Vector Analysis Versus Vector Calculus (Instructor Solution Manual, Solutions) [1 ed.] 9781461421993, 1461421993

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ANTONIO GALBIS AND MANUEL MAESTRE

VECTOR ANALYSIS VERSUS VECTOR CALCULUS

Springer

Solutions to exercises

Solved exercises of Chapter 1 1.3.1. Find the divergence of the vector fields (1) F(x, y) = (sin(x) , ex−y ) . � � (2) F(x, y, z) = sin(y) , cos(z) , z3 . Solution: (1) F = ( f1 , f2 ) where f1 (x, y) = sin(x) ; f2 (x, y) = ex−y . Thus Div F =

∂ f1 ∂ f2 + ∂x ∂y

= cos x − ex−y . (2) Let us put f1 (x, y, z) = sin(y), f2 (x, y, z) = cos(z) and f3 (x, y, z) = z3 . Then the divergence of F is ∂ f1 ∂ f2 ∂ f3 Div F = + + ∂x ∂y ∂z = 3z2 . � 1.3.2. Find the divergence and the curl at the point (1, 1, 0) for the vector field F(x, y, z) = (xyz , y , z) .

1

2

Solutions to exercises

Solution: The components of the vector field are f1 (x, y, z) = xyz, f2 (x, y, z) = y, f3 (x, y, z) = z. Then Div F =

∂ f1 ∂ f2 ∂ f3 + + ∂x ∂y ∂z

= yz + 2. Hence, Div F(1, 1, 0) = 2. We now calculate the curl as � � � e1 e2 e3 � � ∂ ∂ ∂ � Curl F = �� ∂ x ∂ y ∂ z �� = (0, xy, −xz). � xyz y z �

In particular,

(Curl F)(1, 1, 0) = (0, 1, 0). �

1.3.3. Check that Div (Curl F) = 0 for the vector field

� � F(x, y, z) = x2 z , x , 2yz

Solution: The components of the vector field are

f1 (x, y, z) = x2 z, f2 (x, y, z) = x, f3 (x, y, z) = 2yz. Hence

� � � e1 e2 e3 � � ∂ ∂ ∂ � Curl F = �� ∂ x ∂ y ∂ z �� = (2z, x2 , 1). � x2 z x 2yz �

That is, the components of Curl F = (g1 , g2 , g3 ) are given by g1 (x, y, z) = 2z, g2 (x, y, z) = x2 , g3 (x, y, z) = 1. Consequently, Div(Curl F) =

∂ g1 ∂ g2 ∂ g3 + + ∂x ∂y ∂z

= 0.

�

1.3.4. Let F : IR3 → IR3 be a vector field and let g : IR3 → IR be a scalar function, both of class C1 on Rn . Check that ∇g) × F. Curl (gF) = g(Curl F) + (∇

Solutions to exercises

3

Solution: If F = ( f1 , f2 , f3 ) then � � � e1 e2 e3 � � ∂ ∂ ∂ � Curl (gF) = �� ∂ x ∂ y ∂ z �� �gf gf gf � 1 2 3 =

� ∂ (g f ) ∂ (g f ) ∂ (g f ) ∂ (g f ) ∂ (g f ) ∂ (g f ) � 3 2 1 3 2 1 − , − , − ∂y ∂z ∂z ∂x ∂x ∂y

while � � � e1 e2 e3 � � ∂g ∂g ∂g � ∇ g × F = �� ∂ x ∂ y ∂ z �� �f f f � 1 2 3 Consequently,

� ∂g ∂g ∂g ∂g ∂g ∂g� . − f2 , f1 − f3 , f2 − f1 = f3 ∂y ∂z ∂z ∂x ∂x ∂y

�∂ f ∂ f2 ∂ f1 ∂ f3 ∂ f2 ∂ f1 � 3 ∇g × F) = g. Curl (gF) − (∇ − , − , − ∂y ∂z ∂z ∂x ∂x ∂y = g Curl F. � 1.3.5. If F, G : IR3 → IR3 are vector fields of class C1 , prove that Div(F × G) = �Curl F , G� − �F , Curl G� . Solution: We put F = ( f1 , f2 , f3 ) and G = (g1 , g2 , g3 ). According to Definition 1.2.11, we have � � � � � � ∂ f3 ∂ f2 ∂ f1 ∂ f3 ∂ f2 ∂ f1 �Curl F , G� = g1 − + g2 − + g3 − . ∂y ∂z ∂z ∂x ∂x ∂y Analogously, �F , Curl G� = f1

�

∂ g3 ∂ g2 − ∂y ∂z

�

+ f2

�

∂ g1 ∂ g3 − ∂z ∂x

�

+ f3

�

∂ g2 ∂ g1 − ∂x ∂y

Moreover, from Definition 1.1.4 it follows that F × G = (h1 , h2 , h3 ), where h1 = f2 g3 − f3 g2 , h2 = f3 g1 − f1 g3 , h3 = f1 g2 − f2 g1 and from Definition 1.2.10 we have

�

.

4

Solutions to exercises

Div(F × G) =

∂ h1 ∂ h2 ∂ h3 + + . ∂x ∂y ∂z

Finally, it is routine (although a tedious calculation) to check

∂ h1 ∂ h2 ∂ h3 + + = �Curl F , G� − �F , Curl G� . ∂x ∂y ∂z � The previous exercise is solved with full details in Exercise 6.7.7 using the theory of differential forms presented in Chapter 6.

Solved exercises of Chapter 2 2.8.1 Evaluate the length of the path

π γ : [0, ] → IR2 , γ (t) = (et cos(t), et sin(t)). 2 Solution: We first evaluate

γ � (t) = et (cost − sint , sint + cost). Hence, � �1 2 � γ � (t) � = et (cost − sint)2 + (sint + cost)2 √ = et 2

and the length of the path is given by √ � 2

0

π 2

et dt =

� √ � π 2 e2 −1 . �

2.8.2 Obtain a parameterization, counterclockwise, of the ellipse y x ( )2 + ( )2 = 1 ; (a, b > 0). a b Solution: If (x, y) is a point of the ellipse then the point x y ( , ) a b

Solutions to exercises

5

Y

Y

�a cos t , b sen t�

�0,1�

X

�Γ3

Γ2

Γ1 �0,0� (a)

X �1,0�

(b)

Fig. 9.7 (a) Ellipse ; (b) triangle of Exercise 2.8.3.

is on the unit circle, hence �x

a

,

y� = (cost , sint) b

for some t ∈ [0, 2π [. Consequently, a parameterization of the ellipse is given by

γ : [0, 2π ] → IR2 ; γ (t) = (a cost , b sint). � 2.8.3 Integrate the vector field F (x, y) = (y2 , −2xy) along the triangle with vertex (0, 0), (1, 0), (0, 1), oriented counterclockwise. Solution: The triangle is the union of three segments. A parameterization of the segment from (0, 0) to (1, 0) is

γ 1 : [0, 1] → IR2 ; γ 1 (t) = (t, 0). A parameterization of the segment from (1, 0) to (0, 1) is

γ 2 : [0, 1] → IR2 ; γ 2 (t) = (1 − t)(1, 0) + t(0, 1) = (1 − t,t). Now, instead of the segment from (0, 1) to (0, 0) we consider the opposite path (the segment from (0, 0) to (0, 1)), which is easier. This is

γ 3 : [0, 1] → IR2 ; γ 3 (t) = (0,t).

6

Solutions to exercises

Then, the path along which we have to integrate the vector field F is the union of the following three paths: γ = γ 1 ∪ γ 2 ∪ (−γ 3 ) and, consequently, �

γ

We now evaluate

F=

�

γ1

�

F+

γ1

F =

=

�

γ2

F−

�

γ3

F.

�

y2 dx − 2xy dy

� 1

0 dt = 0

γ1

0

and also

�

γ3

Moreover

�

γ2

F =

=

�

y2 dx − 2xy dy

γ2

� 1�

� t 2 (−1) − 2(1 − t)t dt

0

=

F = 0.

� 1 0

(t 2 − 2t) dt =

1 − 1. 3

Finally, �

γ

F=

1 − 1. � 3

2.8.4 (1) Find a path γ whose trajectory is the intersection of the cylinder x2 + y2 = 1 with the plane x + y + z = 1 and with the additional property that the initial (and final) point is (0, −1, 2) and the projection onto the plane XY is oriented counterclockwise. (2) Evaluate �

γ

xy dx + yz dy − x dz.

Solution: (1) The intersection of the cylinder with the plane is an ellipse in IR3 whose projection onto the plane XY is the unit circle. Consequently, we first parameterize the unit circle in such a way that the initial and final point is (0, −1) and it is oriented counterclockwise. That is, we put x = cost ; y = sint (−

π 3π ≤t ≤ ). 2 2

Solutions to exercises

7

Fig. 9.8 Path of Exercise 2.8.4.

Now, from the equation of the plane we deduce z = 1 − x − y = 1 − cost − sint. The path we are looking for is

π 3π γ : [− , ] → IR3 ; γ (t) = (cost , sint , 1 − cost − sint). 2 2 (2) We now denote

ω = xy dx + yz dy − x dz. We could evaluate �

γ

ω=

�

3π 2

− π2

� � ω (γ (t)) γ � (t) dt,

but a more efficient method consists in making a formal substitution in the integral x = cost ; y = sint ; z = 1 − cost − sint and also dx = (− sint) dt ; dy = cost dt ; dz = (sint − cost) dt to obtain �

γ

Moreover,

ω =

�

3π 2

− π2

�

� − 2 sin2 t cost − sint cos2 t + cos2 t dt.

8

Solutions to exercises

�

3π 2

− π2

�

3π 2

− π2

and �

3π 2

− π2

2

cos t dt =

Finally,

sin2 t cost dt =

sin3 t �� 32π = 0, � 3 − π2

− sint cos2 t dt = �

3π 2

− π2

cos3 t �� 32π = 0 � 3 − π2

1 + cos(2t) t sin(2t) �� 32π dt = + � π = π. 2 2 4 −2 �

γ

ω = π. �

2.8.5 Let γ be a simple path whose trajectory is the intersection of the coordinate planes with the portion of the unit sphere in the first octant, oriented according to the sequence (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 0, 0). Evaluate

�

γ

z dx + x dy + y dz.

Fig. 9.9 Path of Exercise 2.8.5.

Solution: The portion of the unit sphere in the first octant is {(x, y, z) ∈ IR3 ; x2 + y2 + z2 = 1 , x ≥ 0 , y ≥ 0 , z ≥ 0} and its intersection with a coordinate plane is an arc of circumference. In particular, the intersection with the plane z = 0 is the trajectory of the path

π γ 1 : [0, ] → IR3 ; γ 1 (t) = (cost, sint, 0), 2

Solutions to exercises

9

while the intersection with the plane x = 0 is the trajectory of

π γ 2 : [0, ] → IR3 ; γ 2 (t) = (0, cost, sint). 2 Also the intersection with y = 0 is the trajectory of

π γ 3 : [0, ] → IR3 ; γ 3 (t) = (sint, 0, cost). 2 The path of integration is

γ = γ 1 ∪ γ 2 ∪ γ 3. We put

ω = z dx + x dy + y dz and we evaluate, bearing in mind that z = 0 on γ 1 , �

γ1

ω =

=

�

γ1 π 2

�

0

x dy =

�

π 2

cos2 t dt

0

1 + cos(2t) π dt = . 2 4

Also, since x = 0 on γ 2 , �

γ2

ω =

�

γ2

y dz =

π 2

�

cos2 t dt =

π . 4

cos2 t dt =

π . 4

0

From the fact that y = 0 along γ 3 we obtain �

γ3

ω =

�

γ3

z dx =

�

π 2

0

Finally �

γ

ω=

3π .� 4

2.8.6 Find a path whose trajectory is the intersection of the upper hemisphere of the sphere with radius 2a x2 + y2 + z2 = 4a2 with the cylinder

x2 + (y − a)2 = a2 .

Solution: If (x, y, z) is a point on that trajectory then (x, y) is on the circle x2 + (y − a)2 = a2 . Hence,

10

Solutions to exercises

Fig. 9.10 Path of Exercise 2.8.6.

x = a cost ; y − a = a sint for some value of t between 0 and 2π . After substituting in the equation of the sphere we obtain a2 cos2 t + (a + a sint)2 + z2 = 4a2 . That is,

z2 = 2a2 (1 − sint)

and from the fact that z ≥ 0 (since we are considering the upper hemisphere), we conclude � z = a 2(1 − sint).

Consequently, the sought-after path is

γ : [0, 2π ] → IR3 defined by

γ (t) =

�

� � a cost , a + a sint , a 2(1 − sint) . �

2.8.7 Determine whether the vector field

is conservative or not.

� � F : IR2 → IR2 ; F (x, y) = x3 y , x

Solution: Le us denote P(x, y) = x3 y ; Q(x, y) = x. Since, in general,

Solutions to exercises

11

∂P ∂Q (x, y) = x3 �= (x, y) = 1 ∂y ∂x we conclude that F is not conservative. An alternative argument is the following. If F were conservative then the integral along a path would depend only on the initial and final point of the path. In particular, F would satisfy �

γ

F =

�

α

F+

�

β

F

where α is the segment from (0, 1) to (0, 0), β is the segment from (0, 0) to(1, 0) and γ is the segment from (0, 1) to (1, 0). Hence, in order to conclude that F is not a conservative field, it is enough to show that the above identity is false. It is easy to compute � � α

F=

β

F = 0.

On the other hand, γ is the path

γ : [0, 1] → IR2 ; γ (t) = (t, 1 − t) which permits us to evaluate �

γ

F =

=

�

γ

x3 y dx + x dy

� 1 0

(t 3 (1 − t) − t) dt = −

9 �= 0. 20

� 2.8.8 Find a potential for the vector field � � F : IR2 → IR2 ; F (x, y) = 2xy , x2 − y2 . Solution: We look for a function f : IR2 → IR of class C1 such that ∇ f (x, y) = F (x, y) for all (x, y) ∈ IR2 . This means that f is a solution of the following equations:

∂f ∂f (x, y) = 2xy ; (x, y) = x2 − y2 . ∂x ∂y

(9.6)

From the first equation we deduce, by integrating with respect to x, f (x, y) = x2 y + ϕ (y).

(9.7)

Here ϕ (y) is the constant of integration with respect to the variable x. More precisely, we observe that � x ∂f 0

∂x

(t, y) dt =

� x 0

2ty dt = x2 y

12

Solutions to exercises

and the Fundamental Theorem of Calculus gives f (x, y) − f (0, y) = x2 y. That is, ϕ (y) = f (0, y). Assuming ϕ is C1 on R, after substituting the expression (9.7) in the second equation (9.6) we get x2 + ϕ � (y) = x2 − y2 , or, equivalently,

ϕ � (y) = −y2 . 3

Hence, after considering ϕ (y) = − y3 , it turns out that f (x, y) := x2 y −

y3 3

is a potential of the vector field F on R. � 2.8.9 For the vector field � � F : IR3 → IR3 ; F (x, y, z) = 2xy , x2 + z2 , 2zy , (1) show that ∇ × F = 0, (2) find a potential for F . Solution: The proof of (1) is left to the reader. Then Poincaré’s Lemma allows us to conclude that F is a conservative vector field on R3 . The potential of F is a C1 function f : IR3 → IR such that ∇ f = F , that is,

∂f ∂f ∂f (x, y, z) = 2xy ; (x, y, z) = x2 + z2 ; (x, y, z) = 2zy. ∂x ∂y ∂z

(9.8)

By integrating the first equation in (9.8) with respect to x we obtain f (x, y, z) = x2 y + ϕ (y, z).

(9.9)

We compute the derivative with respect to y and substitute in the second equation of (9.8) to get ∂ϕ x2 + = x2 + z2 , ∂y that is,

∂ϕ = z2 . ∂y By integrating with respect to y we conclude

Solutions to exercises

13

ϕ (y, z) = z2 y + g(z). We now repeat the process. From the third equation in (9.8) and from (9.9) we get

∂ϕ = 2zy, ∂z which implies

g� (z) = 0.

Finally, we take g(z) as an arbitrary constant, for instance g(z) = 0, then ϕ (y, z) = z2 y and f (x, y, z) = (x2 + z2 )y. � 2.8.10 (1) Show that the vector field F (x, y) = (ex (sin(x + y) + cos(x + y)) + 1 , ex cos(x + y)) is conservative on R2 and find a potential. (2) Evaluate the line integral �

γ

where

F,

� � γ : [0, π ] → IR2 ; γ (t) = sin(π esin(t) ), cos5 (t) .

Solution: (1) Writing P(x, y) = ex (sin(x + y) + cos(x + y)) + 1 and Q(x, y) = ex cos(x + y) it follows that

∂Q ∂P = ex (cos(x + y) − sin(x + y)) = . ∂x ∂y

Hence F is conservative. If f : IR2 → IR is a potential function of F then

and

∂f = ex (sin(x + y) + cos(x + y)) + 1 ∂x

(9.10)

∂f = ex cos(x + y). ∂y

(9.11)

By integrating (9.11) respect to y we obtain

14

Solutions to exercises

f (x, y) = ex sin(x + y) + ϕ (x). We now compute the derivative with respect to x and substitute in (9.10) to get ex (sin(x + y) + cos(x + y)) + ϕ � (x) = ex (sin(x + y) + cos(x + y)) + 1, from where it follows ϕ � (x) = 1. Finally we take ϕ (x) = x and we conclude that f (x, y) = ex sin(x + y) + x is a potential function for F . (2) The computation of the line integral (2) using Definition 2.2.1 could be rather cumbersome due to the expression for the path γ . Fortunately, since the vector field is conservative, we can apply Theorem 2.5.1 to get �

γ

F = f (γ (π )) − f (γ (0)).

Since γ (0) = (0, 1) and γ (π ) = (0, −1) we have �

γ

F = f (0, −1) − f (0, 1) = −2 sin 1. �

2.8.11 Evaluate

�

γ

x dx + y dy + z dz

� � where γ (t) = cos4 (t), sin2 (t) + cos3 (t),t , 0 ≤ t ≤ π . Solution: The vector field F (x, y, z) = (x, y, z) is conservative because 1 f (x, y, z) = (x2 + y2 + z2 ) 2 satisfies ∇ f = F . From γ (π ) = (1, −1, π ) and γ (0) = (1, 1, 0) we obtain, by applying Theorem 2.5.1, �

γ

F = f (1, −1, π ) − f (1, 1, 0) =

π2 . � 2

2.8.12 For the vector field F = ( f1 , f2 ) : IR2 \ {0} → IR2 defined by f1 (x, y) =

−y x ; f2 (x, y) = 2 : x 2 + y2 x + y2

Solutions to exercises

15

(1) Show that

∂ f2 ∂ f1 (x, y) = (x, y) ∂x ∂y for all (x, y) ∈ IR2 \ {0}. (2) Let γ be the unit circle oriented counterclockwise. Show that �

γ

F = 2π .

Is this fact a contradiction to Poincaré’s Lemma? (3) Argue whether this statement is true or not: for every closed and piecewise C1 path α : [a, b] → IR2 \ {(0, 0)} such that α1 (t) ≥ 0 for all t ∈ [a, b] the following holds � F = 0. α

(4) Evaluate

�

γF

where � � π π γ (t) = cos(t), sin7 (t) , − ≤ t ≤ . 2 2

Solution: (1) An easy computation gives

∂ f1 ∂ f2 y2 − x2 (x, y) = (x, y) = 2 ∂y ∂x (x + y2 )2 for every (x, y) ∈ IR2 \ {0}. (2) A parameterization of the unit circle is

γ : [0, 2π ] → IR2 ; γ (t) = (cost , sint). By substituting formally x = cost , y = sint , dx = − sint dt , dy = cost dt we obtain

�

γ

F =

�

γ

f1 (x, y) dx + f2 (x, y) dy

� 2π sin2 t + cos2 t

(9.12)

dt = 2π . sin2 t + cos2 t The fact that the line integral is different from zero is not a contradiction with Poincaré’s Lemma, since the open set is not starlike. =

0

16

Solutions to exercises

(3) The open set � � U := IR2 \ ] − ∞, 0] × {0}

is starlike with respect to the point (1, 0). The set U is obtained by deleting the non-positive part of the X axis. Now, condition (1) and Poincaré’s Lemma permits the conclusion that F is a conservative vector field in U. That is, there exist a C1 function f : U ⊂ IR2 → IR such that ∇ f (x, y) = F (x, y) for all (x, y) ∈ U. If α = (α1 , α2 ) : [a, b] → IR2 \{(0, 0)} is a closed and piecewise C1 path such that α1 (t) ≥ 0 for all t ∈ [a, b] then it follows α ([a, b]) ⊂ U and, because F is conservative in U, we get � α

F = 0.

(4) The trajectory of the path γ is contained in U. Since F is conservative in U, the value of the line integral does not change after replacing γ by any other path contained in U with initial point γ (− π2 ) = (0, −1) and final point γ ( π2 ) = (0, 1). We can consider, for instance, the semicircle

π π β : [− , ] → IR2 ; β (t) = (cost, sint), 2 2 which gives an easier expression for the line integral. In fact, proceeding as in 9.12

Β Γ

Fig. 9.11 The integral of a conservative field along γ coincides with the integral along β .

Solutions to exercises

17

we obtain �

γ

F=

�

β

F=

π 2

�

dt = π . �

− π2

2.8.13 Let γ (t) = (cos(t) , sin(t)) be given and let us consider the paths

α := γ |[−π ,π ] and β := γ |[0,2π ] . (1) Are α and β equivalent paths? (2) Justify why �

α

F=

�

β

F

for any continuous vector field F : U ⊂ IR2 → IR2 defined on the unit circle. Solution: (1) Both paths parameterize the unit circle. However they are not equivalent since they have different initial point. In fact,

α (−π ) = (−1, 0) �= β (0) = (0, 1). (2) We now consider

α 1 := γ |[−π ,0] ; α 2 := γ |[0,π ] and also

β 1 := γ |[0,π ] ; β 2 := γ |[π ,2π ] . Then

�

α

F=

�

F+

F=

�

F+

α1

�

F

�

F.

α2

and analogously �

β

β1

β2

Because α 2 = β 1 , in order to conclude it is enough to prove that α 1 is an equivalent path to β 2 . This is true, since β 2 = α1 ◦ϕ where

ϕ : [π , 2π ] → [−π , 0] is the increasing bijection defined by ϕ (t) = t − 2π . An alternative solution consists of applying Proposition 2.7.2. � 2.8.14 Let γ be the path whose trajectory is the union of the graph of y = x3 from (0, 0) to (1, 1) and the segment from (1, 1) to (0, 0). Using Green’s Theorem, evaluate � (x2 + y2 ) dx + (2xy + x2 ) dy. γ

18

Solutions to exercises

Y �1,1�

Γ2 Γ2 Γ1 K

K X

Γ1

�0,0� ��2,0�

�2,0�

(a)

(b)

Fig. 9.12 (a) Path of Exercise 2.8.14 ; (b) path of Exercise 2.8.15.

Solution: γ = γ 1 ∪ γ 2 is the boundary, oriented counterclockwise, of the region of type I K := {(x, y) ∈ IR2 ; 0 ≤ x ≤ 1 , x3 ≤ y ≤ x}. Writing

P(x, y) = x2 + y2 ; Q(x, y) = 2xy + x2 ,

Green’s Theorem gives �

γ

P(x, y) dx + Q(x, y) dy =

�� � ∂Q K

=

� 1 �� x 0

=

∂x

� 1 0

x3

−

∂P� d(x, y) ∂y

� 2x dy dx

2x(x − x3 ) dx =

4 .� 15

2.8.15 Use Green’s Theorem to evaluate the line integral �

γ

(x2 + 3x2 y2 ) dx + (2x3 y + x2 ) dy

where γ is the boundary of the region lying between the graphs of y = 0 and y = 4 − x2 , oriented counterclockwise. Solution: γ is the boundary of the region of type I K := {(x, y) ∈ IR2 ; −2 ≤ x ≤ 2 , 0 ≤ y ≤ 4 − x2 }. Writing

Solutions to exercises

19

P(x, y) = x2 + 3x2 y2 ; Q(x, y) = 2x3 y + x2 we obtain �

γ

P(x, y) dx + Q(x, y) dy =

�� � ∂Q

∂x

K

=

=

−

� 2 � � 4−x2 −2

� 2

−2

0

∂P� d(x, y) ∂y � 2x dy dx

2x(4 − x2 ) dx = 0. �

2.8.16 Let K be a region of type I or type II and let γ be a path whose trajectory is the boundary of K oriented counterclockwise. Then area (K) =

�

γ

1 (−y dx + x dy) = 2

�

γ

−y dx =

�

γ

x dy.

Solution: The choice P(x, y) = − 2y and Q(x, y) = 2x , produces the following line integral when applying Green’s Theorem �

γ

P dx + Q dy =

�� � ∂Q

∂x

K

=

��

K

−

∂P� d(x, y) ∂y

d(x, y) = area (K). �

2.8.17 Evaluate the area bounded by the cycloid γ : [0, 2π ] → IR2

γ (t) = (at − a sin(t), a − a cos(t)) (a > 0) and the X axis.

Y

�Γ Α �0,0�

X �2Π,0�

Fig. 9.13 Path of Exercise 2.8.17.

Solution: We observe that γ (0) = (0, 0) and γ (2π ) = (a2π , 0). Moreover

20

Solutions to exercises

γ1 (t) := at − a sin(t) is strictly increasing on [0, 2π ], while

γ2 (t) := a − a cos(t) > 0 whenever 0 < t < 2π . Hence the trajectory of the path γ and the X axis limit a region K of type I with boundary, oriented counterclockwise,

α ∪ (−γ ) where α is the segment from (0, 0) to (a2π , 0). We put 1 ω = (−y dx + x dy). 2 From Exercise 2.8.16, the area of K is given by A :=

�

α

ω−

�

γ

ω.

Because y = 0 on the segment from (0, 0) to (a2π , 0), we get �

α

and A=−

1 2

�

ω =0

γ

−y dx + x dy.

In order to evaluate this line integral we substitute x = a(t − sint), dx = a(1 − cost) y = a(1 − cost), dy = a sint. Then 2

A=a

� 2π � 0

Since

� 2π

1 − cost − t

sint � dt. 2

cost dt = 0 and integrating by parts

0

� 2π 0

we conclude

�2π � � −t sint dt = t cost � − 0

2π

cost dt = 2π

0

A = 3π a2 . �

2.8.18 Evaluate the area bounded by the two coordinate axes and the path

Solutions to exercises

21

π γ : [0, ] → IR2 ; γ (t) = (sin4 (t), cos4 (t)). 2 Solution: The function γ1 (t) := sin4 (t) is strictly increasing and γ2 (t) := cos4 (t) is strictly decreasing on [0, π2 ]. We have to evaluate the area of a region of types I and II whose boundary, oriented counterclockwise, is

α ∪ (−γ ) ∪ β where α is the segment from (0, 0) to (1, 0) and β is the segment from (0, 1) to (0, 0). Once again, we consider the 1-form 1 ω = (−y dx + x dy). 2 Because

�

α

ω=

�

1 2

−y dx + x dy.

β

ω = 0,

Green’s Theorem gives A=−

�

γ

We make the formal substitution x = sin4 t, dx = 4 sin3 t cost dt y = cos4 t, dy = −4 cos3 t sint dt to obtain A=2

�

π 2

sin3 t cos3 t (cos2 t + sin2 t) dt = 2

0

�

π 2

sin3 t cos3 t dt.

0

Finally, the change of variables sint = u gives A=2

� 1 0

1 u3 (1 − u2 ) du = . � 6

2.8.19 Evaluate the area limited by the circles � � C1 := (x, y) ∈ R2 : x2 + y2 = a2 and

� � C2 := (x, y) ∈ R2 : x2 + y2 = 2ax (a > 0).

Solution: The circle C1 is centered at the origin and has radius a, while the circle C2 is centered at point (a, 0) and also has radius a. The two circles meet at the solutions of the following system of equations

22

Solutions to exercises

�

x2 + y2 = a2 x2 + y2 = 2ax

(9.13)

√

From a2 = 2ax we deduce x = a2 , hence y = ± 23 a. Consequently, the vertical line x = a2 decompose the region limited by the two circles into two pieces. It is a region

C1

Α

C2

Β

Fig. 9.14 Positively oriented boundary of the region limited by two circles.

of type II whose boundary, oriented counterclockwise, is given by

γ = α ∪β. The trajectory of α is the semi-circle located to the left of the line x = appropriate parameterization is

α :[

a 2

and an

2π 4π , ] → IR2 ; α (t) = (a + a cost, a sint). 3 3

Also, the trajectory of β is the semi-circle located to the right of the line x = 2x , and an appropriate parameterization is

π π β : [− , ] → IR2 ; β (t) = (a cost, a sint). 3 3 Since �

α

−y dx + x dy =

�

4π 3 2π 3

= a2 ( and �

β

−y dx + x dy =

�

a2 (1 + cost) dt

2π √ − 3) 3

π 3

− π3

a2 dt = a2

2π , 3

we conclude that the area is 1 A= 2

√ 3 2π −y dx + x dy = a ( − ). � 3 2 γ

�

2

Solutions to exercises

23

Solved exercises of Chapter 3 3.5.1 1. Show that the cylinder S := {(x, y, z) ∈ IR3 ; x2 + y2 = a2 } is a regular surface. 2. What about the portion of the cylinder {(x, y, z) ∈ S ; 0 < z < 1}? Solution: (1) We consider the mapping of class C∞ f : IR3 → IR ; f (x, y, z) = x2 + y2 . Then

S = {(x, y, z) ∈ IR3 ; f (x, y, z) = a2 }.

Also, whenever surface.

x2

(2) We take

+ y2

∇ f (x, y, z) = (2x , 2y , 0) �= (0, 0, 0)

=

a2 .

Hence, it follows from Example 3.2.2 that S is a regular

U := {(x, y, z) ∈ IR3 ; 0 < z < 1},

which is an open set in IR3 . By Example 3.1.2 we have that S ∩U is a regular surface. � 3.5.2 Show that the cone S := {(x, y, z) ∈ IR3 ; x2 + y2 = z2 , z ≥ 0} is not a regular surface, but S \ {(0, 0, 0)}, the cone without the vertex, is. Solution: Let us assume that S is a regular surface. According to Proposition 3.1.1, there exist an open neighborhood W ⊂ IR2 of (0, 0), an open neighborhood U of (0, 0, 0) in IR3 and a function of class C1 g : W ⊂ IR2 → IR such that (a) S ∩U = {(x, y, g(x, y)) ; (x, y) ∈ W }

24

Solutions to exercises

or (b) S ∩U = {(x, g(x, z), z)) ; (x, z) ∈ W } or (c) S ∩U = {(g(y, z), y, z) ; (y, z) ∈ W }. Since the projections onto the xz- and yz-planes are not injective on S ∩U, the cases (b) and (c) cannot occur. Consequently, S ∩U = {(x, y, g(x, y)) ; (x, y) ∈ W }. In particular g(x, y) =

� x 2 + y2

for every (x, y) ∈ W, hence g does not admit partial derivatives at point (0, 0), which is a contradiction since g is a function of class C1 . Consequently S is not a regular surface. However, S \ {(0, 0, 0)} is a regular surface since it is the graph of the function of class C1 g : IR2 \ {(0, 0)} → IR. � 3.5.3 Determine whether the intersection of the cone x 2 + y2 = z 2 with the plane x+y+z = 1 is a regular curve. Solution. The intersection of the cone with the plane can be described as M := Φ −1 {(0, 0)} where Φ : IR3 → IR2 is the function defined by

Φ (x, y, z) := (x2 + y2 − z2 , x + y + z − 1). Since Φ is a function of class C∞ , it suffices (by Proposition 3.2.1) to check that the Jacobian matrix of Φ has rank 2 at each point (x, y, z) ∈ M. That Jacobian matrix is � � 2x 2y −2z A := , 1 1 1 and it has rank 2 in U := R3 \ {(t,t, −t) : t ∈ R}. Since M is contained in U we conclude that the rank of the matrix is 2 at each point (x, y, z) ∈ M and, consequently, M is a regular curve (regular surface of dimension 1) of class C∞ . �

Solutions to exercises

25

Fig. 9.15 Intersection of a cone and a plane.

3.5.4 Let

γ : [a, b] → U ⊂ R × (0, +∞)

be a path such that γ ([a, b]) is a level curve . Show that the set obtained by rotating γ ([a, b]) around the x-axis is a level surface.

Fig. 9.16 Surface of revolution of Exercise 3.5.4.

Solution: Let f : U ⊂ R2 → IR a C1 function on the open set U, and c ∈ IR be such that γ ([a, b]) = {(x, y) ∈ U ; f (x, y) = c}.

26

Solutions to exercises

If we select an arbitrary point (x0 , y0 ) ∈ γ ([a, b]) and rotate (x0 , y0 , 0) ∈ IR3 through an angle s about the x-axis we get the point with coordinates x = x0 ; y = y0 cos s ; z = y0 sin s. That is, the x-coordinate of the point does not change and we make a rotation of angle s in the yz-plane of the point with coordinates y = y0 , z = 0. Since y0 > 0, the points of the yz-plane of the form (y0 cos s, y0 sin s) are precisely the points (y, z) satisfying y0 = We consider the continuous function

� y2 + z2 .

h : IR3 → IR2 ; h (x, y, z) = (x,

� y2 + z2 )

and V := h −1 (U), which is an open subset of IR3 . Finally we define g := f ◦ h : V ⊂ IR3 → IR. Then, the set obtained after rotating γ ([a, b]) about the x-axis is M = {(x, y, z) ∈ V ; g(x, y, z) = c}. The obtained level surface is known as surface of revolution, and as we have just proven, it is described by the equation � f (x, y2 + z2 ) = c. �

3.5.5 Find a parameterization of the cylinder x2 + y2 = a2 in a neighborhood of (a, 0, 0). Solution: We put

M := {(x, y, z) ; x2 + y2 = a2 }

and try to describe points of M which lie in a neighborhood of (a, 0, 0) in terms of two parameters. To do this, we first observe that for each (x, y, z) ∈ M, the point (x, y) is on the circle with radius a centered at the origin. Hence x = a cost ; y = a sint for some t (our first parameter). The z variable does not appear in these equations, which means that it is independent of (x, y). We now consider as a second parameter the z coordinate. That is z = s. Finally, we define U := {(s,t) ∈ IR2 ; −1 < s < 1 , −π < t < π }

Solutions to exercises

and

27

ϕ : U ⊂ IR2 → M ; ϕ (s,t) = (a cost, a sint, s).

Then ϕ (0, 0) = (a, 0, 0) and (U, ϕ ) is a coordinate system of M. This is easy to prove using Corollary 3.3.1. In fact, it is obvious that ϕ is an injective mapping of class C∞ . Moreover, the Jacobian matrix of ϕ   0 −a sint ϕ � (s,t) =  0 a cost  1 0

always has rank 2 since sint and cost do not vanish simultaneously. �

A

�

Fig. 9.17 A parametrization of the cylinder.

3.5.6 Let ϕ : IR2 → IR3 be the mapping

ϕ (s,t) := (s + t , s − t , 4st) . Show that (IR2 , ϕ ) is a coordinate system of the regular surface with equation z = x 2 − y2 . Solution: The set

M := {(x, y, z) ∈ IR3 ; z = x2 − y2 }

is a regular surface of class C∞ since it is the graph of the function f : IR2 → IR

28

Solutions to exercises

Fig. 9.18 The mapping of Exercise 3.5.6.

defined by f (x, y) = x2 − y2 . We observe that the point with cartesian coordinates x = s + t , y = s − t , z = 4st is in M since Then

x2 − y2 = (s + t)2 − (s − t)2 = 4st = z.

ϕ : IR2 → M ⊂ IR3

is a mapping of class C∞ which is injective since the parameters s and t are uniquely determined by the coordinates x, y, z : 1 1 s = (x + y) , t = (x − y). 2 2 This means that

�1 � 1 (x + y), (x − y) . 2 2 Now, the Jacobian matrix of ϕ at point (s,t),   1 1 ϕ � (s,t) =  1 −1  4t 4s

ϕ −1 (x, y, z) =

has rank two. From Corollary 3.3.1 we conclude that (IR2 , ϕ ) is a coordinate system of M. � 3.5.7 Explain how to obtain a coordinate system of the ellipsoid

Solutions to exercises

29

� x �2 a

+

� y �2 b

+

from a coordinate system of the unit sphere.

� z �2 c

=1

Solution: Let S be the unit sphere and let M the ellipsoid. We consider the linear isomorphism f : IR3 → IR3 defined by f (x, y, z) = (ax , by , cz). The point (x0 , y0 , z0 ) is in S if, and only if, f (x0 , y0 , z0 ) ∈ M. Consequently, if (U, ϕ )

Fig. 9.19 A mapping from the sphere onto the ellipsoid.

is an arbitrary coordinate system of the sphere S then f ◦ ϕ : U ⊂ IR2 → M and (U, f ◦ ϕ ) is a coordinate system of M. For instance, it follows from Example 3.3.2 that (U, ψ ) is a coordinate system of the ellipsoid, where U := {(s,t) : 0 < s < π , 0 < t < 2π } and

ψ (s,t) := (a sin s cost, b sin s sint, c cos s) . � 3.5.8 Let 0 < r < R. (1) Show that the torus obtained by rotation of the circle x2 + (y − R)2 = r2 about the X axis is a regular surface. (2) Find a coordinate system. Solution: (1) We define

30

Solutions to exercises

Fig. 9.20 Torus of Exercise 3.5.8.

f : IR2 → IR ; f (x, y) = x2 + (y − R)2 . Then, the torus is the level surface M = {(x, y, z) ∈ IR3 ; f (x,

�

y2 + z2 ) = r2 }.

In order to conclude that M is a regular surface it suffices to check that g(x, y, z) := x2 +

�� �2 � y2 + z2 − R = x2 + y2 + z2 + R2 − 2R y2 + z2

has gradient different from zero on an open set containing M. Put � � U := R3 \ {(0, y, z) : y2 + z2 = R2 } ∪ {(0, 0, 0)} , which is an open set containing M. Since ∇ g(x, y, z) is the vector � 2Ry 2Rz 2x, 2y − � , 2z − � ), 2 2 y +z y2 + z2

it follows that ∇ g does not vanish on the open set U containing M. (2) We first fix a parameterization of the circle. For instance

γ = (γ1 , γ2 ) :]0, 2π [→ IR2 ; γ (t) = (r cost, R + r sint). If we rotate a point (γ1 (t), γ2 (t), 0) an angle s about the x-axis we obtain the point with coordinates x = γ1 (t) ; y = γ2 (t) cos s ; z = γ2 (t) sin s.

Solutions to exercises

31

This suggests the definition � � ϕ : IR2 → M ⊂ IR3 ; ϕ (s,t) = γ1 (t), γ2 (t) cos s, γ2 (t) sin s .

The mapping ϕ is of class C1 on R2 . If

ϕ (s1 ,t1 ) = ϕ (s2 ,t2 ) then γ1 (t1 ) = γ1 (t2 ) and

γ2 (t1 )(cos s1 , sin s1 ) = γ2 (t2 )(cos s2 , sin s2 ). Evaluating norms in the last identity, we deduce γ2 (t1 ) = γ2 (t2 ). Hence γ (t1 ) = γ (t2 ) and (cos s1 , sin s1 ) = (cos s2 , sin s2 ). From the injectivity of γ and the properties of the functions sin and cos it follows that t1 = t2 and s1 − s2 is multiple of 2π . In particular ϕ is injective on each open set Wc := (c, c + 2π ) × (0, 2π ) ⊆ IR2 (c ∈ IR). We show that � � Wc , ϕ

is a coordinate system. Since ϕ : Wc → ϕ (Wc ) is a continuous bijection it suffices (according to Corollary 3.3.1) to check that ϕ � (s,t) has rank 2, or equivalently,

for every (s,t). Since,

has norm

�∂ϕ ∂ϕ � × (s,t) �= (0, 0, 0) ∂s ∂t

� � �∂ϕ ∂ϕ � × (s,t) = γ2 (t) − γ2� (t), γ1� (t) cos s, γ1� (t) sin s ∂s ∂t

we are done. �

�� �2 � � � � ∂ϕ ∂ϕ � � ∂ s × ∂ t (s,t)� = γ2 (t)2 γ2� (t)2 + γ1� (t)2 � �2 = r2 R + r sint > 0,

3.5.9 Find the tangent plane to the paraboloid z = x2 + y2 at the point ( 12 , 12 , 12 ). Solution: The paraboloid can be described as M = Φ −1 {(0)}

32

Solutions to exercises

Fig. 9.21 Tangent plane to a paraboloid.

for Φ (x, y, z) = z − x2 − y2 . By Proposition 3.4.1, the vector 1 1 1 ∇ Φ ( , , ) = (−1, −1, 1) 2 2 2 is orthogonal to the tangent plane we are asked to find. Hence, the equation of the tangent plane is � � 1 1 1 (x, y, z) − ( , , ) , (−1, −1, 1) = 0, 2 2 2 that is

1 x+y−z = . 2

An alternative approach consists of evaluating a basis for the plane determined by the tangent vectors to the paraboloid. To do this, we consider the following parameterization of the surface

ϕ : IR2 → M ; ϕ (s,t) = (s , t , s2 + t 2 ). Since ϕ ( 12 , 12 ) = ( 12 , 12 , 12 ), it turns out that a basis of the tangent plane is given by the vectors ∂ϕ 1 1 ( , ) = (1, 0, 1) ∂s 2 2 and ∂ϕ 1 1 ( , ) = (0, 1, 1). ∂t 2 2 Consequently, the equation of the tangent plane to M is given by � � �x− 1 y− 1 z− 1 � � 2 2 2� � 1 0 1 �� = 0. � � 0 1 1 � Evaluating the previous determinant we again get

Solutions to exercises

33

1 x+y−z = . � 2 3.5.10 Find a tangent vector at (0, 0, 0) to the curve determined by the equations z = 3 x2 + 5 y2 ; x2 + y2 = 4 x. Solution: We define functions f (x, y, z) = 3 x2 + 5 y2 − z ; g(x, y, z) = x2 + y2 − 4 x and note that the curve is precisely M = {(x, y, z) ∈ IR3 ; f (x, y, z) = g(x, y, z) = 0}. We apply Proposition 3.4.1 to deduce that the vectors ∇ f (0, 0, 0) = (0, 0, −1) , ∇ g(0, 0, 0) = (−4, 0, 0) are normal vectors to M and, consequently, a tangent vector to M is given by the vector product � � � e1 e2 e3 � � � ∇ f (0, 0, 0) × ∇ g(0, 0, 0) = �� 0 0 −1 �� � −4 0 0 � = (0, 4, 0).

Another option consists of finding a coordinate system of the curve in a neighborhood of the origin. We parameterize the circle (x − 2)2 + y2 = 4 as x = 2 + 2 cost , y = 2 sint. Then, from the first defining equation of the curve, we obtain z = 3x2 + 5y2 = 12(1 + cos2 t + 2 cost) + 20 sin2 t. Hence, a parameterization of M in a neighborhood of the origin is

γ : (0, 2π ) → M ⊂ IR3 defined by � � γ (t) = 2 + 2 cost, 2 sint, 12(1 + cos2 t + 2 cost) + 20 sin2 t .

Since (0, 0, 0) = γ (π ), we deduce that a tangent vector to M at the origin is given by

γ � (π ) = (0, −2, 0),

34

Solutions to exercises

a multiple of the tangent vector previously obtained. �

Solved exercises of Chapter 4 4.4.1 Find the area of the triangle in IR3 with vertices P1 = 2 e3 , P2 = e1 − e2 + 2 e3 , P3 = e1 + 3 e3. Solution: The area of the triangle is half the area of the parallelepiped spanned by the vectors v 1 := P 2 − P 1 = (1, −1, 0) , v 2 := P 3 − P 1 = (1, 0, 1).

Fig. 9.22 Triangle of Exercise 4.4.1.

Since

� � � e1 e2 e3 � � � v 1 × v 2 = �� 1 −1 0 �� = (−1, −1, 1), �1 0 1�

we find that the area of the triangle is

Solutions to exercises

35

1 �(−1, −1, 1)� = 2

area =

√

3 .� 2

4.4.2 Let M be a regular surface of class C1 in IR3 and suppose that (W, ϕ ) is a coordinate system of M,

ϕ : W ⊂ IR2 → M ⊂ IR3 , with the property that M \ ϕ (W ) can be decomposed as the union of a finite family of subsets of M each of which is the image of a null set through some parameterization of M. Then area M = area (W, ϕ ). Solution: By hypothesis there is a finite family {(A j , ϕ j ) : j = 1, . . . , m} of coordinate systems of M and, for each j = 1, . . . , m, there is a null set N j ⊂ A j such that � M \ ϕ (W ) = ϕ j (N j ). 1≤ j≤m

Put K1 = N1 and, for 2 ≤ j ≤ m, � � � K j := ϕ −1 ϕ j (N j ) \ ϕ i (Ni ) ⊂ N j . j i< j

It could happen that some K j is empty but after selecting the non empty sets it turns out that {ϕ (W ), ϕ j (K j )} is a partition of M. According to Definition 4.2.3, area M = area (W, ϕ ) + ∑ area (K j , ϕ j ). j

To conclude it suffices to observe that �� � � area (K j , ϕ j ) = det(ϕ �T j (s,t) ◦ ϕ j (s,t))d(s,t) = 0 KJ

since K j is a null set. � � � 4.4.3 Let γ = γ1 , γ2 : (a, b) → IR2 be a parameterization of a regular curve of class C1 , with γ2 ≥ 0. Evaluate the area of the surface of revolution formed by rotating γ (]a, b[) around the x-axis. Solution: Let M denote the surface of revolution. According to Exercises 3.5.4 and 3.5.8, a parameterization of M is given by

ϕ : W ⊂ IR2 → M ⊂ IR3 ,

36

Solutions to exercises

where W = (0, 2π ) × (a, b) and ϕ is defined by � � ϕ (s,t) = γ1 (t), γ2 (t) cos s, γ2 (t) sin s . Since

� � M \ ϕ (W ) = ϕ {0} × (a, b) ,

{0}×(a, b) is a null subset of IR2 and ϕ : (−π , π )×(a, b) → M is a parameterization of M, we conclude area M =

��

W

�

∂ϕ ∂ϕ × (s,t) � d(s,t). ∂s ∂t

Finally, we evaluate

and

Thus

� ∂ϕ � = 0, −γ2 (t) sin s, γ2 (t) cos s ∂s � ∂ϕ � � = γ1 (t), γ2� (t) cos s, γ2� (t) sin s . ∂t

� ∂ϕ ∂ϕ � × = − γ2 (t)γ2� (t), γ2 (t)γ1� (t) cos s , γ2 (t)γ1� (t) sin s ∂s ∂t and consequently �� � b �2 � �2 area M = 2π γ2 (t) γ1� (t) + γ2� (t) dt. � a

4.4.4 Find the area of the part of the plane x y z + + =1 a b c (a, b, c are positive constants) that lies in the first octant. Solution: We have to evaluate the area of x y z + + = 1}. a b c To this end, we first parameterize the surface, taking as parameters M := {(x, y, z) ∈ IR3 ; x > 0, y > 0, z > 0,

s :=

y x , t := . a b

That is, x = as,

y = bt,

z = c(1 − s − t).

Solutions to exercises

37

Fig. 9.23 Triangle of Exercise 4.4.4.

The domain of the parameters is obtained by observing that the three coordinates x, y, z are strictly positive and, consequently, s > 0, t > 0, 1 − s − t > 0. So we consider the coordinate system (U, ϕ ) where U = {(s,t) ∈ IR2 ; s > 0, t > 0, s + t < 1} and

� � ϕ : U ⊂ IR2 → M ; ϕ (s,t) = as, bt, c(1 − s − t) .

Since ϕ (U) = M, it turns out that area M =

��

U

�

∂ϕ ∂ϕ × (s,t) � d(s,t). ∂s ∂t

Finally, evaluate ∂ϕ ∂s

×

∂ϕ ∂ t (s,t)

� � � e1 e2 e3 � � � = �� a 0 −c �� � 0 b −c �

= (cb , ac, ab). If we denote

1

λ := (c2 b2 + a2 c2 + a2 b2 ) 2

we obtain area M = λ

� 1 � � 1−s 0

0

� λ dt ds = . � 2

4.4.5 Evaluate the area of the part of the cone x2 + y2 = z2 that lies above the plane z = 0 and inside the sphere x2 + y2 + z2 = 4ax with a > 0.

38

Solutions to exercises

Solution: The part of the cone above the plane z = 0 and inside the sphere

Fig. 9.24 Part of the cone inside the sphere.

(x − 2a)2 + y2 + z2 = 4a2 is described by M := {(x, y, z) ∈ IR3 ; x2 + y2 = z2 , z > 0 , x2 + y2 + z2 < 4ax}. The equation of the cone suggests a parameterization of the surface, while the two inequalities will determine the domain of the parameters. We note that the intersection of the cone with the horizontal plane z = ρ is a circle with radius ρ whose projection onto the xy-plane admits the following parameterization. x = ρ cos θ , y = ρ sin θ . Now, ρ > 0 and from the inequality x2 + y2 + z2 < 4ax we deduce

ρ < 2a cos θ . In particular cos θ > 0, which means that we have to consider −

π π 0, ∂ρ ∂θ

we conclude that (W, ϕ ) is a coordinate system of M and area M =

�

π 2

− π2

2a cos θ

��

0

√ � = 2a2 2

π 2

− π2

� √ ρ 2 dρ dθ

cos2 θ dθ

√ = π a2 2 . � 4.4.6 Find the area of the part of the cylinder x2 + y2 = 1 bounded by the planes x + y + z = 0 and x + y + z = 1.

Fig. 9.25 Part of a cylinder bounded by two planes.

Solution: We parameterize M := {(x, y, z) ∈ IR3 ; x2 + y2 = 1 , 0 < x + y + z < 1} by choosing as one parameter t := x + y + z. The other parameter will be obtained from the circle x2 + y2 = 1 using polar coordinates. More precisely, we consider

ϕ : IR2 → M ⊂ IR3 defined by

40

Solutions to exercises

and we take

� � ϕ (s,t) = cos s , sin s , t − cos s − sin s Wc := (c, c + 2π ) × (0, 1).

Then (Wc , ϕ ) is a coordinate system of M because ϕ is injective and of class C1 on Wc and also � � ∂ϕ ∂ϕ × (s,t) = cos s, sin s, 0 �= (0, 0, 0) ∂s ∂t for every s ∈ (c, c + 2π ). Since {0} × (0, 1) is a null subset of R2 and � � ϕ (W0 ) = M \ ϕ {0} × (0, 1)

we conclude

area M =

��

W0

�

∂ϕ ∂ϕ × (s,t) � d(s,t) ∂s ∂t

= 2π . � 4.4.7 Determine the flux of the vector field F (x, y, z) = (x, y, 2z) across the part of the sphere

x2 + y2 + z2 = 1

that lies in the first octant. Solution: Begin by writing the portion of the sphere in the first octant as M := {(x, y, z) ∈ IR3 ; x2 + y2 + z2 = 1 , x > 0 , y > 0 , z > 0}.

We consider the coordinate system (U, ϕ ) where

π π ϕ : U = (0, ) × (0, ) → M 2 2 is defined by

ϕ (s,t) = (sin s cost, sin s sint, cos s) and U is chosen so that ϕ (U) = M. First we evaluate � � � e2 e3 �� e1 � ∂ϕ ∂ϕ × (s,t) = �� cos s cost cos s sint − sin s �� ∂s ∂t � − sin s sint sin s cost 0 � = sin s ϕ (s,t).

Using the fact that ϕ (s,t) is a point of the unit sphere, we see

Solutions to exercises

41

Fig. 9.26 Vector field of Exercise 4.4.7.

�

∂ϕ ∂ϕ × (s,t) �= sin s ∂s ∂t

and, for (x, y, z) = ϕ (s,t), N (x, y, z) :=

�

∂ϕ ∂s ∂ϕ ∂s

ϕ

× ∂∂ϕt (s,t) ϕ

× ∂∂ϕt (s,t) �

= (x, y, z)

is a normal vector to M directed outward from the sphere. From F (x, y, z) , N (x, y, z)� = x2 + y2 + 2z2 = 1 + z2 �F and z = cos s we obtain the following expression for the flux of the vector field in the direction of the unit vector N : Flux =

��

U

=

π 2

�

(1 + cos2 s) � π 2

0

∂ϕ ∂ϕ × (s,t) � d(s,t) ∂s ∂t

(1 + cos2 s) sin s ds =

2π . 3

An alternative solution is to consider M as the graph of a function and to take as parameters the coordinates x, y, that is, � x = u, y = v, z = 1 − u2 − v2 . More precisely, we consider the coordinate system (D, ψ ), where

42

Solutions to exercises

D := {(u, v) ∈ IR2 ; u > 0 , v > 0 , u2 + v2 < 1} and We obtain

� � � ψ : D → M ; ψ (u, v) = u, v, 1 − u2 − v2 . Flux =

� � � f1 f2 f3 � � ∂x ∂y ∂z � � � � ∂ u ∂ u ∂ u � d(u, v) � ∂x ∂y ∂z �

��

D

∂v ∂v ∂v

� � √ � 2 2� �� �� u v 2 1 −−uu − v �� � 1 0 √1−u2 −v2 � d(u, v) = � � D � −v � �0 1 √ 2 2 � 1−u −v

2 − u2 − v2 √ d(u, v). D 1 − u2 − v2 The last integral is evaluated after a change to polar coordinates to get =

Flux =

=

��

π 2

� 1 2 − ρ2

π 2

�

0

π 2

� ρ dρ (ρ = sint) 1 − ρ2 (1 + cos2 t) sint dt =

0

2π . 3

We note that the flux just obtained has the same sign as the flux obtained in the previous solution. This is due to the fact that the two coordinate systems define the same field N of unit normal vectors to the sphere. � 4.4.8 Let M be the part of the paraboloid x2 + y2 + z = 4R2 that lies inside the cylinder

(y − R)2 + x2 = R2 .

(1) Find a coordinate system of M such that the associated vector field N of unit normal vectors points upwards. (2) Evaluate the flux of the vector field F (x, y, z) = (x, 0, 0) across M in the direction of N . Solution: (1) We have

Solutions to exercises

43

M = {(x, y, z) ∈ IR3 ; x2 + y2 + z = 4R2 , (y − R)2 + x2 < R2 }. Since the paraboloid is the graph of a function of two variables x and y we can

Fig. 9.27 Part of a paraboloid inside a cylinder.

parameterize it by taking these variables as parameters. That is, we consider the coordinate system (D, ϕ ), where D := {(u, v) ∈ IR2 ; (v − R)2 + u2 < R2 } and

ϕ (u, v) = (u, v, 4R2 − u2 − v2 ).

Now,

� � � e1 e2 e3 � � � ∂ϕ ∂ϕ × (u, v) = �� 1 0 −2u �� ∂u ∂v � 0 1 −2v � = (2u, 2v, 1).

Since the third coordinate is positive, it follows that the vector field N of unit normal vectors obtained from the previous coordinate system points upwards. (2) Using � � ∂ϕ ∂ϕ F (ϕ (u, v)) , × (u, v) = 2u2 , ∂u ∂v we deduce, via the change of coordinates u = ρ cos θ , v = R + ρ sin θ in the double integral, that the flux can be evaluated as Flux =

��

D

2u2 d(u, v) = 2

��

2π 0

cos2 θ dθ

�� �

0

R

� ρ 3 dρ .

44

From

Solutions to exercises

� 2π

cos2 θ dθ = π

0

we finally obtain Flux =

π 4 R .� 2

Solved exercises of Chapter 5 5.3.1 If V is a two dimensional vector space and B1 = {vv1 , v 2 } is a basis of V , does the basis B2 = {vv1 + v 2 , v 1 − v 2 } the same orientation as B1 .? Solution: The change of basis matrix from B2 to B1 is � � 1 1 , 1 −1 and it has negative determinant. Hence B2 reverses the orientation defined by B1 . � 5.3.2 The sets {ee1 − e 2 , e 1 − e 3 } and {ee1 − e 3 , 2ee1 − e 2 − e 3 } span the same vector subspace V of IR3 . Find if the two bases define the same orientation in V. Solution: We put v 1 := e 1 − e 2 , v 2 := e 1 − e 3 and u 1 := e 1 − e 3 , u 2 := 2ee1 − e 2 − e 3 . Then

while

� � � e1 e2 e3 � � � v 1 × v 2 = �� 1 −1 0 �� = (1, 1, 1) � 1 0 −1 �

� � � e1 e2 e3 � � � u1 × u2 = �� 1 0 −1 �� = (−1, −1, −1). � 2 −1 −1 �

Hence the two bases define different orientations in V. �

Solutions to exercises

45

5.3.3 Find if the following coordinate systems of the cylindrical surface S := {(x, y, z) ∈ IR3 ; x2 + y2 = 1 , 0 < z < 1} define the same orientation in the tangent space to S at each common point (x0 , y0 , z0 ) = ϕ (s0 ,t0 ) = ψ (u0 , v0 ) : (1) ϕ : (0, 2π ) × (0, 1) → S ; ϕ (s,t) = (cos s, sin s,t) � √ � (2) ψ : (0, 1) × (0, 1) → S ; ψ (u, v) = u, 1 − u2 , v . Solution: We have to find out if the two vectors

∂ϕ ∂ϕ × (s0 ,t0 ) ∂s ∂t and

∂ψ ∂ψ × (u0 , v0 ) ∂u ∂v

point in the same direction. But, ∂ϕ ∂s

� � � e1 e 2 e 3 �� � × ∂∂ϕt (s0 ,t0 ) = �� − sin s0 cos s0 0 �� � 0 0 1�

= (cos s0 , sin s0 , 0) = (x0 , y0 , 0)

and ∂ψ ∂u

� � � e1 e2 e3 � � � −u0 � � × ∂∂ψv (u0 , v0 ) = � 1 √1−u2 0 � 0 � � �0 0 1�

= ( √−u0 2 , −1, 0) = √−1 2 (x0 , y0 , 0). 1−u0

1−u0

Hence the two coordinate systems define different orientations. � 5.3.4 Find a parameterization of the paraboloid z = 1 − x2 − y2 (0 < z < 1) compatible with the orientation induced by a field of normal vectors pointing inward. Solution: We consider the coordinate system (D, ϕ ) where

46

Solutions to exercises

D := {(u, v) ∈ IR2 ; 0 < u2 + v2 < 1} and

ϕ (u, v) = (u, v, 1 − u2 − v2 ).

According to Exercise 4.4.8,

∂ϕ ∂ϕ × (u, v) = (2u, 2v, 1), ∂u ∂v is a normal vector to the paraboloid and it points to the exterior of the bounded region Ω = {(x, y, z) : 0 ≤ z ≤ 1 − x2 − y2 }. To reverse the orientation we simply consider the coordinate system (D, ψ ) given by

ψ (u, v) = ϕ (v, u) = (v, u, 1 − u2 − v2 ). �

5.3.5 Prove that any level surface M in R3 is orientable. Solution: By hypothesis there is Φ : U ⊂ R3 → R of class C1 on the open set U such that ∇Φ (x) �= 0 for every x ∈ Φ −1 (0) = M. We know that {∇Φ (x)} is a basis of Nx M for every x ∈ M. Hence F(x) :=

∇Φ (x) ||∇Φ (x)||

is a continuous vector field of unit normal vectors. �

Solved exercises of Chapter 6 6.7.1 For

ω = dy ∧ dz − 2 dz ∧ dx + 3 dx ∧ dy ∈ Λ 2 (IR3 )

find ω (uu, v ) where u = (1, −1, 0) and v = (2, 1, 1). Solution:

� � � 1 −2 3 � � � ω (uu, v ) = �� 1 −1 0 �� = 10. � �2 1 1�

Solutions to exercises

47

6.7.2 Let F = (F1 , F2 , F3 ) and G = (G1 , G2 , G3 ) be vector fields on an open set U ⊂ IR3 and let ω , ϕ be the associated differential forms of degree 1. Find the relation between the vector field F × G and the vector field associated to the differential form of degree 2, ω ∧ ϕ . Solution: We have

ω = F1 dx + F2 dy + F3 dz , ϕ = G1 dx + G2 dy + G3 dz, hence

� � � � ω ∧ ϕ = F2 G3 − F3 G2 dy ∧ dz + F3 G1 − F1 G3 dz ∧ dx � � + F1 G2 − F2 G1 dx ∧ dy,

which is the 2-form associated to the vector field F × G . � 6.7.3 We consider in IR2n the differential form of degree 2

ω = dx1 ∧ dx2 + dx3 ∧ dx4 + . . . + dx2n−1 ∧ dx2n . Evaluate the exterior product of n copies of ω . Solution: We observe that

ν := ω ∧ ω ∧ . . . ∧ ω ∈ Λ 2n (IR2n ) and consequently for every x ∈

R2n .

ν (xx) = b(xx) dx1 ∧ dx2 ∧ . . . ∧ dx2n

If µ j = dx2 j−1 ∧ dx2 j (1 ≤ j ≤ n) then

µ j ∧ µ k = µ k ∧ µ j. Hence, b(xx) is constant and it coincides with the number of permutations of a set with n elements. That is,

ν = n! dx1 ∧ dx2 ∧ . . . ∧ dx2n . � 6.7.4 Let ω be the differential form of degree 1 on an open set U in R3 associated to the vector field F and ν the differential form of degree 2 in U associated to F , G � and the the vector field G . Find the relation between the scalar product �F differential form of degree 3, ω ∧ ν . Solution: We have

ω = F1 dx + F2 dy + F3 dz and

48

Solutions to exercises

ν = G1 dy ∧ dz + G2 dz ∧ dx + G3 dx ∧ dy. Then F , G � dx ∧ dy ∧ dz. � ω ∧ ν = �F 6.7.5 Consider the following differential forms in IR3 : (1) ω = dx − zdy (2) ν = (x2 + y2 + z2 ) dx ∧ dz + (xyz) dy ∧ dz. Evaluate dω , ω ∧ dω , dν , ω ∧ ν . Solution: (a) dω = −dz ∧ dy = dy ∧ dz. (b) ω ∧ dω = dx ∧ dy ∧ dz. (c) dν = (yz − 2y) dx ∧ dy ∧ dz. � � (d) ω ∧ ν = xyz + z(x2 + y2 + z2 ) dx ∧ dy ∧ dz. � 6.7.6 For the differential form

ω = dx + (x2 + y2 ) dy − sin x dz, find the associated vector field F and show that ∇ ×F is the vector field associated to the differential form dω . Solution: The vector field F : IR3 → IR3 is F (x, y, z) = (1, x2 + y2 , − sin x). Then

On the other hand,

� � � e1 e2 e 3 �� � ∂ ∂ � ∇ × F = �� ∂ x ∂∂y ∂ z � = (0, cos x, 2x). � 1 x2 + y2 − sin x �

dω = 2x dx ∧ dy − cos x dx ∧ dz = 0 · dy ∧ dz + cos x dz ∧ dx + 2x dx ∧ dy. �

Solutions to exercises

49

6.7.7 Deduce from the previous exercises and the properties of the exterior differential that if F and G are vector fields of class C1 on an open set U ⊂ IR3 then F × G ) = �G G, ∇ × F � − �F F , ∇ × G� . Div(F Solution: Let ω be the differential form of degree 1 associated with the vector field G . Then ω ∧ ϕ is the differential form of degree 2 associated to the vector field F × G (see Exercise 6.7.2) and hence F × G ) dx ∧ dy ∧ dz. d(ω ∧ ϕ ) = Div (F Since dω is the differential form of degree 2 associated to ∇ × F , by Exercise 6.7.4, it follows that G, ∇ × F � dx ∧ dy ∧ dz dω ∧ ϕ = �G and, by the same argument, F , ∇ × G � dx ∧ dy ∧ dz. ω ∧ dϕ = �F From d(ω ∧ ϕ ) = dω ∧ ϕ − ω ∧ dϕ we deduce F × G ) = �G G, ∇ × F � − �F F , ∇ × G� . � Div(F 6.7.8 Let

ω = xz dy − y dx ; ν = x3 dz + dx

and

ϕ (s,t) = (cos s, sin s,t) be given. Evaluate each of

ϕ ∗ (ω ), ϕ ∗ (dω ), ϕ ∗ (ω ∧ ν ) and ϕ ∗ (ω ∧ dν ). Solution: Denote by (s,t) the coordinates of an arbitrary point of IR2 and by (x, y, z) the coordinates of a point of IR3 , so that a basis of Λ 1 (IR2 ) is {ds, dt} while a basis of Λ 1 (IR3 ) is {dx, dy, dz}. If

ϕ = (ϕ1 , ϕ2 , ϕ3 ) then

 ∗  ϕ (dx)(s,t) = dϕ1 (s,t) = − sin s ds ϕ ∗ (dy)(s,t) = dϕ2 (s,t) = cos s ds  ∗ ϕ (dz)(s,t) = dt

Now put f (x, y, z) = xz and g(x, y, z) = −y. Then

50

Solutions to exercises

� � � � ϕ ∗ (ω )(s,t) = f ϕ (s,t) ϕ ∗ (dy) − g ϕ (s,t) ϕ ∗ (dx) = (t cos2 s + sin2 s) ds

and hence

ϕ ∗ (dω )(s,t) = d(ϕ ∗ ω ) = − cos2 s ds ∧ dt.

Analogously

ϕ ∗ (ν )(s,t) = − sin s ds + cos3 s dt,

and we deduce � � ϕ ∗ (ω ∧ ν )(s,t) = ϕ ∗ (ω ) ∧ ϕ ∗ (ν ) (s,t) = (t cos2 s + sin2 s) cos3 s ds ∧ dt.

Finally, we observe that ω ∧ dν ∈ Λ 3 (IR3 ) and, consequently,

ϕ ∗ (ω ∧ dν ) ∈ Λ 3 (IR2 ) = {0}. �

Solved exercises of Chapter 7

7.4.1 Find

�

M

ω where ω = x2 dy ∧ dz − y dx ∧ dz

and M is the portion of the plane x + y + z = 2 whose projection onto the xy-plane is the triangle with vertices (0, 0), (0, 1), (1, 1), oriented according to the normal vector (1, 1, 1). Solution: We consider the coordinate system (W, ϕ ) defined by � � ϕ (u, v) = u, v, 2 − u − v ,

where

W := {(u, v) ∈ IR2 ; 0 < u < 1, u < v < 1}

is the projection of M onto the xy-plane. Since the cross product � � �e e e � ∂ ϕ ∂ ϕ �� 1 2 3 �� × = 1 0 −1 = (1, 1, 1) ∂u ∂ v �� 0 1 −1 ��

it turns out that this coordinate system is compatible with the orientation of M. Also ϕ (W ) = M. Since ω is the differential form of degree 2 associated to the vector field F (x, y, z) = (x2 , y, 0)

Solutions to exercises

51

Y

�1,1�

W

X

(a)

(b)

Fig. 9.28 (a) Triangle of Exercise 7.4.1; (b) projection onto the xy-plane

we get �

ω =

M

W

=

∂ϕ ∂ϕ F (ϕ (u, v)), × ∂u ∂v

�� � ��

�

d(u, v)

(u2 + v) d(u, v)

W

=

� 1 �� 1 0

u

� 5 (u + v) dv du = . � 12 2

7.4.2 Find the height of the center of gravity of the portion of paraboloid 2z = x2 + y2 that is under the plane 2z = 1, assuming that the density (mass per unit area) is constant. Solution: The third coordinate of the center of gravity of a surface M with constant density is given by � 1 z= z dS area(M) M (see for instance Marsden-Tromba [15]). Since M = {(x, y, z) ∈ IR3 ; 2z = x2 + y2 , 2z < 1} is a graph whose projection onto the xy-plane is W = {(u, v) ∈ IR2 ; u2 + v2 < 1}, we consider the coordinate system (W, ϕ ) where

52

Solutions to exercises

� � 1 ϕ (u, v) = u, v, (u2 + v2 ) . 2

We first evaluate the cross product

� � � e1 e2 e3 � � � ∂ϕ ∂ϕ � × = � 1 0 u �� = (−u, −v, 1). ∂u ∂v � 0 1 v �

Then

� � �∂ϕ ∂ϕ � � d(u, v) � (u + v )� × ∂u ∂v � W

1 z dS = 2 M

��

1 2

��

=π

� 1

�

=

2

W

� (u2 + v2 ) 1 + u2 + v2 d(u, v) (polar coordinates)

ρ3

0

2

�

1 + ρ 2 dρ (integrating by parts)

� ρ2

� � � 3 �1 3 1 1 (1 + ρ 2 ) 2 2ρ dρ (1 + ρ 2 ) 2 � − 3 3 0 0 √ = 215π ( 2 + 1).

=π

Also area(M) = =

� � � ∂ ϕ × ∂ ϕ � d(u, v) � ∂v � W ∂u

�� � �

�� �

1 + u2 + v2 d(u, v) (polar coordinates)

W

= 2π

� 1 �

ρ

1 + ρ 2 dρ

0

= Consequently

2π 3 (2

√

2 − 1). √ 1 2+1 √ .� z= 5 2 2−1

7.4.3 Integrate the vector field F (x, y, z) = (0, 0, z) over the ellipsoid of Exercise 3.5.7, oriented according to the coordinate systems describe in that exercise. Solution: We consider the mapping ψ : IR2 → IR3 ,

ψ (s,t) := (a sin s cost, b sin s sint, c cos s) .

Solutions to exercises

53

For each r ∈ IR let Ur denote the open set Ur := {(s,t) : 0 < s < π , r < t < r + 2π }. Then (Ur , ψ ) is a coordinate system of the ellipsoid (see Exercise 3.5.7). Since � � M \ ψ (U0 ) = {(0, 0, c), (0, 0, −c)} ∪ ψ (0, π ) × {0} we deduce

�

M

F · dSS =

Now evaluate

� ∂ψ ∂ψ × , F (ψ (s,t)) d(s,t). ∂s ∂t

� ∂ψ ∂ψ � × = bc sin2 s cost, ac sin2 s sint, ab sin s cos s . ∂s ∂t �

and

U0

�

� ∂ψ � = a cos s cost, b cos s sint, −c sin s , ∂s � ∂ψ � = − a sin s cost, b sin s cost, 0 ∂t

and, finally

Then,

��

�

M

� ∂ψ ∂ψ × , F (ψ (s,t)) = abc sin s cos2 s ∂s ∂t

F · dSS = 2π abc

� π

sin s cos2 s ds =

0

4π abc. � 3

7.4.4 Determine the flux of the vector field F (x, y, z) = (x, y, z) across the surface x2 + y2 + z2 = R2 . Solution: We suppose the sphere is oriented according to the field N of unit normal vectors outward from the center, that is, the vector field N (x, y, z) = R1 (x, y, z) from Example 5.2.1. Then F (x, y, z), N (x, y, z)� = �F and Flux =

�

M

1 2 (x + y2 + z2 ) = R R

F , N � dS = R area(M) = 4π R3 . � �F

54

Solutions to exercises

7.4.5 Find the flux of the vector field F (x, y, z) = (1, xy, z2 ) through the square in the yz-plane defined by 0 < y < 2, −1 < z < 1 and oriented according to the normal vector (−1, 0, 0).

Fig. 9.29 Vector field of Exercise 7.4.5.

Solution: The surface M = {(0, y, z) : 0 < y < 2, −1 < z < 1} is oriented according to the vector field N (x, y, z) = (−1, 0, 0). Since F (x, y, z), N (x, y, z)� = −1, �F the flux is Flux = −area(M) = −4.

Alternatively, we can consider the coordinate system (A, ϕ ) of M defined by A = {(s,t) ∈ R2 : 0 < s < 2, −1 < t < 1} and ϕ (s,t) = (0, s,t). From � � � e1 e2 e3 � � � ∂ϕ ∂ϕ × = �� 0 1 0 �� ∂s ∂t �0 0 1� = (1, 0, 0)

� � N ϕ (s,t) , = −N

we obtain that (A, ϕ ) is not compatible with the chosen orientation of M. Consequently we consider B = {(t, s) ∈ R2 : 0 < s < 2, −1 < t < 1} and

ψ (t, s) = ϕ (s,t) = (0, s,t). From Lemma 5.2.2, (B, ψ ) is a coordinate system of M compatible with the orientation. Hence, keeping in mind that F (ψ (t, s)) = (1, 0,t 2 ), ∂∂ψt = (0, 0, 1) and

Solutions to exercises ∂ψ ∂s

55

= (0, 1, 0), we conclude that the flux is � � �� �� 1 0 t 2 �� � 0 0 1 � d(t, s) Flux = � � B� 01 0� =−

�

B

d(t, s) = −4. �

The negative sign in this calculation of the flux is explained by the fact that the vector field F flows in the opposite direction to that given by the vector (−1, 0, 0).

Solved exercises of Chapter 8

(a)

(b)

Fig. 9.30 (a) Cylinder of Exercise 8.7.1; (b) disk of Exercise 8.7.2.

8.7.1 Show that the portion of the cylinder {(x, y, z) ∈ IR3 : x2 + y2 = a2 , 0 ≤ z < 1} is a regular surface with boundary.

56

Solutions to exercises

Solution: We will apply Theorem 8.3.2. To do this we take S := {(x, y, z) ∈ IR3 ; x2 + y2 = a2 , z < 1} which is a regular surface of class C∞ and define ϕ : IR3 → IR by

ϕ (x, y, z) = −z. If (x, y, z) ∈ S ∩ ϕ −1 ({0}) then ∇ ϕ (x, y, z) = (0, 0, −1) is not orthogonal to the surface S at (x, y, z) since the orthogonal vectors to the cylinder are horizontal. Consequently M := {(x, y, z) ∈ S : ϕ (x, y, z) ≤ 0} is a regular surface with boundary and

∂ M = {(x, y, z) ∈ IR3 ; x2 + y2 = a2 , z = 0}. Alternatively we could apply Criterion 8.3.1 instead of Theorem 8.3.2. Indeed, it suffices to consider U := {(x, y, z) ∈ R3 : z < 1},

which is an open set in R3 and the mappings Φ , f : U ⊂ R3 → R defined by

Φ (x, y, z) = x2 + y2 − a2 ; f (x, y, z) = −z. Then ∇ Φ (x, y, z) �= (0, 0, 0) for every (x, y, z) ∈ Φ −1 ({0}) and � � 2x 2y 0 (Φ , f )� (x, y, z) = 0 0 −1 has rank 2 at each point of M := {(x, y, z) ∈ R3 : Φ (x, y, z) = 0, f (x, y, z) ≤ 0}. By Criterion 8.3.1, M is a regular surface with boundary and

∂ M = {(x, y, z) ∈ IR3 ; Φ (x, y, z) = 0, f (x, y, z) = 0}. � 8.7.2 Let M be the portion of the plane x + y + z = 1 whose projection onto the xy-plane is the closed disk centered at the origin and with radius 1. (1) Show that M is a regular surface with boundary. (2) If M is oriented according to the normal vector (1, 1, 1) and we consider the induced orientation in ∂ M, find

Solutions to exercises

57

�

∂M

x dy + y dz.

Solution: (1) Let S be the plane x + y + z = 1 and ϕ (x, y, z) = x2 + y2 − 1. Then M = {(x, y, z) ∈ S ; ϕ (x, y, z) ≤ 0}. Since ∇ ϕ (x, y, z) = (2x, 2y, 0) is not a multiple of the vector (1, 1, 1) which is orthogonal to the plane S, it follows from Theorem 8.3.2 that M is a regular surface with boundary and

∂ M = {(x, y, z) ∈ IR3 ; x + y + z = 1, x2 + y2 = 1}. (2) Take

S := {(x, y) ∈ IR2 ; x2 + y2 ≤ 1},

which is a regular surface with boundary in IR2 whose boundary is the unit circle. The mapping f : S → M ; f (x, y) = (x, y, 1 − x − y)

is a C1 bijection and since

� � � e1 e2 e3 � � � ∂f ∂f × = �� 1 0 −1 �� = (1, 1, 1), ∂ x ∂ y � 0 1 −1 �

we deduce that the differential d f (x, y) has maximum rank at all points and the restriction of f to the open unit disc defines a coordinate system of M \ ∂ M which is compatible with its orientation. By Criterion 8.6.1, if we define

γ : [0, 2π ] → M ⊂ IR3 as the image under f of the positively oriented circle, � � γ (t) := f (cost, sint) = cost, sint, 1 − cost − sint ,

then

� � (0, 2π ), γ

is a coordinate system of the boundary ∂ M which is compatible with its orientation. Finally, since � � ∂ M = γ [0, 2π ] , we conclude, for ω = x dy + y dz,

58

Solutions to exercises

�

∂M

ω=

�

γ

ω =

� 2π 0

=

�(0, cost, sint), (− sint, cost, sint − cost)� dt

� 2π � 0

� 1 − sint cost dt = 2π .�

8.7.3 Let M be the lower hemisphere of the unit sphere, i.e. the surface specified by � z = − 1 − x 2 − y2 . Suppose the sphere is oriented according to the exterior normal vector. Evaluate �

∂M

−y dx + x dy.

Solution: The boundary of M := {(x, y, z) ∈ IR3 ; x2 + y2 + z2 = 1, z ≤ 0} is the unit circle in the xy-plane

∂ M = {(x, y, z) ∈ IR3 ; x2 + y2 = 1, z = 0}. This circle is oriented so that if we travel along it with our head towards the outside of the sphere then the lower hemisphere is to our left. Or, in other words, ∂ M is oriented clockwise. We can see this more formally as follows. Take the orientable regular 2-surface in R2 S := {(s,t) ;

π ≤ s < π , 0 < t < 2π } 2

and � � ϕ : S → M ; ϕ (s,t) = sin s cost, sin s sint, cos s .

We observe that the oriented boundary of S is the vertical segment from ( π2 , 2π ) to ( π2 , 0), which can be parameterized as t �→ ( π2 , 2π − t). Since the restriction of ϕ to S \ ∂ S defines a coordinate system of M \ ∂ M which is compatible with its orientation, we apply Criterion 8.6.1 to conclude that

π γ (t) = ϕ ( , 2π − t) = (cost, − sint, 0), 0 < t < 2π , 2 defines a coordinate system of ∂ M which is compatible with its orientation. Finally, from ∂ M = γ ([0, 2π ]) we obtain �

∂M

−y dx + x dy =

�

γ

−y dx + x dy = −2π . �

Solutions to exercises

59

Solved exercises of Chapter 9 9.7.1 For the regular surface with boundary M := {(x, y) ∈ IR2 ; a2 ≤ x2 + y2 ≤ b2 } (1) Find a parameterization of the circles x2 + y2 = a2 , x2 + y2 = b2 that is compatible with the orientation of M. (2) Using Green’s formula, evaluate �

∂M

(x3 − y) dx + (cos y + 2x) dy.

Solution: (1) According to the Criterion 8.5.1, we parameterize the circles so that, as we travel along them, we keep the set M to ours left. That is, we consider

Fig. 9.31 Regular surface with boundary of Exercise 9.7.1.

γ 1 : [0, 2π ] → R2 ; γ 1 (t) = (b cost, b sint) and

γ 2 : [0, 2π ] → R2 ; γ 2 (t) = (a cost, −a sint).

Then

∂ M = γ 1 ([0, 2π ]) ∪ γ 2 ([0, 2π ]) and

�

�� � (0, 2π ), γ 1 , (0, 2π ), γ 2

60

Solutions to exercises

are two coordinate systems of ∂ M compatible with the orientation induced by that of M. (2) We apply Green’s formula (Theorem 9.4.5) with P(x, y) = x3 − y, Q(x, y) = cos y + 2x. Accordingly �

∂M

(x3 − y) dx + (cos y + 2x) dy

coincides with �� � M

� �� ∂Q ∂P 3 d(x, y) − d(x, y) = ∂x ∂y M = 3 area(M) = 3π (b2 − a2 ). �

9.7.2 Let M be the part of the plane x + y + z = 1 whose projection onto the xyplane is the closed disc centered at the origen and with radius 1. Suppose that M is oriented according to the normal vector (1, 1, 1). Using Stokes’ Theorem, find �

Solution: Take

∂M

x dy + y dz.

U := {(x, y) ∈ IR2 ; x2 + y2 ≤ 1}

and f : U → M ; f (x, y) = (x, y, 1 − x − y). As we saw in Exercise 8.7.2, �

∂f ∂f × ∂x ∂y

�

� � � e1 e2 e3 � � � (x, y) = �� 1 0 −1 �� = (1, 1, 1), � 0 1 −1 �

and the restriction of f to the open unit disc defines a coordinate system of M \ ∂ M that is compatible with its orientation. We consider the vector field F (x, y, z) = (0, x, y) associated to the differential form of degree 1, ω = x dy + y dz. From Stokes’ Theorem, � � � ω= ∇ × F · dSS = ∇ × F · dSS . ∂M

Moreover, since

M

M\∂ M

Solutions to exercises

it turns out that �

M\∂ M

61

� � e1 � ∇ × F ) (x, y, z) = �� ∂x (∇ �0

∇ × F · dSS = =

e2 ∂y x

� e3 �� ∂z �� = (1, 0, 1) y�

�� �

∇ × F ) ( f (x, y)), (∇

U

��

�

∂f ∂f × ∂x ∂y

�

� (x, y) d(x, y)

2 d(x, y) = 2π .

U

As expected, we obtain the same result as in Exercise 8.7.2. � 9.7.3 Verify Stokes’ Theorem for the lower hemisphere of the unit sphere � z = − 1 − x 2 − y2

and the vector field

F (x, y, z) = (−y, x, z). Solution: Suppose that the sphere is oriented according to the exterior normal vector. We denote M := {(x, y, z) ∈ IR3 ; x2 + y2 + z2 = 1, z ≤ 0}. Consider the coordinate system (D, ψ ) where D := {(u, v) ∈ IR2 ; u2 + v2 < 1} and

� � � ψ : D → M ; ψ (u, v) = u, v, − 1 − u2 − v2 .

Now evaluate the cross product �

∂ψ ∂ψ × ∂u ∂v

�

� � e1 e2 e3 � �1 0 √ u (u, v) = � 1−u2 −v2 � � 0 1 √ v2 2 1−u −v

= √ �

−u , 1−u2 −v2

√

� � � � � � �

−v , 1−u2 −v2

� 1 .

The third coordinate is positive, what means that the normal vector obtained points towards the interior of the sphere. Or, equivalently, the coordinate system (D, ψ ) reverses the orientation of M. From ψ (D) = M \ ∂ M we conclude �

M

∇ × F · dSS = −

�

(D,ψ )

∇ × F · dSS .

62

Solutions to exercises

Since

� � � e1 e2 e3 � � � ∇ × F ) (x, y, z) = �� ∂x ∂y ∂z �� = (0, 0, 2) (∇ � −y x z �

we obtain �

M

�� �

∇ × F · dSS = −

D

��

=−

D

∇ × F ) (ψ (u, v)), (∇

�

∂ψ ∂ψ × ∂u ∂v

�

� (u, v) d(u, v)

2 d(u, v) = −2π .

We already proved in Exercise 8.7.3 that �

∂M

Hence

�

F · dss = �

M

∂M

−y dx + x dy = −2π .

∇ × F · dSS =

�

∂M

F · dss. �

9.7.4 Let M be the sphere with radius a > 0 centered at the origin and F a vector field of class C2 on a neighborhood of the corresponding closed ball. Show that �

M

∇ × F · dSS = 0.

Solution: Suppose that M is oriented according to the normal exterior. By the Divergence Theorem and Theorem 1.2.3, �

M

∇ × F · dSS =

���

B(0,a)

∇ × F )(x, y, z) d(x, y, z) = 0. Div (∇

Alternatively, we can decompose M = M1 ∪ M2 where

M1 := {(x, y, z) ∈ IR3 ; x2 + y2 + z2 = a2 , z ≥ 0}

is the upper hemisphere and M2 := {(x, y, z) ∈ IR3 ; x2 + y2 + z2 = a2 , z ≤ 0} is the lower hemisphere. Note that

∂ M1 = ∂ M2

Solutions to exercises

63

is the circle in the xy-plane with radius a. Moreover, if we put � γ (t) = a cost, a sint, 0)

then

�

(0, 2π ), γ

�

is a coordinate system of ∂ M1 compatible with its orientation but it is a coordinate system of ∂ M2 that reverse the orientation induced by that of M2 in ∂ M2 . By Stokes’ Theorem � � ∇ × F · dSS = F γ

M1

and also

�

M2

�

∇ × F · dSS = −

γ

F.

Consequently �

M

∇ × F · dSS =

�

M1

∇ × F · dSS +

�

M2

∇ × F · dSS = 0. �

9.7.5 Let M be the triangle without vertices; that is {(x, y, z) ∈ IR3 ; x + y + z = 1, x ≥ 0, y ≥ 0, z ≥ 0, } with the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) removed. Verify the validity of Stokes’ Theorem for this surface. Solution: We suppose that the orientation of the plane x + y + z = 1 is defined by the vector (1, 1, 1). Consider the subset D of IR2 consisting of the triangle {(x, y) ∈ IR2 ; x + y ≤ 1, x ≥ 0, y ≥ 0} with the vertices (0, 0), (0, 1), (1, 0) removed. Also take f : IR2 → IR3 to be the mapping defined by f (x, y) = (x, y, 1 − x − y). This is a C∞ function and the rank of f � is 2, which is maximal on R2 . Thus D is a regular 2-surface with boundary in IR2 and M = f (D) is a regular 2-surface with boundary in IR3 . As we proved in Exercise 8.7.2, � � � e1 e2 e3 � � � ∂f ∂f × = �� 1 0 −1 �� = (1, 1, 1), ∂ x ∂ y � 0 1 −1 �

64

Solutions to exercises

Fig. 9.32 Triangle of Exercise 9.7.5.

what means that ( f , D \ ∂ D) is a coordinate system of M \ ∂ M compatible with its orientation. According to Criterion 8.6.1, if γ is a positively oriented parameterization of ∂ D then f ◦ γ is a positively oriented parameterization of ∂ M. Let ω be a differential form of degree 2 and class C1 on a neighborhood of M and denote

ν := f ∗ ω . According to Green’s formula from Chapter 2, �

ν=

�

ν,

∂D

�

dν. D

Since, by 7.1.1 and 6.4.1, �

∂M

we can conclude

ω= �

∂D

∂M

ω=

�

dω =

M

�

M

�

dν D

dω. �

9.7.6 Find the work done in moving a particle counterclockwise around the (topological) boundary of the intersection of the plane x + y + z = 1 with the first octant under the action of the force field F (x, y, z) = (x + xz2 , x, y).

Solutions to exercises

65

Solution: Let M the oriented regular surface of the previous exercise. We are asked to evaluate � F · dss, ∂M

and, to do so, we apply Stokes’ Theorem. First evaluate � � � e1 e2 e3 � � � ∇ × F ) (x, y, z) = �� ∂x ∂y ∂z �� = (1, 1 + 2xz, 1). (∇ � z + xz2 x y � Let D be the triangle

{(u, v) ∈ IR2 ; u + v ≤ 1, u ≥ 0, v ≥ 0} with the vertices (0, 0), (0, 1), (1, 0) removed and let f (u, v) = (u, v, 1 − u − v). As in the previous exercise �

∂M

F · dss = =

�

M

∇ × F · dSS

�� �

∂f ∂f ∇ × F ( f (u, v)), × ∂u ∂v

D

=

�� � D

=

�

d(u, v)

� 3 + 2u(1 − u − v) d(u, v)

� 1 � � 1−u � 0

0

� � 19 3 + 2u(1 − u − v) dv du = . � 12

9.7.7 Verify the Divergence Theorem for the vector field F (x, y, z) = (x, 2y, z) and the closed surface M := M1 ∪ M2 , where M1 := {(x, y, z) ∈ IR3 ; x2 + y2 = z2 , 0 ≤ z ≤ 2} and M2 is the closed disc with radius 2 in the plane z = 2 and centered at (0, 0, 2). Solution: M˜ 1 := M1 \ {(0, 0, 0)} and M2 are regular surfaces with boundary, and we suppose they are oriented according the vector field N of unit normal vectors pointing to the exterior of the region bounded by M. According to Example 9.4.1, a coordinate system of M˜1 \ ∂ M˜ 1 compatible with the orientation is (A, ϕ ) where A = {(u, v) ∈ R2 : 0 < u2 + v2 < 4}

66

and

Solutions to exercises

� � � ϕ : A ⊂ R2 → R3 ; ϕ (u, v) = v, u, u2 + v2 .

On the other hand, a coordinate system of M2 \ ∂ M2 is (B, ψ ), where B := {(u, v) ∈ R2 : u2 + v2 < 4} and Since

ψ : B ⊂ R2 → R3 ; ψ (u, v) = (u, v, 2). �

∂ψ ∂ψ × ∂u ∂v

�

(u, v) = (0, 0, 1) = N (ψ (u, v)),

it turns out that the coordinate system (B, ψ ) of M2 is compatible with the orientation. Note that ϕ (A) = M˜1 \ ∂ M˜ 1 and ψ (B) = M2 \ ∂ M2 . Now evaluate �

M

F , N � dS := �F

�

M˜ 1

F , N � dS + �F

�

M2

F , N � dS. �F

√ Since F (ϕ (u, v)) = (v, 2u, u2 + v2 ) we have � � √ � 2 2� � �� �� v 2u u u+ v �� � 0 1 √u2 +v2 � d(u, v) F , N � dS = �F � � M˜ 1 A� v � �1 0 √ 2 2 � u +v

=

��

√

u2

d(u, v) u2 + v2 �� 2 � � � 2π � 8π 2 2 cos θ dθ = ρ dρ = . 3 0 0 A

Keeping in mind that F (ψ (u, v)), N (ψ (u, v))� = 2 �F we deduce

�

M2

F , N � dS = 2 area(M2 ) = 8π . �F

Consequently, �

M

F , N � dS = �F

32 π. 3

Finally, evaluate F (x, y, z) = 4 DivF and, denoting by

Solutions to exercises

67

Ω := {(x, y, z) ∈ IR3 ; x2 + y2 ≤ z2 , 0 ≤ z ≤ 2} the region bounded by M, we find (using cylindrical coordinates x = ρ cos θ ; y = ρ sin θ ; z = z), � � ��� � 2 �� z �� 2π Div F (x, y, z) d(x, y, z) = 4 ρ dθ dρ dz Ω

0

0

� 2

= 4π

0

z2 dz =

0

32 π. 3

We have verified that �

M

F , N � dS = �F

���

Ω

Div F (x, y, z) d(x, y, z). �

9.7.8 Using Stokes’ Theorem, find the z−coordinate of the center of gravity of the hemisphere M = {(x, y, z) ∈ IR3 ; x2 + y2 + z2 = R2 , z ≥ 0} assuming that the surface density is constant. Solution: Recall that the z−coordinate of the center of gravity is z=

1 area M

�

z dS.

M

Suppose that the sphere is oriented according to the field of normal unit vectors N (x, y, z) = R1 (x, y, z) (see Example 5.2.1). To apply Stokes’ Theorem we need a vector field F such that ∇ × F,N� = z �∇ (9.14) and, to this end, it suffices to have ∇ × F )(x, y, z) = (0, 0, R). (∇ According to the geometrical interpretation of the curl, described following Corollary 9.4.1, the vector field F should represent a rotation around the axes Z. Because ∇ × G )(x, y, z) = (0, 0, 2) of this, we try with G (x, y, z) = (−y, x, 0) and evaluate (∇ whence it follows that the vector field F (x, y, z) :=

R R G (x, y, z) = (−y, x, 0) 2 2

satisfies the identity (9.14). From Stokes’ Theorem 9.4.4,

68

Solutions to exercises

�

z dS =

M

�

M

=

∇ × F , N � dS �∇

�

∂M

R = 2

F · dss

�

γ

(9.15)

−ydx + xdy

= π R3 . where

γ : [0, 2π ] → R3 ; γ (t) = (R cost, R sint, 0)

is a parameterization of the circle of radius R in the xy-plane oriented counterclockwise. Finally, as the area of M is 2π R2 we reach the conclusion z=

R .� 2

9.7.9 Consider the vector field F (x, y, z) = (x, y, z) and let M be the part of the paraboloid z = 4 − x2 − y2 that satisfies z ≥ 0. Assuming that M is oriented so that the third coordinate of the normal vector is positive, find the relation between the flux of the vector field F across M and the flux of F across the disc D in the xy-plane centered at the origin, of radius 2 and oriented according to the normal vector (0, 0, −1). Solution: We proceed as in Example 9.5.2. Consider the compact set K := {(x, y, z) ∈ R3 : 0 ≤ z ≤ 4 − x2 − y2 } whose topological boundary can be decomposed as the union of two regular surfaces with boundary ∂ top (K) = M ∪ D. Define � π π π � f (r, s,t) := 2r sin( s) cos(2π t) , 2r sin( s) sin(2π t) , 4r2 cos2 ( s) , 2 2 2

so that

K = f (I), where I is the closed unit cube. We denote by Flux1 the flux of F across the regular surface with boundary M and by Flux2 the flux of F across the regular surface with boundary D. An analogous argument to that of Example 9.5.2 permits us to conclude that

Solutions to exercises

69

Flux1 + Flux2 =

���

=3

F (x, y, z) d(x, y, z) DivF

K

���

d(x, y, z).

K

To solve this integral, we make the change of variables x = ρ cos θ ; y = ρ sin θ ; z = z to obtain Flux1 + Flux2 = 6π

� 2

� � ρ

= 6π

� 2

ρ (4 − ρ 2 )dρ

0

0

4−ρ 2

dz dρ

0

= 24π . �

�

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