7 4

2

.

2

In fact, since ⎡ ⇀

⇀

∇ × F = det ⎢ ⎢ ⎣

^ ı ı

^ ȷ ȷ

^ k

∂

∂

∂

∂x

∂y

∂z

cos x + y + z

x +z

x +y

⎤ ⎥ ⎥ ⎦

^ = ^ ı ı (1 − 1) − ^ ȷ ȷ (1 − 1) + k(1 − 1) ⇀

= 0 ⇀

⇀

⇀

This F is conservative! (In fact F = ∇( sin x + xy + xz + yz). ) As C is a closed curve, ∮

C

4.4.10

⇀

⇀

F ⋅ d r = 0.

https://math.libretexts.org/@go/page/91912

 Example 4.4.6 Evaluate ∬

S

G⋅n ^ dS

^ where G = (2x) ^ ı ı + (2z − 2x) ^ ȷ ȷ + (2x − 2z) k and 2

S = {(x, y, z)|z = (1 − x

2

3

y

2

− y )(1 − y ) cos x e ,  x

+y

2

≤ 1}

with upward pointing normal Solution 1 The surface S is sketched below. It is a pretty weird surface. About the

only simple thing about it is that its boundary, ∂S, is the circle x + y = 1, z = 0. It is clear that we should not try to evaluate the integral directly 4. In this solution we will combine the divergence theorem with the observation that 2

∂

⇀

2

∂

∇⋅G =

(2x) + ∂x

∂ (2z − 2x) +

∂y

(2x − 2z) = 0 ∂z

to avoid ever having work with the surface S. Here is an outline of what we will do. We first select a simple surface S whose boundary ∂S is also the circle x S , and the surface that we will use, is the disk ′

′

2

+y

2

= 1,

z = 0.

A nice simple choice of

′

S

′

2

= {(x, y, z)| x

+y

2

= 1,  z = 0}

Then we define V to be the solid whose top surface is S and whose bottom surface is S . So the boundary of V is the union of S and S . ′

′

^ ^ =k For S , we will use the upward pointing normal n , which is minus the outward pointing normal to ∂V on S . So the divergence theorem says that ′

′

⇀

∭

∇ ⋅ G dV = ∬

V

^ dS − ∬ G⋅n

S

S

^ dS G⋅n ′

⇀

The left hand side is zero because, as we have already seen, ∇ ⋅ G = 0. So ∬

^ dS = ∬ G⋅n

S

Finally, we compute ∬

S

′

S

^ dS G⋅n ′

^ dS. G⋅n ⇀

⇀

We saw an argument like this (with G = ∇ × F ) in the first remark following the proof of Theorem 4.4.1. So all that we have to do now is compute

4.4.11

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^ dS G⋅n

∬

^ G ⋅ k dS = ∬

^ dS = ∬ G⋅n

=∬

S

S

′

S

′

(2x − 2z) dxdy 2

x +y

2

≤1

z=0

=∬

(2x) dxdy

x2 +y 2 ≤1 z=0

=0

simply because the integrand is odd under x → −x. Solution 2 In this second solution we'll use Stokes' theorem instead of the divergence theorem. To do so, we have to express G in the ⇀

⇀

form ∇ × F. So the first thing to do is to check if G passes the screening test, Theorem 4.1.12, for the existence of vector ⇀

potentials. That is, to check if ∇ ⋅ G = 0. It is. We saw this in Solution 1 above. Next, we have to find a vector potential. In fact, we have already found, in Example 4.1.15, that ⇀

F = (z

2

2

− 2xz) ^ ıı + (x

− 2xz)^ ȷ ȷ

is a vector potential for G, which we can quickly check. Parametrizing ⇀

⇀

2

F( r (t)) = x

by

C

⇀

∣ ^ ȷ ȷ ∣x=cos

r (t) = cos t ^ ıı + sin t ^ ȷ ȷ, 2

= cos

t

0 ≤ t ≤ 2π,

Stokes' theorem gives (recalling that

z =0

on

C

so that

)

t

⇀

∬

^ dS G⋅n

=∬

S

⇀

⇀

^ dS = ∮ ∇×F⋅ n

S

2π ⇀

F ⋅ dr = ∫

C

0

⇀

⇀

⇀

dr

F( r (t)) ⋅

 dt dt

2π 2

=∫

( cos

t)(cos t) dt

0

Of course this integral can be evaluated by using that one antiderivative of the integrand cos t = (1 − sin t) cos t is sin t − sin t and that this antiderivative is zero at t = 0 and at t = 2π. But it is easier to observe that the integral of any odd power of sin t or cos t over any full period is zero. Look, for example, at the graphs of sin x and cos x, below. 3

1

2

3

3

3

3

Either way ^ dS = 0 G⋅n

∬ S

 Example 4.4.7 In this example we compute, in three different ways, ∮

C

⇀

⇀

F ⋅ dr

where

⇀

^ F = (z − y) ^ ı ı − (x + z) ^ ȷ ȷ − (x + y) k

and C is the curve x

2

+y

2

+z

2

= 4,

z =y

oriented counterclockwise when viewed from above.

4.4.12

https://math.libretexts.org/@go/page/91912

Solution 1 Direct Computation: ⇀

In this first computation, we parametrize the curve C and compute ∮ F ⋅ d r directly. The plane z = y passes through the origin, which is the centre of the sphere x + y + z = 4. So C is a circle which, like the sphere, has radius 2 and centre (0, 0, 0). We use a parametrization of the form ⇀

C

2

2

2

⇀

′

′

r (t) = c + ρ cos t ^ ıı + ρ sin t ^ ȷ ȷ

0 ≤ t ≤ 2π

where is the centre of C , is the radius of C and and ^ȷȷ are two vectors that

c = (0, 0, 0) ρ =2 ^ ıı

′

′

1. are unit vectors, 2. are parallel to the plane z = y and 3. are mutually perpendicular.

′

′

The trickiest part is finding suitable vectors ^ ıı and ^ ȷ ȷ : ′

The point (2, 0, 0) satisfies both x + y + z = 4 and z = y and so is on C . We may choose ^ ıı to be the unit vector ^ in the direction from the centre (0, 0, 0) of the circle towards (2, 0, 0). Namely ıı = (1, 0, 0). Since the plane of the circle is z − y = 0, the vector ∇(z − y) = (0, −1, 1) is perpendicular to the plane of C . So 2

2

2

′

⇀

′

^ k =

1 √2

(0, −1, 1)

′

is a unit vector normal to z = y. Then ^ȷȷ ′

′

′ ^ =k ×^ ıı =

1 √2

(0, −1, 1) × (1, 0, 0) =

1 √2

(0, 1, 1)

is a

′

^ ^ unit vector that is perpendicular to ^ ıı and k . Since ^ ȷ ȷ is perpendicular to k , it is parallel to z = y. ′

Substituting in c = (0, 0, 0), ρ = 2,

′

^ ıı = (1, 0, 0)

⇀

′

′

and ^ȷȷ

=

1 √2

(0, 1, 1)

gives

1

r (t) = 2 cos t (1, 0, 0) + 2 sin t

– √2

sin t (0, 1, 1) = 2( cos t,

– √2

sin t ,

– √2

)

0 ≤ t ≤ 2π

To check that this parametrization is correct, note that x +y +z = 4 and z = y. 2

2

At

x = 2 cos t,

– y = √2 sin t,

– z = √2 sin t

satisfies both

2

–

As t increases, z(t) = √2 sin t increases and r (t) moves upwards towards r( This is the desired counterclockwise direction (when viewed from above). Now that we have a parametrization, we can set up the integral. ⇀

t = 0,

π 2

⇀

r (0) = (2, 0, 0).

⇀

– – ) = (0, √2, √2).

4.4.13

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– – r (t) = (2 cos t, √2 sin t, √2 sin t)

⇀

⇀ ′

r

⇀

– – (t) = ( − 2 sin t, √2 cos t, √2 cos t)

⇀

F( r (t))

= (z(t) − y(t), −x(t) − z(t), −x(t) − y(t)) – – – – = (√2 sin t − √2 sin t, −2 cos t − √2 sin t, −2 cos t − √2 sin t) – – = −(0, 2 cos t + √2 sin t, 2 cos t + √2 sin t)

⇀

⇀

⇀ ′

F( r (t)) ⋅ r

– 2 (t) = −[4 √2 cos t + 4 cos t sin t] – – = −[2 √2 cos(2t) + 2 √2 + 2 sin(2t)]

by the double angle formulae sin(2t) = 2 sin t ⇀

∮

cos t 2π

⇀

F ⋅ dr

C

=∫

and cos(2t) = 2 cos

⇀

2

⇀

⇀ ′

F( r (t)) ⋅ r

So

t − 1.

(t) dt

0 2π

=∫

– – −[2 √2 cos(2t) + 2 √2 + 2 sin(2t)] dt

0 2π

– – = −[√2 sin(2t) + 2 √2t − cos(2t)]

0

– = −4 √2π

Oof! Let's do it an easier way. Solution 2 Stokes' Theorem: To apply Stokes' theorem we need to express C as the boundary ∂S of a surface S. As 2

C = {(x, y, z)| x

+y

2

+z

2

= 4,  z = y}

is a closed curve, this is possible. In fact there are many possible choices of S with ∂S = C . Three possible S 's (sketched below) are S S S

′

′′

2

= {(x, y, z)| x

2

= {(x, y, z)| x

2

= {(x, y, z)| x

+y +y +y

2

2

2

+z +z +z

2

2

2

≤ 4,  z = y} = 4,  z ≥ y} = 4,  z ≤ y}

The first of these, which is part of a plane, is likely to lead to simpler computations than the last two, which are parts of a sphere. So we choose what looks like the simpler way. ⇀

⇀

^ dS. For the latter, we apply the formula In preparation for application of Stokes' theorem, we compute ∇ × F and n ^ dS = ±(−f , −f , 1) dxdy (of Equation 3.3.2) to the surface z = f (x, y) = y. We use the + sign to give the normal a n ^ positive k component. x

y

4.4.14

https://math.libretexts.org/@go/page/91912

^ ıı

^ ȷ ȷ

^ k

∂

∂

∂

∂x

∂y

∂z

z−y

−x − z

−x − y

⎡ ⇀

⇀

∇ × F = det ⎢ ⎢ ⎣

⎤ ⎥ ⎥ ⎦

^ = ^ ıı ( − 1 − (−1)) − ^ ȷ ȷ ( − 1 − 1) + k( − 1 − (−1)) =2^ ȷ ȷ ^ dS = (0, −1, 1) dxdy n ⇀

⇀

^ dS ∇×F⋅ n

= (0, 2, 0) ⋅ (0, −1, 1) dxdy = −2 dxdy

The integration variables are x and y and, by definition, the domain of integration is R = {(x, y)|(x, y, z) is in S for some z}

To determine precisely what this domain of integration is, we observe that since z = y on S, x as x + 2y ≤ 4 on S,

2

2

+y

2

+z

2

≤4

is the same

2

2

S = {(x, y, z)| x

+ 2y

2

≤ 4,  z = y}

⟹

2

R = {(x, y)| x

+ 2y

2

≤ 4} –

–

So the domain of integration is an ellipse with semimajor axis a = 2, semiminor axis b = √2 and area πab = 2√2π. The integral is then ⇀

∮

⇀

⇀

C

⇀

– (−2) dxdy = −2 Area (R) = −4 √2π

^ dS = ∬ ∇×F⋅ n

F ⋅ dr = ∬ S

R

Remark (Limits of integration): If the integrand were more complicated, we would have to evaluate the integral over R by expressing it as an iterated integrals with the correct limits of integration. First suppose that we slice up R using thin vertical slices. On each such −−−−−−− − −−−−−−− − slice, x is essentially constant and y runs from −√(4 − x )/2 to √(4 − x )/2 . The leftmost such slice would have x = −2 and the rightmost such slice would have x = 2. So the correct limits with this slicing are 2

2

2

2

∬

f (x, y) dxdy = ∫

R

√(4−x )/2

dx ∫

dy f (x, y) 2

−2

−√(4−x )/2

If, instead, we slice up R using thin horizontal slices, then, on each such slice, y is essentially constant and x runs from − − −−− − − − −−− − – – −√4 − 2y to √4 − 2y . The bottom such slice would have y = −√2 and the top such slice would have y = √2. So the correct limits with this slicing are 2

2

√4−2y 2

√2

∬

f (x, y) dxdy = ∫

R

dy ∫

−√2

−√4−2y

dx f (x, y) 2

Note that the integral with limits √2

∫ −√2

2

dy ∫

dx f (x, y)

−2

4.4.15

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corresponds to a slicing with x running from integration, not what we have here.

to

−2

on {\bf every} slice. This corresponds to a rectangular domain of

2

Stokes' Theorem, Again: Since the integrand is just a constant (after Stoking — not the original integrand) and ⇀

wisely), we can evaluate the integral ∬

S

⇀

⇀

any limits of integration. We already know that ⇀

^ ∇(z − y) = −^ ȷ ȷ +k

^ = is normal to S. So n

⇀

∮ C

Since

is the level surface

S

1

⇀

^ dS = ∬ ∇×F⋅ n

=∬

is so simple (because we chose it z − y = 0,

the gradient

and

√2

⇀

⇀

F ⋅ dr

⇀

∇×F = 2 ^ ȷ ȷ.

^ (−^ ȷ ȷ + k)

1

S

without ever determining dS explicitly and without ever setting up

^ dS ∇×F⋅ n

S

(2 ^ ȷ ȷ) ⋅

S

– √2

^ (−^ ȷ ȷ + k) dS

– – −√2 dS = −√2 Area (S)

=∬ S

As S is a circle of radius 2, ∮

C

⇀

– ⇀ F ⋅ d r = −4 √2π,

yet again.

 Example 4.4.8 ⇀

In Warning 4.1.17, we stated that if a vector field fails to pass the screening test ∇ ⋅ B = 0 at even a single point, for example because the vector field is not defined at that point, then B can fail to have a vector potential. An example is the point source ^ r(x, y, z) B(x, y, z) =

2

r(x, y, z)

of Example 3.4.2. Here, as usual, − −−−−−−−− − 2

r(x, y, z) = √ x

+y

2

+z

2

^ r(x, y, z) =

^ x^ ı ı + y^ ȷ ȷ + zk − −−−−−−−− − 2 2 2 √x +y +z

This vector field is defined on all of R , except for the origin, and its divergence 3

∂

⇀

x

∇⋅B =

( ∂x

2

(x

+y

2

∂ 2

3/2

y

)+

( ∂y

+z )

∂

2

(x

+y

(

2

1

3x −

)

(x2 + y 2 + z 2 )3/2

(x2 + y 2 + z 2 )5/2 1

3y

+(

) (x2 + y 2 + z 2 )5/2

1

3z

+( 2

+y

2

2

3/2

+z )

3(x

+y

2

2

−

)

3/2

2

+z ) 2

3 2

2

− (x

2

2

− (x2 + y 2 + z 2 )3/2

+y

)

)

=(

2

3/2

(x2 + y 2 + z 2 )3/2

∂z

(x

2

+z )

z

+

=

2

(x

+y

2

(x

+y

2

2

5/2

+z )

2

+z ) 2

5/2

+z )

is zero everywhere except at the origin, where it is not defined. This 3

R

vector

field

cannot

have

a

vector potential on its domain of definition, i.e. on To see this, suppose to the contrary that it did have a vector potential A.

∖ {(0, 0, 0)} = {(x, y, z)|(x, y, z) ≠ (0, 0, 0)} .

4.4.16

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Then its flux through any closed surface 5 (i.e. surface without a boundary) S would be ⇀

∬

^ dS = ∬ B⋅n

S

⇀

^ dS = ∮ ∇×A ⋅ n

S

A ⋅ dr = 0

∂S

by Stokes' theorem, since ∂S is empty. But we found in Example 3.4.2, with centred on the origin is 4π.

m = 1,

that the flux of

B

through any sphere

The Interpretation of Div and Curl Revisited In sections 4.1.4 and 4.1.5 we derived interpretations of the divergence and of the curl. Now that we have the divergence theorem and Stokes' theorem, we can simplify those derivations a lot.

Divergence Let ε > 0 be a tiny positive number, and then let 2

2

Bε (x0 , y0 , z0 ) = {(x, y, z)|(x − x0 )

be a tiny ball of radius ε centred on the point (x

0,

y0 , z0 ).

+ (y − y0 )

2

⇀

ε (x0 ,

y0 , z0 )

2

2

+ (y − y0 )

+ (z − z0 )

is really small, ∇ ⋅ v is essentially constant in B ⇀

∭

⇀

ε (x0 ,

⇀

⇀

2

0

ε

(x0 , y0 , z0 )

ε

0

0

0

0

0

0

ε

ε

ε

ε

Because D

ε (x0 ,

⇀

y0 , z0 )

ε

is really small, ∇× v is essentially constant on D ⇀

⇀

ε (x0 ,

⇀

⇀

^  dS ∇× v ⋅ n

∬

y0 , z0 )

and we essentially have

⇀

^  Area(Dε (x0 , y0 , z0 )) = ∇× v (x0 , y0 , z0 ) ⋅ n

Dε ( x0 , y0 , z0 ) 2

⇀

⇀

^ = π ε  ∇ ×v (x0 , y0 , z0 ) ⋅ n

Again, this is really an approximate statement which gets better and better as ε → 0. A more precise statement is ⇀

∬

⇀

⇀

^ = lim ∇ ×v (x0 , y0 , z0 ) ⋅ n

Dε ( x0 , y0 , z0 )

πε

ε→0

⇀

^  dS ∇× v ⋅ n 2

By Stokes' theorem, we also have ⇀

⇀

Dε ( x0 , y0 , z0 )

Again, think of the vector field

⇀

v

⇀

⇀

^  dS = ∮ ∇× v ⋅ n

∬

v ⋅ dr

Cε ( x0 , y0 , z0 )

as the velocity of a moving fluid. Then ∮

⇀

⇀

Cε

v ⋅ dr

is called the circulation of

⇀

v

around C

ε.

To measure the circulation experimentally, place a small paddle wheel in the fluid, with the axle of the paddle wheel pointing along n ^ and each of the paddles perpendicular to C and centred on C . ε

ε

Each paddle moves tangentially to C . It would like to move with the same speed as the tangential speed v ⋅ ^t (where ^t is the forward pointing unit tangent vector to C at the location of the paddle) of the fluid at its location. But all paddles are rigidly fixed to the axle of the paddle wheel and so must all move with the same speed. That common speed will be the average value of v ⋅ ^t around C . If ds represents an element of arc length of C , the average value of v ⋅ ^t around C is ⇀

ε

ε

⇀

⇀

ε

ε

¯¯¯¯ ¯ vT

1 =

∮ 2πε

ε

⇀

v ⋅^ t  ds =

1 ∮ 2πε

Cε

⇀

⇀

v ⋅ dr

Cε

since d r has direction ^t and length ds so that d r = ^tds, and since 2πε is the circumference of C . If the paddle wheel rotates at Ω radians per unit time, each paddle travels a distance Ωε per unit time (remember that ε is the radius of C ). That is, v = Ωε. Combining all this information, ⇀

⇀

ε

ε

4.4.18

¯¯¯¯ ¯ T

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⇀ ⇀

∬

^ = lim ∇ ×v (x0 , y0 , z0 ) ⋅ n

πε2

ε→0

⇀

∮

Cε

= lim ε→0

⇀

v ⋅ dr

πε

= lim

⇀

∇× v ⋅ n ^  dS

Dε ( x0 , y0 , z0 )

⇀

2

¯¯¯¯ ¯ 2πε v T

πε

ε→0

2

2πε (Ωε) = lim ε→0

πε

2

= 2Ω

so that 1 Ω = 2 ⇀

⇀

⇀

^ ∇ ×v (x0 , y0 , z0 ) ⋅ n

^ is twice the rate at which the paddle wheel turns when it is put into the The component of ∇ ×v (x , y , z ) in any direction n ^ . The direction of ∇ ×v (x , y , z ) is the axle direction which gives fluid at (x , y , z ) with its axle pointing in the direction n maximum rate of rotation and the magnitude of ∇ ×v (x , y , z ) is twice that maximum rate of rotation. For this reason, ∇ × v is called the “vorticity”. ⇀

0

0

0

⇀

0

0

⇀

0

0

⇀

0

0

⇀

⇀

0

0

⇀

0

Optional — An Application of Stokes' Theorem — Faraday's Law Magnetic induction refers to a physical process whereby an electric voltage is created (“induced”) by a time varying magnetic field. This process is exploited in many applications, including electric generators, induction motors, induction cooking, induction welding and inductive charging. Michael Faraday 6 is generally credited with the discovery of magnetic induction. Faraday's law is the following. Let S be an oriented surface with boundary C . Let E and B be the (time dependent) electric and magnetic fields and define ⇀

∮

E ⋅ d r = voltage around C

C

∬

^ dS = magnetic flux through S B⋅n

S

Then the voltage around C is the negative of the rate of change of the magnetic flux through S. As an equation, Faraday's Law is ∂

⇀

∮

E ⋅ dr = −

∬ ∂t

C

^ dS B⋅n

S

We can reformulate this as a partial differential equation. By Stokes' Theorem, ⇀

⇀

∮

^ dS (∇ × E) ⋅ n

E ⋅ dr = ∬

C

S

so Faraday's law becomes ∂B

⇀

∬

(∇ × E +

^ dS = 0 )⋅n

∂t

S

This is true for all surfaces S. So the integrand, assuming that it is continuous, must be zero. ⇀

To see this, let G = (∇ × E +

∂B ∂t

).

Suppose that

G(x0 ) ≠ 0.

Pick a unit vector n ^ in the direction of

^ (the vector we picked). Then G(x small flat disk centered on x with normal n

0)

0

4.4.19

^ >0 ⋅n

G(x0 ).

Let S be a very

^ > 0 for all and, by continuity, G(x) ⋅ n

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x

on

S,

if we have picked

S

small enough. Then

⇀

∬

S

(∇ × E +

∂B ∂t

^ dS > 0, )⋅n

which is a contradiction. So

⇀

G = 0

everywhere and we conclude that ∂B

⇀

∇×E+

=0 ∂t

This is one of Maxwell's electromagnetic field equations 7.

Exercises Stage 1  1 Each of the figures below contains a sketch of a surface S and its boundary ∂S. Stokes' theorem says that ^ dS ^ is a correctly oriented unit normal vector to S. Add to each sketch a typical such normal ∮ F ⋅ dr = ∬ ∇ × F ⋅ n if n vector. ⇀

⇀

⇀

∂S

⇀

S

(b)

(a)

(c)

 2 Let be a finite region in the xy-plane, the boundary, C , of R consist of a single piecewise smooth, simple closed curve R

that is oriented (i.e. an arrow is put on C ) consistently with R in the sense that if you walk along C in the direction of the arrow, then R is on your left

F1 (x, y)

and F

2 (x,

y)

have continuous first partial derivatives at every point of R.

Use Stokes' theorem to show that ∮

[ F1 (x, y) dx + F2 (x, y) dy] = ∬

C

R

(

∂F2 ∂x

−

∂F1

) dxdy

∂y

i.e. to show Green's theorem.

 3 ⇀

⇀

Verify the identity   ∮ ϕ∇ψ ⋅ d r = − ∮ ψ∇ϕ ⋅ d r   for any continuously differentiable scalar fields that is the boundary of a piecewise smooth surface. C

⇀

⇀

C

4.4.20

ϕ

and

ψ

and curve

C

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Stage 2  4 Let C be the curve of intersection of the cylinder x + y direction as seen from (0, 0, 100). Let F = (x − y , y + x , 2

⇀

2

2

2

=1

and the surface z = y Calculate ∮ F ⋅ d r ⇀

1).

oriented in the counterclockwise

2

⇀

C

1. by direct evaluation 2. by using Stokes' Theorem.

 5 Evaluate ∮

C

⇀

⇀

where F = y e

⇀

F ⋅ dr

x

x

^ ı ı + (x + e ) ^ ȷ ȷ +z

2

and C is the curve

^ k

^ r (t) = (1 + cos t) ^ ı ı + (1 + sin t) ^ ȷ ȷ + (1 − sin t − cos t) k

⇀

0 ≤ t ≤ 2π

 6✳ Find the value of ∬

⇀

S

⇀

^ dS ∇×F⋅ n

⇀

where F = (z − y ,

x , −x) 3

2

{(x, y, z) ∈ R | x

+y

and S is the hemisphere 2

+z

2

= 4,  z ≥ 0}

oriented so the surface normals point away from the centre of the hemisphere.

 7✳ Let S be the part of the surface z = 16 − (x

2

2

2

+y )

⇀

which lies above the xy-plane. Let F be the vector field

⇀

^ F = x ln(1 + z) ^ ı ı + x(3 + y) ^ ȷ ȷ + y cos z k

Calculate ⇀

∬

⇀

^ dS ∇×F⋅ n

S

^ is the upward normal on S. where n

 8✳ Let C be the intersection of the paraboloid z = 4 − x − y with the cylinder x + (y − 1) ^ when viewed from high on the z -axis. Let F = xz ^ ı ı +x ^ ȷ ȷ + yz k. Find ∮ F ⋅ d r . 2

2

2

⇀

⇀

2

= 1,

oriented counterclockwise

⇀

C

 9 ⇀

^ Let F = −y e ^ ı ı + x cos z ^ ȷ ȷ + z sin(xy) k, and let S be the part of the surface z = (1 − x square −1 ≤ x ≤ 1, −1 ≤ y ≤ 1 in the xy-plane. Find the flux of ∇ × F upward through S. z

3

2

⇀

2

)(1 − y )

that lies above the

⇀

 10 ⇀

⇀

Evaluate the integral ∮ F ⋅ d r , in which F = (e (0, 1, 0) to (0, 0, 1) to (1, 0, 0). C

⇀

2

x

− yz , sin y − yz , xz + 2y)

4.4.21

and C is the triangular path from (1, 0, 0) to

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 11 ✳ Let

⇀

^ F(x, y, z) = −z ^ ı ı +x ^ ȷ ȷ +y k

be a vector field. Use Stokes' theorem to evaluate the line integral

the intersection of the plane z = y and the ellipsoid the z -axis.

2

x

+

4

y

2

+

2

z

2

2

= 1,

⇀

∮

C

⇀

F ⋅ dr

where

C

is

oriented counter-clockwise when viewed from high on

 12 ✳ ⇀

Consider the vector field F(x, y, z) = z

2

2

^ ı ı +x

⇀

^ ȷ ȷ +y

2

in R

^ k

3

.

1. Compute the line integral I = ∫ F ⋅ d r where C is the curve consisting of three line segments, L from (2, 0, 0) to (0, 2, 0), then L from (0, 2, 0) to (0, 0, 2), finally L from (0, 0, 2) to (2, 0, 0). 2. A simple closed curve C lies on the plane E : x + y + z = 2, enclosing a region R on the plane of area 3, and oriented in a counterclockwise direction as observed from the positive x-axis. Compute the line integral I = ∫ F ⋅ d r . 1

⇀

1

C1

2

1

3

2

⇀

2

⇀

C2

 13 ✳ Let C

= C1 + C2 + C3

be the curve given by the union of the three parameterized curves ⇀

r 1 (t) = (2 cos t, 2 sin t, 0),

0 ≤ t ≤ π/2

⇀ r 2 (t)

= (0, 2 cos t, 2 sin t),

0 ≤ t ≤ π/2

⇀ r 3 (t)

= (2 sin t, 0, 2 cos t),

0 ≤ t ≤ π/2

1. Draw a picture of C . Clearly mark each of the curves C , C , and C and indicate the orientations given by the parameterizations. 2. Find and parameterize an oriented surface S whose boundary is C (with the given orientations). 3. Compute the line integral ∫ F ⋅ d r where 1

⇀

2

3

⇀

C

⇀

2

2

F = (y + sin(x ) , z − 3x + ln(1 + y ) , y + e

z

2

)

 14 ✳ − −− −− −

We consider the cone with equation z = √x + y . Note that its tip, or vertex, is located at the origin (0, 0, 0). The cone is oriented in such a way that the normal vectors point downwards (and away from the z axis). In the parts below, both S and S are oriented this way. 2

2

1

2

⇀

Let F = ( − zy, zx, xy cos(yz)). 1. Let S be the part of the cone that lies between the planes z = 0 and z = 4. Note that S does not include any part of the plane z = 4. Use Stokes' theorem to determine the value of 1

1

⇀

⇀

^ dS ∇×F⋅ n

∬ S1

Make a sketch indicating the orientations of S and of the contour(s) of integration. 2. Let S be the part of the cone that lies below the plane z = 4 and above z = 1. Note that S does not include any part of 1

2

2

⇀

⇀

the planes z = 1 and z = 4. Determine the flux of ∇ × F across S . Justify your answer, including a sketch indicating the orientations of S and of the contour(s) of integration. 2

2

 15 ✳ Consider the curve C that is the intersection of the plane z = x + 4 and the cylinder x so that it is traversed clockwise as seen from above.

2

⇀

Let F(x, y, z) = (x

3

+y

2

= 4,

and suppose

C

is oriented

2

+ 2y , sin(y) + z , x + sin(z )).

4.4.22

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Use Stokes' Theorem to evaluate the line integral ∮

⇀

C

⇀

F ⋅ dr .

 16 ✳ ⇀

⇀

1. Consider the vector field F(x, y, z) = (z , x , y ) in R . Compute the line integral ∮ F ⋅ d r , where C is the curve consisting of the three line segments, L from (2, 0, 0) to (0, 2, 0), then L from (0, 2, 0) to (0, 0, 2), and finally L from (0, 0, 2) to (2, 0, 0). 2. A simple closed curve C lies in the plane x + y + z = 2. The surface this curve C surrounds inside the plane x + y + z = 2 has area 3. The curve C is oriented in a counterclockwise direction as observed from the positive x-axis. Compute the line integral ∮ F ⋅ d r , where F is as in (a). 2

2

3

2

⇀

C

1

⇀

2

3

⇀

⇀

C

 17 ✳ Evaluate the line integral 1 ∫

(z +

x ) dx + xz dy + (3xy −

1 +z

C

(z + 1)2

) dz

where C is the curve parameterized by ⇀

2

r (t) = ( cos t , sin t , 1 − cos

t sin t)

0 ≤ t ≤ 2π

 18 ✳ A simple closed curve C lies in the plane x + y + z = 1. The surface this curve C surrounds inside the plane x + y + z = 1 has area 5. The curve C is oriented in a clockwise direction as observed from the positive z -axis looking down at the plane x + y + z = 1. ⇀

Compute the line integral of F(x, y, z) = (z

2

2

2

,x ,y )

around C .

 19 ✳ Let C be the oriented curve consisting of the 5 line segments which form the paths from (0, 0, 0) to (0, 1, 1), from (0, 1, 1) to (0, 1, 2), from (0, 1, 2) to (0, 2, 0), from (0, 2, 0) to (2, 2, 0), and from (2, 2, 0) to (0, 0, 0). Let ⇀

F = (−y + e

Evaluate the integral ∫

C

⇀

x

sin x) ^ ı ı +y

4

^ ^ ȷ ȷ + √z tan z k

⇀

F ⋅ dr .

 20 ✳ Suppose the curve C is the intersection of the cylinder x + y = 1 with the surface z = x y from the positive z-axis, i.e. viewed “from above”. Evaluate the line integral 2

∫

2

3

(z + sin z) dx + (x

2

,

traversed clockwise if viewed

2

− x y) dy + (x cos z − y) dz

C

 21 ✳ Evaluate ∬ normal, and

S

⇀

⇀

^ dS ∇×F⋅ n

where S is that part of the sphere x

2

⇀

F(x, y, z) = −y

2

3

^ ı ı +x

+y

2

+z

^ ȷ ȷ + (e

4.4.23

x

2

=2

+e

y

above the plane z = 1,

^ n

is the upward unit

^ + z) k

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 22 ✳ Let ⇀

F = x sin y ^ ı ı − y sin x ^ ȷ ȷ + (x − y)z

2

^ k

Use Stokes' theorem to evaluate ⇀

∫

⇀

F ⋅ dr

C

along the path consisting of the straight line segments successively joining the points P = (0, 0, 0) to P = (π/2, 0, 1) to P = (0, 0, 1) to P = (0, π/2, 1) to P = (0, π/2, 0), and back to (0, 0, 0). 0

2

3

4

P1 = (π/2, 0, 0)

to

5

 23 ✳ Let 2z

⇀

3z

2

F =(

2

+ sin(x ) , 1 +y

+ sin(y ) , 5(x + 1)(y + 2)) 1 +x

Let C be the oriented curve consisting of four line segments from (0, 0, 0) to (2, 0, 0), from (2, 0, 0) to (0, 0, 2), from (0, 0, 2) to (0, 3, 0), and from (0, 3, 0) to (0, 0, 0). 1. Draw a picture of C . Clearly indicate the orientation on each line segment. 2. Compute the work integral ∫ F ⋅ d r . ⇀

⇀

C

 24 ✳ Evaluate

⇀

∬

⇀

∇×F⋅ n ^ dS

where

⇀

and

^ F =y ^ ı ı + 2z ^ ȷ ȷ + 3x k

S

is the surface

− −−−−−−− − z = √1 − x2 − y 2 ,

z ≥0

and

n ^

is a unit

S

^ ^ ⋅k normal to S obeying n ≥ 0.

 25 ✳ Let S be the curved surface below, oriented by the outward normal: 2

x

+y

2

2

+ 2(z − 1 )

= 6,

z ≥ 0.

^ (E.g., at the high point of the surface, the unit normal is k .)

Define ⇀

G = ∇ × F,

Find ∬

S

⇀

where

F = (xz − y

3

3

cos z) ^ ı ı +x e

z

2

^ ȷ ȷ + xyze

x +y

2

+z

2

^ k.

^ dS. G⋅n

 26 ✳ Let C be a circle of radius R lying in the plane x + y + z = 3. Use Stokes' Theorem to calculate the value of ⇀

∮

⇀

F ⋅ dr

C ⇀

where F = z

2

2 2^ ^ ı ı +x ^ ȷ ȷ + y k.

(You may use either orientation of the circle.)

4.4.24

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 27 Let

S

be the oriented surface consisting of the top and four sides of the cube whose vertices are ⇀

outward. If F(x, y, z) = (xyz, x y

2

⇀

2

, x yz),

(±1, ±1, ±1),

oriented

⇀

find the flux of ∇ × F through S.

 28 Let S denote the part of the spiral ramp (that is helicoidal surface) parametrized by x = u cos v,  y = u sin v,  z = v

0 ≤ u ≤ 1,  0 ≤ v ≤ 2π

Let C denote the boundary of S with orientation specified by the upward pointing normal on S. Find ∫

y dx − x dy + xy dz

C

Stage 3  29 Let C be the intersection of x + 2y − z = 7 and viewed from high on the z -axis. Let ⇀

2

F = (e

Evaluate ∮

C

⇀

x

2

x

− 2x + 4 y

2

The curve

= 15.

C

is oriented counterclockwise when

2 2 2 ^ + yz) ^ ı ı + ( cos(y ) − x ) ^ ȷ ȷ + ( sin(z ) + xy) k

⇀

F ⋅ dr .

 30 ✳ ⇀

1. Find the curl of the vector field F = (2 + x + z , 0 , 3 + x z). 2. Let C be the curve in R from the point (0, 0, 0) to the point (2, 0, 0), consisting of three consecutive line segments connecting the points (0, 0, 0) to (0, 0, 3), (0, 0, 3) to (0, 1, 0), and (0, 1, 0) to (2, 0, 0). Evaluate the line integral 2

2

3

⇀

∫

⇀

F ⋅ dr

C ⇀

where F is the vector field from (a).

 31 ✳ 1. Let S be the bucket shaped surface consisting of the cylindrical surface y + z = 9 between x = 0 and x = 5, and the disc inside the yz-plane of radius 3 centered at the origin. (The bucket S has a bottom, but no lid.) Orient S in such a way 2

2

⇀

that the unit normal points outward. Compute the flux of the vector field ∇ × G through S, where G = (x, −z, y). 2. Compute the flux of the vector field F = (2 + z, x z , x cos y) through S, where S is as in (a). ⇀

2

 32 ✳ Let ⇀

y

F(x, y, z) = (

2

1+x

+x

2

, x

−y

1+y

2

5

, cos (ln z))

x ⇀

1. Write down the domain D of F. 2. Circle the correct statement(s): 1. D is connected. 2. D is simply connected. 3. D is disconnected.

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⇀

⇀

3. Compute ∇ × F. 4. Let C be the square with corners (3 ± 1, 3 ± 1) in the plane z = 2, oriented clockwise (viewed from above, i.e. down z axis). Compute ⇀

∫

⇀

F ⋅ dr

C ⇀

5. Is F conservative?

 33 ✳ ⇀

⇀

A physicist studies a vector field F(x, y, z). From experiments, it is known that F is of the form ⇀

y

F(x, y, z) = xz ^ ı ı + (ax e z + byz) ^ ȷ ȷ + (y ⇀

2

y 2 ^ − xe z ) k

⇀

for some real numbers a and b. It is further known that F = ∇ × G for some differentiable vector field G. 1. Determine a and b. 2. Evaluate the surface integral ⇀

^ dS F⋅n

∬ S

where S is the part of the ellipsoid x component.

2

+y

2

+

1 4

z

2

=1

for which z ≥ 0, oriented so that its normal vector has a positive z -

 34 ✳ Let C be the curve in the xy-plane from the point (0, 0) to the point (5, 5) consisting of the ten line segments consecutively connecting the points (0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3), (3, 3), (3, 4), (4, 4), (4, 5), (5, 5). Evaluate the line integral ⇀

∫

⇀

F ⋅ dr

C

where ⇀

F =y ^ ı ı + (2x − 10) ^ ȷ ȷ

 35 ✳ ⇀

⇀

Let F = ( sin x , xz , z ). Evaluate ∮ F ⋅ d r around the curve C of intersection of the cylinder surface z = x , traversed counter clockwise as viewed from high on the z -axis. 2

2

⇀

C

2

x

+y

2

=4

with the

2

 36 ✳ Explain how one deduces the differential form 1 ∂H

⇀

∇×E = − c

∂t

of Faraday's law from its integral form ∮ C

⇀

1

E ⋅ dr = −

d  

c

∬ dt

4.4.26

^ dS H⋅n

S

https://math.libretexts.org/@go/page/91912

 37 ✳ Let C be the curve given by the parametric equations: – x = cos t,  y = √2 sin t,  z = cos t,  0 ≤ t ≤ 2π

and let ⇀

3

F =z ^ ı ı +x ^ ȷ ȷ +y z

3

^ k

Use Stokes' theorem to evaluate ⇀

∮

⇀

F ⋅ dr

C

 38 ✳ Use Stokes' theorem to evaluate ∮

z dx + x dy − y dz

C

where C is the closed curve which is the intersection of the plane x + y + z = 1 with the sphere x that C is oriented clockwise as viewed from the origin.

2

+y

2

+z

2

= 1.

Assume

 39 ✳ Let S be the part of the half cone − −− −− − 2

z = √x

+y

2

,

y ≥ 0,

that lies below the plane z = 1. 1. Find a parametrization for S. 2. Calculate the flux of the velocity field ^ v =x ^ ı ı +y ^ ȷ ȷ − 2z k

⇀

downward through S. ^ 3. A vector field F has curl ∇ × F = x ^ ı ı +y ^ ȷ ȷ − 2z k. On the xz-plane, the vector field F is constant with F(x, 0, z) = ^ȷȷ . Given this information, calculate ⇀

⇀

⇀

⇀

⇀

∫

⇀

⇀

F ⋅ dr ,

C

where C is the half circle 2

x

+y

2

= 1,  z = 1,  y ≥ 0

oriented from (−1, 0, 1) to (1, 0, 1).

 40 ⇀

⇀

^ dS ^ is the Consider ∬ (∇ × F) ⋅ n where S is the portion of the sphere x + y + z = 1 that obeys x + y + z ≥ 1, n ^ upward pointing normal to the sphere and F = (y − z) ^ ı ı + (z − x)^ ȷ ȷ + (x − y)k. Find another surface S with the property 2

2

2

S

⇀

that ∬

S

⇀

⇀

⇀

^ dS = ∬ (∇ × F) ⋅ n

S

′

⇀

^ dS (∇ × F) ⋅ n

′

and evaluate ∬

S

⇀ ′

⇀

^ dS. (∇ × F) ⋅ n

1. Sir George Gabriel Stokes (1819–1903) was an Irish physicist and mathematician. In addition to Stokes' theorem, he is known for the Navier-Stokes equations of fluid dynamics and for his work on the wave theory of light. He gave evidence to the Royal

4.4.27

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Commission on the Use of Iron in Railway Structures after the Dee bridge disaster of 1847. 2. Otherwise, decompose S into simpler pieces, analogously to what we did in the proof of the divergence theorem. 3. By Salvador Dali? 4. That way lies pain. 5. If you are uncomfortable with the surface not having a boundary, poke a very small hole in the surface, giving it a very small boundary. Then take the limit as the hole tends to zero. 6. Michael Faraday (1791–1867) was an English physicist and chemist. He ended up being an extremely influential scientist despite having only the most basic of formal educations. 7. For the others, see Example 4.1.2. This page titled 4.4: Stokes' Theorem is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

4.4.28

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4.5: Optional — Which Vector Fields Obey ∇ × F = 0 ⇀

⇀

⇀

We already know that if a vector field F passes the screening test ∇ × F = 0 on all of R or R , then there is a function φ with 1 F = ∇φ. That is, F is conservative. We are now going to take a first look at what happens when F passes the screening test ∇ × F = 0 only on some proper subset D of R , n = 2 or 3. We will just scratch the surface of this topic — there is a whole subbranch of Mathematics (cohomology theory, which is part of algebraic topology) concerned with a general form of this ⇀

⇀

⇀

⇀

2

3

⇀

⇀

n

⇀

question. We shall imagine that we are given a vector field F that is only defined on D and we shall assume that D is a connected, open subset of R with n = 2 or n = 3 (see Definition 4.5.1, below) that all first order derivatives of all vector fields and functions that we consider are continuous and that all curves we consider are piecewise smooth. A curve is piecewise smooth if it is a union of a finite number of smooth curves C , C , ⋯ , C with the end point of C being the beginning point of C for each 1 ≤ i < m. A curve is smooth 2 if it has a parametrization r (t), a ≤ t ≤ b, whose first derivative r (t) exists, is continuous and is nonzero everywhere. n

1

2

m

i

i+1

⇀′

⇀

 Definition 4.5.1 Let n ≥ 1 be an integer. 1. Let a ∈ R and ε > 0. The open ball of radius ε centred on a is n

n

Bε (a) = { x ∈ R  ∣ ∣ |x − a| < ε }

Note the strict inequality in |x − a| < ε. 2. A subset O ⊂ R is said to be an “open subset of R ” if, for each point a ∈ O, there is an ε > 0 such that B (a) ⊂ O. Equivalently, O is open if and only if it is a union of open balls. 3. A subset D ⊂ R is said to be (pathwise) connected if every pair of points in D can be joined by a piecewise smooth curve in D. n

n

ε

n

Here are some examples to help explain this definition.

 Example 4.5.2 1. The open rectangle O = { (x, y) ∈ R  ∣∣ 0 < x < 1,  0 < y < 1 } is an open subset of R because any point a = (x , y ) ∈ O is a nonzero distance, namely d = min { x , 1 − x , y , 1 − y } away from the boundary of O. So the open ball B (a) is contained in O. 2. The closed rectangle C = { (x, y) ∈ R  ∣∣ 0 ≤ x ≤ 1,  0 ≤ y ≤ 1 } is not an open subset of R . For example, 0 = (0, 0) 2

0

2

0

0

0

0

0

d/2

2

⇀

2

is a point in C. No matter what ε > 0 we pick, the open ball B (− , 0), which is not in C.

⇀

ε (0)

is not contained in C because B

⇀

ε (0)

contains the point

ε

2

3. The x-axis, X = { (x, y) ∈ R  ∣∣ y = 0 }, in R is not an open subset of R because for any point (x ε > 0, the ball B ((x , 0)) contains points with nonzero y -coordinates and so is not contained in X . 4. The union of open balls 2

ε

2

2

0,

0) ∈ X

and any

0

B1 ((0, 0)) ∪ B1 ((2, 0)) 2

2

= {(x, y) ∈ R ∣ ∣x

+y

2

2

< 1 or (x−2 )

+y

2

< 1}

is not connected, since any continuous path from, for example, (2, 0) to (0, 0) must leave the union. In the figure on the left below, an “empty disk” has been sketched at (1, 0) just to emphasise that the point (1, 0) is not in the union.

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5. On the other hand the union of “closed balls” 2

2

{(x, y) ∈ R ∣ ∣x

+y

2

2

≤ 1 or (x − 2 )

+y

2

≤ 1}

is connected. For example, the straight line segment from (2, 0) to (0, 0) remains in the the union.

Many, but not all, of the basic facts that we developed, in §2.4.1, about conservative fields in proofs) to fields on D.

also applies (with the same

n

R

 Theorem 4.5.3 ⇀

For a vector field F on D ⊂ R

n

,

⇀

⇀

F is conservative on D

⇀

⟺

F = ∇φ on D, for some function φ

⟺

for each P0 , P1 ∈ D, the integral  ∫

⇀

⇀

F ⋅ d r  takes

C

the same value for all curves C from P0  to P1 ⇀

⟺

∮

⇀

F ⋅ d r = 0  for all closed curves C in D

C ⇀

⇀

  ⟹ ∇ × F = 0  on D

Note that the last line of this theorem contains only a one way implication. Combining this with Stokes' Theorem 4.4.1 (when consequences.

n = 3,

or Green's Theorem 4.3.2 when

n =2

) gives us the following two

 Theorem 4.5.4 1. If D has the property that every closed curve C in D is the boundary of a bounded oriented surface, S, in D

then

⇀

⇀

⇀

F is conservative on D ⟺

∇ × F = 0 on D

⇀

⇀

⇀

2. For any D, if ∇ × F = 0 on D, then F is locally conservative. This means that for each point x and a function φ such that F = ∇φ on B (x ).

0

⇀

∈ D,

there is an ε > 0

⇀

ε

0

Proof ⇀

⇀

(a) This is simply because if ∇ × F = 0 on D and if the curve C = ∂S, with S an oriented surface in D, then Stokes' theorem gives ⇀

∫ C

⇀

⇀

F ⋅ dr = ∫

⇀

⇀

F ⋅ dr = ∬

∂S

⇀

^  dS = 0 ∇×F⋅ n

S

⇀

So F is conservative by Theorem 4.5.3. (b) This is true simply because B

ε (x0 )

satisfies property (H).

4.5.2

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 Example 4.5.5 Here are some examples of D's that violate (H). When D = D = { (x, y) ∈ R  ∣∣ 0 < |(x, y)| < 3} (an open ball with its centre removed), then the circle x + y = 4 is a curve in D that is not the boundary of a surface in D. The circle x + y = 4 is the boundary of the disk x + y < 4, but the disk x + y < 4 is not contained in D because the point (0, 0) is in the disk and not in D. See the figure on the left below. When D = D = { (x, y, z) ∈ R  ∣∣ |(x, y, z)| < 2, |(x, y)| > 0} (an open ball with the z -axis removed), then the circle x + y = 1, z = 0 is a curve in D that is not the boundary of a surface in D. The circle is the boundary of many different surfaces in R , but each contains a point on the z -axis and so is not contained in D. See the figure in the centre below. 2

2

2

1

2

2

2

2

2

2

3

2

2

2

3

On the other hand, here is an example which does satisfy (H). Let D = D

3

2

x

+y

2

(an open ball with its centre removed). For example the circle is the boundary of the surface { (x, y, z) ∈ R  ∣∣ x + y + z = 1, z > 0} ⊂ D. See the figure on 3

= { (x, y, z) ∈ R  ∣ ∣ 0 < |(x, y, z)| < 2}

= 1, z = 0

3

2

2

2

the right above. This leaves the question of what happens when (H) is violated. We'll just look at one example, which however gives the flavour of the general theory. The punctured disk is

2

D = { (x, y) ∈ R  ∣ ∣ 0 < |(x, y)| < 1 }

We'll start by looking at one particular vector field, which passes the screening test, but which cannot possibly be conservative. The field, which we saw in Example 2.3.14, is y

⇀

Th = −

2

x

+y

2

^ ı ı +

x 2

x

+y

2

^ ȷ ȷ

with domain of definition D. We'll first check that it passes the screening test: ⇀

∂

⇀

∇ × Th

x

={

( ∂x

2

x

∂

+y

2

2

x

+y

(− ∂y

2

x

+y

2

2

1 = {(

y

)−

2x 2

−

2

(x

^ )}k

1 2

2

)+(

+y )

2

x

2y

+y

2

−

2

(x

2

2

2

^ )}k

+y )

=0

Next we'll check that it cannot be conservative. Denote by C the circle Parametrize C by r (θ) = ε cos θ ^ ı ı + ε sin θ ^ ȷ ȷ with 0 ≤ θ ≤ 2π. Then ε

2

x

+y

2

2

=ε ,

with counterclockwise orientation.

⇀

ε

4.5.3

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2π

⇀

∫

⇀

⇀

Th ⋅ d r = ∫

Cε

⇀

dr

⇀

Th( r (θ)) ⋅

2π

=∫

1 (−

sin θ ^ ı ı +

ε

0

(θ) dθ dθ

0

1

cos θ ^ ȷ ȷ ) ⋅ ( − ε sin θ ^ ı ı + ε cos θ ^ ȷ ȷ ) dθ

(E1)

ε

2π

=∫

dθ

0

= 2π ⇀

is not zero. By Theorem 4.5.3, Th cannot be conservative on the punctured disk since the integral ∫ curve C is nonzero.

Cε

⇀

⇀

Th ⋅ d r

around the closed

ε

⇀

Next we'll check that it is locally conservative. That is, it can be written in the form ∇θ(x, y) near any point (x , y ) in its domain. Define θ(x, y) to be the polar angle of (x, y) with, for example, −π < θ < π. This θ is defined on all of D, except for the negative real axis. The domain of definition, D , is sketched on the left below. 0

0

π

If (x , y ) happens to lie on the negative real axis, just replace −π < θ < π by a different interval of length 2π, like 0 < θ < 2π. The domain of definition of θ would then change to the D , sketched on the right above. 0

0

0

⇀

⇀

It's now a simple matter to check that ∇θ(x, y) = Th(x, y) on the domain of definition of θ. If x ≠ 0, then, from the figure below,

we have that tan θ(x, y) =

y x

,

and cos θ(x, y) = ∂

x √x2 +y 2

,

so that

y tan θ(x, y) = −

∂x

∂

2

⟹

[

x

θ(x, y)]  sec

⟹

θ(x, y) = −

2

2

x

[

θ(x, y)]  sec

⟹

+y

2

1 θ(x, y) = x

1

2

θ(x, y) =

cos

∂y

θ(x, y)

x 2

1 =

x  

x y ≠0

2

2

x

∂y ∂

If x = 0, then we must have and arrive at the same result.

y =−

2

+y

∂ ⟹

θ(x, y)

x  

x

x

cos

2

y

tan θ(x, y) =

2

2

x =−

1

2

x y

∂x

∂y

θ(x, y) = −

∂x ∂

∂

y

2

2

x

x

+y

2

=

2

x

+y

2

(since (0, 0) is not in the domain of definition to θ ), and we can use

So far we have just looked at one vector field on D. We are now ready to consider any vector field screening test ∇ × F = 0 on D. We claim that there is a function φ on D such that ⇀

cot θ(x, y) =

⇀

F

on

D

x y

instead

that passes the

⇀

⇀

⇀ F

1

⇀

F = α⇀ Th + ∇φ

where

4.5.4

α⇀ = F

⇀

∮ 2π

⇀

F ⋅ dr

Cε

https://math.libretexts.org/@go/page/92316

The significance of this claim is that it says that if a vector field on D passes the screening test on D, then, either it is conservative (that's the case if and only if α = 0 ) or, if it fails to be conservative, then it differs from a conservative field (namely ∇φ) only ⇀

⇀ F

⇀

by a constant (namely α ) times the fixed vector field Th. That is, there is only one nonconservative vector field on D that passes the screening test, up to multiplication by constants and addition of conservative fields. This is a nice simple surprise. ⇀ F

Observe that in the definition of α , we did not specify the radius ε of the circle C to be used for the integration curve. That's because the answer to the integral does not depend on the choice of ε. To see this, take any 0 < ε < ε < 1 and consider the surface S = { (x, y) ∈ R  ∣∣ ε < |(x, y)| < ε}. ⇀ F

ε

′

2

′

It is completely contained in D. The boundary of S consists of two parts. The outside part is C , oriented counterclockwise as usual. The inside part is C , but oriented clockwise. It is usually denoted −C . So, by Stokes' theorem, ε

ε

′

ε

⇀

∮

⇀

⇀

F ⋅ dr − ∮

Cε

⇀

⇀

F ⋅ dr

=∮

Cε′

′

⇀

⇀

F ⋅ dr + ∮

Cε

−Cε′ ⇀

⇀

⇀

F ⋅ dr = ∮

⇀

F ⋅ dr

∂S

⇀

^  dS = 0 ∇×F⋅ n

=∬ S

⇀

⇀

Finally to verify the claim (E2), we check that the vector field G = F − α Th is conservative on D. To do so, it suffices to check that ∮ G ⋅ d r = 0 for any closed curve C in D. In fact we can restrict our attention to curves C that are simple, closed, counterclockwise oriented curves on D. A curve is called simple if it does not cross itself. Closed curves which are not simple can be split up into simple closed subcurves. And changing the orientation of C just changes the sign of ∮ G ⋅ d r = 0, which does not affect whether it is zero or not. ⇀ F

⇀

C

⇀

C

So let C be a simple, closed, counterclockwise oriented curve in D. We need to verify that ∮ G ⋅ d r = 0. Any simple closed curve in R divides R into three mutually disjoint subsets 3 — C itself, the set of points inside C and the set of points outside C. Since (0, 0) is not on C, it must be either outside C, as in the figure of the left below, or inside C as in the figure on the right below. ⇀

C

2

2

Case 1: (0, 0) outside C. In this case C is the boundary of a set, S, which is completely contained in D, namely all of the points inside C. So, by Stokes' theorem, ∮

⇀

G ⋅ dr

⇀

=∮

C

⇀

⇀

(F − α⇀ Th) ⋅ d r F

∂S ⇀

=∬ S

⇀

⇀

^  dS − α⇀ ∬ ∇×F⋅ n F

S

⇀

^  dS = 0 − α⇀ 0 ∇ × Th ⋅ n F

=0

Case 2: (0, 0) inside C. Since (0, 0) is not on C, we can choose ε small enough that the circle C lies completely inside C. Then the curve C − C is the boundary of a set, S, which is completely contained in D, namely the part of D that is between C and C. So, by Stokes' theorem, ε

ε

ε

4.5.5

https://math.libretexts.org/@go/page/92316

⇀

∮

⇀

G ⋅ dr − ∮

C

G ⋅ dr

⇀

=∮

Cε

C−Cε

⇀

⇀

G ⋅ dr = ∮

^  dS ∇×G⋅ n

G ⋅ dr = ∬

∂S

S

=0 ⇀

⇀

⇀

⇀

since ∇ × G = ∇ × F − α

⇀

⇀∇ F

∮

× Th = 0

⇀

G ⋅ dr = ∮

C

Cε

by the definition, (E2), of α

⇀ F

on D. Hence ⇀

⇀

G ⋅ dr = ∮

⇀

⇀

F ⋅ d r − α⇀ ∮ F

Cε

⇀

Th ⋅ d r = 2π α⇀ − α⇀ (2π) = 0 F

Cε

F

and (E1).

⇀

So G is conservative on D and F is of the form (E2) on D. The ideas that we have explored here can be generalised quite a bit. For example, if we had a disk with n > 1 punctures, we could use arguments like those above to show that any vector field F that passes the screening test has to be of the form ⇀

n ⇀

⇀

⇀

F = ∇φ + ∑ αℓ Thℓ ℓ=1 ⇀

⇀

with Th simply being the above Th translated so as to be centered on the ℓ

th

ℓ

puncture.

1. Russell Crowe posed a related question in the movie A Beautiful Mind. The movie is based on the life of the American mathematician John Nash, who won a Nobel Prize in Economics. 2. The word “smooth” does not have a universal meaning in mathematics. It is used with different meanings in different contexts. We are here using one of the standard definitions. Another standard definition requires that all derivatives of all orders are continuous. 3. This, intuitively obvious, but hard to prove, result is called the Jordan curve theorem. It is named after the French mathematician Camille Jordan (1838–1922), who first proved it. This page titled 4.5: Optional — Which Vector Fields Obey ∇ × F = 0 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

4.5.6

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4.6: Really Optional — More Interpretation of Div and Curl We are now going to determine, in much more detail than before 1, what the divergence and curl of a vector field tells us about the flow of that vector field. Consider a (possibly compressible) fluid with velocity field that, at time t , it is a cube with corners at

Pick any time

⇀

v (x, t).

t0

and a really tiny piece of the fluid; assume

0

(1)

^ {x0 + n1 εe

(2)

^ + n2 εe

(3)

^ + n3 εe

∣ ∣n1 , n2 , n3 ∈ {0, 1}}

^ ^ ^ Here ε > 0 is the length of each edge of the cube and is assumed to be really small. The vectors e ,  e and e are three mutually perpendicular unit vectors that give the orientation of the edges of the cube. The vectors from the corner x to its three ^ ^ ^ nearest neighbour corners are εe ,  εe and εe . (1)

(2)

(3)

0

(1)

(2)

(3)

As time progresses, the chunk of fluid moves. In particular, the corners move. Let us denote by εb (t) the vector, at time t, joining the n = n = n = 0 corner to the n = 1,  n = n = 0 corner. Define εb (t) and εb (t) similarly. For times very close to t we can think of our chunk of fluid as being essentially a parallelepiped with edges εb (t). (1)

(2)

1

2

3

1

2

(3)

3

(k)

0

By concentrating on the edges εb (t) of the chunk of fluid, rather than the corners, we are ignoring any translations that the chunk of fluid might have undergone. We want, instead, to determine how the size and orientation of the parallelepiped changes as t increases. (k)

At time t

0,

(k)

b

(k)

^ =e

.

The velocities of the corners of the chunk of fluid at time t are 0

⇀

(1)

(2)

^ v (x0 + n1 εe

In particular, at time t , the tail of εb (using a Taylor approximation),

(k)

0

has velocity

^ + n2 εe

⇀

v (x0 , t0 )

(k)

3 (k)

⇀

^ (t0 ) = v (x0 + εe

dt

, t0 )

and the head of εb

(k)

db ε

(3)

^ + n3 εe

⇀

(k)

⇀

^ v (x0 + εe

, t0 ).

Consequently

⇀

∂v

, t0 ) − v (x0 , t0 ) = ∑ ε j=1

has velocity

∂xj

(k)

^ (x0 , t0 )e

j

2

+ O(ε )

and so (k)

db

dt

The notation O(ε

n

j

th

db

dt

(t0 )

j=1

⇀

∂v

∂xj

(k)

^ (x0 , t0 )e

j

+ O(ε)

represents a function that is bounded by a constant times ε for all sufficiently small ε. That is, we are saying n

)

(k)

that

3

(t0 ) = ∑

3

⇀ ∂ v

j=1

∂xj

is ∑

(k)

^ (x0 , t0 )e

j

^ component of the vector e

(k)

(k)

^ plus a small error that is bounded by a constant time ε. The notation e

j

just refers to the

.

4.6.1

https://math.libretexts.org/@go/page/92317

Denote by V the 3 × 3 matrix whose (i, j) matrix element is ⇀

∂ vi

Vi,j =

∂xj

(x0 , t0 )

1 ≤ i, j ≤ 3

Then we can write the above more compactly: (k)

db

(k)

dt

(t0 ) = V b

(t0 ) + O(ε)

Here V b (t ) is the product of the 3 × 3 matrix V and the 3 × 1 column vector b (t ). We study the behaviour of small ε and t close to t , by studying the behaviour of the solutions to the initial value problems (k)

(k)

0

0

(k)

b

(t)

for

0

(k)

db

(k)

(t) = V b

(k)

(t)

b

dt

(k)

^ (t0 ) = e

To warm up, we first look at two two-dimensional examples. In both examples, the velocity field Consequently, in these examples, coincides with the exact example.

 Example 4.6.1.

(k)

b

(t).

⇀

v (x0

(k)

^ + εe

⇀

, t0 ) − v (xx0 , t0 )

is exactly

3

∑

j=1

⇀ ∂ v

ε

∂xj

Following each example, we discuss a broad class of

V

(k)

^ (x0 , t0 )e

j

⇀

v (x, y)

is linear in

(x, y).

and the solution to (IVP)

's that generate behaviour similar to that

⇀

v (x, y) = 2x ^ ı + 3y ^ ȷ

In this example 2

0

0

3

V =[

]

The solution to the initial value problem ′

b (t) = Vb(t)

b(0) = [

β1

′

]

b1 (t) = 2 b1 (t)

or equivalently

b1 (0) = β1

′

β2

b (t) = 3 b2 (t)

b2 (0) = β2

2

is b1 (t) = e b2 (t) = e

If one chooses

(1)

^ e

= ^ ı ı

^ and e

(2)

=^ ȷ ȷ,

2t

3t

β1

or equivalently

b(t) = [

β2

the edges,

e

2t

0 (1)

b

(t) = e

2t

(1)

^ e

and b

(2)

(t) = e

0 e 3t

3t

] b(0)

(2)

^ e

,

of the chunk of fluid never change (k)

direction. But their lengths do change. The relative rate of change of length per unit time,

db

(k)

|

(t)|/| b

(t)|,

is 2 for

(1)

b

dt

and 3 for b . In the figure below, the darker rectangle is the initial square. That is, the square with edges b lighter rectangle is that with edges b (t) for some t a bit bigger than t . (2)

(k)

(k)

^ (t0 ) = e

.

The

(k)

0

As time increases the initial cube becomes a larger and larger rectangle.

4.6.2

https://math.libretexts.org/@go/page/92317

 Example 4.6.2. Example 4.6.1, generalized The behaviour of Example 4.6.1 is typical of V 's that are symmetric matrices, i.e. that obey 2 V symmetric matrix 3 (with real entries)

= Vj,i

i,j

for all i, j. Any d × d

has d real eigenvalues has d mutually orthogonal real unit eigenvectors. Denote by (k)

Ve ^

the eigenvalues of for all 1 ≤ k ≤ d. Then

λk ,  1 ≤ k ≤ d, (k)

= λk e ^

and choose

V

(k)

b

(t) = e

mutually perpendicular real unit vectors,

d

λk (t−t0 )

(k)

e ^

that obey

,

(k)

^  e

obeys (k)

db

(t) = λk e

dt

So b

(k)

(t) = e

λk (t−t0 )

(k)

^  e

λk (t−t0 )

(k)

^  e

=e

λk (t−t0 )

(k)

(k)

^  V e

= Vb

(t)

and

(k)

b

(k)

^ (t0 ) = e

satisfies (IVP) for all t and 1 ≤ k ≤ d. (k)

^ If we start, at time t , with a cube whose edges, e , are eigenvectors of D, then as time progresses the edges, b (t), of the chunk of fluid never change direction. But their lengths change with the relative rate of change of length per unit time being λ for edge number k. This rate of change may be positive (the edge grows with time) or negative (the edge shrinks in time) depending on the sign of λ . (k)

0

k

k

The volume of the chunk of fluid at time t is V (t) = e ⋯e . The relative rate of change of volume per unit time is V (t)/V (t) = λ ⋯ + λ , the sum of the d eigenvalues. The sum of the eigenvalues of any d × d matrix V is given by its trace ∑ V . For the matrix (M) λ1 (t−t0 )

λd (t−t0 )

′

1

d

d

i=1

i,i

V

′

d

(t0 )

=∑ V (t0 )

So, at least when the matrix

V

i=1

∂vi

⇀

⇀

(x0 , t0 ) = ∇ ⋅ v (x0 , t0 )

∂xi

defined in (M) is symmetric, the divergence

volume per unit time for our tiny chunk of fluid at time t and position particular, this is the case when the fluid is incompressible.

⇀

x0 .

0

⇀

∇ ⋅ v (x0 , t0 )

Thus when

gives the relative rate of change of ⇀

⇀

∇⋅ v = 0

the volume is fixed. In

In fact we can relax the symmetry condition.

 Example 4.6.3. Example 4.6.1, generalized yet again For any d × d matrix V, the solution of ′

b (t) = Vb(t)

b(t0 ) = e

is b(t) = e

V (t−t0 )

e

where the exponential of a d × d matrix B is defined by the power series e

B

1 = 1 +B+

2

B

∞

1 +

2

3

B 3!

with 1 denoting the d × d identity matrix. This sum converges

1

+⋯ = ∑ n=0

4

for all d × d matrices B. Furthermore it easy to check, using the power series, that e is the identity matrix when t = t

0.

So b(t) = e

V (t−t0 )

Pick any d vectors e ,  1 ≤ k ≤ d, and define e and E(t) be the d × d matrix whose k (k)

(k)

(k)

b

th

e

n

B n!

V (t−t0 )

obeys

really does obey b (t) = Vb(t) and b(t

(t) = e

′

V (t−t0 )

column is

(k)

b

4.6.3

(k)

e

(t).

.

0)

d e

V (t−t0 )

= Ve

V (t−t0 )

and

dt = e.

Also let E be the d × d matrix whose k column is Then the volume of the parallelepiped with edges th

https://math.libretexts.org/@go/page/92317

(k)

e

,  1 ≤ k ≤ d,

is V (t

0)

= det E

and the volume of the parallelepiped with edges b

(k)

V (t) = det E(t) = det (e

V (t−t0 )

E) = det (e

V (t−t0 )

(t),  1 ≤ k ≤ d,

) det E = det (e

V (t−t0 )

is

)V (t0 )

Of course now we have to compute the determinant of the exponential of a matrix. Luckily, there is an easy way to do this. For any d × d matrix B, we have 5 det e = e , where trB, called the trace of the matrix B, is the sum of the diagonal matrix elements of B. So B

trB

V (t) = e

(t−t0 )trV

V V (t0 )

′

d

(t0 )

⇒

= trV = ∑ Vi,i

V (t0 )

i=1 ⇀

(k)

^ So, for any matrix V defined as in (M) and any choice of e ,  1 ≤ k ≤ d, the divergence ∇ ⋅ v (x of change of volume per unit time for our tiny chunk of fluid at time t and position x . 0

 Example 4.6.4.

⇀

0,

t0 )

gives the relative rate

0

⇀

v (x, y) = −y ^ ı + x^ ȷ

In this example 0

−1

1

0

V =[

]

The solution 6 to ′

b (t) = Vb(t)

b(0) = [

β1

′

b (t) = −b2 (t) ]

1

or equivalently

′

β2

b (t) = b1 (t) 2

b1 (0) = β1 b2 (0) = β2

is b1 (t) = β1 cos t − β2 sin t

or equivalently

b(t) = [

b2 (t) = β1 sin t + β2 cos t

cos t

− sin t

sin t

cos t

] b(0)

Consequently the vector b(t) has the same length as b(0). The angle between b(t) and b(0) is just t radians. So, in this example, no matter what direction vectors e ^ we pick, the chunk of fluid just rotates at one radian per unit time. In the figure below, the outlined rectangle is the initial square. That is, the square with edges b (t ) = e ^ . The shaded rectangle is that with edges b (t) for some t a bit bigger than t . (k)

(k)

(k)

0

(k)

0

 Example 4.6.5. Example 4.6.4, generalized The behaviour of Example 4.6.4 is typical of V 's that are antisymmetric matrices, i.e. that obey V = −V have already observed, for any d × d matrix V, the solution of  b (t) = Vb(t),  b(0) = e  is  b(t) = e that if V is a 3 × 3 antisymmetric matrix, then e is a rotation. i,j

′

j,i Vt

for all i, j. As we e. We now show

Vt

Assuming that V is not the zero matrix (in which case e is the identity matrix for all t ), we can find a number ^ unit vector k = (k , k , k ) (not necessarily the standard unit vector parallel to the z -axis) such that Vt

1

2

Ω >0

and a

3

⎡ V =⎢ ⎣

0

−Ωk3

Ωk3

0

−Ωk2

Ωk1

4.6.4

Ωk2

⎤

−Ωk1 ⎥ 0

⎦

https://math.libretexts.org/@go/page/92317

This is easy. Because V is antisymmetric, all of the entries on its diagonal must be zero. Define Ω to be

−−−−−−−−−−−− − √V

2

1,2

+V

2

1,3

+V

2

2,3

^ and k = −V /Ω, k = V /Ω, k = −V /Ω. Also, let ^ ı ı be any unit vector orthogonal to k (again, not necessarily the ^ ^ standard one) and ^ȷȷ = k × ^ ı ı . So ^ ı ı ,  ^ ȷ ȷ ,  k is a right-handed system of three mutually perpendicular unit vectors. 1

2,3

2

1,3

3

Observe that, for any vector e = (e

1,

e2 , e3 )

0

⎡ Ve = ⎢ ⎣

1,2

−Ωk3

Ωk3

0

−Ωk2

Ωk1

Ωk2

⎤⎡

e1

−Ωk1 ⎥ ⎢ e2 ⎦⎣

0

e3

k2 e3 − k3 e2 ⎡ ⎤ ^ = Ω ⎥ ⎢ k3 e1 − k1 e3 ⎥ = Ω k × e ⎤

⎦

⎣

k1 e2 − k2 e1

⎦

In particular, ^ V^ ı ı = Ωk × ^ ı ı = Ω^ ȷ ȷ 2

V

2

^ ı ı = ΩV ^ ȷ ȷ = −Ω ^ ı ı

3

V

4

V

2

^ ı ı = ΩV

3

^ ı ı = ΩV

⇀

^ V^ ȷ ȷ = Ωk × ^ ȷ ȷ = −Ω ^ ı ı V

3

^ ȷ ȷ = −Ω ^ ȷ ȷ

V

4

^ ȷ ȷ =Ω ^ ı ı

V

2

3

4

^ ^ ^ Vk = Ω k × k = 0

2

^ ȷ ȷ = −ΩV ^ ı ı = −Ω ^ ȷ ȷ ^ ȷ ȷ = −Ω V ^ ȷ ȷ = −Ω V

2

3

V

3

^ ı ı =Ω ^ ı ı

V

4

^ ı ı =Ω ^ ȷ ȷ

V

⇀

⇀

2

^ k = V0 = 0

3

2 ^ k =V 0 = 0

4

⇀

⇀

⇀

⇀

3 ^ k =V 0 = 0

and so on. For all odd n ≥ 1, V

n

(n−1)/2

^ ı ı = (−1 )

n

Ω ^ ȷ ȷ

V

n

(n−1)/2

^ ȷ ȷ = −(−1 )

n

Ω

^ ı ı

V

n

⇀

^ k = 0

and all even n ≥ 2, V

n

n/2

^ ı ı = (−1 )

n

Ω

^ ı ı

V

n

n/2

^ ȷ ȷ = (−1 )

n

Ω ^ ȷ ȷ

n

V

⇀

^ k = 0

Hence we can write ∞

e

Vt

^ ı ı =∑ n=0

∞ Vt

n

(Vt)

(n−1)/2

(−1)

n

^ ı ı = ∑

n!

(Ωt)

(−1)

n!

n even

n

^ ı ı + ∑

(Ωt) ^ ȷ ȷ n!

n odd

cos(Ωt) ^ ı ı + sin(Ωt) ^ ȷ ȷ

=

e

n/2

1

^ ȷ ȷ =∑ n=0

n/2

1 n!

(n−1)/2

(−1)

n

(Vt) ^ ȷ ȷ = ∑

(−1)

n

n

(Ωt) ^ ȷ ȷ  − ∑ n!

n even

(Ωt)

^ ı ı

n!

n odd

= − sin(Ωt) ^ ı ı + cos(Ωt) ^ ȷ ȷ ∞

e

Vt

^ k =∑ n=0

1

n^ (Vt) k

n!

^ =k

So e

Vt

^ is rotation by an angle Ωt about the axis k .

 Example 4.6.6. Example 4.6.5, continued Whether or not the matrix V defined in (M) is antisymmetric, the related matrix with entries 1 Ai,j =

2

(Vi,j − Vj,i )

is. When V is antisymmetric, A and V coincide. The matrix A is (to write it out explicitly) ⎡ ⎢ ⎢ 1⎢ − ⎢ ⎢ 2⎢ ⎢ ⎢

⇀ ∂ v1

0 ⇀ ∂ v1 ∂x2

(x0 , t0 )+

∂x2 ⇀ ∂ v2 ∂x1

(x0 , t0 )−

(x0 , t0 )

⇀ ∂ v2 ∂x1

(x0 , t0 )

∂v1 ∂x3

(x0 , t0 )− ⇀ ∂ v2

0

∂x3

⇀ ∂ v3

(x0 , t0 )−

textbf x0 , t0 ) ⇀ ∂ v1

⎣−

∂x3

(x0 , t0 )+

⇀ ∂ v3 ∂x1

(x0 , t0 )

⇀ ∂ v2

−

∂x3

(x0 , t0 )+

⇀ ∂ v3 ∂x2

(x0 , t0 )

∂x1

0

(x0 , t0 ) ⎤ ⇀ ∂ v3 ∂x2

(

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Comparing this with (R), we see that

4.6.5

https://math.libretexts.org/@go/page/92317

^ Ωk =

1 2

⇀

∇ × v (x0 , t0 )

So, at least when the matrix V defined in (M) is antisymmetric, our tiny cube rotates about the axis with ∇ × v (x ⇀

1 2

0,

t0 )

at rate

⇀

∣ ∣∇ × v (x0 , t0 )∣ ∣.

 Remark 4.6.7 In the generalization, Example 4.6.5, of Example 4.6.4, we only considered dimension 3. It is a nice exercise in eigenvalues and eigenvectors to handle general dimension. Here are the main facts about antisymmetric matrices with real entries that are used. All eigenvalues of antisymmetric matrices are either zero or pure imaginary. For antisymmetric matrices with real entries, the nonzero eigenvalues come in complex conjugate pairs. The corresponding eigenvectors may also be chosen to be complex conjugates. ^ Choose as basis vectors (like ^ ı ı ,  ^ ȷ ȷ ,  k above) ^ the eigenvectors of eigenvalue 0 (they act like k above) the real and imaginary parts of each complex conjugate pair of eigenvectors (they act like ^ ı ı ,  ^ ȷ ȷ above)

Resumé so far: We have now seen that (k)

^ when the matrix V defined in (M) is symmetric and the direction vectors e of the cube are eigenvectors of V, then, at time t the chunk of fluid is not changing orientation but is changing volume at instantaneous relative rate ∇ ⋅ v (x , t ) and when the matrix V defined in (M) is antisymmetric, then, at time t , the chunk of fluid is not changing shape or size but is rotating about the axis ∇ × v (x , t ) at rate ∣∣∇ × v (x , t )∣∣. For this reason, ∇ × v is often referred to as a “vorticity” meter.

0,

⇀

0

0

0

⇀

1

0

0

⇀

⇀

0

2

0

These agree with our earlier interpretations of divergence and curl. The general case: Now consider a general matrix V. It can always be written as the sum V = S +A

of a symmetric matrix S and an antisymmetric matrix A. Just define 1 Si,j =

2

1 (Vi,j + Vj,i )

Ai,j =

2

(Vi,j − Vj,i )

As we have already observed, the solution of ′

b (t) = Vb(t)

b(0) = e

is b(t) = e

Vt

e =e

If S and A were ordinary numbers, we would have e =e 7 happen to commute . For arbitrary matrices, it is still true that (A+S)t

e

(A+S)t

At

e

St

= lim [e

(A+S)t

.

e

But for matrices this need not be the case, unless

At/n

e

St/n

S

and A

n

]

n→∞

8

This is called the Lie product formula. It shows that our tiny chunk of fluid mixes together the behaviours of A and S, scaling a bit, then rotating a bit, then scaling a bit and so on.

4.6.6

https://math.libretexts.org/@go/page/92317

 Example 4.6.8.  v (x, y) = 2y ^ ı ⇀

In this example V =[

0

2

0

0

] = S +A

with

S =[

0

1

1

0

]

0

1

−1

0

A =[

]

The solution to the full flow ′

b (t) = Vb(t)

b(0) = [

β1

′

b (t) = 2 b2 (t) ]

1

or equivalently

′

β2

b (t) = 0

b1 (0) = β1 b2 (0) = β2

2

is b1 (t) = β1 + 2 β2 t or equivalently

1

2t

0

1

b(t) = [

b2 (t) = β2

] b(0)

The solution to the S part of the flow ′

b (t) = Sb(t)

b(0) = [

β1

′

b (t) = b2 (t) ]

b1 (0) = β1

1

or equivalently

′

β2

b (t) = b1 (t)

b2 (0) = β2

2

is 9 b1 (t) = β1 cosh t + β2 sinh t

or equivalently

b(t) = [

b2 (t) = β1 sinh t + β2 cosh t

cosh t

sinh t

sinh t

cosh t

] b(0)

The eigenvectors of S are (1)

^ e

1 =

– √2

[

1

]

(2)

^ e

1 =

– √2

1

1

[

]

−1

The corresponding eigenvalues are +1 and −1. The eigenvectors obey e

e

St

St

(1)

^ e

cosh t

sinh t

sinh t

cosh t

cosh t

sinh t

sinh t

cosh t

=[

(2)

^ e

=[

(1)

^ ]e

(2)

^ ]e

t

(1)

^ =e e

=e

−t

(2)

^ e

Under the S part of the flow e ^ scales by a factor of e , which is bigger than one for t > 0 and e ^ which is smaller than one for t > 0. (1)

(2)

t

scales by a factor of e

−t

,

The solution to the A part of the flow ′

b (t) = Ab(t)

b(0) = [

β1

′

]

b (t) = b2 (t)

b1 (0) = β1

′

b2 (0) = β2

1

or equivalently

β2

b (t) = −b1 (t) 2

is b1 (t) = β1 cos t + β2 sin t or equivalently

cos t

sin t

− sin t

cos t

b(t) = [

b2 (t) = −β1 sin t + β2 cos t

] b(0)

The A part of the flow rotates clockwise about the origin at one radian per unit time. Here are some figures to help us visualize this. (1)

(2)

^ ^ The first shows a square with edges e ,  e and its image under the full flow t = 0.4 later. Under this full flow the vector ^ ^ ^ e → e e . The darkly shaded parallelogram has edges e e . ^ ^ The second shows its image under 0.4 time units of the S -flow (that is, e → e e ). The lightly shaded rectangle has ^ edges e e . The third applies 0.4 time units of the A -flow to the shaded rectangle of the second figure. So the lightly shaded rectangle ^ ^ of the third figure has edges e e and the darkly shaded rectangle has edges e e e . (k)

0.4V

(k)

0.4V

(k)

(k)

0.4S

0.4S

(k)

(k)

0.4S

(k)

0.4A

4.6.7

0.4S

(k)

https://math.libretexts.org/@go/page/92317

Of course e

0.4(A+S)

e

(k)

^ e

(as in the darkly shaded rectangle of the third figure) is not a very good approximation for ^ (as in the darkly shaded parallelogram of the first figure). It is much better to take [e e ] e with n

0.4A

e

0.4S

(k)

^ e

0.4A/n

large. Each of the following figures shows two parallelograms. In each, the shaded region has edges e ^ and the outlined region has edges [ e e ] e . 0.4A/n

So we can see that, as n increases, [e

0.4S/n

0.4A/n

e

n

0.4S/n

0.4V

0.4S/n

(k)

^ e

n

=e

(k)

0.4(A+S)

(k)

^ e

(k)

n

(k)

^ ] e

becomes a better and better approximation to e

0.4(A+S)

(k)

^ e

.

1. We'll also use some more mathematics than before. In this section, we'll use matrix eigenvalues and eigenvectors and solve some simple systems of ordinary differential equations. We'll also need to use a lot of subscripts and superscripts. It only looks intimidating. ⇀ ∂ vi

2. In terms of our original vector field, this condition is that ⇀

∂xj

(x0 , t0 ) =

⇀ ∂ vj ∂xi

(x0 , t0 ).

So, in three dimensions, it comes down

to the requirement that ∇ × v be zero at the point (x , t ). 3. This was proven by the French mathematician and physicist Augustin-Louis Cauchy (1789–1857) in 1829. 4. The proof is not so hard, though we'll only outline it. Just denote by β the magnitude of the largest matrix element of B. Then use the definition of the matrix product to prove that the largest matrix element of B has magnitude at most (dβ ) . 5. Again, we won't prove this. But for a diagonal matrix, it is easy — just compute both sides. So for a diagonalizable matrix it is also easy — diagonalize. 6. You can find the solution either by guessing, or by using eigenvalues and eigenvectors. 7. By definition, the matrices S and A commute when AS = SA. 8. This formula is named after the Norwegian mathematician Marius Sophus Lie (1842–1899). In 1870, he was arrested and held in prison in France for a month, because he was suspected of being a German spy. His mathematics notes were thought to be top secret coded messages. 9. Recall that sinh t = (e − e ) and cosh t = (e + e ). ⇀

0

0

n

1 2

t

−t

1

t

n

−t

2

This page titled 4.6: Really Optional — More Interpretation of Div and Curl is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

4.6.8

https://math.libretexts.org/@go/page/92317

4.7: Optional — A Generalized Stokes' Theorem As we have seen, the fundamental theorem of calculus, the divergence theorem, Greens' theorem and Stokes' theorem share a number of common features. There is in fact a single framework which encompasses and generalizes all of them, and there is a single theorem of which they are all special cases. We now give a bare bones introduction to this framework and theorem. A proper treatment typically takes up a good part of a full course. Here is an outline of what we shall do: First, we will define differential forms. To try and keep things as simple and concrete as possible, we'll only define 1 differential forms on R — all of our functions will be defined on R . Very roughly speaking, a k -form is what you write after the integral sign of an integral over a k dimensional object. Here k is one of 0, 1, 2, 3. As a example, a 1-form is an expression of the form F (x, y, z) dx + F (x, y, z) dy + F (x, y, z) dz. For k = 0, think of a point as a zero dimensional object and think of evaluating a function at a point as “integrating the function over the point”. Then we will define some operations on differential forms, so that we can add them, multiply them, differentiate them and, eventually, integrate them. The derivative of a k -form ω is a (k + 1) -form that is denoted dω. It will turn out that 3

1

3

2

3

differentiating a 0-form amounts to taking a gradient, differentiating a 1-form amounts to taking a curl, and differentiating a 2-form amounts to taking a divergence. Finally we will get to the generalized Stokes' theorem which says that, if ω is a k -form (with k = 0, 1, 2) and D is a (k + 1) dimensional domain of integration, then ∫

dω = ∫

D

ω

∂D

It will turn out that when k = 0, this is just the fundamental theorem of calculus and when k = 1, this is both Green's theorem and our Stokes' theorem, and when k = 2, this is the divergence theorem. Now let's get to work. For simplicity, we will assume throughout this section that all derivatives of all functions exist and are continuous. Our first task to define differential forms. As we said above we will define a 1-form as an expression of the form F (x, y, z) dx + F (x, y, z) dy + F (x, y, z) dz. When you learned the definition of the integral the symbol “dx” was not given any mathematical meaning by itself. A meaning was given only to collections of symbols like the indefinite integral “∫ f (x) dx ” and the definite integral “∫ f (x) dx ”. Later in this section, we will give a meaning to dx. We will, in Definition 4.7.9, define a differentiation operator that we will call \(\text{d}}\text{.\) Then dx will be that differentiation operator applied to the function f (x) = x. However, until then we will have to treat dx and dy and dz just as symbols. Their sole role in F (x, y, z) dx + F (x, y, z) dy + F (x, y, z) dz is to allow us to distinguish 2 F (x, y, z), F (x, y, z) and F (x, y, z). 1

2

3

b

a

1

1

2

2

3

3

Similarly, we will define a 2-form as an expression of the form F (x, y, z) dy ∧ dz + F (x, y, z) dz ∧ dx + F (x, y, z) dx ∧ dy. Once again there is a symbol, namely “∧”, that we have not yet given a meaning to. We will, in Definition 4.7.3, define a product, called the wedge product, with ∧ as the multiplication symbol. Then dx ∧ dy will be the wedge product of dx and dy. Until then we will have to treat dy ∧ dz, dz ∧ dx and dx ∧ dy just as three more meaningless symbols. 1

2

3

Finally here is the definition.

 Definition 4.7.1 1. A 0-form is a function f (x, y, z). 2. A 1-form is an expression of the form F1 (x, y, z) dx + F2 (x, y, z) dy + F3 (x, y, z) dz

with F (x, y, z), F (x, y, z) and F (x, y, z) being functions of three variables. 3. A 2-form is an expression of the form 1

2

3

F1 (x, y, z) dy ∧ dz + F2 (x, y, z) dz ∧ dx + F3 (x, y, z) dx ∧ dy

4.7.1

https://math.libretexts.org/@go/page/92318

with F (x, y, z), F (x, y, z) and F (x, y, z) being functions of three variables. 4. A 3-form is an expression of the form f (x, y, z) dx ∧ dy ∧ dz, with f (x, y, z) being a function of three variables. 1

2

3

At this stage (there'll be more later), just think of “dx”, “dy ”, “dz ”, “dx ∧ dy ”, and so on, as symbols. Do not yet attempt to attach any significance to them. There are four operations involving differential forms — addition, multiplication (∧), differentiation (d) and integration. Here are their definitions. First, addition is defined, and works, just the way that you would expect it to.

 Definition 4.7.2. Addition of differential forms 1. The sum of the 0-forms f and g is the 0-form f + g. 2. The sum of two 1-forms is the 1-form [ F1 dx + F2 dy + F3 dz] + [ G1 dx + G2 dy + G3 dz] = (F1 + G1 ) dx + (F2 + G2 ) dy + (F3 + G3 ) dz

3. The sum of two 2-forms is the 2-form [ F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy] + [ G1 dy ∧ dz + G2 dz ∧ dx + G3 dx ∧ dy] =

(F1 + G1 ) dy ∧ dz + (F2 + G2 ) dz ∧ dx + (F3 + G3 ) dx ∧ dy

4. The sum of two 3-forms is the 3-form f dx ∧ dy ∧ dz  +  g dx ∧ dy ∧ dz  =  (f + g) dx ∧ dy ∧ dz

There is one wrinkle in multiplication. It is not commutative, meaning that α ∧ β need not be the same as β ∧ α. You have already seen some noncommutative products. If a and b are two vectors in R , then a × b = −b × a. Also, if A and B are two n × n matrices, the matrix product AB need not be the same as BA. 3

 Definition 4.7.3. Multiplication of differential forms We now define a multiplication rule for differential forms. If ω is a k -form and ω is a k -form then the product will be a (k + k ) -form and will be denoted ω ∧ ω (read “omega wedge omega prime”). It is determined by the following properties. ′

′

′

′

1. If f is a function (i.e. a 0-form), then f [ F1 dx + F2 dy + F3 dz]

= (f F1 ) dx+(f F2 ) dy+(f F3 ) dz

f [ F1 dy ∧ dz+F2 dz ∧ dx+F3 dx ∧ dy]

= (f F1 ) dy ∧ dz + (f F2 ) dz ∧ dx + (f F3 ) dx ∧ dy

f [g dx ∧ dy ∧ dz]

= (f g) dx ∧ dy ∧ dz

Traditionally, the ∧ is not written when multiplying a differential form by a function (i.e. a 0-form). 2. ω ∧ ω is linear in ω and in ω . This means that if ω = f ω + f ω , where f ,f are functions and ω ′

′

1

1

2

2

1

′

1,

2

′

ω2

are forms, then

′

(f1 ω1 + f2 ω2 ) ∧ ω = f1 (ω1 ∧ ω ) + f2 (ω2 ∧ ω )

Similarly, ′

ω ∧ (f1 ω

1

′

′

′

2

1

2

+ f2 ω ) = f1 (ω ∧ ω ) + f2 (ω ∧ ω )

3. If ω is a k -form and ω is a k -form then ′

′

′

′

kk

ω ∧ ω = (−1 )

′

ω ∧ω

That is, if at least one of k and k is even, then ′

′

′

ω∧ω = ω ∧ω

4.7.2

https://math.libretexts.org/@go/page/92318

(so that the wedge product is commutative) and if k and k are both odd then ′

′

′

ω ∧ ω = −ω ∧ ω

(so that the wedge product is anticommutative). In particular, if ω is a d -form with d odd ω∧ω = 0

4. The wedge product is associative. This means that ′

′′

(ω ∧ ω ) ∧ ω

′

′′

= ω ∧ (ω ∧ ω )

So the wedge product obeys most of the usual multiplication rules, with the one big exception that if form with k and k both odd then ω ∧ ω = −ω ∧ ω. ′

′

ω

is k -form and

′

ω

is a

′

k

-

′

The best way to get a handle on the wedge product is to work through some examples, like these.

 Example 4.7.4 Let ω = F

1

dx + F2 dy + F3 dz ′

ω∧ω

and ω

′

= G1 dx + G2 dy + G3 dz

be any two 1-forms. Their product is

= [ F1 dx + F2 dy + F3 dz] ∧ [ G1 dx + G2 dy + G3 dz] =  (F1 dx) ∧ (G1 dx) + (F1 dx) ∧ (G2 dy) + (F1 dx) ∧ (G3 dz) + (F2 dy) ∧ (G1 dx) + (F2 dy) ∧ (G2 dy) + (F2 dy) ∧ (G3 dz) + (F3 dz) ∧ (G1 dx) + (F3 dz) ∧ (G2 dy) + (F3 dz) ∧ (G3 dz) (by linearity, i.e. by part (b) of the last Definition) =  F1 G1 dx ∧ dx + F1 G2 dx ∧ dy + F1 G3 dx ∧ dz + F2 G1 dy ∧ dx + F2 G2 dy ∧ dy + F2 G3 dy ∧ dz + F3 G1 dz ∧ dx + F3 G2 dz ∧ dy + F3 G3 dz ∧ dz = (F1 G2 − F2 G1 ) dx ∧ dy + (F3 G1 − F1 G3 ) dz ∧ dx + (F2 G3 − F3 G2 ) dy ∧ dz

because dx ∧ dx = dy ∧ dy = dz ∧ dz = 0

and dx ∧ dy = −dy ∧ dx

Note that, looking at the last example, if we view simply as

dx ∧ dz = −dz ∧ dx

⇀

F = (F1 , F2 , F3 )

dz ∧ dy = −dy ∧ dz

and G = (G

1,

G2 , G3 )

as vectors, we can write the product

 Equation 4.7.5 [ F1 dx + F2 dy + F3 dz] ∧ [ G1 dx + G2 dy + G3 dz] ⇀

⇀

⇀

= (F × G)1 dy ∧ dz + (F × G)2 dz ∧ dx + (F × G)3 dx ∧ dy

⇀

where we are using (F × G) to denote the ℓ have

th

ℓ

⇀

component of the cross product F × G. In the special case that F

3

= G3 = 0,

we

 Equation 4.7.6 [ F1 dx + F2 dy] ∧ [ G1 dx + G2 dy]

= (F1 G2 − F2 G1 ) dx ∧ dy

= det [

4.7.3

F1

F2

G1

G2

] dx ∧ dy

https://math.libretexts.org/@go/page/92318

We can now see why in the Definition 4.7.1.c of 2-forms there were no dx ∧ dx or dy ∧ dy or dz ∧ dz terms — they are all zero and there were no dy ∧ dx or dz ∧ dy or dx ∧ dz terms — they can all be rewritten using dx ∧ dy, vice versa).

dy ∧ dz

and dz ∧ dx terms (or

The reason that we chose to write the Definition 4.7.1.c as F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy

as opposed to in the form, for example, f1 dx ∧ dy + f2 dx ∧ dz + f3 dy ∧ dz

was to make formulae like 4.7.5 work. The easy way to remember F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy

is to rename (in your head) x, y, z to x

1,

x2 , x3 .

Then the subscripts in the three terms of

F1 dx2 ∧ dx3 + F2 dx3 ∧ dx1 + F3 dx1 ∧ dx2

are just 1, 2, 3 and 2, 3, 1 and 3, 1, 2 — the three cyclic permutations of 1, 2, 3.

 Example 4.7.7 The

product

of

the

(general)

1

′

-form ω = F dx + F dy + F dz and the (general) (again note the numbering of the coefficients in the 2-form) is 1

2

3

2

-form

ω = [ G1 dy ∧ dz + G2 dz ∧ dx + G3 dx ∧ dy] ′

ω∧ω

= [ F1 dx + F2 dy + F3 dz] ∧ [ G1 dy ∧ dz + G2 dz ∧ dx + G3 dx ∧ dy] =  F1 G1 dx ∧ dy ∧ dz + F2 G2 dy ∧ dz ∧ dx + F3 G3 dz ∧ dx ∧ dy = (F1 G1 + F2 G2 + F3 G3 ) dx ∧ dy ∧ dz

Here we have used that, for 1-forms, α ∧ β = −β ∧ α, so that dy ∧ dz ∧ dx

= −dy ∧ dx ∧ dz = dx ∧ dy ∧ dz

dz ∧ dx ∧ dy

= −dx ∧ dz ∧ dy = dx ∧ dy ∧ dz

We have also used that any wedge product of three same is zero. For example

d{x or y\text{ or }z}

's with at least two of the coordinates being the

dx ∧ dz ∧ dx = −dx ∧ dx ∧ dz = 0

So [ F1 dx + F2 dy + F3 dz] ∧ [ G1 dy ∧ dz + G2 dz ∧ dx + G3 dx ∧ dy] ⇀

= F ⋅ G dx ∧ dy ∧ dz

 Example 4.7.8 Combining Examples 4.7.4 and 4.7.7, we have the wedge product of any three (general) 1-forms F G dx + G dy + G dz and H dx + H dy + H dz is

1

1

2

3

1

2

dx + F2 dy + F3 dz

and

3

[ F1 dx+F2 dy+F3 dz] ∧ [ G1 dx+G2 dy+G3 dz] ∧ [ H1 dx + H2 dy + H3 dz] = [ F1 dx + F2 dy + F3 dz]∧ [(G × H)1 dy ∧ dz + (G × H)2 dz ∧ dx + (G × H)3 dx ∧ dy] = { F1 (G × H)1 + F2 (G × H)2 + F3 (G × H)3 }dx ∧ dy ∧ dz = { F1 (G2 H3 −G3 H2 )+F2 (G3 H1 −G1 H3 )+F3 (G1 H2 −G2 H1 }dx ∧ dy ∧ dz

This can be expressed cleanly in terms of determinants. Recalling the rule for expanding a determinant along its top row

4.7.4

https://math.libretexts.org/@go/page/92318

[ F1 dx+F2 dy+F3 dz] ∧ [ G1 dx+G2 dy+G3 dz] ∧ [ H1 dx+H2 dy+H3 dz] F1

F2

F3

= det ⎢ G1

G2

G3 ⎥ dx ∧ dy ∧ dz

H2

H3

⎡

⎣

H1

⎤

⎦

Our next operation is a differential operator which unifies and generalizes gradient, curl and divergence.

 Definition 4.7.9. Differentiation of differential forms If ω is a k -form, then dω is a k + 1 -form, with d being the unique 3 such operator that obeys 1. d is linear. That is, if ω

1,

ω2

are k -forms and a

1,

then

a2 ∈ R,

d(a1 ω1 + a2 ω2 ) = a1 dω1 + a2 dω2

2. d obeys a “graded product rule”. Precisely, if ω

(k)

( ℓ)

(k)∧ω

d(ω

)

is a k -form and ω

(ℓ)

(k)

= (dω

(ℓ)

)∧ω

is an ℓ -form, then k

(k)

+ (−1 ) ω

(ℓ)

∧ (dω

)

3. If f (x, y, z) is a 0-form, then ∂f df

=

∂f (x, y, z) dx +

∂x ⇀

∂f (x, y, z) dy +

∂y

(x, y, z) dz ∂z

⇀ ^ where d r = dx ^ ı ı + dy ^ ȷ ȷ + dz k

⇀

= ∇f (x, y, z) ⋅ d r

4. For any differential form ω, d(dω) = 0

 Example 4.7.10 1. If f (x, y, z) = x, then ∂x df =

∂x (x, y, z) dx +

∂x

∂x (x, y, z) dy +

∂y

(x, y, z) dz = dx ∂z

That is, dx really is the operator d applied to the function x. Similarly, dy really is the operator d applied to the function y and dz really is the operator d applied to the function z. 2. For any k -form ω d[ω ∧ dx]

k

= dω ∧ dx + (−1 ) ω ∧ d(dx) = dω ∧ dx

Similarly d[ω ∧ dy] = dω ∧ dy

d[ω ∧ dz] = dω ∧ dz

3. For any 1-form

4.7.5

https://math.libretexts.org/@go/page/92318

d[ F1 dx + F2 dy + F3 dz] = dF1 ∧ dx + dF2 ∧ dy + dF3 ∧ dz

=(

∂F1

∂F1

 dx +

∂x

∂F1

 dy +

∂y

 dz) ∧ dx

∂z

+(

∂F2

∂F2

 dx +

∂x +(

∂F3

∂F3

∂F2

−

∂y

∂z ∂F2

⇀

∂F3

−

−

∂F1

 dz) ∧ dz

∂z ∂F3

∂z

∂x ⇀

 dy +

∂y ∂F1

 dz) ∧ dy

∂z

∂F3

 dx +

) dy ∧ dz + (

+(

∂F2

∂y

∂x

=(

 dy +

) dz ∧ dx

∂x ) dx ∧ dy

∂y

⇀

⇀

⇀

⇀

= (∇ × F)1 dy ∧ dz + (∇ × F)2 dz ∧ dx + (∇ × F)3 dx ∧ dy

4. For any 2-form d[ F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy] = dF1 ∧ dy ∧ dz + dF2 ∧ dz ∧ dx + dF3 ∧ dx ∧ dy =(

∂F1

∂F1

 dx +

∂x

 dy +

∂F1

∂y

 dz) ∧ dy ∧ dz

∂z +(

∂F2

 dx +

∂x +(

∂F3

∂F1 ∂x

⇀

+

∂F2

+

∂F3

∂y

 dy +

∂y  dx +

∂x =(

∂F2

∂F2

 dz) ∧ dz ∧ dx

∂z

∂F3

 dy +

∂y

∂F3

 dz) ∧ dx ∧ dy

∂z

) dx ∧ dy ∧ dz

∂z

⇀

= ∇ ⋅ F dx ∧ dy ∧ dz

5. For any 3-form ∂f d[f dx ∧ dy ∧ dz]

∂f

=(

 dx + ∂x

∂f  dy +

∂y

 dz) ∧ dx ∧ dy ∧ dz ∂z

=0

 Example 4.7.11 In Definition 4.7.9.c, we defined, for any function f (x, y, z) of three variables ∂f df

=

∂f (x, y, z) dx +

∂x

∂f (x, y, z) dy +

∂y

(x, y, z) dz ∂z

The analogous formulae 4 for functions of one or two variables also apply. df df (t) =

(t) dt dt ∂f

df (u, v) =

∂f (u, v) du +

∂u

1. Let F

1 (x,

z = z(t),

(u, v) dv ∂v

be a 1-form. Suppose that we substitute x = x(t), y = y(t) and so that we are restricting our 1-form to a parametrized curve. Then, writing r (t) = (x(t), y(t), z(t)), y, z) dx + F2 (x, y, z) dy + F3 (x, y, z) dz

⇀

F1 (x(t), y(t), z(t)) dx(t) + F2 (x(t), y(t), z(t)) dy(t) + F3 (x(t), y(t), z(t)) dz(t) ⇀

dx

= F1 ( r (t))

⇀

⇀

dy

(t) dt + F2 ( r (t)) dt

⇀

dz

(t) dt + F3 ( r (t)) dt

(t) dt dt

⇀

⇀

dr

= F( r (t)) ⋅

(t) dt dt

4.7.6

https://math.libretexts.org/@go/page/92318

2. Let F

1 (x,

be a 2-form. Suppose that we substitute x = x(u, v), and z = z(u, v), so that we are restricting our 2-form to a parametrized surface. Then, writing

y, z) dy ∧ dz + F2 (x, y, z) dz ∧ dx + F3 (x, y, z) dx ∧ dy

y = y(u, v) ⇀

r (u, v) = (x(u, v), y(u, v), z(u, v)),

F1 (x(u, v), y(u, v), z(u, v)) dy(u, v) ∧ dz(u, v) + F2 (x(u, v), y(u, v), z(u, v)) dz(u, v) ∧ dx(u, v) + F3 (x(u, v), y(u, v), z(u, v)) dx(u, v) ∧ dy(u, v) ∂y

⇀

= F1 ( r (u, v)) (

∂y

∂z

du + ∂u

dv) ∧ ( ∂v ∂z

⇀

+ F2 ( r (u, v)) (

∂y ∂z

⇀

⇀

⇀

dv) ∂v

∂z ∂x

⇀

∂z ∂x −

∂u ∂v

) ∂v ∂u

∂x ∂y −

∂u ∂v

)]du ∧ dv ∂v ∂u

⇀

∂r

⇀

∂y du +

∂u

) + F2 ( r (u, v)) (

∂x ∂y

⇀

dv) ∂v

∂y dv) ∧ (

∂v

∂v ∂u

∂x du +

∂u

∂x du +

∂u

∂y ∂z

+ F3 ( r (u, v)) (

= [F( r (u, v)) ⋅

∂x dv) ∧ (

∂v

− ∂u ∂v

dv) ∂v

∂z du +

∂u ∂x

⇀

+ F3 ( r (u, v)) (

= [F1 ( r (u, v)) (

∂z du +

∂u

∂r (u, v) ×

∂u

(u, v)]du ∧ dv ∂v

Let us summarize what we have seen in the Example 4.7.10.

 Lemma 4.7.12 1. For any 0-form ⇀

⇀

df = ∇f (x, y, z) ⋅ d r

2. For any 1-form d[ F1 dx + F2 dy + F3 dz] ⇀

⇀

⇀

⇀

⇀

⇀

= (∇ × F)1 dy ∧ dz + (∇ × F)2 dz ∧ dx + (∇ × F)3 dx ∧ dy

3. For any 2-form ⇀

⇀

d[ F1 dy ∧ dz + F2 dz ∧ dx + F3 dx ∧ dy] = ∇ ⋅ F dx ∧ dy ∧ dz

4. For any 3-form d[f dx ∧ dy ∧ dz] = 0

Our final operation is integration of differential forms.

 Definition 4.7.13. Integration of differential forms 1. Let f (x, y, z) be a 0-form and P

3

= (x0 , y0 , z0 ) ∈ R

∫

be a point. Then f = f (x0 , y0 , z0 )

P

More generally if, for each 1 ≤ i ≤ ℓ,

3

Pi = (xi , yi , zi ) ∈ R

is a point and n is an integer, then i

ℓ

∫

f = ∑ ni f (xi , yi , zi ) ℓ

Σi=1 ni Pi ⇀

2. Let ω = F( r ) ⋅ d r = F by r (t) = (x(t) , y(t) , ⇀

⇀

⇀

1 (x,

i=1

be a 1-form. Let C be a curve that is parametrized Then, motivated by Example 4.7.11.a above,

y, z) dx + F2 (x, y, z) dy + F3 (x, y, z) dz

z(t)), a ≤ t ≤ b.

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b

∫

ω =∫

C

⇀

⇀

dr

⇀

F( r (t)) ⋅

⇀

(t) dt = ∫ dt

a

⇀

F ⋅ dr

C

3. Let ω = F (x, y, z) dy ∧ dz + F (x, y, z) dz ∧ dx + F (x, y, z) dx ∧ dy be a 2-form. Let S be an oriented surface that is parametrized by r (u, v) = (x(u, v) , y(u, v) , z(u, v)), with (u, v) running over a region R in the uv-plane. Assume that 1

2

3

⇀

⇀

r (u, v)

^ dS = + is orientation preserving in the sense that n

ω

=∬

S

du dv.

Then, motivated by Example 4.7.11.b above,

∂r

⇀

(u, v) × ∂u

R

∂v

⇀

∂r

⇀

[F( r (u, v)) ⋅

⇀ ∂ r

×

∂u

⇀

⇀

∫

⇀ ∂ r

^ dS F⋅n

(u, v)]du ∧ dv = ∬ ∂v

S

4. Let ω = f (x, y, z) dx ∧ dy ∧ dz be a 3-form. Let V be a solid in R . Then 3

∫

ω =∭

V

f (x, y, z) dxdydz

V

Finally, after all of these definitions, we have a very compact theorem that simultaneously covers the fundamental theorem of calculus, Green's theorem. Stokes' theorem and the divergence theorem. Had we given all of our definitions in n dimensions, rather than just three dimensions, it would cover a lot more. This general theorem is also called Stokes' theorem.

 Theorem 4.7.14. Stokes' Theorem If ω is a k -form (with k = 0, 1, 2) and D is a (k + 1) -dimensional domain of integration, then ∫

dω = ∫

D

ω

∂D

Here ∂D is the boundary of D (suitably oriented). To see the connection between the general Stokes' theorem 4.7.14 and the Stokes' and divergence theorems of the earlier part of this chapter, here are the k = 1 and k = 2 cases of Theorem 4.7.14 again. Let ω = F dx + F dy + F dz be a 1-form and let S be a piecewise smooth oriented surface as in (our original) Stokes' theorem 4.4.1. Then, by Lemma 4.7.12.b, 1

2

3

⇀

⇀

⇀

⇀

⇀

⇀

dω = (∇ × F)1 dy ∧ dz + (∇ × F)2 dz ∧ dx + (∇ × F)3 dx ∧ dy ⇀

⇀

⇀

So, by parts (c) (but with F replaced by ∇ × F ) and (b) of Definition 4.7.13, the conclusion ∫ Stokes' theorem 4.7.14 is

D

⇀

⇀

⇀

^ dS = ∫ ∇×F⋅ n

∬ S

dω = ∫

S

ω =∫

∂S

dω = ∫

∂D

ω

of (the general)

⇀

F ⋅ dr

∂S

which is the conclusion of (our original) Stokes' theorem 4.4.1. be a 2-form and let V be a solid as in the divergence

ω = F1 (x, y, z) dy ∧ dz + F2 (x, y, z) dz ∧ dx + F3 (x, y, z) dx ∧ dy

theorem 4.2.2. Then, by Lemma 4.7.12.c, ⇀

⇀

dω = ∇ ⋅ F dx ∧ dy ∧ dz

So, by parts (d) (with f 4.7.14 is

⇀

⇀

= ∇⋅F

) and (c) of Definition 4.7.13, the conclusion ∫

D

⇀

∭

dω = ∫

⇀

ω

of (the general) Stokes' theorem

⇀

∇ ⋅ F dxdydz = ∫

V

∂D

dω = ∫

V

∂V

ω =∬

F⋅n ^ dS

∂V

which is the conclusion of the divergence theorem 4.2.2.

1. In general, a differential form is defined on a manifold, which is an abstract generalization of a multi-dimensional surface, like a sphere or a torus.

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2. We could also define, for example, a 1-form as an ordered list (F (x, y, z) , F (x, y, z) , F (x, y, z))of three functions and just view F (x, y, z) dx + F (x, y, z) dy + F (x, y, z) dz as another notation for (F (x, y, z) , F (x, y, z) , F (x, y, z)). 3. That d is unique just means that the action of d on any differential form is completely determined by the four rules (a), (b), (c), (d). We will see in Example 4.7.10.c,d,e, that this is indeed the case. 4. Indeed, you can view f (t) as a function of three variables that happens to be independent of two of the three variables. Similarly you can view f (u, v) as a function of three variables that happens to be independent of one of the three variables. 1

1

2

3

2

3

1

2

3

This page titled 4.7: Optional — A Generalized Stokes' Theorem is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

4.7.9

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CHAPTER OVERVIEW 5: True/False and Other Short Questions 5.2: Exercises

This page titled 5: True/False and Other Short Questions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

5.1

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5.2: Exercises  1✳ True or false? ⇀

1. ∇ ⋅ (a × r ) = 0, where a is a constant vector in R , and ⇀

⇀

⇀

⇀

⇀

3

⇀

r

is the vector field

⇀

r = (x, y, z).

2. ∇ × (∇f ) = 0 for all scalar fields f on R with continuous second partial derivatives. 3. ∇ ⋅ (f F) = ∇(f ) ⋅ F + f ∇ ⋅ F, for every vector field F in R with continuous partial derivatives, and every scalar function f in R with continuous partial derivatives. 3

⇀

⇀

⇀

⇀

⇀

3

3

⇀

4. Suppose F is a vector field with continuous partial derivatives in the region D, where D is R without the origin. If ∇ ⋅ F > 0 throughout D, then the flux of F through the sphere of radius 5 with center at the origin is positive. 5. If a vector field F is defined and has continuous partial derivatives everywhere in R , and it satisfies ∇ ⋅ F = 0, everywhere, then, for every sphere, the flux out of one hemisphere is equal to the flux into the opposite hemisphere. 6. If r (t) is a twice continuously differentiable path in R with constant curvature κ, then r (t) parametrizes part of a circle of radius 1/κ. ⇀

3

⇀

⇀

⇀

3

⇀

2

⇀

7. The vector field F = (−

y 2

x +y

2

,

⇀

x 2

x +y

2

)

⇀

⇀

is conservative in its domain, which is R , without the origin. 2

⇀

8. If a vector field F = (P , Q) in R has Q = 0 everywhere in R , then the line integral ∮ F ⋅ d r is zero, for every simple closed curve in R . 9. If the acceleration and the speed of a moving particle in R are constant, then the motion is taking place along a spiral. 2

2

⇀

2

3

 2✳ True or false? ⇀

1. ∇ × (a × r ) = 0, where a is a constant vector in R , and r is the vector field r = (x, y, z). 2. ∇ ⋅ (∇f ) = 0 for all scalar fields f on R with continuous second partial derivatives. ⇀

⇀

⇀

⇀

⇀

3

⇀

⇀

3

⇀

⇀

3. ∇(∇ ⋅ F) = 0 for every vector field F on R with continuous second partial derivatives. 4. Suppose F is a vector field with continuous partial derivatives in the region D, where D is R without the origin. If ∇ ⋅ F = 0, then the flux of F through the sphere of radius 5 with center at the origin is 0. 5. Suppose F is a vector field with continuous partial derivatives in the region D, where D is R without the origin. If 3

⇀

⇀

3

⇀

⇀

⇀

⇀

⇀

3

⇀

⇀

∇ × F = 0 then ∮ F ⋅ d r is zero, for every simple and smooth closed curve C in R which avoids the origin. 6. If a vector field F is defined and has continuous partial derivatives everywhere in R , and it satisfies ∇ ⋅ F > 0, everywhere, then, for every sphere, the flux out of one hemisphere is larger than the flux into the opposite hemisphere. 7. If r (t) is a path in R with constant curvature κ, then r (t) parametrizes part of a circle of radius 1/κ. ⇀

3

C

⇀

⇀

⇀

3

3

⇀

⇀

8. The vector field F = (−

⇀

y 2

x +y

2

,

x 2

x +y

2

, z)

is conservative in its domain, which is R , without the z -axis. 3

9. If all flow lines of a vector field in R are parallel to the z -axis, then the circulation of the vector field around every closed curve is 0. 10. If the speed of a moving particle is constant, then its acceleration is orthogonal to its velocity. 3

 3✳ ⇀′′

1. True or false? If r (t) is the position at time t of an object moving in R , and r (t) is twice differentiable, then | r (t)| is the tangential component of its acceleration. ^ ^ ^ 2. Let r (t) is a smooth curve in R with unit tangent, normal and binormal vectors T (t), N(t), B(t). Which two of these vectors span the plane normal to the curve at r (t)? ^ 3. True or false? If F = P ^ ı ı + Q^ ȷ ȷ + Rk is a vector field on R such that P , Q, R have continuous first order derivatives, and if ∇ × F = 0 everywhere on R , then F is conservative. ^ 4. True or false? If F = P ^ ı ı + Q^ ȷ ȷ + Rk is a vector field on R such that P , Q, R have continuous second order derivatives, ⇀

⇀

3

⇀

3

⇀

⇀

⇀

⇀

⇀

⇀

⇀

3

3

⇀

3

⇀

then ∇ × (∇ ⋅ F ) = 0.

5.2.1

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⇀

⇀

5. True or false? If F is a vector field on R such that |F(x, y, z)| = 1 for all x, y, z, and if S is the sphere 3

⇀

^ dS = 4π. then ∬ F ⋅ n 6. True or false? Every closed surface S in R is orientable. (Recall that S is closed if it is the boundary of a solid region E.) 2

x

+y

2

+z

2

= 1,

S

3

 4✳ 1. In the curve shown below (a helix lying in the surface of a cone), is the curvature increasing, decreasing, or constant as z increases?

2. Of the two functions shown below, one is a function f (x) and one is its curvature κ(x). Which is which?

3. Let C be the curve of intersection of the cylinder x + z = 1 and the saddle xz = y. Parametrise C . (Be sure to specify the domain of your parametrisation.) 4. Let H be the helical ramp (also known as a helicoid) which revolves around the z -axis in a clockwise direction viewed from above, beginning at the y-axis when z = 0, and rising 2π units each time it makes a full revolution. Let S be the the portion of H which lies outside the cylinder x + y = 4, above the z = 0 plane and below the z = 5 plane. Choose one of the following functions and give the domain on which the function you have chosen parametrizes S. (Hint: Only one of the following functions is possible.) 2

2

1. 2. 3. 4.

2

2

⇀

r (u, v) = (u cos v, u sin v, u)

⇀

r (u, v) = (u cos v, u sin v, v)

⇀

r (u, v) = (u sin v, u cos v, u)

⇀

r (u, v) = (u sin v, u cos v, v)

5. Write down a parametrized curve of zero curvature and arclength 1. (Be sure to specify the domain of your parametrisation.) 6. If ∇ ⋅ F is a constant C on all of R , and S is a cube of unit volume such that the flux outward through each side of S is 1, what is C ? 7. Let ⇀

⇀

3

⇀

F(x, y) = (ax + by , cx + dy) ⇀

Give the full set of a, b, c and d such that F is conservative. 8. If r (s) has been parametrized by arclength (i.e. s is arclength), what is the arclength of r (s) between s = 3 and s = 5? 9. Let F be a 2D vector field which is defined everywhere except at the points marked P and Q. Suppose that ∇ × F = 0 everywhere on the domain of F. Consider the five curves R, S, T , U , and V shown in the picture. ⇀

⇀

⇀

⇀

⇀

⇀

Which of the following is necessarily true?

5.2.2

https://math.libretexts.org/@go/page/92321

⇀

1.

∫

2.

∫

3.

∫

4.

∫

⇀

⇀

F ⋅ dr = ∫

S

T ⇀

⇀

⇀

F ⋅ dr = ∫

R

S ⇀

⇀

F ⋅ dr

U ⇀

⇀

F ⋅ dr + ∫

R ⇀

⇀

⇀

F ⋅ dr = ∫

T ⇀

⇀

⇀

F ⋅ dr = 0

U ⇀

⇀

F ⋅ dr + ∫

F ⋅ dr = ∫

U

⇀

⇀

F ⋅ dr = ∫

T ⇀

⇀

S ⇀

∫

⇀

⇀

F ⋅ dr = ∫

F ⋅ dr + ∫

R

5.

⇀

F ⋅ dr

⇀

F ⋅ dr

S

⇀

F ⋅ dr = 0

V ⇀

10. Write down a 3D vector field F such that for all closed surfaces S, the volume enclosed by S is equal to ⇀

^ dS F⋅n

∬ S

th

⇀

^ 11. Consider the vector field F in the xy-plane shown below. Is the k

⇀

⇀

component of ∇ × F at P positive, negative or zero?

 5✳ Say whether the following statements are true or false. ⇀

1. If F is a 3D vector field defined on all of R , and S and S are two surfaces with the same boundary, but ∬ F⋅n ^ dS ≠ ∬ F⋅n ^ dS, then ∇ ⋅ F is not zero anywhere. 3

1

⇀

S1

⇀

⇀

2

⇀

S2

⇀

⇀

⇀

⇀

2. If F is a vector field satisfying ∇ × F = 0 whose domain is not simply-connected, then F is not conservative. 3. The osculating circle of a curve C at a point has the same unit tangent vector, unit normal vector, and curvature as C at that point. 4. A planet orbiting a sun has period proportional to the cube of the major axis of the orbit. ⇀

⇀

⇀

⇀

5. For any 3D vector field F, ∇ ⋅ (∇ × F) = 0. 6. A field whose divergence is zero everywhere in its domain has closed surfaces S in its domain. 7. The gravitational force field is conservative. 8. If F is a field defined on all of R such that ∫ F ⋅ d r = 3 for some curve C , then ∇ × F is non-zero at some point. 9. The normal component of acceleration for a curve of constant curvature is constant. 10. The curve defined by ⇀

⇀

3

⇀

⇀

⇀

C

5.2.3

https://math.libretexts.org/@go/page/92321

⇀

4

4

r 1 (t) = cos(t ) ^ ı ı + 3t ^ ȷ ȷ,

−∞ < t < ∞,

is the same as the curve defined by ⇀ r 2 (t)

^ = cos t ^ ı ı + 3t k,

−∞ < t < ∞

 6✳ Which of the following statements are true (T) and which are false (F)? ⇀

All real valued functions f (x, y, z) and all vector fields F(x, y, z) have domain R unless specified otherwise. 3

1. If f is a continuous real valued function and S a smooth oriented surface, then ∬

f dS = − ∬

S

f dS

−S

where `−S ' denotes the surface S but with the opposite orientation. ⇀

⇀

⇀

^ dS = 0 for every 2. Suppose the components of the vector field F have continuous partial derivatives. If ∬ ∇ × F ⋅ n closed smooth surface, then F is conservative. 3. Suppose S is a smooth surface bounded by a smooth simple closed curve C . The orientation of C is determined by that of S as in Stokes' theorem. Suppose the real valued function f has continuous partial derivatives. Then S

⇀

∂f ∫

f dx = ∬

C

(

∂f

∂z

S

^ ^ dS k) ⋅ n

^ ȷ ȷ − ∂y

4. Suppose the real valued function f (x, y, z) has continuous second order partial derivatives. Then ⇀

⇀

⇀

⇀

(∇f ) × (∇f ) = ∇ × (∇f )

5. The curve parameterized by ⇀

3

r (t) = (2 + 4 t

has curvature κ(t) = 0 for all t. 6. If a smooth curve is parameterized by

⇀

r (s)

3

, −t

3

, 1 − 2t )

−∞ < t < ∞

where s is arc length, then its tangent vector satisfies ⇀′

| r (s)| = 1

7. If S is the sphere x

2

+y

2

+z

2

⇀

=1

and F is a constant vector field, then ∬

S

⇀

⇀

F⋅n ^ dS = 0.

8. There exists a vector field F whose components have continuous second order partial derivatives such that ⇀

⇀

∇ × F = (x, y, z).

 7✳ ⇀

The vector field F = P (x, y) ^ ı ı + Q(x, y) ^ ȷ ȷ is plotted below.

5.2.4

https://math.libretexts.org/@go/page/92321

In the following questions, give the answer that is best supported by the plot. 1. The derivative P at the point labelled A is (a) positive, (b) negative, (c) zero, (d) there is not enough information to tell. 2. The derivative Q at the point labelled A is (a) positive, (b) negative, (c) zero, (d) there is not enough information to tell. ^ ^ 3. The curl of F at the point labelled A is (a) in the direction of +k (b) in the direction of −k (c) zero (d) there is not enough information to tell. 4. The work done by the vector field on a particle travelling from point B to point C along the curve C is (a) positive (b) negative (c) zero (d) there is not enough information to tell. 5. The work done by the vector field on a particle travelling from point B to point C along the curve C is (a) positive (b) negative (c) zero (d) there is not enough information to tell. 6. The vector field F is (a) the gradient of some function f (b) the curl of some vector field G (c) not conservative (d) divergence free. y

x

⇀

1

2

⇀

 8✳ Which of the following statements are true (T) and which are false (F)? 1. The curve defined by ⇀

2

2

2

r 1 (t) = cos(t ) ^ ı ı + sin(t ) ^ ȷ ȷ + 2t

^ k,

−∞ < t < ∞

is the same as the curve defined by ⇀ r 2 (t)

^ = cos t ^ ı ı + sin t ^ ȷ ȷ + 2t k,

−∞ < t < ∞

2. The curve defined by ⇀ r 1 (t)

2

2

2

= cos(t ) ^ ı ı + sin(t ) ^ ȷ ȷ + 2t

^ k,

0 ≤t ≤1

is the same as the curve defined by ⇀ r 2 (t)

3. If a smooth curve is parameterized by

⇀

r (s)

^ = cos t ^ ı ı + sin t ^ ȷ ȷ + 2t k,

0 ≤t ≤1

where s is arc length, then its tangent vector satisfies

5.2.5

https://math.libretexts.org/@go/page/92321

⇀′

| r (s)| = 1

4. If r (t) defines a smooth curve C in space that has constant curvature κ > 0, then C is part of a circle with radius 1/κ. 5. If the speed of a moving object is constant, then its acceleration is orthogonal to its velocity. 6. The vector field ⇀

−y

⇀

F(x, y, z) =

2

x

x

+y

^ ı ı +

2

2

x

+y

2

^ ^ ȷ ȷ + zk

is conservative. ⇀

7. Suppose the vector field F(x, y, z) is defined on an open domain and its components have continuous partial derivatives. If ∇ × F = 0, then F is conservative. 8. The region D = {(x, y)∣∣Bx + y > 1} is simply connected. 9. The region D = {(x, y)∣∣By − x > 0} is simply connected. ⇀

⇀

⇀

2

2

2

⇀

10. If F is a vector field whose components have two continuous partial derivatives, then ⇀

∬

⇀

^ dS = 0 ∇×F⋅ n

S

when S is the boundary of a solid region E in R

3

.

 9✳ Which of the following statements are true (T) and which are false (F)? 1. If a smooth curve C is parameterized by r (s) where s is arc length, then the tangent vector r (s) satisfies | r (s)| = 1. 2. If r (t) defines a smooth curve C in space that has constant curvature κ > 0, then C is part of a circle with radius 1/κ. ⇀′

⇀

⇀′

⇀

⇀

3. Suppose F is a continuous vector field with open domain D. If ⇀

∫

⇀

F ⋅ dr = 0

C ⇀

for every piecewise smooth closed curve C in D, then F is conservative. ⇀

⇀

4. Suppose F is a vector field with open domain D, and the components of F have continuous partial derivatives. If ∇ × F = 0 everywhere on D, then F is conservative. 5. The curve defined by ⇀

⇀

⇀

⇀

2

2

2

r 1 (t) = cos(t ) ^ ı ı + sin(t ) ^ ȷ ȷ + 2t

^ k,

−∞ < t < ∞

is the same as the curve defined by ⇀

^ r 2 (t) = cos t ^ ı ı + sin t ^ ȷ ȷ + 2t k,

−∞ < t < ∞

6. The curve defined by ⇀

2

2

2

r 1 (t) = cos(t ) ^ ı ı + sin(t ) ^ ȷ ȷ + 2t

^ k,

0 ≤t ≤1

is the same as the curve defined by ⇀ r 2 (t)

^ = cos t ^ ı ı + sin t ^ ȷ ȷ + 2t k,

0 ≤t ≤1

⇀

7. Suppose F(x, y, z) is a vector field whose components have continuous second order partial derivatives. Then ⇀

⇀

∇ ⋅ (∇ × F ) = 0. ⇀

⇀

8. Suppose the real valued function f (x, y, z) has continuous second order partial derivatives. Then ∇ ⋅ (∇f ) = 0. 9. The region D = {(x, y)∣∣x + y > 1} is simply connected. 10. The region D = {(x, y)∣∣y − x > 0} is simply connected. 2

2

2

5.2.6

https://math.libretexts.org/@go/page/92321

 10 ✳ ⇀

⇀

Let F, G be vector fields, and f , g be scalar fields. Assume F, G, f , g are defined on all of derivatives of all orders everywhere. Mark each of the following as True (T) or False (F). ⇀

3

R

and have continuous partial

⇀

1. If C is a closed curve and ∇f = 0 , then ∫ f ds = 0. 2. If r (t) is a parametrization of a smooth curve C and the binormal B(t) is constant then C is a straight line. 3. If r (t) is the position of a particle which travels with constant speed, then r (t) ⋅ r (t) = 0. 4. If C is a path from points A to B, then the line integral ∫ (F × G) ⋅ d r is independent of the path C . 5. The line integral ∫ f ds does not depend of the orientation of the curve C . 6. If S is a parametric surface r (u, v) then a normal to S is given by C

⇀

⇀′

⇀

⇀

⇀′′

⇀

C

C

⇀

⇀

⇀

∂r

∂r ×

∂u

7. The surface area of the parametric surface S given by ∬

∂u

⇀

^ r (u, v) = x(u, v) ^ ı ı + y(u, v) ^ ȷ ȷ + z(u, v) k,

(1 + (

D ⇀

∂z ∂u

⇀

2

)

+(

∂z ∂v

2

(u, v) ∈ D,

is given by

1/2

) )

dudv

⇀

8. If F is the velocity field of an incompressible fluid then ∇ ⋅ F = 0. ⇀

⇀

⇀

⇀

⇀

⇀

9. ∇ ⋅ (F × G) = (∇ ⋅ F)G + (∇ ⋅ G)F

 11 ✳ Say whether the following statements are true (T) or false (F). You may assume that all functions and vector fields are defined everywhere and have derivatives of all orders everywhere. ⇀

⇀

⇀

1. The divergence of ∇ × F is zero, for every F. 2. In a simply connected region, ∫ F ⋅ d r depends only on the endpoints of C . ⇀

⇀

C

⇀

3. If ∇f = 0, then f is a constant function. 4. If ∇ × F = 0 , then F is a constant vector field. ^ dS = 0 for every closed surface S. 5. If ∇ ⋅ F = 0, then ∬ F ⋅ n ⇀

⇀

⇀

⇀

⇀

⇀

⇀

S

⇀

⇀

⇀

6. If ∫ F ⋅ d r = 0 for every closed curve C , then ∇ × F = 0. 7. If r (t) is a path in three space with constant speed | v (t)|, then the acceleration is perpendicular to the tangent vector, i.e. ⇀

C

⇀

⇀

^ a ⋅ T = 0.

8. If r (t) is a path in three space with constant curvature κ, then r (t) parameterizes part of a circle of radius 1/κ. 9. Let F be a vector field and suppose that S and S are oriented surfaces with the same boundary curve C , and C is given ⇀

⇀

⇀

1

2

the direction that is compatible with the orientations of S and S . Then ∬ 1

2

S1

⇀

⇀

^ dS = ∬ F⋅n

10. Let A(t) be the area swept out by the trajectory of a planet from time t = 0 to time t. The

S2

^ dS. F⋅n

dA dt

is constant.

 12 ✳ ⇀

Find the correct identity, if f is a function and G and F are vector fields. Select the true statement. ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

1. ∇ ⋅ (f F) = f ∇ × (F) + (∇f ) × F 2. ∇ ⋅ (f F) = f ∇ ⋅ (F) + F ⋅ ∇f ⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

⇀

3. ∇ × (f F) = f ∇ ⋅ (F) + F ⋅ ∇f 4. None of the above are true.

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 13 ✳ ⇀

True or False. Consider vector fields F and scalar functions f and g which are defined and smooth in all of three-dimensional space. Let r = (x, y, z) represent a variable point in space, and let ω = (ω , ω , ω ) be a constant vector. Let Ω be a smoothly ^ . Which of the following are identites, always valid under these assumptions? bounded domain with outer normal n ⇀

1

⇀

2

3

⇀

1. ∇ ⋅ ∇f ⇀

=0

⇀

⇀

2. F × ∇f ⇀2

⇀

= f ∇⋅F ⇀

⇀

3. ∇ f = ∇(∇ ⋅ f ) 4. ∇ × ∇f = 0 5. (∇ × f ) + (∇ × g) = ∇f × ∇g 6. ∇ ⋅ ∇ × F = 0 ⇀

⇀

⇀

⇀

⇀

⇀

⇀

7. ∇ ⋅

⇀

⇀

⇀ r

⇀

⇀

⇀ 2 | r |

=0

⇀

for

⇀

⇀

r ≠ 0

⇀

8. ∇ × (ω × r ) = 0 ⇀

⇀

9. ∭

⇀

⇀

f ∇ ⋅ F dV = − ∭

Ω

10.

⇀

⇀

∇f ⋅ F dV + ∬

Ω

^ dS fF ⋅ n

∂Ω

⇀

^ dS = − ∭ fn

∬ ∂Ω

∇f dV

Ω

 14 ✳ Determine if the given statements are True or False. Provide a reason or a counterexample. 1. A constant vector field is conservative on R . 2. If ∇ ⋅ F = 0 for all points in the domain of F then F is a constant vector field. 3

⇀

⇀

⇀

⇀

⇀

3. Let

⇀

r (t)

be a parametrization of a curve C in R . If 3

⇀

r (t)

and

dr dt

are orthogonal at all points of the curve C , then C lies

on the surface of a sphere x + y + z = a for some a > 0. 4. The curvature κ at a point on a curve depends on the orientation of the curve. 5. The domain of a conservative vector field must be simply connected. 2

2

2

2

 15 ✳ Provide a short answer to each question. ⇀

^ 1. Compute ∇ ⋅ (x y ^ ı ı + e sin x ^ ȷ ȷ +e k) ^ 2. Compute ∇ × (cos x ^ ı ı −y z^ ȷ ȷ + xz k) 3. Let ⇀

2

y

2

zx

3

x

⇀

F =

2

x

+y

y 2

^ ı ı +

2

x

+y

2

^ ȷ ȷ +z

2

^ k

⇀

and let D be the domain of F. Consider the following four statments. 1. 2. 3. 4.

is connected is disconnected D is simply connected D is not simply connected D D

Choose one of the following: 1. 2. 3. 4. 5.

(II) and (III) are true (I) and (III) are true (I) and (IV) are true (II) and (IV) are true Not enough information to determine

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4. True or False? If the speed of a particle is constant then the acceleration of the particle is zero. If your answer is True, provide a reason. If your answer is False, provide a counter example.

 16 ✳ Are each of the following statements True or False? Recall that f are continuous. ⇀

⇀

⇀

∈ C

k

means that all derivatives of f up to order k exist and

1. ∇ × (f ∇f ) = 0 for all C scalar functions f in R . 2. ∇ ⋅ (f F) = ∇f ⋅ F + f ∇ ⋅ F for all C scalar functions f and C vector fields F in R . 3. A smooth space curve C with constant curvature κ = 0 must be a part of a straight line. 4. A smooth space curve C with constant curvature κ ≠ 0 must be part of a circle of radius 1/κ. 5. If f is any smooth function defined in R and if C is any circle, then ∫ ∇f ⋅ d r = 0. ⇀

⇀

⇀

3

2

⇀

⇀

⇀

1

⇀

1

⇀

3

⇀

⇀

3

⇀

C

⇀

6. Suppose F is a smooth vector field in R and ∇ ⋅ F = 0 everywhere. Then, for every sphere, the flux out of one hemisphere is equal to the flux into the opposite hemisphere. 7. Let F(x, y, z) be a continuously differentiable vector field which is defined for every (x, y, z). Then, 3

⇀

⇀

∬

S

⇀

for any closed surface S. (A closed surface is a surface that is the boundary of a solid region.)

^ dS = 0 ∇×F⋅ n

 17 ✳ True or false (reasons must be given): 1. If a smooth vector field on R is curl free and divergence free, then its potential is harmonic. By definition, ϕ(x, y, z) is 3

harmonic if (

∂

2

2

∂x

+

∂

2

∂y

2

+

∂

2

∂z

2

)ϕ(x, y, z) = 0.

⇀

2. If F is a smooth conservative vector field on R , then its flux through any smooth closed surface is zero. 3

 18 ✳ The following statements may be true or false. Decide which. If true, give a proof. If false, provide a counter-example. 1. If f is any smooth function defined in R and if C is any circle, then ∫ ^ 2. There is a vector field F that obeys ∇ × F = x ^ ı ı +y ^ ȷ ȷ + z k. 3

C

⇀

⇀

⇀

⇀

∇f ⋅ d r = 0.

⇀

 19 ✳ Short answers: ⇀

1. Let S be the level surface f (x, y, z) = 0. Why is ∫ ∇f ⋅ d r = 0 for any curve C on S? 2. A point moving in space with position r (t) at time t satisfies the condition a(t) = f (t) r (t) for all t for some real valued function f . Why is v × r a constant vector? 3. Why is the trajectory of the point in (b) contained in a plane? ^ ^ and unit normal 4. Is the binormal vector, B , of a particle moving in space, always orthogonal to the unit tangent vector T ⇀

C

⇀

⇀

⇀

⇀

^ N?

5. If the curvature of the path of a particle moving in space is constant, is the acceleration zero when maximum speed occurs?

 20 ✳ A region R is bounded by a simple closed curve C. The curve C is oriented such that R lies to the left of C when walking along C in the direction of C. Determine whether or not each of the following expressions is equal to the area of R. You must justify your conclusions. 1.

1 ∫ 2

−y dx + x dy

C

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2.

1 ∫ 2

−x dx + y dy

C

3. ∫

y dx

C

4. ∫

3y dx + 4x dy

C

 21 ✳ Say whether each of the following statements is true or false and explain why. 1. A moving particle has velocity and acceleration vectors that satisfy | v | = 1 and |a| = 1 at all times. Then the curvature of this particle's path is a constant. 2. If F is any smooth vector field defined in R and if S is any sphere, then ⇀

⇀

3

⇀

∬

⇀

^ dS = 0 ∇×F⋅ n

S

Here n ^ is the outward normal to S. ⇀

3. If F and G are smooth vector fields in R and if ∮ 3

⇀

⇀

⇀

F ⋅ dr = ∮

C

G ⋅ dr

⇀

for every circle C , then F = G.

C

 22 ✳ Three quickies: 1. A moving particle with position

⇀

r (t) = (x(t), y(t), z(t))

satisfies ⇀

⇀

⇀

a = f ( r , v) r

for some scalar-valued function f . Prove that r × v is constant. ^ 2. Calculate ∬ (x ^ ı ı −y ^ ȷ ȷ + z k) ⋅ n ^ dS, where S is the boundary of any solid right circular cylinder of radius b with one base in the plane z = 1 and the other base in the plane z = 3. ^ dS, 3. Let F and G be smooth vector fields defined in R . Suppose that, for every circle C , we have ∮ F ⋅ d r = ∬ G ⋅ n ⇀

⇀

2

S

⇀

⇀

3

⇀

⇀

C

⇀

S

where S is the oriented disk with boundary C . Prove that G = ∇ × F. This page titled 5.2: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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CHAPTER OVERVIEW Appendices A: Appendices A.1: Trigonometry A.2: Powers and Logarithms A.3: Table of Derivatives A.4: Table of Integrals A.5: Table of Taylor Expansions A.6: 3d Coordinate Systems A.7: ISO Coordinate System Notation A.8: Conic Sections and Quadric Surfaces A.9: Review of Linear Ordinary Differential Equations B: Hints for Exercises C: Answers to Exercises D: Solutions to Exercises

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1

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A.1

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A.1: Trigonometry A.1.1 Trigonometry — Graphs

sin θ

cos θ

tan θ

A.1.2 Trigonometry — Special Triangles

From the above pair of special triangles we have π sin

1 =

4

– √2

π cos

π sin

1 =

4

– √2

=1

2

π

– √3

= 3

cos 2

3

π

1

π

= 6

tan

– √3

1 =

6 tan

2

π

=

4

– √3

π sin

6

cos

π tan

1 =

2 – = √3

3

A.1.3 Trigonometry — Simple Identities Periodicity sin(θ + 2π) = sin(θ)

cos(θ + 2π) = cos(θ)

Reflection sin(−θ) = − sin(θ)

cos(−θ) = cos(θ)

Reflection around π/4 sin(

π 2

− θ) = cos θ

cos(

π 2

− θ) = sin θ

Reflection around π/2 sin(π − θ) = sin θ

cos(π − θ) = − cos θ

Rotation by π sin(θ + π) = − sin θ

cos(θ + π) = − cos θ

Pythagoras

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2

sin

2

θ + cos 2

tan

θ + 1 = sec 2

1 + cot sin

θ =1

θ = csc

2

2

θ θ

and cos building blocks sin θ

1

tan θ =

1

csc θ = cos θ

cos θ

sec θ = sin θ

cot θ = cos θ

1 =

sin θ

tan θ

A.1.4 Trigonometry — Add and Subtract Angles Sine sin(α ± β) = sin(α) cos(β) ± cos(α) sin(β)

Cosine cos(α ± β) = cos(α) cos(β) ∓ sin(α) sin(β)

Tangent tan α + tan β tan(α + β) = 1 − tan α tan β tan α − tan β tan(α − β) = 1 + tan α tan β

Double angle sin(2θ) = 2 sin(θ) cos(θ) 2

2

cos(2θ) = cos (θ) − sin (θ) 2

= 2 cos (θ) − 1 2

= 1 − 2 sin (θ) 2 tan(θ) tan(2θ) =

2

1 − tan 2

cos

θ

1 + cos(2θ) θ = 2

2

sin

1 − cos(2θ) θ = 2

2

tan

1 − cos(2θ) θ = 1 + cos(2θ)

Products to sums sin(α + β) + sin(α − β) sin(α) cos(β) = 2 cos(α − β) − cos(α + β) sin(α) sin(β) = 2 cos(α − β) + cos(α + β) cos(α) cos(β) = 2

Sums to products α +β sin α + sin β = 2 sin

α −β cos

2

2

α +β sin α − sin β = 2 cos

α −β sin

2

2

α +β cos α + cos β = 2 cos

α −β cos

2

2

α +β cos α − cos β = −2 sin

sin 2

A.1.2

α −β 2

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A.1.5 Inverse Trigonometric Functions

arcsin x

Domain: −1 ≤ x ≤ 1 Range: −

π 2

arctan x

arccos x

Domain: all real numbers

Domain: −1 ≤ x ≤ 1

≤ arcsin x ≤

π 2

Range: −

Range: 0 ≤ arccos x ≤ π

π 2

< arctan x
1, is similar.

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A.2.2

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A.3: Table of Derivatives Throughout this table, a and b are constants, independent of x. dF

′

F (x)

F (x) =

dx ′

′

af (x) + bg(x)

af (x) + bg (x)

f (x) + g(x)

f (x) + g (x)

f (x) − g(x)

f (x) − g (x)

af (x)

af (x)

f (x)g(x)

f (x)g(x) + f (x)g (x)

f (x)g(x)h(x)

f (x)g(x)h(x) + f (x)g (x)h(x) + f (x)g(x)h (x)

f(x)

f (x)g(x)−f(x)g (x)

′

′

′

′

′

′

′

′

′

′

′

′

2

g(x)

g(x) ′

g (x)

1

−

g(x)

2

g(x)

′

f (g(x))

′

f (g(x))g (x)

dF

′

F (x)

F (x) = dx

a

0

a

a−1

x

ax a

a−1

g(x)

′

ag(x )

g (x)

sin x

cosx

sin g(x)

g (x) cosg(x)

cosx

− sin x

cosg(x)

−g (x) sin g(x)

′

′

tan x

sec

2

x

csc x

− csc x cot x

sec x

sec x tan x

cot x

− csc

x

2

x

x

e

e

g(x)

′

e

g(x)

g (x)e

x

x

a

(ln a) a

dF

′

F (x)

F (x) = dx 1

ln x

x ′

g (x)

ln g(x)

log

a

g(x) 1

x

x ln a 1

arcsin x

√1−x2

A.3.1

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′

g (x)

arcsin g(x)

2

√1−g(x)

−

arccosx

1 √1−x2

1

arctan x

1+x2 ′

g (x)

arctan g(x)

2

1+g(x)

−

arccsc x

1 |x|√x2 −1

1

arcsec x

|x|√x2 −1

−

arccot x

1 1+x2

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A.3.2

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A.4: Table of Integrals Throughout this table, a and b are given constants, independent of x and C is an arbitrary constant. f(x)

F (x) = ∫ f(x) dx

af (x) + bg(x)

a ∫ f (x) dx + b ∫ g(x) dx  +  C

f (x) + g(x)

∫ f (x) dx + ∫ g(x) dx  +  C

f (x) − g(x)

∫ f (x) dx − ∫ g(x) dx  +  C

af (x)

a ∫ f (x) dx  +  C

′

′

u(x)v (x)

u(x)v(x) − ∫ u (x)v(x) dx  +  C

′

f (y(x))y (x)

F (y(x)) where F (y) = ∫ f (y) dy

a

ax + C a+ 1

x

a

x

a+1

1

+ C if a ≠ −1

ln |x| + C

x

a+ 1

a

g(x)

′

g(x ) g (x)

a+1

f(x)

+ C if a ≠ −1

F (x) = ∫ f(x) dx

sin x

− cosx + C

g (x) sin g(x)

− cosg(x) + C

cosx

sin x + C

tan x

ln | sec x| + C

csc x

ln | csc x − cot x| + C

sec x

ln | sec x + tan x| + C

cot x

ln | sin x| + C

′

2

sec

2

csc

x

tan x + C

x

− cot x + C

sec x tan x

sec x + C

csc x cot x

− csc x + C

f(x)

F (x) = ∫ f(x) dx

x

x

e g(x)

e

e

′

+C

g(x)

g (x)

e 1

ax

e

a

a

ln a

ln x

ax

 e

1

x

+C

x

 a

+C

+C

x ln x − x + C

1

arcsin x + C

√1−x2

′

g (x)

arcsin g(x) + C

2

√1−g(x)

A.4.1

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1

arcsin

√a2 −x2

1

x a

+C

arctan x + C

1+x2 ′

g (x)

arctan g(x) + C

2

1+g(x)

1

1

a2 +x2

a

arctan

1

arcsecx + C

x √x2 −1

x a

+C

\quad(x

)

> 1

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A.4.2

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A.5: Table of Taylor Expansions Let n ≥ be an integer. Then if the function f has n + 1 derivatives on an interval that contains both x and x, we have the Taylor expansion 0

1

′

f (x)

= f (x0 ) + f (x0 ) (x − x0 ) + 1 +

(n+1)

f

f 2!

′′

1

2

(x0 ) (x − x0 )

n+1

(c) (x − x0 )

+⋯ +

f n!

(n)

n

(x0 ) (x − x0 )

for some c between x0  and x

(n + 1)!

The limit as n → ∞ gives the Taylor series ∞

(n)

f

f (x) = ∑ n=0

for f . When x

0

=0

(x0 )

n!

n

(x − x0 )

this is also called the Maclaurin series for f . Here are Taylor series expansions of some important functions. ∞

e

x

1

n

=∑ n=0

x

for  − ∞ < x < ∞

n! 1

1

2

= 1 +x +

x

∞

1

3

+

x

2

n

+⋯ +

3!

x

+⋯

n!

n

(−1)

2n+1

sin x = ∑

x

n=0

1

n

1

3

=x−

x

(−1)

5

+

x

3! ∞

for  − ∞ < x < ∞

(2n + 1)!

2n+1

−⋯ +

x

5!

+⋯

(2n + 1)!

n

(−1)

2n

cos x = ∑

x

for  − ∞ < x < ∞

(2n)!

n=0

1

n

1

2

=1−

x

4

+

x

2!

(−1)

2n

−⋯ +

x

4!

+⋯

(2n)!

∞

1

n

= ∑x 1 −x

for  − 1 ≤ x < 1

n=0 2

3

= 1 +x +x

+x

n

+⋯ +x

+⋯

∞

1

n

n

= ∑(−1 ) x 1 +x

for  − 1 < x ≤ 1

n=0 2

3

= 1 −x +x ∞

1

for  − 1 ≤ x < 1

1

2

x

−

2

3

x

1 −⋯ −

3

n

x

−⋯

n

n

(−1)

ln(1 + x) = − ∑

1 =x−

n

x

for  − 1 < x ≤ 1

n

n=1

2

x 2

p

+⋯

x

1

(1 + x)

n

n

= −x − ∞

n

+ ⋯ + (−1 ) x

n

ln(1 − x) = − ∑ n=1

−x

n

1 +

3

x

(−1) −⋯ −

3

n

x

−⋯

n

p(p − 1) = 1 + px +

2

x

p(p − 1)(p − 2) +

2

3

x

+⋯

3!

p(p − 1)(p − 2) ⋯ (p − n + 1) +

n

x

+⋯

n!

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A.5.1

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A.6: 3d Coordinate Systems A.6.1 Cartesian Coordinates Here is a figure showing the definitions of the three Cartesian coordinates (x, y, z)

and here are three figures showing a surface of constant x, a surface of constant x, and a surface of constant z.

Finally here is a figure showing the volume element dV in cartesian coordinates.

A.6.2 Cylindrical Coordinates Here is a figure showing the definitions of the three cylindrical coordinates r =  distance from (0, 0, 0) to (x, y, 0) θ =  angle between the the x axis and the line joining (x, y, 0) to (0, 0, 0) z =  signed distance from (x, y, z) to the xy-plane

The cartesian and cylindrical coordinates are related by x = r cos θ

y = r sin θ

− −− −− − 2

r = √x

+y

2

z =z y

θ = arctan

z =z x

Here are three figures showing a surface of constant r, a surface of constant θ, and a surface of constant z.

A.6.1

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Finally here is a figure showing the volume element dV in cylindrical coordinates.

A.6.3 Spherical Coordinates Here is a figure showing the definitions of the three spherical coordinates ρ =  distance from (0, 0, 0) to (x, y, z) φ =  angle between the z axis and the line joining (x, y, z) to (0, 0, 0) θ =  angle between the x axis and the line joining (x, y, 0) to (0, 0, 0)

and here are two more figures giving the side and top views of the previous figure.

The cartesian and spherical coordinates are related by x = ρ sin φ cos θ

y = ρ sin φ sin θ

− −−−−−−−− − 2

ρ = √x

+y

2

+z

2

z = ρ cos φ − −− −− − √ x2 + y 2

y θ = arctan

φ = arctan x

z

Here are three figures showing a surface of constant ρ, a surface of constant θ, and a surface of constant φ.

A.6.2

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Here is a figure showing the surface element dS in spherical coordinates

and two extracts of the above figure to make it easier to see how the factors ρ dφ and ρ sin φ dθ arise.

Finally, here is a figure showing the volume element dV in spherical coordinates

A.6.3

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A.6.4

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A.7: ISO Coordinate System Notation In this text we have chosen symbols for the various polar, cylindrical and spherical coordinates that are standard for mathematics. There is another, different, set of symbols that are commonly used in the physical sciences and engineering. Indeed, there is an international convention, called ISO 80000-2, that specifies those symbols 1. In this appendix, we summarize the definitions and standard properties of the polar, cylindrical and spherical coordinate systems using the ISO symbols.

A.7.1 Polar Coordinates In the ISO convention the symbols ρ and ϕ are used (instead of r and θ ) for polar coordinates. ρ =  the distance from (0, 0) to (x, y) ϕ =  the (counter-clockwise) angle between the x-axis   and the line joining (x, y) to (0, 0)

Cartesian and polar coordinates are related by x = ρ cos ϕ

y = ρ sin ϕ

− −− −− − 2

ρ = √x

+y

2

y ϕ = arctan x

The following two figures show a number of lines of constant ϕ, on the left, and curves of constant ρ, on the right.

Note that the polar angle ϕ is only defined up to integer multiples of 2π. For example, the point (1, 0) on the x-axis could have ϕ = 0, but could also have ϕ = 2π or ϕ = 4π. It is sometimes convenient to assign ϕ negative values. When ϕ < 0, the counterclockwise angle ϕ refers to the clockwise angle |ϕ|. For example, the point (0, −1) on the negative y -axis can have ϕ = − and can also have ϕ = . π 2

3π 2

It is also sometimes convenient to extend the above definitions by saying that x = ρ cos ϕ and negative. For example, the following figure shows (x, y) for ρ = 1, ϕ = and for ρ = −1, ϕ = .

A.7.1

π

π

4

4

y = ρ sin ϕ

even when

ρ

is

https://math.libretexts.org/@go/page/92439

Both points lie on the line through the origin that makes an angle of 45 with the x-axis and both are a distance one from the origin. But they are on opposite sides of the the origin. ∘

The area element in polar coordinates is dA = ρ dρ dϕ

A.7.2 Cylindrical Coordinates In the ISO convention the symbols ρ, ϕ and z are used (instead of r, θ and z ) for cylindrical coordinates. ρ =  distance from (0, 0, 0) to (x, y, 0) ϕ =  angle between the the x axis and the line joining (x, y, 0) to (0, 0, 0) z =  signed distance from (x, y, z) to the xy-plane

The cartesian and cylindrical coordinates are related by x = ρ cos ϕ

y = ρ sin ϕ

− −− −− − 2

ρ = √x

+y

2

z =z y

ϕ = arctan

z =z x

Here are three figures showing a surface of constant ρ, a surface of constant ϕ, and a surface of constant z.

Finally here is a figure showing the volume element dV in cylindrical coordinates.

A.7.2

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A.7.3 Spherical Coordinates In the ISO convention the symbols r (instead of ρ), ϕ (instead of θ ) and θ (instead of ϕ ) are used for spherical coordinates. r =  distance from (0, 0, 0) to (x, y, z) θ =  angle between the z axis and the line joining (x, y, z) to (0, 0, 0) ϕ =  angle between the x axis and the line joining (x, y, 0) to (0, 0, 0)

Here are two more figures giving the side and top views of the previous figure.

The cartesian and spherical coordinates are related by x = r sin θ cos ϕ

y = r sin θ sin ϕ

− −−−−−−−− − 2

r = √x

+y

2

+z

2

z = r cos θ − −− −− − √ x2 + y 2

y ϕ = arctan

θ = arctan x

z

Here are three figures showing a surface of constant r, a surface of constant ϕ, and a surface of constant θ.

Finally, here is a figure showing the volume element dV in spherical coordinates

A.7.3

https://math.libretexts.org/@go/page/92439

and two extracts of the above figure to make it easier to see how the factors r dθ and r sin θ dϕ arise.

1. It specifies more than just those symbols. See https://en.Wikipedia.org/wiki/ISO_31-11 and https://en.Wikipedia.org/wiki/ISO/IEC_80000 The full ISO 80000-2 is available at https://www.iso.org/standard/64973.html — for $$. This page titled A.7: ISO Coordinate System Notation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

A.7.4

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A.8: Conic Sections and Quadric Surfaces A conic section is the curve of intersection of a cone and a plane that does not pass through the vertex of the cone. This is illustrated in the figures below.

An equivalent 1 (and often used) definition is that a conic section is the set of all points in the xy-plane that obey Q(x, y) = 0 with 2

Q(x, y) = Ax

+ By

2

+ C xy + Dx + Ey + F = 0

being a polynomial of degree two 2. By rotating and translating our coordinate system the equation of the conic section can be brought into one of the forms 3 This statement can be justified using a linear algebra eigenvalue/eigenvector analysis. It is beyond what we can cover here, but is not too difficult for a standard linear algeba course. with α, \be, γ > 0, which is an ellipse (or a circle), α x − β y = γ with α, β > 0, γ ≠ 0, which is a hyperbola, x = δy, with δ ≠ 0 which is a parabola. 2

αx

2

+ βy

2

=γ

2

2

The three dimensional analogs of conic sections, surfaces in three dimensions given by quadratic equations, are called quadrics. An example is the sphere x + y + z = 1. 2

2

2

Here are some tables giving all of the quadric surfaces. name equation in standard form

2

x

a2

+

parabolic cylinder

2

y

2

= 1

hyperbolic cylinder 2

x

2

y = ax

a2

b

−

sphere

2

y

2

= 1

2

two lines

one line

two lines

circle

constant cross-section

two lines

two lines

two lines

circle

constant cross-section

ellipse

parabola

hyperbola

circle

y =

2

x +y +z

2

2

= r

b

constant cross-section

x =

z =

elliptic cylinder

sketch

Figure A.8.1. Table of conic sections

A.8.1

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name equation in standard form x =

constant cross-section

ellipsoid 2

x

a2

y

+

elliptic paraboloid

2

2

+

b

z

2

c2

2

= 1

ellipse

x

a2

y

+

elliptic cone

2

2

=

b

2

z

x

c

a2

2

y

+

2

=

b

z

two lines if x

parabola

2

c2

= 0,

hyperbola if

x ≠ 0

y =

constant cross-section

ellipse

two lines if y

parabola

hyperbola if

= 0,

y ≠ 0

z =

constant cross-section

ellipse

ellipse

ellipse

sketch

Figure A.8.2. Table of quadric surfaces-1

name equation in standard form

hyperboloid of one sheet 2

x

a2

+

2

y

2

−

b

z

2

c2

hyperboloid of two sheets 2

= 1

x

a2

+

2

y

2

−

b

z

2

c2

= −1

hyperbolic paraboloid 2

y

2

2

−

b

x

a2

x =

constant cross-section

hyperbola

hyperbola

parabola

y =

constant cross-section

hyperbola

hyperbola

parabola

z =

constant cross-section

ellipse

ellipse

=

z c

two lines if z = 0, hyperbola if z ≠ 0

sketch

Figure A.8.3. Table of quadric surfaces-2 It is outside our scope to prove this equivalence. Technically, we should also require that the constants A, B, C , D, E, F , are real numbers, that A, B, C are not all zero, that Q(x, y) = 0 has more than one real solution, and that the polynomial can't be factored into the product of two polynomials of degree one. This page titled A.8: Conic Sections and Quadric Surfaces is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

A.8.2

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A.9: Review of Linear Ordinary Differential Equations Definition A.9.1. 1. A differential equation is an equation for an unknown function that contains the derivatives of that unknown function. For example y (t) + y(t) = 0 is a differential equation for the unknown function y(t). 2. A differential equation is called an ordinary differential equation (often shortened to “ODE”) if only ordinary derivatives appear. That is, if the unknown function has only a single independent variable. A differential equation is called a partial differential equation (often shortened to “PDE”) if partial derivatives appear. That is, if the unknown function has more than ′′

one independent variable. For example y

′′

is an ODE while

(t) + y(t) = 0

∂

2

u 2

2 ∂

(x, t) = c

∂ t

2

u

∂ x2

(x, t)

3. The order of a differential equation is the order of the highest derivative that appears. For example y second order ODE. 4. An ordinary differential equation that is of the form a0 (t)y

(n)

(t) + a1 (t)y

(n−1)

is a PDE. ′′

is a

(t) + y(t) = 0

(t) + ⋯ + an (t)y(t) = F (t)

(A.9.1)

with given coefficient functions a (t), ⋯ , a (t) and F (t) is said to be linear. Otherwise, the ODE is said to be nonlinear. For example, y (t) + y(t) = 0, y (t)y (t) + y(t) = 0 and y (t) = e are all nonlinear. 0

′

2

n

′

′′

′

y(t)

5. The ODE (A.9.1) is said to have constant coefficients if the coefficients a (t), a (t), ⋯ , a (t) are all constants. Otherwise, it is said to have variable coefficients. For example, the ODE y (t) + 7y(t) = sin t is constant coefficient, while y (t) + ty(t) = sin t is variable coefficient. 6. The ODE (A.9.1) is said to be homogeneous if F (t) is identically zero. Otherwise, it is said to be inhomogeneous or nonhomogeneous. For example, the ODE y (t) + 7y(t) = 0 is homogeneous, while y (t) + 7y(t) = sin t is inhomogeneous. A homogeneous ODE always has the trivial solution y(t) = 0. 7. An initial value problem is a problem in which one is to find an unknown function y(t) that satisfies both a given ODE and given initial conditions, like y(t ) = 1, y (t ) = 0. Note that all of the conditions involve the function y(t) (or its derivatives) evaluated at a single time t = t . 8. A boundary value problem is a problem in which one is to find an unknown function y(t) that satisfies both a given ODE and given boundary conditions, like y(t ) = 0, y(t ) = 0. Note that the conditions involve the function y(t) (or its derivatives) evaluated at two different times. 0

1

n

′′

′′

′′

′′

′

0

0

0

0

1

The following theorem gives the form of solutions to the ODE (A.9.1).

 Theorem A.9.2 Assume that the coefficients a

0 (t),

a1 (t), ⋯ , an−1 (t), an (t)

and F (t) are continuous functions and that a

0 (t)

is not zero.

1. The general solution to the ODE (A.9.1) is of the form y(t) = yp (t) + C1 y1 (t) + C2 y2 (t) + ⋯ + Cn yn (t)

(A.9.2)

where is the order of (A.9.1) is any solution to (A.9.1) C , C , ⋯ , C are arbitrary constants y , y , ⋯,y are n independent solutions to the homogenous equation n

yp (t) 1

1

2

n

2

n

a0 (t)y

(n)

(t) + a1 (t)y

(n−1)

′

(t) + ⋯ + an−1 (t)y (t) + an (t)y(t) = 0

associated to (A.9.1). “Independent” just means that no y can be written as a linear combination of the other y 's. For example, y (t) cannot be expressed in the form b y (t) + ⋯ + b y (t). i

1

In (A.9.2), solution”.

yp

2

is called the “particular solution” and

2. Given any constants b

0,

⋯ , bn−1

j

2

n

n

C1 y1 (t) + C2 y2 (t) + ⋯ + Cn yn (t)

is called the “complementary

there is exactly one function y(t) that obeys the ODE (A.9.1) and the initial conditions

y(0) = b0

′

y (0) = b1

A.9.1

⋯

y

(n−1)

(0) = bn−1

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 Example A.9.3. RLC circuit As an example of the most commonly used techniques for solving linear, constant coefficient ODE's, we consider the RLC circuit, which is the electrical circuit consisting of a resistor of resistance R, a coil (or solenoid) of inductance L, a capacitor of capacitance C and a voltage source arranged in series, as shown below. Here R, L and C are all nonnegative constants.

We're going to think of the voltage x(t) as an input signal, and the voltage y(t) as an output signal. The goal is to determine the output signal produced by a given input signal. If i(t) is the current flowing at time t in the loop as shown and q(t) is the charge on the capacitor, then the voltages across

R, L

and C , respectively, at time

are

t

di Ri(t), L

(t) dt

1

and

q(t)

y(t) =

C

.

By

the Kirchhoff's law that says that the voltage between any two points has to be independent of the path used to travel between the two points, these three voltages must add up to x(t) so that q(t)

di Ri(t) + L

(t) +

= x(t)

dt

(A.9.3)

C

Assuming that R,  L,  C and x(t) are known, this is still one differential equation in two unknowns, i(t) and q(t). Fortunately, there is a relationship between the two. Namely dq i(t) =

′

(t) = C y (t)

(A.9.4)

dt

This just says that the capacitor cannot create or destroy charge on its own; all charging of the capacitor must come from the current. Substituting (A.9.4) into (A.9.3) gives LC y

′′

′

(t) + RC y (t) + y(t) = x(t)

As a concrete example, we'll take an ac voltage source and choose the origin of time so that Then the differential equation becomes LC y

′′

x(0) = 0, x(t) = E0 sin(ωt).

′

(t) + RC y (t) + y(t) = E0 sin(ωt)

(A.9.5)

This is a second order, linear, constant coefficient ODE. So we know, from Theorem A.9.2, that the general solution is of the form y (t) + C y (t) + C y (t), where p

1

1

2

2

the particular solution, is any one solution to (A.9.5), C , C are arbitrary constants and y (t), y (t) are any two independent solutions of the corresponding homogeneous equation yp (t), 1

1

2

2

LC y

′′

′

(t) + RC y (t) + y(t) = 0

So to find the general solution to (A.9.5), we need to find three functions: y

1 (t),

(A.9.6)

and y

y2 (t)

p (t).

Finding y (t) and y (t): The best way to find y and y is to guess them. Any solution, y (t), of (A.9.6) has to have the property that y (t), RC y (t) and LC y (t) cancel each other out for all t. We choose our guess so that y (t), y (t) and y (t) are all proportional to a single function of t. Then it will be easy to see if y (t), RC y (t) and LC y (t) all cancel. All derivatives of the function e are again proportional to e . Hence we try y (t) = e , with the constant r to the determined. This guess is a solution of (A.9.6) if and only if 1

2

h

1

′

′′

h

h

2

h

h

′′

h

h

rt

rt

2

LC r e

rt

+ RC re

rt

+e

rt

=0

2

LC r

′′

h

h

′

h

rt

h

⟺

′

+ RC r + 1 = 0

−−−−−−−−− − 2 2 −RC ± √ R C − 4LC ⟺

r = 2LC

How we proceed depends on the sign of R

2

(A.9.7) ≡ r1,2 − −

C

2

− 4LC .

That is, whether R > 2√

A.9.2

L

C

− −

or R < 2√

L

C

− −

or R = 2√

L

C

.

https://math.libretexts.org/@go/page/92441

Finding y

1 (t)

1 (t)

1 (t)

when R > 2√

2 (t),

may take y Finding y

− −

and y =e

and y

r1 t

2 (t)

and y

=e

Then R

2

:

C

2

and r and r are two different real numbers. We

− 4LC > 0,

1

2

so that the complimentary solution is C

r2 t

1 y1 (t)

− −

when R < 2√

2 (t),

L C

L

C

Then R

2

:

C

2

+ C2 y2 (t) = C1 e

r1 t

+ C2 e

r2 t

.

and r and r are the two different complex

− 4LC < 0

1

2

numbers −ρ ± iν , where −−−−−−−−− − √ 4LC − R2 C 2

R ρ =

and

ν =

2L

2LC

We may again take C e + C e as the complimentray solution. However we can also rewrite terms of real valued functions by using that e = cos θ ± i sin θ: 1

r1 t

2

r2 t

C1 e

r1 t

+ C2 e

r2 t

in

±iθ

C1 e

r1 t

+ C2 e =e =e

where 2 D = C + C ,  D complementary solution. 1

1

2

2

r2 t

−ρt

−ρt

=e

−ρt

[ C1 e

iν t

+ C2 e

−iν t

]

[ C1 { cos(ν t) + i sin(ν t)} + C2 { cos(ν t) − i sin(ν t)}]

[ D1 cos(ν t) + D2 sin(ν t)]

So we may also take

= i(C1 − C2 ).

y1 (t) = e

−ρt

cos(ν t), y2 (t) = e

−ρt

sin(ν t)

in the

There is yet a third useful way to write the complementary solution. Think of (D , D ) as a point in the xy-plane. Call the polar coordinates of that point A and θ so that D = A cos θ and D = A sin θ. Then, using the trig identity cos(α + β) = cos α cos β − sin α sin β, with α = ν t and β = −θ, 1

1

e

−ρt

2

2

[ D1 cos(ν t) + D2 sin(ν t)] =e

−ρt

= Ae

(A.9.8)

[A cos(ν t) cos θ + A sin(ν t) sin θ]

−ρt

cos(ν t − θ)

We have, in effect, replaced the two arbitrary constants D and D , whose values would normally be determined by initial conditions, by two other arbitrary constants, R and θ, whose values would also normally be determined by initial conditions. 1

Finding y

1 (t)

− −

and y

when R = 2√

2 (t),

L C

Then R

2

:

C

2

2

so that r

− 4LC = 0

1

= r2 .

We may take y

1

=e

r1 t

but

,

=e is certainly not a second independent solution. So we still need to find y . Here is a trick (called reduction of order 3) for finding the other solutions: look for solutions of the form v(t)e . Here e is the solution we have already found and v(t) is to be determined. To save writing, set ρ = so that r = r = ρ. To save writing also e

r2 t

r1 t

2

−r1 t

−r1 t

R

1

2L

divide ((A.9.5) ) by LC and substitute that

R

h

= 2ρ

L

and

1 LC

2

R

=

2

2

4L

So ((A.9.5) ) is equivalent to

2

(Recall that we are assuming that R

2

=ρ .

=

4L C

.

)

h

y

′′

h

2

′

(t) + 2ρ y (t) + ρ h

yh (t) = 0

Substitute in yh (t) =    v(t)e

−ρt

′

y (t) = −ρv(t)e

−ρt

h

y

′′

h

So when y

h (t)

= v(t)e

−ρt

y

2

(t) =

ρ v(t)e

′

+

−ρt

v (t)e ′

−ρt

− 2ρv (t)e

−ρt

′′

+ v (t)e

−ρt

,

′′

h

2

′

(t) + 2ρ y (t) + ρ h

2

2

yh (t) 2

= [ ρ −2 ρ +ρ ]v(t)e ′′

= v (t)e

−ρt

′

+ [ − 2ρ+2ρ] v (t)e

−ρt

′′

+ v (t)e

−ρt

−ρt

Thus v(t)e is a solution of ((A.9.5) ) whenever the function v (t) = 0 for all t. But, for any values of the constants C and C , v(t) = C + C t has vanishing second derivative so (C + C t)e = (C + C t)e solves ((A.9.5) ). This is of the form C y (t) + C y (t) with y (t) = e , the solution that we found first, and y (t) = te , a second independent solution. So we may take y (t) = te . −ρt

1

2

′′

h

1

2

1

2

−ρt

1

2

−r1 t

−r1 t

h

1

1

2

2

1

−r1 t

2

r1 t

2

A.9.3

https://math.libretexts.org/@go/page/92441

Finding y (t): The best way to find y is to guess it. We guess that the circuit responds to an oscillating input voltage with an output voltage that oscillates at the same frequency. So we try y (t) = A sin(ωt − φ) with the amplitude A and phase φ to be determined. p

p

p

For y

p (t)

to be a solution, we need ′′

′

LC yp (t) + RC yp (t) + yp (t) = E0 sin(ωt)

or 2

− LC ω A sin(ωt − φ) + RC ωA cos(ωt − φ) + A sin(ωt − φ) = E0 sin(ωt) = E0 sin(ωt − φ + φ)

and hence, applying sin(A + B) = sin A cos B + cos A sin B with A = ωt − φ and B = φ, 2

(1 − LC ω )A sin(ωt − φ) + RC ωA cos(ωt − φ) = E0 cos(φ) sin(ωt − φ) + E0 sin(φ) cos(ωt − φ)

Matching coefficients of sin(ωt − φ) and cos(ωt − φ) on the left and right hand sides gives 2

(1 − LC ω )A = E0 cos(φ)

(A.9.9)

RC ωA = E0 sin(φ)

(A.9.10)

It is now easy to solve for A and φ (A.9.10)

RC ω ⟹

tan(φ) =

⟹

φ = arctan

(A.9.9)

1 − LC ω2 RC ω

2

2

√ (A.9.9) + (A.9.10)

⟹

⟹

Naturally, different input frequencies parameters held fixed.

ω

2

1 − LC ω − −−−−−−−−−−−−−−−− −

− −−−−−−−−−−−−−− −

2

2

√ (1−LC ω )

A =

2

+R C

2

2

ω

 A = E0

E0 − −−−−−−−−−−−−−−−− − √ (1−LC ω2 )2 + R2 C 2 ω2

give different output amplitudes

A.

Here is a graph of

A

against

ω,

with all other

Note that there is a small range of frequencies that give a large amplitude response. This is the phenomenon of resonance. It is exploited in the design of radio and television tuning circuitry. It has also been dramatically illustrated in, for example, the collapse 4 of the Tacoma narrows bridge.

 Example A.9.4. Boundary Value Problems By part (b) of Theorem A.9.2, an initial value problem consisting of an n order linear ODE with reasonable 5 coefficients and n initial conditions always has exactly one solution. We shall now see that a boundary value problem may have no solutions at all. Or it may have exactly one solution. Or it may have infinitely many solutions. We shall start by finding all solutions to the ODE th

′

y +y = 0

A.9.4

(A.9.11)

https://math.libretexts.org/@go/page/92441

We shall then impose various boundary conditions and see what happens. The function y(t) = e

rt

is a solution to (A.9.11) if and only if 2

r e

rt

+e

rt

=0

⟺

2

r

+1 = 0

⟺

r = ±i

where i (which electrical engineers often denote 6 j ) is a square root of −1. Thus the general solution to the second order linear ODE (A.9.11) is y(t) = C e + C e , with C and C arbitrary constants. We may rewrite this general solution in terms of sin t and cos t by substituting in ′

it

−it

′

1

2

e

it

′

′

1

2

= cos t + i sin t

e

−it

= cos t − i sin t

This gives ′

′

1

2

y(t) = C ( cos t + i sin t) + C (cos t − i sin t) = C1 cos t + C2 sin t

where C = C + C , and C So there is nothing stopping C 1

′

′

1

2

2 1

= i(C

′

1

=C

′

1

′

Note that there is nothing stopping C and C from being complex numbers. and C = i(C − C ) from being real numbers.

− C ). 2

′

+C , 2

2

′

′

1

2

′

′

1

2

1. Now consider the boundary value problem ′

y +y = 0

y(0) = 0

y(2π) = 1

(A.9.12)

The function y(t) satisfies the ODE if and only if it is of the form y(t) = C1 cos t + C2 sin t

for some constants C and C . A function of this form satisfies the boundary condition y(0) = 0 if and only if 1

2

0 = y(0) = C1 cos 0 + C2 sin 0 = C1

A function of this form satisfies the boundary condition y(2π) = 1 if and only if 1 = y(2π) = C1 cos 2π + C2 sin 2π = C1

The two requirements C = 0 and C = 1 are incompatible. So the boundary value problem (A.9.12) has no solution at all. 2. Next consider the boundary value problem 1

1

π

′

y +y = 0

y(0) = 0

y(

) =0

(A.9.13)

2

The function y(t) satisfies the ODE if and only if it is of the form y(t) = C1 cos t + C2 sin t

for some constants C and C . A function of this form satisfies the boundary condition y(0) = 0 if and only if 1

2

0 = y(0) = C1 cos 0 + C2 sin 0 = C1

A function of this form satisfies the boundary condition y(

π 2

) =0

π 0 = y( 2

So we have a solution if and only if C = C namely y(t) = 0, which is a bit dull. 3. Finally consider the boundary value problem 1

2

′

if and only if

π ) = C1 cos (

=0

2

π ) + C2 sin (

2

) = C2

and the boundary value problem (A.9.13) has exactly one solution,

y +y = 0

y(0) = 0

y(2π) = 0

(A.9.14)

The function y(t) satisfies the ODE if and only if it is of the form y(t) = C1 cos t + C2 sin t

for some constants C and C . A function of this form satisfies the boundary condition y(0) = 0 if and only if 1

2

0 = y(0) = C1 cos 0 + C2 sin 0 = C1

A.9.5

https://math.libretexts.org/@go/page/92441

A function of this form satisfies the boundary condition y(2π) = 0 if and only if 0 = y(2π) = C1 cos(2π) + C2 sin(2π) = C1

So we have a solution if and only if C = 0 and the boundary value problem (A.9.14) has infinitely many solutions, namely y(t) = C sin t with C being an arbitrary constant. 1

2

2

1. Gustav Robert Kirchhoff (1824--1887) was a German physicist. 2. Don't make the mistake of thinking that C and C have to be real numbers, forcing D to be pure imaginary. In most applications, D and D will be pure real and C and C will be complex. 3. The modern method of reduction of order was created by the French mathematician, physicist and music theorist Jean le Rond d'Alembert (1717-1783). The interested reader can easily search out more about his life. 4. There are videos of the collapse on the web. 5. For example, continuous. 6. This is to avoid confusion with the current, which is typically called i. 1

1

2

2

1

2

2

This page titled A.9: Review of Linear Ordinary Differential Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

A.9.6

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B: Hints for Exercises This page titled B: Hints for Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

B.1

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C: Answers to Exercises This page titled C: Answers to Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

C.1

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D: Solutions to Exercises This page titled D: Solutions to Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

D.1

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Index C

L

Conservative Vector Fields

line integrals

2.3: Conservative Vector Fields

curl 2.3: Conservative Vector Fields

T tangent planes

2.4: Line Integrals

3.2: Tangent Planes

N

V

normal vector

E

vector field

3.2: Tangent Planes

equipotential surface 2.3: Conservative Vector Fields

2.1: Definitions and First Examples

P

W

pendulum

F

work

2.5: Optional — The Pendulum

field lines 2.2: Optional — Field Lines

flux integrals 3.3: Surface Integrals 3.4: Interpretation of Flux Integrals

2.4: Line Integrals

S surface integrals 3.3: Surface Integrals

1

https://math.libretexts.org/@go/page/91917

Glossary absolute convergence | if the series \displaystyle

\sum^∞_{n=1}|a_n| converges, the series \displaystyle \sum^∞_{n=1}a_n is said to converge absolutely

absolute error | if B is an estimate of some

quantity having an actual value of A, then the absolute error is given by |A−B|

absolute extremum | if f has an absolute maximum or absolute minimum at c, we say f has an absolute extremum at c

absolute maximum | if f(c)≥f(x) for all x in the domain of f, we say f has an absolute maximum at c

absolute minimum | if f(c)≤f(x) for all x in the domain of f, we say f has an absolute minimum at c

absolute value function | f(x)=\begin{cases}−x, & \text{if } x0 such that for any value c∈(k−ε,k+ε) the solution to the initial-value problem y′=f(x,y),y(x_0)=c approaches k as x approaches infinity

asymptotically unstable solution | y=k if there

exists ε>0 such that for any value c∈(k−ε,k+ε) the solution to the initial-value problem y′=f(x,y),y(x_0)=c never approaches k as xapproaches infinity

autonomous differential equation

perimeter of a circle that is rolling around a fixed circle of the same radius; the equation of a cardioid is r=a(1+\sin θ) or r=a(1+\cos θ)

carrying capacity | the maximum population of an |

| an equation in which the right-hand side is a function of y alone

closed curve | a curve for which there exists a

parameterization \vecs r(t), a≤t≤b, such that \vecs r(a)=\vecs r(b), and the curve is traversed exactly once

closed set | a set S that contains all its boundary

(1+x)^r; it is given by (1+x)^r=\sum_{n=0}^∞(^r_n)x^n=1+rx+\dfrac{r(r−1) }{2!}x^2+⋯+\dfrac{r(r−1)⋯(r−n+1)}{n!}x^n+⋯ for |x|0 over I, then f is concave up over I; if f''< over I, then f is concave down over I

conditional convergence

| if the series \displaystyle \sum^∞_{n=1}a_n converges, but the series \displaystyle \sum^∞_{n=1}|a_n| diverges, the series \displaystyle \sum^∞_{n=1}a_n is said to converge conditionally conic section | a conic section is any curve formed

by the intersection of a plane with a cone of two nappes

connected region | a region in which any two points can be connected by a path with a trace contained entirely inside the region

connected set | an open set S that cannot be

represented as the union of two or more disjoint, nonempty open subsets

conservative field | a vector field for which there exists a scalar function f such that \vecs ∇f=\vecs{F}

1

https://math.libretexts.org/@go/page/91918

constant multiple law for limits | the limit law \lim_{x→a}cf(x)=c⋅\lim_{x→a}f(x)=cL \nonumber

constant multiple rule | the derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative: \dfrac{d} {dx}\big(cf(x)\big)=cf′(x)

constant rule | the derivative of a constant function is zero: \dfrac{d}{dx}(c)=0, where c is a constant

constraint | an inequality or equation involving one

or more variables that is used in an optimization problem; the constraint enforces a limit on the possible solutions for the problem

continuity at a point | A function f(x) is

continuous at a point a if and only if the following three conditions are satisfied: (1) f(a) is defined, (2) \displaystyle \lim_{x→a}f(x) exists, and (3) \displaystyle \lim{x→a}f(x)=f(a)

continuity from the left | A function is

continuous from the left at b if \displaystyle \lim_{x→b^−}f(x)=f(b)

continuity from the right | A function is continuous from the right at a if \displaystyle \lim_{x→a^+}f(x)=f(a)

continuity over an interval | a function that can

be traced with a pencil without lifting the pencil; a function is continuous over an open interval if it is continuous at every point in the interval; a function f(x) is continuous over a closed interval of the form [a,b] if it is continuous at every point in (a,b), and it is continuous from the right at a and from the left at b

contour map | a plot of the various level curves of a given function f(x,y)

convergence of a series | a series converges if the sequence of partial sums for that series converges

convergent sequence | a convergent sequence is a sequence \displaystyle {a_n} for which there exists a real number \displaystyle L such that \displaystyle a_n is arbitrarily close to \displaystyle L as long as \displaystyle n is sufficiently large

coordinate plane | a plane containing two of the

three coordinate axes in the three-dimensional coordinate system, named by the axes it contains: the xy-plane, xz-plane, or the yz-plane

critical point | if f'(c)=0 or f'(c) is undefined, we say that c is a critical point of f

critical point of a function of two variables |

the point (x_0,y_0) is called a critical point of f(x,y) if one of the two following conditions holds: 1. f_x(x_0,y_0)=f_y(x_0,y_0)=0 2. At least one of f_x(x_0,y_0) and f_y(x_0,y_0) do not exist

cross

product | \vecs u×\vecs v= (u_2v_3−u_3v_2)\mathbf{\hat i}− (u_1v_3−u_3v_1)\mathbf{\hat j}+ (u_1v_2−u_2v_1)\mathbf{\hat k}, where \vecs u=⟨u_1,u_2,u_3⟩ and \vecs v=⟨v_1,v_2,v_3⟩ determinant a real number associated with a square matrix parallelepiped a three-dimensional prism with six faces that are parallelograms torque the effect of a force that causes an object to rotate triple scalar product the dot product of a vector with the cross product of two other vectors: \vecs u⋅(\vecs v×\vecs w) vector product the cross product of two vectors. cross-section | the intersection of a plane and a solid object

cubic function | a polynomial of degree 3; that is, a

function of the form f(x)=ax^3+bx^2+cx+d, where a≠0

curl | the curl of vector field \vecs{F}=⟨P,Q,R⟩, denoted \vecs ∇× \vecs{F} is the “determinant” of the matrix \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \ \dfrac{\partial} {\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \ P & Q & R \end{vmatrix}. \nonumber and is given by the expression (R_y−Q_z)\,\mathbf{\hat i} + (P_z−R_x)\,\mathbf{\hat j} +(Q_x−P_y)\,\mathbf{\hat k} ; it measures the tendency of particles at a point to rotate about the axis that points in the direction of the curl at the point

curvature | the derivative of the unit tangent vector with respect to the arc-length parameter

cusp | a pointed end or part where two curves meet cycloid | the curve traced by a point on the rim of a

circular wheel as the wheel rolls along a straight line without slippage

cylinder | a set of lines parallel to a given line passing through a given curve

cylindrical coordinate system | a way to

describe a location in space with an ordered triple (r,θ,z), where (r,θ) represents the polar coordinates of the point’s projection in the xy-plane, and z represents the point’s projection onto the z-axis

decreasing on the interval I | a function

differentiable | a function f(x,y) is differentiable at

(x_0,y_0) if f(x,y) can be expressed in the form f(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x−x_0)+f_y(x_0,y_0) (y−y_0)+E(x,y), where the error term E(x,y) satisfies \lim_{(x,y)→(x_0,y_0)}\dfrac{E(x,y)} {\sqrt{(x−x_0)^2+(y−y_0)^2}}=0

differentiable at a | a function for which f'(a) exists is differentiable at a

differentiable function | a function for which f'(x) exists is a differentiable function

differentiable on S | a function for which f'(x) exists for each x in the open set S is differentiable on S

differential | the differential dx is an independent

variable that can be assigned any nonzero real number; the differential dy is defined to be dy=f'(x)\,dx

differential calculus | the field of calculus

concerned with the study of derivatives and their applications

differential equation | an equation involving a function y=y(x) and one or more of its derivatives

differential form | given a differentiable function y=f'(x), the equation dy=f'(x)\,dx is the differential form of the derivative of y with respect to x

differentiation | the process of taking a derivative

all

direction angles | the angles formed by a nonzero

definite integral | a primary operation of calculus;

direction cosines | the cosines of the angles formed

decreasing on the interval I if, x_1,\,x_2∈I,\;f(x_1)≥f(x_2) if x_10, then f has a local minimum at c; if f''(c)