Calculus BLUE Multivariable Calculus Vol III Integrals [3, 3 ed.] 9781944655051

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CALCULUS BLUE MULTIVARIABLE VOLUME 3 : INTEGRALS ROBERT GHRIST 3rd edition, kindle format Copyright © 2019 Robert Ghrist All rights reserved worldwide Agenbyte Press, Jenkintown PA, USA ISBN 978-1-944655-05-1 1st edition © 2016 Robert Ghrist 2nd edition © 2017 Robert Ghrist

prologue chapter 1: defining integrals chapter 2: the fubini theorem chapter 3: double integrals chapter 4: triple integrals chapter 5: averages chapter 6: centroids & centers chapter 7: moments of inertia chapter 8: the inertia matrix chapter 9: solid body mechanics chapter 10: probability & integration

chapter 11: multiple random variables chapter 12: covariance matrices chapter 13: cylindrical coordinates chapter 14: spherical coordinates chapter 15: changes of variables chapter 16: choosing coordinates chapter 17: surface integrals chapter 18: gaussians, redux chapter 19: data and dimension epilogue foreshadowing: calculus of fields

enjoy learning! use your full imagination & read joyfully… this material may seem easy, but it’s not! it takes hard work to learn mathematics well… work with a teacher, tutor, or friends and discuss what you are learning. this text is meant to teach you the big ideas and how they are useful in modern applications; it’s not rigorous, and it’s not comprehensive, but it should inspire you to do things with math… exercises at chapter ends are for you to practice. don’t be too discouraged if some are hard… keep working! keep learning!

thy fearful symmetry

to put the pieces together

but you may recall… defining integrals in single-variable calculus was definitely nontrivial!

just like (definite) integrals in single-variable calculus

the integral of a function over a region is a limit of sums of function values on “cubes” filling up the region, weighted by the volumes of the cubes

why do multivariate integrals?

integrals are at the heart of so very many applications of mathematics, including…

we will (finally!) learn how to compute the surface area of objects that are not symmetric about some axis…

likewise with volumes of irregular shapes…

this is useful in MULTIPLE dimensions…

f =

@

f dx R

@

dx R

…SUCH AS CENTROIDS AND…

center of mass and moments of inertia are multiple integrals based on mass density

the inertia matrix encodes moments about the three coordinate axes

[I] =

Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz

inertia matrices help clarify the complexities of rotating bodies in 3-d

[I] =

Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz

but there's more than 3-d when we get to…

a Gaussian density on  is: (x-μ)2 - 2σ2 1

2πσ2

e

these applications will motivate us to learn some new coordinate systems…

which will in turn reveal new perspectives on derivatives and integrals

we’ll use integrals to explore volumes of spheres & apply this to Gaussian probability densities

in single variable calculus, it was “easy”

NOTATION:

@

b x=a

f(x) dx

notation:

@ f(x) dx

IT’S A NUMBER

IT’S A CLASS OF FUNCTIONS (SAME UP TO A CONSTANT)

defined in terms of a limit of Riemann sums (ouch!)

defined in terms of an antiderivative (yay!)

used to compute area, volume, work, force, etc…

used to compute definite integrals

we will restrict attention to n integrals of f :  _ 

d dx

f:_ f' :  _ 

n

∫-dx

f: _ D

?

[Df] the derivative is not the same type of function… so you can’t invert?

the definite integral

∆x xi

recall that for a function f on the interval [ a , b ] one defines…

@

b

f(x) dx =  + # f(xi) ∆x

x=a

∆x_0

f(x)

i

a if the limit exists, then we say that the integrand is “Riemann-integrable”

b x

how do you do “area under a curve” in higher dimensions?

do not think in terms of area under the graph of the function – in fact, don’t think in terms of the graph at all… do think of the integrand as a (potentially negative) density function; the integral is the total mass so, then, how would you think about computing mass of an object?

n

for f :  _ , the integral is a limit of “Riemann sums” over a grid of “cubes” of dimension n & side lengths ∆x n=1

n=2

@

n=3

n

n

f(x) dx =  +# f(xi) (∆x) ∆x_0 i

but that definition seems a little… complicated

how do you integrate over a bounded region?

to integrate over a domain… fill the domain with cubes and sample there…

fill the domain with cubes and sample there…

to integrate over a domain…

then decrease the grid size unto convergence!

n

for f : R ⊂  _, the integral is denoted…

@

R

f dx

or equivalently

@ @@ f dx dx …

R

1

2

this is the volume element on n

… dxn

in terms of coordinates x = ( x1, x2 , … , xn ) b

@ f dx a

double and triple integrals are used in physics to compute total charge based on surface/volumetric charge density. also, magnetic flux density is integrated over a surface to compute total flux…

the physics of motion of a solid body requires use of mass density and 3-d integrals to compute: center of mass, moment of inertia, angular momentum, torque, kinetic energy, and more…

a vector of n random variables determines a probability density in n-dimensions. various statistical features (mean, variance, etc.) are expressed in terms of integrals in n-d with respect to this density…

but how do we compute integrals…?

a few simple integrals ? compute the integral…

compute the integral…

@ 5 dx = @@@ 5 dx dy dz = @ 5 dV = 5V

@ 4 dx = @@ 4 dx dy = @ 4 dA = 4A

R

R

R

where R is the unit ball in 3

R

this is the volume element dy dx

4 20 = 5 3π = 3 π

dz

where R⊂2 is -2 ≤ x, y ≤ 1

R R

= 36

this is the area element dy dx

we have not discussed the many complexities of defining the integral…

for example, when do the Riemann sums converge?

what happens if the integrand is highly discontinuous or oscillatory?

worse still, what happens if the domain of integration has a weird, fractal shape?

if you take a course in real analysis you will spend a lot of time learning integration theory of the many types of integrals one can define, the lebesgue integral has the best combination of power and generality

the big picture

multivariate integrals are all definite integrals: think in terms of mass & density

1

Consider f : 2 _  as shown in the contour map, with superimposed mesh: ------->

0 1 2

3

4

5

6

6

write Riemann sum approximations to the integral over this mesh, assuming mesh square size of length 1/10 and…; a) using the maximal value on each square b) using the minimal value on each square

2 3 4

8

7

6

5 4 3 2 1

if you visualize the double integral of f(x, y) = x+y over R = { 0 ≤ x, y ≤ 2 } as the “volume under the graph”, you can compute the integral via geometry. try it! use the method of the last problem to compute the integral of f(x, y) = (x2+y2)1/2 over the unit disc R = { (x2+y2)1/2 ≤ 1 } in the plane. think geometrically! what is the integral of a function on 0? Recall, 0 is simply a single point, zero. begin by answering what a function f:0_  looks like & think!

Integration theory is not for the weak in spirit. For the curious, here is an overview. One begins with a notion of measure, μ, of subsets of Rn, that is non-negative and additive, in the sense that: ∩

μ(A B) = μ(A) + μ(B) – μ(A∩B) The first difficulty is this: some sets have measure zero, even though they are not empty (nor even all that “small”). For example, the rational numbers have measure zero in R1, even though they are dense in the reals. (!)

The second difficulty is more subtle: not all sets are measureable. This unfortunate immeasurability is at the heart of various curious facts, like the Banach-Tarski paradox… Take comfort: any set you will ever see or use in practice is measurable. The Lebesgue integral ∫X f dμ (pronounced “luh-BAYG”) is defined in terms of a limit (big surprise!), but not like the Riemann integral, though it agrees with it when both are computable.

You may want to take a course that covers the Lebesgue integral: that course is called “REAL ANALYSIS” and it usually covers two semesters. By the end of that year of study, you will have carefully defined and derived all its (seemingly obvious) properties. And that is just the start… Don’t worry about the details if you are just starting out. Just check the box and move on… But know that much more awaits!

I have read the above completely and agree to abide by these terms

WE KNOW WHAT AN INTEGRAL IS

the definition of an integral in terms of Riemann sums is… daunting. is there hope?

consider the analogous case of multivariate differentiation… how did we handle that?

we differentiated multivariate functions?

we simply worked one variable at-a-time

AS A “PARTIAL INTEGRAL”

start off with a matrix of numerical entries… the double integral is the sum of all the entries but you could add along the rows… or along columns rows-then-columns is equal to columns-then-rows

n

for f :  → a “sufficiently integrable” function, The fine print? don’t even ask…

@

n

f dx =

@ (@ ( (@ …

 



) ))

f dx1 dx2 … dxn

concerning order see below…

each integral with respect to dxi assumes all other xj constant

it will take some time to get an intuition for this, but don’t despair... this is so helpful!

partial integration is the way to go

in general, the limits of integration are over the entire real axis; but in practice, one restricts to limits that may or may not be constants

@

2

x=1

2xyz - 3y2 dx

= x2yz - 3y2x

2 x=1

@

1 y=-1

2xyz - 3y2 dy

= xy2z – y3

1 y=-1

@

y

z=x2

2xyz - 3y2 dz

= xyz2 - 3y2z

y z=x2

= (4-1)yz - 3y2(2-1)

= ( 1 –1 ) xz - ( 1 – (-1) )

= xy(y2-x4) - 3y2(y-x2)

= 3yz - 3y2

= 2

= xy3-x5y-3y3+3x2y2

for numerical limits of integration, the integral is a function that depends only on the other variables

for symbolic limits, things can get more complicated…

a simple double integral y

@@

R

this is easy to evaluate because we are integrating over a rectangle

6x2y dA

over the rectangle 1≤x≤4 & 0≤y≤3 4

=

@ @ x=1

3 y=0

6x2y dy dx =

@

4

x=1

3

3x2y2 dx = y=0

@

dA = dx dy = dy dx

dy dx

x

4

4

27x2 dx = 9x3

x=1

x=1

= 9(64-1) = 567

does the order matter? no. no, it does not. 3

=

@ @

4

y=0 x=1

6x2y dx dy =

@

3

y=0

4

2x3y

dy =

x=1

@

3

3

126 y dy = 63y2 = 63(9-0) = 567

y=0

y=0

Back to area-between-curves

the area element is an “infinitesimal rectangle” dy dx

A=

f

y

dA = dx dy = dy dx

@ dA = @@ 1 dx dy = @@ 1 dy dx A =

b

@ @

f(x)

x=a y=g(x)

dy dx =

@

b x=a

y

f(x)

g

a

dx =

y=g(x)

@

b

f(x)-g(x) dx

x=a

b

x

You have to pay attention to limits! In general, limits will be non-constant And functions of the as-yet-unintegrated variables

In theory, the order doesn’t matter… In practice, well, you can guess what happens… you have to invest time & effort practicing multiple integrals: that comes next!

well, yes & no. the hard part (for us) is not in the integration itself…

the big picture

the fubini theorem lets you do “partial integration” one variable at a time in any order you like

1

compute the following “partial integrals” with numerical limits A) D)

@ @

3

4xy – 2y2 dx

b)

xy + x2 dy

e)

x=0 0

y=4

@ @

1 t=0 7

 (st) dt

c)

y2 – z3 dx

f)

x=-3

@

1

z=-1 8

@

s=8

(x+y+z) dz 5st ds

note that all your answers should be functions of the other variables…

2

compute the following “partial integrals” with variable limits A) D)

@

y

z=0 3s

@

t=s

3y + 2z dz 2

2t-3st dt

b) e)

@

0

x=y y

@

z=x

(1-xy) dx

c)

e-xyz dz

f)

@

v2

u=v 1

@

z=-1

(u+v)2 du (x+y+z) dz

3

compute the following integrals over rectangular domains A)

3/2 1/2

@ @

x=0 x=0

x+2y dx dy

b)

1

@ @

1

 (s+t) ds dt

c)

s=0 t=0

3

@ @

1

u=1 v=-1

(u+v)2 dv du

if you feel like it, reverse the orders of integration & verify the answers

4

explain exactly what is wrong with the following definite integrals A)

2y

@ @

x

x=1 y=0

x+2y dy dx

b)

2

@ @

z

z=1 y=0

(y-z) dz dy

c)

2

z

@ @ @

x

x=1 y=x z=y

(xyz)3 dz dy dx

remember: in a definite integral, the final answer has to be numerical!

5

true or false: given a double integral over a rectangular region in the plane, you don’t have to worry about which variable to integrate with respect to first. either way will be essentially the same computation.

a double integral is given by

@@ dA =

dy dx

R

f dA =

= dx dy

@@ f(x, y) dx dy R

WHEN SETTING UP LIMITS

area of an ellipse x2 y2 + = 1 9 4

compute the area bounded by the ellipse y

y 3

x

x

3

A =

-3

@ @

4-4x2/9

x=-3 y=- 4-4x2/9

dy dx

y

y

2

x

x -2

2

A =

@ @

9-9y2/4

y=-2 x=- 9-9y2/4

dx dy

A simple mass computation compute the mass of a 2-d plate given by

0 ≤ y ≤ 4 - x2

=

if the density is

ρ(x, y) = 1 + x2 + y y=4-x2

y

=

@@ @ @

dM =

2

x=-2 2 x=-2

4

dy dx

-2

M =

2

2

@ @

4-x2

x=-2 y=0

y + yx2 + 21 y2

4-x2 y=0

dx

(4-x2) + (4x2 – x4) + (8 – 4x2 + 1 x4 ) dx 2

12 - x2 – 21 x4

= 12x - 31 x3 - 101 x5 x

1 + x2 + y dy dx

2

x=-2

= 544 15

would it have been more difficult horizontally first? try it!

The order does matter

Order-of-integration matters!

@@ =

R

over the triangular region

3 x y e dA

(1,2)

dy

2

@ @ y=0

y

y=2x

1

dx

3 x y e dx dy =

Integrating this seems impossible!

x=y/2

(0,0)

(1,0)

But if we reverse the order of integration…

=

1

@ @

2x

x3

ye

dy dx

=

x=0 y=0

let u=x3 then du=3x2 dx

=

@ @

1

x=0 1

u=0

1 2 y e 2

x3

2x

dx =

y=0

2 u 2 eu = e du 3 3

@

1

x=0 1

u=0

2x2

3 x e dx

= 2 (e-1) 3

x

Leads to complicated limits

Additivity & integration domains

@@

R

this would seem to require several integrals to break up the domain into components…

x2y dA

y

x but integration is “additive” – you can subtract

=

5

@ @

5

x2y dy dx -

x=1 y=0 3 5 2 5

x = 3

x=1

y 2

y=0

-

5

@ @

2

x2y dy dx -

x=2 y=1 5 2 3 2 x y

3

x=2 2

y=1

-

4 3 x

3

4

@ @

4

x2y dy dx

x=2 y=3 4 2 y

x=2 2

y=3

=

this is not hard! just antidifferentiate…

2357 = 6

unbounded/singular integrals are a double threat

improper double integrals

@@ =

R

∞

@ @ y=0

= =

∞

@

y=0 ∞

@

y=0

over the first quadrant in the plane

dA x2+y2

this has both an unbounded domain and a singular integrand at the origin

∞

dx dy 2+y2 x x=0 1 y

@

∞ u=0

y

=

∞

@ @

y=0 x=0

du dy = 2 1+u

1 π dy y 2

∞

∞

@

y=0

dx dy 2 2 2 y (1+x /y ) 1 y (u)

∞

x dy

u=0

let u = x/y then du = 1/y dx

that improper integrals never show up… right? hello? I mean, it can’t be all that…

we will investigate gaussians later when we cover basic multivariate probability…

a 2-d Gaussian centered at the origin, symmetric, with standard deviation σ>0, & amplitude C>0 is given by the function

-(x2+y2)/2σ2

f( x, y ) = Ce

one of the more important results concerns the integral of a Gaussian over the plane

@@

2

f dA =

∞

@ @

∞

-(x2+y2)/2σ2

Ce

x=-∞ y=-∞

2 = 2πCσ dx dy

the big picture

double integrals are dicey! pay attention to bounds & order-of-integration to avoid getting stuck

1

Compute the following double integrals A) b) c) d)

e)

@@ @@ @@ @@ @@

R

R

R

R

R

u – v2 du dv

where R = { 0 ≤ u ≤ 4 & -2 ≤ v ≤ 2 }

s2 + t2 ds dt

where R = region in the plane between

x2 exy dx dy

where R = triangle defined (0,0), (0,4), & (4,4) by the points

eax + by dx dy

where R = triangle bounded by

t = s2 & t = 2s

x = 0, y = 0, & ax + by = 1

(this answer will depend on a and b…)

2 – u + v du dv where R = region in the plane given by

1 ≤ u2 + v2 ≤ 3

(think before integrating: what does the domain look like? notice any symmetry?)

2

reverse the order of integration of the following double integrals. be very careful with the limits… you may have to think!@ a)

π/2

@ @

u

u=0 v=0

b)

2

@ @ s=0

C)

a

@ @

s2 t=0

evaluate this once you have reversed limits

s2 - 2t dt ds

evaluate this once you have reversed limits

a- a2-x2

x=0 y=0

3

 u dv du

xey dy dx 2 (y-a)

do not evaluate this: it does not converge…

compute the following improper integrals by taking appropriate limits a)

∞

@ @

∞

x=0 y=0

e

-2x-3y

dx dy

b)

∞

@ @

∞

x=0 y=0

2(x+y) dx dy

this one is a challenge!

a triple integral can be reduced

integrating out one variable takes you from a triple to a double integral

then, integrate the next variable to go from a double to a single integral

& a single integral is a simple integral!

is ordering the variables

limits on a 3-dimensional simplex what are all the ways to set up

over the 3-simplex given by

@@@ f dV

6x + 3y + 2z ≤ 6 x,y,z≥0

z

(0,0,3)

6x+3y+2z=6

(0,2,0) y (0,0,0) (1,0,0)

y

z (0,3)

(0,2) (0,0)

6x+3y=6

(1,0)

x

(0,0)

z (0,3)

6x+2z=6

(1,0)

x

(0,0)

x

3y+2z=6

(2,0)

y

limits on a 3-dimensional simplex 3

2-2z/3

@ @ z=0 1

y=0

@ @

2-2x

x=0 y=0

@

1-y/2-z/3

@

3-3y/2-3x

f dx dy dz

to integrate x first, fix y and z constant, then solve for x limits via

f dz dy dx

for y next, fix z constant, then solve for y limits via the y-z projection

x=0

(0,2,0) y (0,0,0)

z 6x+3y=6

(1,0)

6x+3y+2z=6

(1,0,0) (0,3)

(0,0)

(0,0,3)

z=0

y (0,2)

z

x

(0,0)

z (0,3)

6x+2z=6

(1,0)

x

(0,0)

x

3y+2z=6

(2,0)

y

is a delightful challenge

fill in the blanks! inference based on partial information is a good test of how well you “get” integrals since these limits have y and z in them, they must refer to the x variable… reconstruct the y-z plane… 2

y

4

@@ @ 4

z

@@ @

y+z

0 z/2 0 z

@@ @

y=z/2

y+z

0 z/2 0

4

y= z

z

z

y+z

0 z/2 0

4

reverse the limits on y and z, being careful of direction!

4

z

@@ @

y+z

0 z/2 0

f d d d

f dx d d f d x dy d f d x dy d

=

= = =

@ @ @ @ @ @

f dx dz dy y+z 0

2y

y+z

y2

0

2

2y

y+z

0

y2

0

@ @ @ @ @ @

f dx dz dy f dx dz dy f dx dz dy

to infer the shape of the domain from the limits

visualizing a 3-dimensional domain z 1

1

@ @ @

1- x

(0,1)

f dz dx dy

y=-1 x=y2 z=0

the innermost limits implicate only x and z, so plot in the x-z plane

z=1- x

(0,0)

this is something you need to know if you want to, say, change the order of integration variables…

the remaining limits are for the x and y variables, so plot this in the x-y plane

y x=y2 (0,0)

(1,0)

x

(1,1) x (1,-1)

visualizing a 3-dimensional domain

1

1

@ @ @

1- x

f dz dx dy

seeing the big picture like this can be tough, but it makes reversing the order of integration easy…

z (0,1) z=1- x

y=-1 x=y2 z=0

(0,0) y x=y2 (0,0)

(1,0)

x

(1,1) x (1,-1)

it's not so easy to see in 3-d with 2-d images you have to try!

limits of integration for triple integrals can be very challenging! try reasoning from planar projections, as best you can

simply cannot be represented by one integral using planar projections

the big picture

triple integrals are tricky & can easily break your visual intuition! use careful projections in order to get correct limits

1

Compute the volumes of the following regions using triple integrals A) where x, y, z ≥ 0 and between the planes x + y + z = 1 and x + 2y + 3z = 1 b) where x, y, z ≥ 0 and between the regions x + y + z = a>0 and x + y + z = b>a C) where x, y, z ≥ 0 and between z = x2 + y2 – 9 (top) and y = 4 – x2 (side) D) bounded between x = z2 + y2 – 4 and x = 2 – (y2 + z2)/2

2

compute the following triple integrals (do you need to change order?) a)

2

u

@ @ @

v

u=0 v=0 w=u

c)

1

π/2 1

@ @ @

y=0 x=0 z=-1

uvw dw dv du x  (xy) dz dx dy

b)

2π

π/2 R

@ @ @

θ=0 φ=0 ρ=0

d)

1

πz

@ @ @

z

ρ2  φ dρ dφ dθ  (y/x) dx dy dz

z=0 y=0 x=y/π

this one is a true challenge!

3

change the order of integration of the following as specified (do not evaluate) a) b)

1

1

@ @ @

u2

v=0 u=0 w=0 1 1 1-y

@ @ @

x=0 y= x z=0

c)

2

2a

@ @ @

a

a=0 c=0 b=0

4

?

?

f(u, v, w) dw du dv =

@ @ @

f(x, y, z) dz dy dx

@ @ @

f(a, b, c) db dc da

?

f(u, v, w) du dv dw

w=? v=? u=? ? ? ?

=

f(x, y, z) dx dy dz

z=? y=? x=? ?

?

@ @ @

=

c=?

?

f(a, b, c) da db dc

b=? a=?

draw projections of the regions defined by these integrals into the x-y plane a)

1

3

@ @ @

y2

y=0 x=3y z=-1

f(x, y, z) dz dx dy

b)

1

2

@ @ @

y

y=0 x=2y z=0

f(x, y, z) dz dx dy

Are a great source of integration problems

You learned about averages?

let’s recall the classical setting

Classical cases of averages use integrals For f : { 1, 2, …, n } _ 

f= f

1 n

n

= # f i i=1

1 n

For f : [a, b] _ 

@

n

f i=1

f = f

1 b-a

@

b a

f

the AVERAGE OF A FUNCTION OVER A REGION

the average of a function f(x) on a n region R in  is defined to be…

f =

@

f dx R

@

dx R

=

1 Voln(R)

@

R

f dx

n

Where voln denotes the volume in 

single variable calculus

average vs. extremal temperatures y 0≤x≤a

With radially quadratic temperature distribution

0≤y≤b

T(x, y) = C(x2+y2) + A b

T

=

a

How close is the average temperature on the plate to the maximal/minimal temperature?

@ T dA @ @ C(x +y ) + A dx dy = @ 1 dA @ @ 1 dx dy R

2

y=0 x=0 b

R

3y+xy3 x = C ab 3

a

2

C = ab

b

@ @

(a, b)

x

(0, 0)

a

(x2+y2)

y=0 x=0

Aab dx dy + ab

y=0 x=0

a x=0

b

3

2

C(a3b+ab ) C(a2+b ) +A = +A = +A 3ab 3 y=0

average vs. extremal temperatures y With radially quadratic temperature distribution

T(x, y) = C(x2+y2) + A

T

2

C(a2+b ) = +A 3

0≤x≤a 0≤y≤b

Tmax

How close is the average temperature on the plate to the maximal/minimal temperature?

(a, b)

x

(0, 0)

= C(a2+b2) + A

Tmin T

=

Tmin +

= A

Tmax - Tmin 3

when computing averages…

average over the right domain what is the average area of all parallelograms in the plane satisfying…

y

0≤ q ≤ p ≤ 1 0≤ s ≤ t ≤ 1

t s

the area is computed via a determinant

A = | pt - qs | = pt - qs but the integral is 4-dimensional… integrate over the domain

0≤ q ≤ p ≤ 1 &

q

since pt > qs

1

q

p 1

0≤ s ≤ t ≤ 1

p

these are independent conditions 1

x

t s 1

average over the right domain 1

@ A dx @ 1 dx R

A =

=

R

=4 =

1

1

t

@ @ @ pqt0 0 0

1

@@

1

0 0

1

t p

@ @ @ @ pt-qs dq ds dp dt @ @ @ @ 1 dq ds dp dt

1

0 0 0 0 1 1 t p

q

this equals 1/4 = (1/2)2

p

0 0 0 0

1

2

q2s

4p2st-p2s2

t

p

ds dp dt

q=0

dp dt

s=0

=

@

1

0

2 3 pt

1 3 = 3t

1

1

1

dt

1

t

p=0

1 = 3 t=0

s 1

are commonly used in applications…

for f : R _  a function n on region R in 

fRMS =

2

f

high-dimensional cubes & corners… xn what’s the root-meansquare distance to origin?

0 ≤ xi ≤ 1

the square distance from the origin to a point is

i = 1…n

2

d =

n

on the unit cube in 

2 x1 +

2 x2

+ …+

2 xn

1

x1

@ d dV … = @ @ @ @ 1 dV 1 1 2

2

d =

R

R

1

1

1

0

0 0

x21 +

x22

+ …+

x2n

1 n … = 3+ 3+ +3 = 3

dx

dRMS

=

n 3

so, in dimension three, the RMS distance equals one!

the big picture

averages are defined In terms of integrals & provide great motivation For higher-dimensional Integration domains

1

Compute the following averages of the function (xy)1/2 A) over the square 0 ≤ x, y ≤ a. how does this grow as a function of a? B) over the region x, y ≥ 0 and x+y ≤ a. how does this compare with what you found in the previous average as a function of a>0? c) can you compute the average of this over the region 0 ≤ x, y and x2+y2 ≤ a2 ? that seems hard… what would you guess the behavior is as a increases?

2

compute the average of the function (x1) (x2) … (xn) over the n-cube defined by { 0 ≤ xi ≤ π : i =1…n }. what happens to this average as n_∞ ?

3

compute both the average and root-mean-square average of the following A) ex+y over the square 0 ≤ x, y ≤ 1 in 2. b) yz (x) over the cube -π ≤ x, y, z ≤ π in 3.

c) x + y over the region between y = x2 & y = 2x in 2.

6

prove that the average of a constant function is precisely that constant. be sure to invoke the definition of the average!

7

consider the cube C in n defined by -1 ≤ xi ≤ 1 for i=1…n. Compute the average of the function f(x) = exp(-x1-x2-…-xn) on C. in particular, consider what happens to this average as n_∞. does it go to zero? one? infinity? guess before solving.

8

challenge: in a certain room, a light flashes exactly once every 10 seconds. you walk into the room and open your eyes. what is the average time you wait until the light flashes? figure out the [obvious] answer using an integral. now, assume that there are N lights in the room, each flashing once every 10 seconds, but independently and unsynchronized. now, what is your average wait time until the first light flash? to solve this problem, coordinatize the N-cube with side length 10 and average the minimum-of-coordinates function. that may be tough! perhaps you should do the 2-d and 3-d case for starters… can you now say what the average time to see all N lights is? think!

are called “centroids”

you may recall formulae for centroids of regions between graphs b

x

=

@ x(f(x)-g(x)) dx x=a b

@ f(x)-g(x) dx

x=a b

y

=

=

@ (f(x) -g(x) )/2 dx @ f(x)-g(x) dx x=a b

x=a

2

1

A

2

=

@@

R

y

x dA y

1

A @@

f

R

y dA

x a

x

g

b

you may recall formulae for centroids of regions between graphs

x

=

1

A

@@

R

x dA =

1

A

@

b

y

f

x=a f(x)

y

=

1

A @@

R

y dA =

1

A

@

b

x=a

y y=g(x)

x a

x

g

b

in 3-d, a body has a centroid with three coordinates

& these tell the average x, y, & z positions in the body…

the centroid may or may not be a point within the body!

centroid of a 3-d shape z

V= =

1

1

@ @ @

1- x

dz dx dy

y=-1 x=y2 z=0 1 1

@ @

1 - x dx dy

y=-1 x=y2

= =

@

1

y=-1 1

@

x=y2

first, compute the volume fortunately, we set up this integral in chapter 5

z

y

z=1- x

(0,0) y x=y2

dy

1 – y2 - 32 + 32 |y|3 dy

y=-1

= 31

x - 32 x3/2

1

(0,1)

x danger! you need to use the absolute value here…

(0,0)

(1,0)

x

(1,1) x (1,-1)

centroid of a 3-d shape clearly, the centroid has Y coordinate equal to zero by symmetry…

1

@ V@ @

x = 1

= 3

1- x

x dz dx dy

y=-1 x=y2 z=0 1 1

@ @

x – x3/2 dx dy

y=-1 x=y2

x = 52

= 3

y = 0

= 3

1 z = 5

1

@

1

y=-1 1

@

y=-1

1 2 2 5/2 x - 5x 2

1 x=y2

dy

1 1 4 2 5 y + 5 |y| dy 10 2

2 = 2 x = 3* 15 5

1

1

@ V@ @

z = 1

= 3

1- x

z dz dx dy

y=-1 x=y2 z=0 1 1

@ @

y=-1 x=y

= 3

@

= 3

@

1

y=-1 1

1 – x + x dx dy 2 2 2

x 2 3/2 1 2 - x +4x 2 3

1

dy

x=y2

5 1 2 2 3 1 4 - y + 3 |y| - 4 y dy 6 2

y=-1

z = 3* 151 = 51

is a density-weighted centroid

n

the center of mass of a region R in  with mass density ρ(x) is the point x with coordinates…

xi =

@

R

@

xi ρ(x) dx R

ρ(x) dx

=

@

R

@

xi dM dM R

1 = M

@

R

xi dM

Where dM denotes the mass element & M is the mass

center of mass compute the center of mass of the “crane arm” with density

ρ – ρ0 ρ0 + 1 x L M =

y ρ0

@ @ x=0

h

ρ1

h

ρ1 – ρ0 ρ0 + x dy dx = L y=2hx/3L

L

density varies linearly in x…

L ρ1 – ρ0 ( ρ0 + L x x=0

@

L

)( h -

2 3h

x

2hx dx ) 3L

good: if both density (ρ1 – ρ0)h L2 ρ0 2h L2 (ρ1 – ρ0)2h L3 constants are the same, this = ρ0hL + 2 2 3 is the mass as per area L 3L 2 3L 1 1 2 ρ 1 2 ρ 7 ρ 5 ρ = hL ( 1 - - + ) 0 + ( - ) 1 = hL ( 0 + 1) 2 3 9 2 9 18 18

(

)

center of mass y ρ0

compute the center of mass of the “crane arm” with density

1 x = M

L

ρ – ρ0 2 1 ρ0x + 1 x dy dx = M L y=2hx/3L

@ @ x=0

ρ1

h

ρ – ρ0 ρ0 + 1 x L L

density varies linearly in x…

h

2

3

3

1 2 = hL M

((

ρ1 – ρ0 2 2hx ρ x + x h dx (0 ) ( ) 3L L x=0

@

1 1 2 1 ρ 1 1 ρ - - + ) 0+ ( - ) 1 2 3 9 6 3 6

substitute the previous result for mass…

x

L

4

(ρ1 – ρ0)h L ρ0 2h L (ρ1 – ρ0)2h L 1 L ρ0h + = M 2 3 4 L 3L 3 3L2

(

2 3h

)

=

) 2ρ0 + 3ρ1 = L 7ρ0 + 5ρ1 it’s the correct units!

center of mass compute the center of mass of the “crane arm” with density

ρ – ρ0 ρ0 + 1 x L 1 y = M

L

density varies linearly in x…

ρ1

h

L

L ρ1 – ρ0 ρ1 – ρ0 1 ρ0 y + ρ0 + x xy dy dx = ( 2M L L y=2hx/3L x=0

@ @ x=0

y ρ0

h

@

2 3h

2hx 2 ) ( h - 3L ) dx

this integral requires a bit more work & simplification, but it’s just a polynomial…

this gives the coordinates of the center of mass

2ρ0 + 3ρ1 x = L 7ρ0 + 5ρ1

17ρ0 + 9ρ1 y = h 21ρ0 + 15ρ1

you can verify the details of this integral…

x

integrals are additive

center of mass, by parts compute the center of mass of the three plates with density

ρ=x+y+z

all same by symmetry

1 1

@@

Mxy =

0

1 1 x + y + z dx dy = + =1 2 2 0

0

z y x unit squares we can compute the center of mass of each plate, then take their average location

x = y = z = 0

1 1

@@ 0

1 1 7 x(x + y + z) dx dy = + = 3 4 12 0

0

on the x-y plate

take the mass-weighted average of these three centers to obtain the net center of mass…

x = y = z = 1

7 7 + +0 3 12 12

(

)

7 = 18

here’s a problem about centroids with a surprise: let’s say you have a solid, convex, uniform-density object & you set it on a flat surface…

well, of course, it rocks until it comes to rest

at a point on the surface which minimizes the distance to the centroid stable equilibrium

question: is there…

it’s easy to find a monostatic solid if we vary the density or make it nonconvex

watermelons mangoes AVocados samosas black beans jelly beans koopas cupcakes pawpaws quenepas

if you have constant density & convexity… most object shapes have multiple stable equilibria

the big picture

centroids & centers of mass are simply the average position with respect to mass density

1

Compute the centroids of the regions defined by the following: A) in 2, the set x, y ≥ 0 and x + y3 ≤ 8 b) in 2, the set bounded by y = x2 and x = y2 c) in 3, the set bounded by 0 ≤ x, y, z ≤ 1 and x + y ≥ 1. draw a picture! d) in 3, the set z ≥ 0 and x2 + y2 + z2 ≤ 9. don’t forget to use symmetry! e) in 3, the set 0 ≤ z ≤ 4 and x2 + y2 ≤ z2. don’t forget to use symmetry!

2

compute the centroid of the tetrahedron (or “3-simplex”) in 3 spanned by the points (0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0, c) for a, b, c > 0. hint: if you argue using symmetry, you should be able to solve this with a single triple integral & a permutation of the coordinates… think!

3

repeat the previous exercise for an “n-simplex” in n spanned by the origin and points ci > 0 along the n positive axes. Don’t compute n integrals, though: think!

4

compute the center of mass of the following regions with densities as given: A) in 2, the set x = |y| and x ≤ 2 with ρ = (x + y)2 b) in 2, the set 0 ≤ x ≤ 1 and 0 ≤ y ≤ ex with ρ = y c) in 3, the set 0 ≤ x ≤ 1 , 0 ≤ y ≤ 2 , and 0 ≤ z ≤ xy with ρ = xyz

5

consider A solid region R in n, and assume its centroid is at a point u ∈ n. suppose that you transform R to A(R) by applying a linear transformation A: n _ n. what can you say about the centroid of A(R) ? use an integral!

6

compute the center of mass of a wire tracing out the unit circle in 2 with density function ρ = 2 + x + y. hint: parametrize by an angle (& think).

7

find the coordinates of the centroid of the region In n defined by min(xi) ≥ 1 and max(xi) ≤ 2 for all i. what does this region look like? try to draw some pictures for n=2 & 3. hint: break the domain up into equal-sized cubes.

you did moment of inertia? whether you have seen it or not…

we will use our new understanding of multivariate integrals to clarify this subject

let R ⊂ n be a solid body to be rotated about an axis

r dM

I=

@

R

dI

moment of inertia measures the resistance to rotation

dI = r2 dM

where r is the distance to the axis & dM is the mass element

among objects of identical mass, those whose mass is concentrated away from the axis have the higher moment of inertia…

can be easy or difficult to compute depending on the SHAPE OF THE DOMAIN...

planar moment of inertia moment of inertia of a square plate about the origin if the density varies as…

dM = (x + y) dA dI = r2 dM = (x2 + y2) dM = (x2 + y2) (x + y) dA M=∫

l

∫

l

x=0 y=0

3

=l

x+y dy dx

y

∫

I=

∫ ∫ x=0

4

dx

(0, l)

I = dI l

dy

r

(0,0)

(l, 0)

x

l

3 + xy2 + x2y + y3 dy dx x y=0 2

3

3

2

4

5

l l l l l l 5l 5 2 = 4 l + 2 3 + 3 2 + 4 l = 6 = 6 Ml

are what one often cares about...

moment of inertia of a solid cone z dM = ρ dV dI = r2 dM = (x2 + y2) dM

moment of inertia of a solid cone of uniform density about the central axis

h

I=

R

@ @

R2-x2

@

h-h( x2+y2 )/R

x=-R y=- R2-x2 z=0

to solve this, we could either…

r

y

(x2 + y2) ρ dz dy dx

these limits are not so nice to work with…

x

R

use I for a disc of thickness dz & integrate with respect to z use a “polar” coordinate system to simplify [coming soon!]

for more complicated shapes

the parallel axis theorem

if I0 is the moment of inertia of an object about an axis through its center-of-mass then the moment of inertia about a parallel axis distance D away equals 2

ID = I0 + MD D

@ = @ (r + D) dM = @ r dM I + 2D @ r dM MD + @ D dM

I0 = ID

r2 dM

2

2

0

2

this term vanishes since the centroid has “AVERAGE” RADIAL COORDINATE EQUAL TO ZERO

2

rotating cubes & parallel axes place a coordinate frame at the centroid

s s s

s/2 s/2 s/2

@ @ @

2 M s (x2 + y2) ρ dz dy dx = I0 = 6 -s/2 -s/2 -s/2

s2 s2 5 M s2 I= M ( + ) = 6 4 12 s2 s2 2 M s2 I= M ( + ) = 6 2 3

can be handled using double integrals

spinning a shell y

moment of inertia of a cubical shell of uniform density about a vertical axis through center

x

the top & bottom faces each have the same moment…

s s s

I=

s/2

∫ ∫

s/2

-s/2 -s/2

(x2 + y2) ρ dx dy

3

3

x y = ρ 3 y + ρx 3 we can do this! one face at a time…

4

2

s = ρ s6 = M 36

s/2

s/2

x=-s/2 y=-s/2

spinning a shell moment of inertia of a cubical shell of uniform density about a vertical axis through center

s

for the side faces, rotate one about its center

I=

s/2

∫ ∫

s/2

-s/2 -s/2

3

s s

then translate via parallel axis

x = ρ3z 4

z

x2 ρ dx dz s/2

x

s/2

x=-s/2 z=-s/2

2

s s = ρ 12 = M 72

s2 M s2 M s2 I=M + = 72 6 4 18

5 M s2 I = 18

and how it is distributed spatially

let R ⊂ n be a solid body to be rotated about an axis

rg

if all the Mass were concentrated at a single point…

M

I =

Mr2 g

I rg = M

how far away from the axis would that point have to be… to have the same moment of inertia?

s

Ms2 6

1 s 6

s

5Ms2 18

5s 18

2Ms2 3

2s 3

s

s

l

r

2M r2 3

2 r 3

r

2M r2 5

2 r 5

r

M 2 2 ( 3r + l ) 12

3r2

2

+l 12

the big picture

moment of inertia tells of the distribution of mass about an axis of rotation… the RADIUS OF GYRATION GIVES A “MASSIVE DISTANCE” TO THE AXIS

1

compute the following moments of inertia of bodies in 3 about the z-axis (assume a uniform density ρ and total mass M.) a) the set bounded by -2 ≤ x, y ≤ 2, z ≥ 0, and x + y + z ≤ 5 b) challenge! the set given by 1 ≤ |x| ≤ 2, 1 ≤ |y| ≤ 2, & 1 ≤ |z| ≤ 2, what is that shape?!

2

compute the following moments of inertia of bodies in 3 about the z-axis (with non-uniform density ρ as given and total mass M.) a) the cube 0 ≤ x, y, z ≤ 1 with ρ = 1 + 2x + 3y + 4z b) the set bounded by x, y, z ≥ 1 and x y z ≤ 6 with ρ = 1 / (x2 + y2)

3

try to prove that the radius of gyration of a body cannot exceed the maximal distance from the axis to a point on that body. (recall, density is non-negative!)

4

assume that for a uniform planar disc of radius r and mass M, the moment of inertia about the center (an axis orthogonal to the disc) equals Mr2/2. integrate this to compute I for the cone problem earlier in this chapter.

6

compute the moment of inertia & radius of gyration of an n-dimensional cube of unit density and side length s about an axis parallel to an edge & passing through the center. was this one easier or harder than you thought?

7

which do you think has larger moment of inertia: a solid cone of height h and base diameter D about the vertical (symmetric) axis? or a thin plate shaped as an isosceles triangle with height h and base D about the vertical (symmetric) axis? Assume both have uniform density and identical mass M. Guess, then prove.

8

compute the moment of inertia I of a cylindrical shell (with end caps) of radius r, height h, uniform density, and mass M. you may be able to solve this problem without any integrals at all, if you use previous problems/results…

9

compute I for a spherical shell of radius R and mass M about an axis (say, the z-axis) through the center. hint: do it first for a planar circular hoop of radius r with orthogonal axis of rotation, then integrate with respect to z.

Are a good data structure for storing information

given an x, y, & z coordinate frame

[I] =

Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz

the inertia matrix encodes moments about the three axes along its diagonal

@ = @ (x + z ) dM = @ (x + y ) dM

Ixx = Iyy Izz

(y2 + z2) dM

@ = @ -yz dM = @ -xz dM

Ixy = -xy dM = Iyx

2

2

Iyz

2

2

Ixz

= Izy = Izx

the off-diagonal terms are called “mixed moments” or “products of inertia”

inertia matrix of a prism

z

a/2 b/2 c/2

Izz =

@ @ @

(x2 + y2) ρ dz dy dx

-a/2 -b/2 -c/2

c x

b

= cρ = cρ

a

@ @ a/2

bx2

-a/2

3

3

b + dx 12 2

a ba = bcρ + 12 12

(

2

2

x2 + y2 dy dx

-a/2 -b/2

@

2

b +c Ixx = M 12

a/2 b/2

y

2

)

you can do this… 2

2

a +b = M 12

a +c Iyy = M 12

computing the rest of the inertia matrix is not so hard…

inertia matrix of a prism

z

a/2 b/2 c/2

Ixy =

@ @ @

-xy ρ dz dy dx

-a/2 -b/2 -c/2 a/2 b/2

y

c x

b

= cρ

@ @

-xy dy dx

-a/2 -b/2

2 a/2

a

= -cρ x 2

2 b/2

y 2 x=-a/2

= 0 = Ixz = Iyz

y=-b/2

Ixx Ixy Ixz [ I ] = Iyx Iyy Iyz Izx Izy Izz M = 12

b2+c2 0

0

0 a2+c2 0 0

0 a2+b2

the mixed moments are non-zero?

mixed moments & symmetry y

y

x

Ixy = 0

y

x

Ixy = 0

y

x

Ixy ≠ 0

x

Ixy ≠ 0

why bother with inertia matrices?

yes, yes, matrices are data structures… but in order to be truly useful, the matrix algebra should have physical meaning!

rotation about a “skew” axis?

the moment of inertia about an axis through the center with direction vector

if u = i, then

Iu = i * [I] i = Ixx moment about the x-axis

u if the body is spherically symmetric unit vector

Iu = uT [I] u = u * [I] u

[I]

is a constant times the identity by symmetry thus Iu is constant, as expected

rotating cubes & skew axes

M s2 Izz = 6 = Ixx = Iyy

s s

s

Ixy = 0 = Ixz = Iyz

Ixx Ixy Ixz 1 0 0 Ms2 [ I ] = Iyx Iyy Iyz = 6 0 1 0 Izx Izy Izz 0 0 1 if | u| = 1, then

Ms2 Ms2 Iu = u * [I] u = 6 u * u = 6

the constant diagonal matrix indicates a symmetry of mass about the center

what is the moment of inertia of a uniform density rectangular prism along its longest diagonal?

z

y

the direction of a diagonal axis is…

c x

b

u= a

1 a2+b2+c2

±a ±b ±c

M [ I ] = 12

b2+c2 0

0

0 a2+c2 0 0

0 a2+b2

the moment of inertia along such a diagonal is…

Iu = u * [I] u M (a2b2 + b2c2 + a2c2) = 6 ( a2+b2+c2 )

check that it makes sense when a = b = c !!!

rotating a prism along the diagonal

these are called the principal axes it’s a fact that principal axes always exist… there is some coordinate frame in which the inertia matrix is diagonal & all mixed moments vanish!

you will learn about these specialized directions have as examples principal axes of the inertia matrix

the big picture

the inertia matrix bundles together all the information about how mass is distributed with respect to a given coordinate frame

1

compute [I] for the following flat bodies in the plane with (x,y) coordinates, with densities, ρ, as indicated. what structure do all these matrices share in common? A) the set -1 ≤ x ≤ 2 and 1 ≤ y ≤ 3 with ρ = 2y b) the set 0 ≤ x, y ≤ 4 and x + y ≤ 6 with ρ = x + y c) challenge: the set 0 ≤ x, y ≤ ∞ with ρ = e-(x + y)

2

compute the inertia matrix for an axis-aligned cube with side length s>0 and constant density ρ about its corner (that’s where the origin is). show that you get 2/3 Ms2 on the diagonals and –1/4 Ms2 off the diagonals. use this result about [I] to compute the moment of inertia about the axis passing through the origin (0, 0, 0) and the far corner (s, s, s).

3

compute the inertia matrix for an axis-aligned cube with side length s>0 about its center, but this time assume that the density ρ is not constant, but rather is proportional to the square of the distance to the origin: ρ = C(x2+y2+z2).

4

compute the inertia matrix, [I], of a rigid rod of mass M, constant density ρ, and length L, that is centered at the origin in 3-d but points along the diagonal vector i+j+k. if you use integrals to compute this, they should be 1-d integrals. are there other ways to compute this [I] ? what do you observe?

5

compute the inertia matrix for an axis-aligned cube-shell with side length s>0 about its center, assuming uniform density ρ and mass M. use this to compute the moment of inertia of this hollow shell rotated about the diagonal.

6

compute the inertia matrix for a circular hoop of mass M, radius R, and constant density ρ with a coordinate frame having as its origin the center of the hoop, and the hoop contained within the (x, y) plane. challenge: figure out how this changes if the origin is at a point on the hoop…

7

use the inertia matrix to argue that for a rectangular prism, the minimal moment of inertia possible is realizaed about the center through the longest axis.

is the basis of a lot of great applications in solid-body mechanics

let’s define & derive some facts about the physics of rotation…

is framed in terms of vectors

consider a mass element rotating about an axis in 3-d

choose a coordinate frame with origin on the axis

position vector to dM unit vector along axis velocity of mass element Angle of mass-axis plane

its length is the rotational speed & its direction is along the axis for a solid body, all elements have the same angular velocity

angular velocity of a rotating body in 3-d is the vector, ω, satisfying

ω =

dφ n dt

v = ω×r

is also a vector

angular momentum of a mass element is the vector

dL = r × v dM integrate over the body to get the vector L in other words, it’s the cross product of position with the linear momentum element

to relate to angular velocity…

dL = r × v dM = r × ( ω × r ) dM = (( r * r ) ω – ( r * ω ) r) dM this last step follows from the identity

a×(b×c) = = (a*b)c–(a*c)b

x r= y z let’s expand this element

dL = dL =

ωx ω = ωy ωz

dL = r × v dM = r × ( ω × r ) dM = (( r * r ) ω – ( r * ω ) r) dM

(x2 + y2 + z2) ωx - (x ωx + y ωy + z ωz ) x (x2 + y2 + z2) ωy - (x ωx + y ωy + z ωz ) y dM (x2 + y2 + z2) ωz - (x ωx + y ωy + z ωz ) z (y2 + z2) ωx - xy ωy - xz ωz - xy ωx + (x2 + z2) ωy - yz ωz dM - xz ωx - yz ωy + (x2 + y2) ωz

dL = [dI] ω L = [I] ω

inertia matrix of a prism

z

[I]= 2*3*4

y

4

= 12

x

3

2

[I]=

32+42 0

0

0 22+42 0 0

0 22+32

50

0

0

0

40

0

0

0

26

If we rotate this body about an axis through the center of mass, then…

L = [I] ω =

50ωx 40ωy 26ωz

Unless you rotate about a principal axis, then…

why is angular momentum so weird ?

dL dω τ= = [I] dt dt

assuming constant inertia

this is the “angular” version of newton’s 2nd: F = ma torque is parallel to angular momentum, L, not angular acceleration, ω

L = [I] ω when a twirling dancer decreases moment of inertia, angular velocity increases

L

when a cat falls to the ground, it rotates in opposite directions to keep angular momentum zero

in the absence of torque angular momentum does not change

L

once you learn how to think

a mass element, dM, has kinetic energy, dK, given by the following 2 1 dK = 2 | v | dM

what is this for a rotating body?

kinetic energy of rotating body

K =

1 2

ω * [I] ω

dK = = = =

1 2 1 2

( v * v ) dM

1 2 1 2

ω * ( r × v ) dM

v * ( ω × r ) dM ω * dL

expand symmetry of the scalar triple product ah ha !

we’ve skimmed the following results for the mechanics of a solid rotating body in 3-d

angular velocity is a vector, ω, satisfying

v = ω×r

angular momentum is a vector, L, satisfying

L = [I] ω torque is a vector, τ, satisfying dL dω τ = dt = [I] dt kinetic energy is a scalar, K, satisfying

K=

1 2

ω * [I] ω

the big picture

the inertia matrix makes it possible to frame solid-body mechanics in the language of vectors: angular velocity, angular momentum, torque, & more

1

consider a rectangular thin plate in the (x, y) plane, of constant density ρ, and dimensions h– by– w , spinning about its diagonal axis with angular velocity ω. what is the angle between its angular velocity & its angular momentum, L ? (your answer had better depend on h and w…) A) does your answer depend on the size of the plate? if you double the dimensions, does this angle change? B) what happens in the case where h = w ?

2

consider the following moments of inertia of bodies in 3-d & in each case find: 3

0

0

2

-1

0

3

0

2

(1) [ I ] = M 0

2

0

( 2 ) [ I ] = M -1

2

0

(3) [ I ] = M 0

4

1

0

0

5

0

0

1

2

1

5

a) about which axis is kinetic energy maximized by unit-speed rotation? b) what angular velocity is required to obtain an angular momentum L = (1, 1, 1)T ?

3

given: a unit-density rectangular prism of sides 0 < a < b < c rotating about its diagonal axis (through opposite corners) with angular velocity, ω = (a, b, c)T. A) What is the angular momentum, L ? [use results from chapter 8 to get [I]…] b) assume that you have the ability to instantly modify the side lengths of the prism (but you cannot change the volume) so that the inertia matrix is suddenly different. assume also that the angular velocity does not change. what happens to the angular momentum? is it possible to change dimensions so that the new angular momentum points in any direction you like? this is interesting.

4

fact: the inertia matrix is symmetric & positive definite: v * [I] v > 0 for all v > 0. (a) what does this fact tell you about kinetic energy? (b) using what you know about the inertia matrix, can it happen that L ⊥ ω ? (c) challenge: spin a top so that it precesses (look it up!) argue that in this case, torque is perpendicular to angular momentum. you may have to resort to pictures & physical reasoning for this, since you don’t have explicit equations.

is a vital application of integrals

this is a probabilistic function characterized by a probability density on , ρ, satisfying

ρ≥0

&

@  ρ(x) dx = 1

ρ

to compute the probability P that X lies in A⊂ , integrate the probability element over A

dP = ρ(x) dx P(X ∈ A) = “LIES IN”

@ AdP

=

@ A ρ(x) dx

A

leads to asking about how density is distributed

expectation (or mean), is an average value

(X) =

@  x dP

variance is a measure of “spread” about mean

(X) =

@  ( x – (X) )

2

dP

standard deviation is a dimensionless variance

σ(X) =

(X)

σ

σ 

mass & mass distributions

mass density, ρ mass,

@

A

ρ(x) dx

probability density, ρ probability,

@

A

ρ(x) dx

center of mass, x

expectation [mean], (X)

moment of inertia about the center of mass, I

variance, (X)

radius of gyration, rg

standard deviation, σ(X)

simple 1-d probability ρ:α=1;β=2

consider the following probability density

2

βαβ ρ(x) = (x+α)β+1 on [0,∞) for α, β > 0

ρ(x) = (x+1)3 x

in the case α = 1, β = 2 , compute (X)

use partial fractions to compute

@

∞

2x dx = (X) = 3 (x+1) 0 -2 2 = (x+1) -2(x+1)2

@

∞

2 2 dx 2 3 (x+1) (x+1) 0 ∞

= 2-1 = 1 0

in this case, compute P( X< (X) )

isn’t it obviously 50% ? let’s check…

@

1

2 2 dx P(X2 for that)

the Gaussian density on  has:

1 -x2/2 e ρ= 2π

a general Gaussian on  has: (x-μ)2 - 2σ2 1 ρ=

2πσ2

e

Cauchy density Chi-squared density frechet density Gamma density levY density Log-normal density pareto density Pikachu density you don’t need to know all superthese… mariobut density you may want to familiar with some names donkeybekong density

what happened to all the multiple integrals?

n

consider a random X taking values in 

ρ : n _ 

net “mass” = 1

this likewise has a probability density, ρ : n _ , with ρ satisfying

ρ≥0

&

@

n

ρ(x) dx = 1

n

 to compute the probability P(X ∈ A) for A⊂ n, integrate the probability element on A

dP = ρ(x) dx P( X ∈ A) =

@AdP

=

n

@A ρ(x) dx

98%

75%

62%

40%

for example…

with what probability is a randomly chosen point within a subset?

consider a random X taking values in n

ρ

(X)

the expectation, (X) is simply the ρ-weighted centroid with coordinates:

(X) =

@

n

x ρ(x) dx

 the variance (X) and standard deviation σ(X) are numbers computed via

(X) = σ(X) =

@

2

| x – (X)|| ρ(x) dx

n

(X)

n

(X) expectation is “like” the center of mass of the probability density variance measures how hard it is to “rotate the mass” about an axis orthogonal to the expectation

consider the location of an autonomous car or robot as a state (position/bearing/velocity/etc) various (noisy!) sensors allow the controller to estimate the object’s state… but only in terms of a probability density additional sensor readings (gps, lidar, visual cameras, wireless, etc) allow one to update the density, hopefully reducing the variance and improving the estimated position

various (noisy!) sensors allow the controller to estimate the object’s state… but only in terms of a probability density

2-d probability given a probability density

ρ =

ax2

+ xy

on the unit square 0 ≤ x, y ≤ 1 for some constant a>0 compute the constant a 1

1

@@ a + y dy =@ 3 2

1=

ax2 + xy dx dy

0 0 1

compute the expectation (X) = ( (X) ,  (Y) ) 1

(X) = =

0

@@

1

0 0 1

=

a = 9 4

@ @ 1

( X) =

2

0

@@

4

9 y + 1 y2 = 16 6

y ( 94 x2 + xy ) dx dy =

@

1

0

1 y=0

0 0

1

dy

x=0

= 35 48

9 x3y + 1 x2y2 12 2

3 y + 1 y2 dy = 3 y2 + 1 y3 4 2 8 6 1

3

0

9 + 1 y dy 16 3

1

(Y) =

1

@ @ x ( 94 x + xy ) dx dy = @ 169 x + 31 x y 0 0 1

0

a = 3 + 41

1

1 y=0

1

dy

x=0

= 13 24

13 2 9 2 35 2 ( (x- 48 ) + (y- 24 ) )( 4 x + xy ) dx dy

=

1439 11520

2-d probability given a probability density

9 2 ρ = 4 x + xy

on the unit square 0 ≤ x, y ≤ 1 and some constant a>0

compute the probability that a random (X,Y) satisfies X ≥ 23 1 1 9 2 2 x + xy dx dy P(X≥ 3 ) = dP = 4 2 2 y=0 x=

@@ = @

1

@ @

x≥ 3

x2

3 3 x + y 4 2 y=0

1 x= 23

3

dy =

@

1

19 5 + y dy = 2 36 18 3 y=0

compute the constant a 1

1

@@ a + y dy =@ 3 2

1=

ax2 + xy dx dy

0 0 1

@@ = @

P(X 0 R = { 0 ≤ x2 + y2 ≤ 1 }

2

show that the surface areas of the following surfaces in 3-D are equal 2 A) a2x2 + b y2 = z elliptic paraboloid now go look at the graphs (x, y) ∈ R 2 2 [blue vol 1, ch 2] - surprised? 2 2 b) a x - b y = z hyperbolic paraboloid

3

set up (& try to solve?) the integrals to compute the centroid (in x, y, & z) of a shell in the shape of a hemisphere of radius R in the halfspace z≥0. hint: two of these integrals are easy to solve via symmetry… but what about the 3rd?

with all the tools of integration at hand…

the standard Gaussian on  is:

1 2π

-x2/2

e

these are very common probability densities… gaussians have some remarkable properties, as we shall see…

a general Gaussian on  is (x-μ)2 - 2 1

@

∞

-∞

1 -x2/2 e dx 2π

= 1

e 2π

higher dimensional Gaussians?

compare to the 2-d case in chapter 13…

n

the standard gaussian density on  with mean 0 and covariance matrix [] = I (identity matrix) is given by

1 (2π)

@

1 n



n

(2π)

n

1 -2 | x|

e

2

@ = =

- -21 | x|

e

2

dx

=1

=

1 n (2π)

1 (2π)

n

(2π)

n

1 1

(2π)

n

n

1 | x| 2 --2 dx

e

--21 (x1 +…+xn )

@@ @ e ( @e …



(

2

2

1 ©2 n -2 d©

2π

n

)

) = 1

dx1…dxn expand fubini

the general gaussian density on n with mean μ and covariance matrix [] is given by the equation

1 (2π)

n

T

-1

1 -2 (x-μ) [] (x-μ)

e | [] |

(x-μ)2 - 2

1 e 2π

depends only on the covariance matrix (& the mean)

1 0 0 1 standard gaussian

2 1 1 2

2 -1 -1 2

3 1 1 1/2

level sets of a Gaussian are ellipses

makes gaussians ideal for data fusion

prediction is augmented by measurement recall from chapter 12 what happens to expectation and variance under a predictive motion model

f

(Y) = f ((X)) [(Y)] = [Df][(X)][Df]

[Df] assume that some sensors give you a measured mean & covariance matrix

one method of merging predicted & measured densities is called…

T

there are lots of ways to perform data fusion… we will gloss a popular method known as the…

f [Df] assume that some sensors give you a measured mean & covariance matrix

this often reduces the uncertainty

the (rescaled) product of the predicted & sensed densities is the best choice for a fused density why the product? well, it’s like an “and” operation… we want both predicted and measured states to be true!

the product of any two gaussians is again (up to scaling) a gaussian & since gaussians are determined by their mean & covariance matrix, everything should be simple!

predict. measure. fuse. keep going… & that’s a track…

About products of gaussians

the product of two gaussians is a Gaussian, up to rescaling up to rescaling, the density

--21 (ax2 -2bx + c)

e

with

b = a

is a 1-d Gaussian &  = -a1

a(x2 -2(b/a)x + c/a) take the exponent 2 x –2(b/a)x + c/a = 1/a complete 2 the square (x - (b/a)) = +C 1/a constant 



(rescaling!)

the product of two gaussians in 1-d with means 0, 1 and variances 0, 1, is, up to rescaling…

--21 (x - 0)2/ 0

e

--21 (x - 1)2/ 1

*e

--21 ( ( 1 + 1 ) x2 -2 ( 0 + 1 ) x + C ) 0 1 0 1 = e a

b

this is a quadratic exponent, and thus Gaussian by the lemma, with…

0 1 1 = 1 1 = 0 +1 ( + ) 0

1

add the exponents expand & factor

ignore the constant terms (ugly!)

0 1 (  +  )   +   = 10 1 1 = 1 0 0 1 0 +1 ( + ) 0 1

of course, this proof is only valid in 1-d… you’ll hopefully trust me for the matrix version

the product of two gaussians is a Gaussian, up to rescaling the product of two gaussians with mean vectors 0, & 1 and covariance matrices [0], [1], is the Gaussian given by… -1

 = [1] ( [0] + [1] ) 0 -1 + [0] ( [0] + [1] ) 1 -1

[] = [1] ( [0] + [1] ) [0]

the product of two gaussians in 1-d with means 0, 1 and variances 0, 1, is, up to rescaling…

--21 (x - 0)2/ 0

e

--21 (x - 1)2/ 1

*e

--21 ( ( 1 + 1 ) x2 -2 ( 0 + 1 ) x + C ) 0 1 0 1 = e a

b

this is a quadratic exponent, and thus Gaussian by the lemma, with…

0 1 1 = 1 1 = 0 +1 ( + ) 0

1

add the exponents expand & factor

ignore the constant terms (ugly!)

0 1 (  +  )   +   = 10 1 1 = 1 0 0 1 0 +1 ( + ) 0 1

this, finally, tells us how to set up the kalman filer for doing iterative data fusion

is the basis for the Kalman filter

the kalman filter predicts, measures, then fuses using the derived properties of product Gaussian densities…

f [Df]

P & [P] predict this

o & [o]

-1

F = [M] ( [P] + [M] ) P -1 + [P] ( [P] + [M] ) M -1

start here

M & [M] sense that

[F] = [M] ( [P] + [M] ) [P] multiply & rescale to “fuse” feedback & continual updating is the key to how the filter performs…

usually, the kalman filter is done in measurement space, which makes the formulae more complicated if you’re in engineering, you’ll likely see (& use!) kalman filters. don’t be intimidated: it’s simple! (in fact, it’s rather too simple for many problems…)

gaussians are not all there is…

the big picture

gaussians are great! & ubiquitous in applications! but to work with gaussians, you need a lot of (matrix) algebra…

1

given the following covariance matrices, draw pictures of the ellipse(-oid)s on which the zero-mean Gaussian is constant (i.e., level sets of the Gaussian) 2 -1 0 2 0 2 1 2 -1 A) b) c) d) -1 1 0 0 1 1 3 -1 3 0 0 1/2

2

show that the normalization constant in front of a general Gaussian equals  ((2π [] )-1 ), where the product with 2π is scalar-matrix multiplication.

3

consider two 1-d Gaussian functions with identical variances. show that their product has mean equal to the average of the means of the two inputs.

4

show that for an n-dimensional random variable X with Gaussian density & covariance matrix [] equal to a constant times the identity, [] = σ2 I, the expected value of X * X (the distance-squared function) equals nσ2.

5

fuse! perform the fusion step on the following means/covariances and check (by plotting level sets of the gaussians, if you like) that it “works”. 3 2 0 4 3 1 P = [P] = M = [M] = 4 0 3 6 1 2

6

show that for the fusion of two 1-d gaussians given in terms of (0 , 0) & (1 , 1), the fused Gaussian can be described as an update to (0 , 0) in the form of: 1 = 0 + K(1 - 0) 1 = 0 - K0 for some scalar K, called the “kalman gain”. solve for K in terms of 0 & 1

7 8

repeat the previous problem for arbitrary dimensions, solving for the kalman gain as a certain matrix used to do the update. not a surprise, eh ? the Gaussian product formulae become simpler when working with []-1 , the precision matrix. show that for Gaussian densities with precisions [0]-1 & [1]-1, -1 -1 the product Gaussian has precision matrix equal to the sum [0] + [1] .

but they are not always so intuitive

the majority of the probability mass lies within one standard deviation of the mean…

σ

σ

σ σ

almost all probability mass lies within three standard deviations…

σ μ

σ

what happens for higher dimensional gaussians?

consider the 2-d case of a Gaussian with unit variance and identity matrix as covariance…

hey! these numbers for mass within radius 1, 2, & 3 are not the same as in 1-d

μ

but, still, most all of the mass is within 3 of the mean

can get both interesting & nonintuitive

compute the volume Bn and “surface area” Ωn of the unit-radius ball in n

ρn-1dΩn

time to remember high-dimensional spherical coordinates (chapter 14)

dVn = ρn-1 dρ dΩn take advantage of spherical coordinates…

dVn = ρ = dΩn =

volume element radial coordinate solid angle element

we will need to use the gamma function…

Γ(x) =

@

∞

-t x-1

e t

t=0

dt = (x-1) !

fubini

@e

-||x|

n



2

dx =

see chapters 13, & 18

( @e

-©

n

2



n/2

d© ) = π

n/2

π

π

∞

@@ = @ dΩ @ = Ω @ =

-ρ2 n-1 e ρ dρ dΩ

ρ=0

n

n/2

nota bene: this volume is really more like “surface area” as in the 3-d case

convert to spherical coordinates…

2π Ωn = n Γ( 2 )

t =

ρ2

∞

n

=

1 2

∞

n

-ρ2 n-1 e ρ dρ

ρ=0

t=0

1 -t 2 e

n Ωn Γ( 2

)

(n/2)-1

t

dt

the unit n-dimensional ball has volume

@@ = @

Bn =

1

= =

1

ρ=0

ρ=0 ρn 1

n

1 n

ρn-1 dρ dΩ

0

@

n

ρn-1 dρ dΩn Ωn

Ωn

n/2

2π Bn = n n Γ( 2 )

R

n

n/2

2π Ωn = n Γ( 2 )

R

n-1

about volumes of balls & spheres

most of the points lie very close to the sphere at the boundary!

most of the points lie near the equator… for any choice of equator!

most pairs of points have nearly the same distance between them

that thing about Gaussian densities having most of the mass near the mean?

pick a really large dimension, n, and consider the unit-variance standard Gaussian the probability of being within distance R of the mean is given by the integral volume

@

ball of radius R

1 (2π)

--21 | x|

n e

2

dx ≤

1

(2π)n

V(

ball of radius R

the probability is…

n

R = ) n/2 2 (n/2) !

this is ≤ 1 mark it zero, dude

that specifies where the mass is hiding…

so, even though there is a lot of probability density near the mean, there’s not enough volume nearby to matter! the critical sphere is where density and volume balance…

there’s so much more to learn!

the big picture

when it comes to statistics be both curious & skeptical! there are many ways in which high-dimensional data are mysterious: calculus is key to understanding!

1

let’s recall a few facts about the gamma function, in case you forgot… A) show, using integration by parts, that Γ(x+1) = xΓ(x).

Γ(x) =

@

∞

e-t tx-1 dt = (x-1) !

t=0`

B) show that Γ( 1 ) = 1 and conclude that Γ(n) = (n-1) ! for n a positive integer. C) show that Γ(1/2) = π by manipulation to the integral of a Gaussian.

2

compute how much probability mass lies within one, two, and three units of the mean of a standard (unit-variance) gaussian on 3. use decimal percentages.

3

recall from chapter 15 exercises the “spherical coordinates” in 4 given by { (ρ, θ, φ, ψ) } with 0 ≤ θ ≤ 2π & 0 ≤ φ, ψ ≤ π . x1 = ρ  θ  φ  ψ x2 = ρ  θ  φ  ψ A) using this, argue that dΩ4 =  φ 2 ψ dψ dφ dθ x3 = ρ  φ  ψ b) integrate dΩ4 to compute that Ω4 = 2π2. x4 = ρ  ψ verify that this agrees with our formula.

6

let’s unpack some of the unusual claims about the distribution of points in ndimensional balls as n_∞. a) show that the fraction of n-volume of the unit n-ball within distance (1-ε) of the center is O(e-εn). “most points are near the boundary” b) challenge: show that the fraction of n-volume of the unit n-ball satisfying |x1| < 3/ n is >99% as n_∞. “most points are near the equator… any equator!” HINT: relate the volume of this slice to the integral of a standard Gaussian on .

7

using the results of the previous problem, show that for “most” choices of a pair of points in an n-dimensional unit ball, the vectors formed by these two points are nearly orthogonal. hint: let the first vector be along the x1 axis. use this to compute the expected distance between the points.

8

given the formulae for volumes of balls and spheres, compute the n-volume of a cylinder in n defined by the inequalities { | (x1, x2, …, xn-1) | 2 ≤ r , 0 ≤ xn ≤ h }. what is the (n-1)-dimensional “surface area” of this cylinder?

modern applications of integrals

you should learn…

and integration theory

ok, fine, there’s more to integration than just definitions

n

given f:  _  you convert it to a different function… by integrating against a “kernel” K n+n K:  _ 

(Tf)(y) =

@

n

K(x ,y) f(x) dx

y K

which kernel is best? ah, that’s the question!

x

functions f : (+)n _ + = [0, ∞)

K(x ,y ) = e-x*y (Lf)( y ) =

@

-x*y e f(x) dx +n

( )

used in control theory, this converts systems of differential equations to systems of algebraic equations…

functions f : n _ 

K(x ,y ) = e-2πi x*y (Ff)( y ) =

@

-2πi x*y e f(x) dx n



used in everything. yeah, seriously. this converts to a “frequency space” that makes it easy to work with waves

in practice, one uses discrete versions of these [sums]

actually computing integrals?

n

approximating the integral of f:  _  n over a domain R ⊂  can be done discretely sample the integrand f on a mesh ( say, a uniform grid of points {xi} ) & compute an approximation

@

R

f(x) dx ≅

voln(R) C

#i wi f(xi)

f

R

these depend on the method used & derive from a taylor expansion of the integrand

1 4 1 2 1 2 1 2 1 4

1 2

1 2

1 2

1 4 1 2 1 2 1 2 1 4

1 1 1

1 1 1

1 1 1

1 2

1 2

1 2

second-order error

for a double integral on a rectangle with uniform mesh, two common methods are trapezoid & simpson’s

these weights can be derived by using the 1-d versions and following the fubini theorem… try it! this is a fun & useful subject… just be sure to remember your taylor series & big-O notation!

1 9

4 9

2 9

4 9

1 9

4 9 2 9 4 9 1 9

16 9 8 9 16 9 4 9

8 9 4 9 8 9 2 9

16 9 8 9 16 9 4 9

4 9 2 9 4 9 1 9

fourth-order error

the number of sample points you need to compute these approximations grows

suppose you are trying to price the investment value of a device that depends on a 30 year variable-rate mortgage that adjusts monthly… model each month’s rate as a random variable… computing the expected value (or variance) would require computing an integral over a domain of dimension equal to 12*30 = 360

but for a cube in n, with each axis sampled κ times, you need κn total points

but it uses randomness to aid in computation

approximate the integral of f: n _  over a bounded domain R ⊂ n

for a random set of samples { xi } on R average the sampled values of f

for X a uniform random variable on R (that is, the density is constant on R )

(f(X)) =

@

R

if this can be approximated by sampling…

f(x) dP =

1 voln(R)

& we assume we know this…

@ f(x) dx

(f(X)) ≅

1

N

N

# f(xi) i=1

convert this to an approximate integral of f

R

then we can compute this!

@

R

f(x) dx ≅

voln(R) N

N

# f(xi) i=1

with work & some computations involving variance, one can show that

error = O(N

-1/2

)

this is independent of the dimension! although not as good as, say, Simpson’s method in 1-d (which is O(N-4), this is a huge improvement in high dimensions

@

R

f(x) dx ≅

voln(R) N

N

# f(xi) i=1

probability makes use of integrals, but returns the favor in high dimensions…

are SIMPLY A name for a type of function

a © - field on a domain D means there is a © at every point in D

IS WHAT WE’VE BEEN DOING all along…

planar vector fields to draw a planar vector field, try evaluating the field at various points… don’t forget to look for where the field “vanishes” (evaluates to zero) pay attention to what happens along the axes: that’s helpful

 = -x i + y j

with practice, you can recognize patterns & types of vector fields

 = -y i + x j

IT IS PERHAPS IRONIC…

INTEGRATING A VECTOR FIELD

IS NOT that HARD TO DEFINE…

OR TO COMPUTE!

the scalar path integral adds up the values of the integrand along the path

given a parametrization of the path, “pull back” the scalar values & integrate along the parametrization…

ARE MORE INVOLVED FOR VECTOR FIELDS

recall: the gradient of a vector field is the vector form of the partial derivatives in the coordinates

∇f =

^f i ^x

+

^f j ^y

+

^f k ^z n

gradient makes sense for vector fields on any 

the divergence of a vector field is the sum of the partial derivatives in the coordinates

∇* =

^Vx ^x

+

^Vy ^y

+

^Vz ^z n

divergence, like gradient, makes sense for vector fields on any 

curl measures local rotation of volume elements under the vector field’s flow

∇× =

^Vz ^Vy ^y ^z

i+

^Vx ^Vz ^z ^x

j+

^Vy ^Vx k ^x ^y 3

important!!! curl, like cross-product, is well-defined only on 

CULMINATES IN A SET OF THEOREMS

@  * dx = @ @ (∇× ) * k dA @  * dx = @@ (∇× ) * n dS @ @  * n dS = @@@ ∇*  dV ^D

D

^D

D

^D

D

REALLY, REALLY COMPLICATED

IT LEADS TO A DRAMATIC CONCLUSION

@α ^D

=

@ dα D

we meet again our old friend

the big picture the familiar calculus of scalar fields leads naturally to a calculus of vector fields this, in turn, generalizes to a wonderful & unifying calculus of differential forms

bY

Robert ghrist Is the andrea Mitchell professor Of mathematics and Electrical & systems engineering at the university of pennsylvania He’s an award-winning researcher, teacher, writer, & speaker HIS RESEARCH SPECIALTY IS APPLIed TOPOLOGY

Good textbook on calculus that does integration in more than simply 2-d & 3-d: Hubbard, J. and Hubbard, B. B., Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach, 5th ed., Matrix Editions, 2015.

More advanced undergraduate-level text on the mathematics of integration & more: Spivak, M., Calculus on Manifolds, 5th ed. CRC Press, 1971. Good introduction to applied probability: Venkatesh, S., The Theory of Probability: Explorations & Applications, Cambridge, 2012. Good book (in progress) on the mathematics of data: Blum, A., Hopcroft, J. and Kannan, R., Foundations of Data Science, in progress.

all writing, design, drawing, & layout by prof/g [Robert ghrist] prof/g acknowledges the support of andrea Mitchell & the fantastic engineering students at the university of pennsylvania during the writing of calculus blue, prof/g’s research was generously supported by the united states department of defense through the ASDR&E vannevar bush faculty fellowship