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MATH2202 Notebook 3 Fall 2015/2016 prepared by Professor Jenny Baglivo Contents 3 MATH2202 Notebook 3 3 3.1 Reminder: Riemann Sums, Definite Integrals, Applications ................ 3 3.2 Riemann Sums, Double Integrals and Iterated Integrals .................. 9 3.3 Double Integrals Over General Regions ............................ 13 3.4 Riemann Sums, Double Integrals In Polar Coordinates ................... 22 3.5 Riemann Sums, Triple Integrals and Iterated Integrals ................... 28 3.6 Triple Integrals Over General Regions ............................ 33 3.7 Triple Integrals in Cylindrical Coordinates .......................... 38 3.8 Timeout: Spherical Coordinates ................................ 41 3.9 Triple Integrals in Spherical Coordinates ........................... 44 3.10 Determinants, Areas and Volumes Revisited ......................... 48 3.11 Change of Variables in Double Integrals ........................... 52 3.12 Change of Variables in Triple Integrals ............................ 57 c Copyright 2008-2016 by Jenny A. Baglivo. All Rights Reserved. 1

Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

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Page 1: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

MATH2202 Notebook 3 Fall 2015/2016

prepared by Professor Jenny Baglivo

Contents

3 MATH2202 Notebook 3 3

3.1 Reminder: Riemann Sums, Definite Integrals, Applications . . . . . . . . . . . . . . . . 3

3.2 Riemann Sums, Double Integrals and Iterated Integrals . . . . . . . . . . . . . . . . . . 9

3.3 Double Integrals Over General Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.4 Riemann Sums, Double Integrals In Polar Coordinates . . . . . . . . . . . . . . . . . . . 22

3.5 Riemann Sums, Triple Integrals and Iterated Integrals . . . . . . . . . . . . . . . . . . . 28

3.6 Triple Integrals Over General Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.7 Triple Integrals in Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.8 Timeout: Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.9 Triple Integrals in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.10 Determinants, Areas and Volumes Revisited . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.11 Change of Variables in Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.12 Change of Variables in Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

c© Copyright 2008-2016 by Jenny A. Baglivo. All Rights Reserved.

1

Page 2: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

2

Page 3: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3 MATH2202 Notebook 3

This notebook is concerned with topics from Chapter 12 of the Stewart textbook.

3.1 Reminder: Riemann Sums, Definite Integrals, Applications

Definite integral for functions of 1 variable. Let f be a function of 1 variable whosedomain D contains the closed interval [a, b], and let n be a positive integer.

Let 4x = (b− a)/n and xi = a+ i4x for i = 0, 1, 2, . . . , n. Then the n+ 1 numbers

a = x0 < x1 < x2 < · · · < xn = b

partition (or subdivide) the interval [a, b] into n subintervals of equal length. Lastly, let x∗i bethe midpoint of the ith subinterval: x∗i = (xi−1 + xi)/2.

The definite integral of f from a to b is∫ b

af(x) dx = lim

n→∞

n∑i=1

f(x∗i )4x,

if this limit exists. The sum on the right is called a Riemann sum, and f is said to be integrableif the limit of Riemann sums exists.

Note: Continuous functions are always integrable. Further, it is possible to show that the limitexists even if f has a finite number of “jump discontinuities” on [a, b].

Example. Consider the function with rule

f(x) =x3

2− 3x2 + 4x+ 6

on the interval [0, 5]. Since f is continuous and non-negative, the definite integral represents the areaunder the curve and above the interval. Using 5subintervals, the Riemann sum is

(f(0.5) + f(1.5) + · · ·+ f(4.5))×∆x = 32.8125, � � � � ��

��

where ∆x = . The exact area under the curve and above the interval is

3

Page 4: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Example. Consider the function with rule

f(x) = x3 − 4x

on the interval [0, 3]. Since f is partly above andpartly below the x-axis on this interval, the definiteintegral can be interpreted as the difference betweentwo areas. Using 9 subintervals, the Riemann sumis

(f(1/6) + f(3/6) + · · ·+ f(17/6))×∆x = 2.125,

� � ��

-�

��

��

where ∆x = . The exact difference between the areas is

Average values. Let f be a continuous function on the interval [a, b]. The average value off on this interval can be computed using the formula

Average(f) =1

(b− a)

∫ b

af(x) dx

Note: The proof of the average value formula uses a limit of Riemann sums.

(i) Let f be the average of f at the midpoints of the n subintervals:

f =1

n

n∑i=1

f(x∗i ).

(ii) Multiply and divide by the length of [a, b], rearrange terms, and calculate the limit:

f =(b− a)

(b− a)

(1

n

n∑i=1

f(x∗i )

)=

1

(b− a)

n∑i=1

f(x∗i )4x

=⇒ limn→∞

f =1

(b− a)

(limn→∞

n∑i=1

f(x∗i )4x

)=

1

(b− a)

∫ b

af(x) dx.

4

Page 5: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Example. Consider the function with rule

f(x) = 225e0.08x

on the interval [0, 35]. Using 5 subintervals to ap-proximate the average, we get

1

5(f(3.5) + f(10.5) + . . .+ f(31.5)) ≈ 1225.02.

The exact average is� �� �� �� ��

���

����

����

����

Volumes using the method of slices. Let f be a continuous function of 2 variables whosedomain D contains the rectangular region

R = [a, b]× [c, d] = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d},

and assume that the values of f are nonnegative on the rectangle R.

The method of slices (also called Cavalieri’s principle) can be used to find the volume of thesolid bounded above by the graph of f , below by the rectangle R, and with sides parallel tothe xz-plane and the yz-plane.

As a first step, let

1. Ax(x) be the area under the partial function z = f(x, y) for fixed x and y ∈ [c, d].

2. Ay(y) be the area under the partial function z = f(x, y) for fixed y and x ∈ [a, b].

5

Page 6: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let f(x, y) = 3x2 + y + 2 on the rectangle [0, 4]× [0, 6]. The plots below show thesolid bounded above by the graph of f and below by the rectangle, and typical slices throughthe solid for fixed x (left) and for fixed y (right).

Find general formulas for

(a) Ax(x) (the area of a slice of the solid when x is fixed and y varies), and

(b) Ay(y) (the area of a slice of the solid when y is fixed and x varies).

6

Page 7: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

The volume of the solid is obtained by integrating one of the area functions. That is,

V olume =

∫ b

aAx(x) dx =

∫ d

cAy(y) dy

Exercise, continued. Let f(x, y) = 3x2 + y + 2 on the rectangle [0, 4]× [0, 6]. Find the volumeunder the graph of f and above the rectangle by integrating Ax(x) and by integrating Ay(y).

Notes: The proof that volume can be computed as the integral of area, no matter which waywe choose to slice the solid, was given by the mathematician Cavalieri in the 1600’s. Since thearea functions are the values of definite integrals,

Ax(x) =

∫ d

cf(x, y) dy and Ay(y) =

∫ b

af(x, y) dx,

another way to write the volume formula is

V olume =

∫ b

a

∫ d

cf(x, y) dy dx =

∫ d

c

∫ b

af(x, y) dx dy

The integrals on the right are called iterated integrals. Iterated integrals are evaluated byevaluating the innermost integral first.

7

Page 8: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let f(x, y) =100

(x+2y)2 on the rectangle [3, 4]× [0, 1].

Use the method of slices to find the volume of the solid bounded above by the graph of f andbelow by the rectangle.

8

Page 9: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.2 Riemann Sums, Double Integrals and Iterated Integrals

Double integrals over rectangles. Let f be a function of 2 variables whose domain containsthe rectangle R = [a, b]× [c, d], and let m and n be positive integers.

1. Partition in the x-Direction: Let 4x = (b− a)/m and xi = a+ i4x for i = 0, 1, . . . ,m.

Then the (m+ 1) numbers

a = x0 < x1 < x2 < · · · < xm = b

partition the interval [a, b] into m subintervals of equal length with midpoints

x∗i = (xi−1 + xi)/2 for i = 1, 2, . . . ,m.

The length of each subinterval is 4x.

2. Partition in the y-Direction: Let 4y = (d− c)/n and yj = c+ j4y for j = 0, 1, . . . , n.

Then the (n+ 1) numbers

c = y0 < y1 < y2 < · · · < yn = d

partition the interval [c, d] into n subintervals of equal length with midpoints

y∗j = (yj−1 + yj)/2 for j = 1, 2, . . . , n.

The length of each subinterval is 4y.

3. Partition in the Plane: The collection of mn subrectangles,

Ri,j = [xi−1, xi]× [yj−1, yj ], for i = 1, 2, . . . ,m, j = 1, 2, . . . , n,

form an m-by-n partition of the rectangle R.

The midpoint of Ri,j is (x∗i , y∗j ). Each subrectangle has area 4A = (4x) (4y).

4. Double Integral of f over R: The double integral of f over R is

∫∫Rf dA = lim

m,n→∞

m∑i=1

n∑j=1

f(x∗i , y∗j )4A,

when this limit exists. The sum on the right is called a Riemann sum, and f is said tobe integrable if the limit of Riemann sums exists.

Note: Continuous functions are always integrable. Further, it is possible to show that the limitexists when the values of f are bounded, and the set of discontinuities of f has area zero.

9

Page 10: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

For example, let f(x, y) = 3x2 + y + 2 on the rectangle [0, 4]× [0, 6].

2-by-3 Partition 8-by-12 Partition

The following list of approximate values suggests that the limit is 504.

m 2 4 8 16 32 64 128 256

n 3 6 12 24 48 96 192 384

Sum: 498.0 502.5 503.625 503.906 503.977 503.994 503.999

Double and iterated integrals. The following theorem, discovered by Fubini in the early1900’s, says that double integrals can often be evaluated using iterated integrals.

Fubini’s Theorem. Let f is a continuous function of 2 variables whose domaincontains the rectangle R = [a, b]× [c, d]. Then∫∫

Rf dA =

∫ d

c

∫ b

af(x, y) dx dy =

∫ b

a

∫ d

cf(x, y) dy dx .

Note: Fubini’s theorem generalizes Cavalieri’s principle to all continuous functions.

Fubini’s theorem remains true if f is a bounded function on R satisfying (1) the set of dis-continuities of f has area 0, and (2) each slice (for fixed x or for fixed y) intersects the set ofdiscontinuities in at most a finite number of points.

10

Page 11: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Volumes and difference in volumes. If f is nonnegative on R, then the double integralover R can be interpreted as the volume under the graph of f and above the rectangle. If fis sometimes positive and sometimes negative, then the double integral can be interpreted asthe difference in volumes.

Exercise. Consider the function with rule

f(x, y) =20(x+ 1)

(2y + 3)2

on the rectangle R = [−2, 1]× [0, 1].

(1) Find the double integral of f over R.

(2) Find the volume of the solid bounded by

z = f(x, y), z = 0, x = −2, x = 1, y = 0, y = 1.

11

Page 12: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Average values. Let f be a continuous function of 2 variables whose domain contains therectangle R = [a, b] × [c, d]. The average value of f on this rectangle can be computed usingthe formula

Average(f) =1

(b− a)(d− c)

∫∫Rf dA

Note: The proof of the average formula uses a limit of Riemann sums.

(i) Let f be the average of f at the midpoints of the mn sub-rectangles:

f =1

mn

∑i,j

f(x∗i , y∗j ).

and let A(R) = (b− a)(d− c) be the area of the rectangle.

(ii) Multiply and divide by A(R), rearrange terms, and calculate the limit:

f =A(R)

A(R)

1

mn

∑i,j

f(x∗i , y∗j )

=1

A(R)

∑i,j

f(x∗i , y∗j ) 4A

=⇒ limm,n→∞

f =1

A(R)

limm,n→∞

∑i,j

f(x∗i , y∗j ) 4A

=1

A(R)

∫∫Rf dA.

Example, continued. Let f(x, y) = 3x2+y+2 on the rectangle [0, 4]× [0, 6]. The approximate

average based on the midpoints of the 2-by-3 partition is .

The exact average of f on the rectangle is .

Exercise, continued. Consider the function with rule f(x, y) =20(x+1)(2y+3)2 .

(3) Find the average value of f on the rectangle [−2, 1]× [0, 1].

12

Page 13: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.3 Double Integrals Over General Regions

Double and iterated integrals can be evaluated over more general domains. Specifically,

1. Domain of Type 1: Let

D = {(x, y) : a ≤ x ≤ b, `(x) ≤ y ≤ u(x)},

where `(x) and u(x) are continuous on [a, b], and assumef is continuous on a domain containing D. Then∫∫

Df dA =

∫ b

a

∫ u(x)

`(x)f(x, y) dy dx .

2. Domain of Type 2: Let

D = {(x, y) : c ≤ y ≤ d, `(y) ≤ x ≤ u(y)},

where `(y) and u(y) are continuous on [c, d], and assumef is continuous on a domain containing D. Then∫∫

Df dA =

∫ d

c

∫ u(y)

`(y)f(x, y) dx dy .

Exercise. Let D be the triangular region with corners

(0, 1), (1, 0), (1, 2).

Evaluate

∫∫D

70xy2 dA.

���� ��� ���� ���

���

��

���

��

13

Page 14: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let D be the region bounded by

x = y and x =√y.

Evaluate

∫∫D

70xy2 dA.

���� ��� ���� ���

����

���

����

��

Average values over general domains. Let f be a continuous function of 2 variableswhose domain contains a region D of positive area, and let A(D) be the area of D. Then theaverage value of f on D can be computed using the formula

Average(f) =1

A(D)

∫∫Df dA

Note that the area of D can be computed as follows: A(D) =∫∫D 1 dA.

Exercise, continued. Find the average value of f(x, y) = 70xy2 on the region bounded by thecurves x = y and x =

√y.

14

Page 15: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let D be the region satisfying

x ≥ 0 and x2 + y2 ≤ 9.

Find the average value of f(x, y) = x over D.

� � ��

-�

-�

-�

15

Page 16: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Set up the iterated integrals that you would need to find the volume of the solidbetween the planes

z = 3x+ 2y + 1 and z = x+ y

and above the triangular region with corners (1, 0), (0, 1) and (2, 2), as illustrated below.

� ��

16

Page 17: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise: Changing the order of integration. In each case, your goal is to

1. Identify and sketch the domain of integration.

2. Change the order of integration (that is, write an equivalent iterated integralwith “dx dy” replaced by “dy dx” or vice versa).

3. Evaluate the iterated integral obtained in the second step.

(a)∫ 10

∫ 1y e

x2 dx dy

17

Page 18: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

(b)∫ 20

∫ 1x/2 y

2 cos(xy) dy dx

18

Page 19: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

(c)∫ 20

∫ 51+y2 y e

(x−1)2 dx dy

19

Page 20: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Application: probability density functions. Researchers often use nonnegative integrablefunctions to study probability. Let f be a function of 2 variables with domain D, where

1. (x, y) represents a pair of quantities of interest and

2. D represents the set of all possible pairs.

Then f is a probability density function for this pair of quantities if f is a nonnegative integrablefunction that has been scaled so that

∫∫D f dA = 1, and so that∫∫

Df dA = Proportion of individuals with (x, y) ∈ D, when D ⊆ D.

Note: The ideas studied here can be generalized to studying any number of quantities ofinterest. For example, economists might consider the following 3 quantities: total incomeearned, percentage of income saved, and percentage of income donated to charity.

Example (Blood Fats). Cholesterol and triglyceridesare examples of blood fats. For adults living inthe United States, the average cholesterol level is200 milligrams per deciliter (mg/dL) and the aver-age triglycerides level is 150 mg/dL.

Letting x represent cholesterol and y representtriglycerides, a simple probability density functionfor (x, y)-pairs is

f(x, y) = C x9 y9e−x/20−y/15

where C is chosen so that the volume under thesurface for x, y ≥ 0 is 1.

An individual whose cholesterol level is 240 mg/dL or more is classified as having high choles-terol, and an individual whose triglycerides level is 200 mg/dL or more is classified as havinghigh triglycerides. Since

1−∫ 200

0

∫ 240

0f(x, y) dx dy ≈ 0.3522,

this simple model predicts that about 35.22% of adults have at least one of these problems.

20

Page 21: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise (Insurance). A company offers basic insurance to its employees. Employees may alsochoose to purchase a supplemental plan. Let x be the proportion of employees who choose topurchase at least the basic plan, let y be the proportion who choose to purchase both the basicand supplemental plans, and assume that

f(x, y) = C(x+ y), where 0 ≤ y ≤ x ≤ 1,

is a probability density function for these characteristics.

(a) Find the value of C.

(b) Use your answer to part (a) to find the probability that at most 50% of employeespurchase insurance (either the basic plan or the basic plus supplemental plans).

21

Page 22: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.4 Riemann Sums, Double Integrals In Polar Coordinates

Our goal is to compute∫∫D f(x, y) dA using an important change of variables:

x = r cos(θ) and y = r sin(θ),

where r is the distance from (x, y) to the origin, and θ is the angle measured counterclockwisefrom the positive x-axis to the ray containing the origin and (x, y).

Polar rectangles. A polar rectangle R is a setdescribed in polar coordinates as follows:

R = {(r, θ) : a ≤ r ≤ b, α ≤ θ ≤ β}.

An example of a polar rectangle where both

0 < α < π/2 and 0 < β < π/2

is shown to the right.

The area of R can be found as follows:

θ=α

θ=β

� ��

A(R) = (Area of Outer Wedge)− (Area of Inner Wedge)

= 12(∆θ)b2 − 1

2(∆θ)a2 where ∆θ = β − α

= 12(b2 − a2)(∆θ) = 1

2(b+ a)(b− a)(∆θ)

= r∗(∆r)(∆θ) where ∆r = b− a and r∗ = (a+b)2 .

Double integrals over polar rectangles. Let f be a function of 2 variables whose domaincontains the polar rectangle R, and let m and n be positive integers.

1. Partition in the r-Direction: Let 4r = (b− a)/m and ri = a+ i4r for i = 0, 1, . . . ,m.

Then the (m+ 1) numbers

a = r0 < r1 < r2 < · · · < rm = b

partition the interval [a, b] into m subintervals of equal length with midpoints

r∗i = (ri−1 + ri)/2 for i = 1, 2, . . . ,m.

The length of each subinterval is 4r.

22

Page 23: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

2. Partition in the θ-Direction: Let 4θ = (β − α)/n and θj = α+ j4θ for j = 0, 1, . . . , n.

Then the (n+ 1) numbers

α = θ0 < θ1 < θ2 < · · · < θn = β

partition the interval [α, β] into n subintervals of equal length with midpoints

θ∗j = (θj−1 + θj)/2 for j = 1, 2, . . . , n.

The length of each subinterval is 4θ.

3. Partition in the Plane: Let Ri,j be the polar rectangle with r ∈ [ri−1, ri] and θ ∈ [θi−1, θi],and let 4Ai,j be its area. From the work above, we know that

4Ai,j = r∗i (4r)(4θ), for i = 1, 2, . . . ,m, j = 1, 2, . . . , n.

4. Double Integral of f over R: The double integral of f over R is

∫∫R f dA = lim

m,n→∞

m∑i=1

n∑j=1

f(r∗i cos(θ∗j ), r∗i sin(θ∗j ))4Ai,j ,

= limm,n→∞

m∑i=1

n∑j=1

f(r∗i cos(θ∗j ), r∗i sin(θ∗j )) r

∗i (4r)(4θ),

when this limit exists. The sum on the right is called a Riemann sum, and f is said tobe integrable if the limit of Riemann sums exists.

When the limit exists, its value can be found using the iterated integral∫∫Rf(x, y) dA =

∫ β

α

∫ b

af(r cos(θ), r sin(θ)) r dr dθ.

Double integrals over general polar domains. The ideas above can be extended to moregeneral domains. For example, if the domain D can be described as

D = {(r, θ) : α ≤ θ ≤ β, `(θ) ≤ r ≤ u(θ)}, where `(θ) and u(θ) are continuous,

then ∫∫Df(x, y) dA =

∫ β

α

∫ u(θ)

`(θ)f(r cos(θ), r sin(θ)) r dr dθ.

Notes: (1) When evaluating double integrals in polar coordinates, we let

x = r cos(θ), y = r sin(θ) and dA = r dr dθ.

The method is useful when a trig substitution would simplify integration.

23

Page 24: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

(2) You may find the following trig identities useful:

1. Double angle formulas for working with squares:

sin2(θ) =1

2(1− cos(2θ)) and cos2(θ) =

1

2(1 + cos(2θ))

2. Simplifying cubes:

sin3(θ) = (1− cos2(θ)) sin(θ) and cos3(θ) = (1− sin2(θ)) cos(θ)

Exercise. Evaluate∫∫D y dA, where D is the region

satisfying

y ≥ x, y ≥ −x, x2 + y2 ≤ 9.

-� -� -� � � ��

���

���

���

���

���

���

24

Page 25: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Evaluate∫∫D cos(x2 + y2) dA, where D is

the region satisfying

y ≥ 0, y ≥√

3x, x2 + y2 ≤ 1.

-��� -��� ��� ����

���

���

���

���

���

25

Page 26: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Evaluate∫∫D 5x dA, where D is the region

satisfying

x2 + (y − 2)2 ≤ 4 and x ≥ 0.

� � � ��

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Page 27: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Find the volume of the solid lying belowthe surface

z = 9− x2 − y2

and above the surface

z = 3x2 + 3y2 − 16.

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Page 28: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.5 Riemann Sums, Triple Integrals and Iterated Integrals

Triple integrals over boxes. Let f be a function of 3 variables whose domain contains thebox B = [a, b]× [c, d]× [p, q], and let `, m and n be positive integers.

1. Partition in the x-Direction: Let 4x = (b− a)/` and xi = a+ i4x for i = 0, 1, . . . , `.

Then the (`+ 1) numbers

a = x0 < x1 < x2 < · · · < x` = b

partition the interval [a, b] into ` subintervals of equal length with midpoints

x∗i = (xi−1 + xi)/2 for i = 1, 2, . . . , `.

The length of each subinterval is 4x.

2. Partition in the y-Direction: Let 4y = (d− c)/m and yj = c+ j4y for j = 0, 1, . . . ,m.

Then the (m+ 1) numbers

c = y0 < y1 < y2 < · · · < ym = d

partition the interval [c, d] into m subintervals of equal length with midpoints

y∗j = (yj−1 + yj)/2 for j = 1, 2, . . . ,m.

The length of each subinterval is 4y.

3. Partition in the z-Direction: Let 4z = (q − p)/n and zk = c+ k4z for k = 0, 1, . . . , n.

Then the (n+ 1) numbers

p = z0 < z1 < z2 < · · · < zn = q

partition the interval [p, q] into n subintervals of equal length with midpoints

z∗k = (zk−1 + zk)/2 for k = 1, 2, . . . , n.

The length of each subinterval is 4z.

4. Partition in 3-Space: The collection of `mn subboxes,

Bi,j,k = [xi−1, xi]× [yj−1, yj ]× [zk−1, zk], for i = 1, 2, . . . , `, j = 1, 2, . . . ,m, k = 1, 2, . . . , n,

form an `-by-m-by-n partition of the box B. The midpoint of Bi,j,k is (x∗i , y∗j , z∗k), and

the volume of each subbox is 4V = (4x) (4y) (4z).

5. Triple Integral of f over B: The triple integral of f over B is∫∫∫Bf dV = lim

`,m,n→∞

∑̀i=1

m∑j=1

n∑k=1

f(x∗i , y∗j , z∗k)4V,

when this limit exists. The sum on the right is called a Riemann sum, and f is said tobe integrable if the limit of Riemann sums exists.

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Page 29: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Note: Continuous functions are always integrable. Further, it is possible to show that the limitexists if the values of f are bounded and the set of discontinuities of f on B has volume zero.

Exercise 1. Consider the function with rule

f(x, y, z) = 3x2(y − z)

on the box B = [−2, 2]× [−1, 3]× [0, 4].

(a) Use a 2-by-2-by-2 midpoint partition to estimate∫∫∫Bf dV.

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Page 30: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Triple and iterated integrals. The following theorem, discovered by Fubini in the early1900’s, says that triple integrals can often be evaluated using iterated integrals.

Fubini’s Theorem. Let f be a continuous function of 3 variables whose domaincontains the box B = [a, b]× [c, d]× [p, q]. Then the triple integral can be computedusing iterated integrals, where the iteration can be done in any one of six orders.In particular, ∫∫∫

Bf(x, y, z) dV =

∫ b

a

∫ d

c

∫ q

pf(x, y, z) dz dy dx .

Note: This theorem generalizes Fubini’s theorem for functions of 2 variables.

Fubini’s theorem remains true if f is a bounded function on B satisfying (1) the set of discon-tinuities of f has volume 0, and (2) each line parallel to one of the coordinate axes intersectsthe set of discontinuities of f in at most a finite number of points.

Exercise 1, continued. (b) Evaluate∫∫∫

B 3x2(y − z) dV , where B = [−2, 2]× [−1, 3]× [0, 4].

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Page 31: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Average values. Let f be a continuous function of 3 variables whose domain contains thebox B = [a, b] × [c, d] × [p, q]. The average value of f on this box can be computed using theformula

Average(f) =1

(b− a)(d− c)(q − p)

∫∫∫Bf dV

Note: The proof of the average formula uses a limit of Riemann sums.

(i) Let f be the average of f at the midpoints of the sub-boxes:

f =1

`mn

∑i,j,k

f(x∗i , y∗j , z∗k),

and let V (B) = (b− a)(d− c)(q − p) be the volume of the box.

(ii) Multiply and divide by V (B), rearrange terms, and calculate the limit.

f =V (B)

V (B)

1

`mn

∑i,j,k

f(x∗i , y∗j , z∗k)

=1

V (B)

∑i,j,k

f(x∗i , y∗j , z∗k)4V

=⇒ lim`,m,n→∞

f =1

V (B)

lim`,m,n→∞

∑i,j,k

f(x∗i , y∗j , z∗k) 4V

=1

V (B)

∫∫∫Bf dV.

Exercise 1, continued. Let f(x, y, z) = 3x2(y − z) on the box B = [−2, 2]× [−1, 3]× [0, 4].

(c) Find the approximate average of f on B based on the 2-by-2-by-2 partition.

(d) Find the exact average of f on B.

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Page 32: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise 2. Let B = [−1, 1]× [0, 2]× [1, e].

(a) Evaluate∫∫∫

B(y3 − x/z) dV .

(b) Find the average of (y3 − x/z) on B.

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Page 33: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.6 Triple Integrals Over General Regions

Triple integrals can also be computed over more general domains. There are six basic types,depending on the order of integration. For example, let

E = {(x, y, z) : a ≤ x ≤ b, `1(x) ≤ y ≤ u1(x), `2(x, y) ≤ z ≤ u2(x, y)}, where

1. `1(x) and u1(x) are continuous on [a, b] and

2. `2(x, y) and u2(x, y) are continuous on the domain

D = {(x, y) : a ≤ x ≤ b, `1(x) ≤ y ≤ u1(x)} ⊂ R2,

and assume that f is continuous on a domain containing E. Then∫∫∫Ef dV =

∫ b

a

∫ u1(x)

`1(x)

∫ u2(x,y)

`2(x,y)f(x, y, z) dz dy dx .

Exercise. Let E be the region bounded by

x = 0, x = 2, y = 0, y = 1, z = 0, z = 4 + x+ y.

Evaluate∫∫∫

E xy dV .

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Page 34: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let E be the region bounded by

y = 0, y = x, x+ y = 20, z = 0, z = 25− y.

Evaluate∫∫∫

E 3x dV

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Page 35: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let E be the region bounded by

z = −1, z = 1, x = z2, x = 1, y = 0, y = 1− x.

Evaluate∫∫∫

E 35x dV .

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Page 36: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Average values over general domains. Let f be a continuous function of 3 variableswhose domain contains a region E of positive volume, and let V (E) be the volume of E. Thenthe average value of f on E can be computed using the formula

Average(f) =1

V (E)

∫∫∫Ef(x, y, z) dV

Note that the volume of E can be computed as follows: V (E) =∫∫∫

E 1 dV .

Exercise. Let E be the region bounded by

x = 0, y =√x, y = 1, z = 0, z = 1− y.

(a) Find the volume of E.

36

Page 37: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

(b) Find the average value of 40y on E.

Applications: density functions. Researchers often use nonnegative integrable functionsto study density and probability. Specifically,

1. Density and Mass: Let E be a physical object in 3-space, P (x, y, z) be an arbitrary pointin E, and let

f(x, y, z) be mass per unit volume at P (i.e., the density of the object at P ).

Then, the triple integral can be interpreted as the total mass of the object, and theaverage of f on E can be interpreted as the average density of the object.

2. Probability: Let f be a function of 3 variables with domain D, where

(a) (x, y, z) represents a triple of quantities of interest, such as height, weight andsystolic blood pressure; and

(b) D represents the set of all possible triples for individuals in a certain population.

Then f is a probability density function for this triple if f is a nonnegative integrablefunction that has been scaled so that

∫∫∫D f dV = 1, and so that∫∫∫

Ef dV = Proportion of individuals with (x, y, z) ∈ E, when E ⊆ D.

37

Page 38: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.7 Triple Integrals in Cylindrical Coordinates

Many triple integration problems can be simplified by transforming either the xy-plane or theyz-plane or the xz-plane to a “polar plane.” In the usual setup, we let

x = r cos(θ), y = r sin(θ), z = z,

where r is the distance from (x, y, 0) to the origin, and θ is the angle measured counterclockwisefrom the positive x-axis to the ray containing the origin and (x, y, 0).

Cylindrical boxes. In the usual set up, a cylindricalbox is a set of the form

B = {(r, θ, z) : a ≤ r ≤ b, α ≤ θ ≤ β, p ≤ z ≤ q}.

An example where both

0 < α < π/2 and 0 < β < π/2

is shown to the right.

The volume of B can be found as follows:

V (B) = (Area of Base)(Height)

=[r∗(4r)(4θ)

](4z)

where 4r = b− a, 4θ = β − α, 4z = q − p and r∗ = (a+ b)/2.

Triple integrals over cylindrical boxes. Let f be a continuous function of 3 variableswhose domain contains the cylindrical box B. The triple integral of f over B can be computedusing the following iterated integral:∫∫∫

Bf dV =

∫ q

p

∫ β

α

∫ b

af(r cos(θ), r sin(θ), z) r dr dθ dz.

Notes: When evaluating triple integrals in cylindrical coordinates in the usual form, we let

x = r cos(θ), y = r sin(θ), z = z, dV = r dr dθ dz.

The method is useful in situations where trig substitution would simplify computations.

Triple integrals in cylindrical coordinates are defined as limits of Riemann sums, followingthe patterns of the definitions we have seen in earlier sections. In addition, triple integrals incylindrical coordinates can be defined over more general regions.

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Page 39: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let E be the region bounded by

x2 + y2 + z2 ≤ 25, x ≥ 0, y ≥ 0, z ≥ 0.

Evaluate ∫∫∫Exz dV.

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Page 40: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let E be the region bounded by

x2 + y2 = 1, x2 + y2 = 4, z = 0, z = x+ 3.

Evaluate ∫∫∫Ex dV.

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Page 41: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.8 Timeout: Spherical Coordinates

Each P (x, y, z) can be written in terms of three parameters, (ρ, θ, φ), where

1. Distance: ρ represents the distance from the origin to P :

ρ =√x2 + y2 + z2.

2. Longitude: θ represents longitude usually measured from 0 to 2π.

3. Latitude: φ represents latitude measured from 0 (positive z-axis) to π (negative z-axis).

The transformation to spherical coordinates is as follows:

x = ρ sin(φ) cos(θ), y = ρ sin(φ) sin(θ), z = ρ cos(φ).

To visualize the transformation,

ϕ

1. Left plot: A fixed value of ρ corresponds to a sphere of radius ρ centered at the origin,and a fixed value of θ corresponds to a half-plane starting from the z-axis. The half-planewill cut a sphere of radius ρ in a half-circle, as illustrated in the left plot.

2. Right plot: The z-coordinate of P is z = ρ cos(φ), and

r =√x2 + y2 = ρ sin(φ)

is the distance from the point (x, y, 0) to the origin, as illustrated in the right plot. Now,

x = r cos(θ) = ρ sin(φ) cos(θ) and y = r sin(θ) = ρ sin(φ) sin(θ).

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Page 42: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Fixed values of φ. As stated above, φ = 0 corresponds to the positive z-axis and φ = πcorresponds to the negative z-axis. In addition,

φ = π/2 corresonds to the xy-plane (where z = 0).

Given any other fixed value of φ,

tan(φ) =r

zor z = cot(φ)r = cot(φ)

√x2 + y2

is a cone with vertex at the origin.

Exercise. In each case, find a simplified expression for the cone in x, y and z only.

(a) φ = π/6 (b) φ = 3π/4

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Page 43: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise: Describing subsets of a sphere. Consider the sphere of radius 3 centered atthe origin (ρ = 3). In each case, give bounds on θ and φ.

(a) The subset with x ≥ 0 corresponds to

(b) The subset with y ≥ 0 corresponds to

(c) The subset with z ≥ 0 corresponds to

(d) The subset with |z| ≤ 1 corresponds to

43

Page 44: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.9 Triple Integrals in Spherical Coordinates

Many triple integral problems can be simplified by using spherical coordinates:

x = ρ sin(φ) cos(θ), y = ρ sin(φ) sin(θ) and z = ρ cos(φ).

Spherical boxes. A spherical box is a set of the form

B = {(ρ, θ, φ) : a ≤ ρ ≤ b, α ≤ θ ≤ β, γ ≤ φ ≤ δ}.

An example where all angles are between 0 and π/2 isshown to the right.

If 4ρ = (b − a), 4θ = (β − α), and 4φ = (δ − γ) aresmall enough, then the spherical box is approximately arectangular box with approximate volume

V (B) ≈ (Length)(Width)(Height)

= (4ρ)(r∗ 4θ)(ρ∗ 4φ) where ρ∗ = (a+ b)/2 and φ∗ = (γ + δ)/2

= (ρ∗)2 sin(φ∗)(∆ρ)(∆θ)(∆φ) where r∗ = ρ∗ sin(φ∗)

Triple integrals over spherical boxes. Let f be a continuous function of 3 variables whosedomain contains the spherical box B. The triple integral of f over B can be computed usingthe following iterated integral:∫∫∫

Bf dV =

∫ δ

γ

∫ β

α

∫ b

af(ρ cos(θ) sin(φ), ρ sin(θ) sin(φ), ρ cos(θ)) ρ2 sin(φ) dρ dθ dφ.

Notes: When evaluating triple integrals in spherical coordinates, we let

x = ρ sin(φ) cos(θ), y = ρ sin(φ) sin(θ), z = ρ cos(φ), dV = ρ2 sin(φ) dρ dθ dφ.

The method is useful in situations where trig substitution would simplify computations.

Triple integrals in spherical coordinates are defined as limits of Riemann sums, following thepatterns of definitions we have seen in earlier sections. In addition, triple integrals in sphericalcoordinates can be defined over more general regions.

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Page 45: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let E be the region bounded by

4 ≤ x2 + y2 + z2 ≤ 9, y ≥ 0, z ≥ 0.

Evaluate ∫∫∫Ex2 dV.

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Page 46: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let E be the region bounded by

x2 + y2 + z2 ≤ 25 and z ≥√

3x2 + 3y2.

Evaluate ∫∫∫E

(x2 + y2 + z2) dV.

46

Page 47: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let E be the region bounded by

x2 + y2 + (z − 1)2 ≤ 1 and z ≥√x2 + y2.

Evaluate ∫∫∫Ez dV.

47

Page 48: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.10 Determinants, Areas and Volumes Revisited

Areas of parallelograms. Recall that the area of a par-allelogram in the plane can be computed as the absolutevalue of the determinant of the matrix

A =

[a1 a2b1 b2

]whose rows are the coordinates of the vectors a and b asshown in the plot to the right.

Exercise. Find the area of the parallelogram whose corners are (1, 1), (2, 5), (8, 7), (7, 3) asyou go around in the order given.

48

Page 49: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Polar rectangles. Let

x = r cos(θ), y = r sin(θ),

and let R be the polar rectangle whose

1. lower corner is (r, θ) and

2. upper corner is (r +4r, θ +4θ),

as illustrated in the plot to the right, where R lies entirely inthe first quadrant, 4r > 0 and 4θ > 0. �

If 4r and 4θ are small enough, then the area of the polar rectangle R is close to the area ofthe parallelogram with adjacent vectors (4r)a and (4θ)b, where

a = 〈xr, yr〉, b = 〈xθ, yθ〉.

Note: The determinant of the matrix whose rows are the vectors a and b is often written as

∂(x, y)

∂(r, θ)=

∣∣∣∣ xr yrxθ yθ

∣∣∣∣ ,and is called the Jacobian of the polar transformation. The area of the parallelogram thatapproximates R is the absolute value of the Jacobian times the product of 4r and 4θ.

Exercise. Demonstrate that the area of the parallelogram described above is r(4r)(4θ).

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Page 50: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Volumes of parallelepipeds. Recall that the volume ofa parallelepiped in 3-space can be computed as the abso-lute value of the determinant of the matrix

A =

a1 a2 a3b1 b2 b3c1 c2 c3

whose rows are the coordinates of the vectors a, b and c as shown in the plot above.

Spherical boxes. Let

x = ρ sin(φ) cos(θ), y = ρ sin(φ) sin(θ), z = ρ cos(φ)

and let B be the spherical box whose

1. lower corner is (ρ, θ, φ) and

2. upper corner is (ρ+4ρ, θ +4θ, φ+4φ),

as illustrated in the plot to the right, where B lies entirely inthe first octant.

If 4ρ, 4θ and 4φ are small enough, then the volume of B is close to the volume of theparallelepiped with adjacent vectors (4ρ)a, (4θ)b and (4φ)c, where

a = 〈xρ, yρ, zρ〉, b = 〈xθ, yθ, zθ〉, c = 〈xφ, yφ, zφ〉.

Note: The determinant of the matrix whose rows are the vectors a, b and c is often written as

∂(x, y, z)

∂(ρ, θ, φ)=

∣∣∣∣∣∣xρ yρ zρxθ yθ zθxφ yφ zφ

∣∣∣∣∣∣ ,and is called the Jacobian of the spherical transformation.

The volume of the parallelepiped that approximates B is the absolute value of the Jacobiantimes the product of 4ρ, 4θ and 4φ.

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Page 51: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Demonstrate that the volume of the parallelepiped described above is ρ2 sin(φ)(4ρ)(4θ)(4φ).

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Page 52: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.11 Change of Variables in Double Integrals

Many double integral problems can be simplified by using a substitution method (that is, bychanging variables to a different coordinate system). In this section,

1. f is a continuous function of 2 variables whose domain contains D,

2. x = x(u, v), y = y(u, v) is a parametric description of D,

3. D can be thought of as the image of D∗,

D = {(x, y) = (x(u, v), y(u, v)) : (u, v) ∈ D∗},

where each (x, y) ∈ D is the image of a unique (u, v) ∈ D∗,

4. x(u, v), y(u, v) have continuous partial derivatives,

5. the determinant of the matrix of partial derivatives of x(u, v), y(u, v) is denoted by

∂(x, y)

∂(u, v)=

∣∣∣∣ xu xvyu yv

∣∣∣∣and is called the Jacobian of the transformation.

Then

Change of Variables Theorem. Under the conditions above,∫∫Df(x, y) dA =

∫∫D∗f(x(u, v), y(u, v))

∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ dA∗,where dA = dx dy and dA∗ = du dv are area elements in the two systems.

Notes:

1. Substitutions are made for each component in the double integral of f over D, including

dA =

∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ dA∗.Thus, the absolute value of the Jacobian can be interpreted as a stretching or shrinkingfactor relating the area elements in the two coordinate systems.

2. The parametric description of D must be a one-to-one transformation from D∗ to D.

The theorem remains true, however, as long as each point in the interior of D is theimage of a unique point in the interior of D∗.

3. Substitutions in double integrals are made to simplify f , to simplify D, or both.

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Page 53: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let D be the polygonal region with corners

(0, 0), (1, 1), (2, 0), (1,−1).

Evaluate ∫∫D

sin(x− y) cos(x+ y) dA.

� ��

-�

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Page 54: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let D be the polygonal region with corners

(0, 0), (2, 0), (0, 2).

Evaluate ∫∫D

cos

(x− yx+ y

)dA.

� ��

54

Page 55: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let D be the region satisfying

x2 + 4y2 ≤ 4

(or(x

2

)2+ y2 ≤ 1

).

Evaluate ∫∫D

(x2 + y2) dA.

-� -� � ��

-�

55

Page 56: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Exercise. Let D be the region satisfying

1 ≤ xy ≤ 3 andx

2≤ y ≤ 2x,

Evaluate ∫∫Dxy dA.

��� ��� ����

���

���

���

56

Page 57: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

3.12 Change of Variables in Triple Integrals

Many triple integral problems can be simplified by using a substitution method (that is, bychanging variables to a different coordinate system). In this section,

1. f is a continuous function of 3 variables whose domain contains E,

2. x = x(u, v, w), y = y(u, v, w), z = z(u, v, w) is a parametric description of E,

3. E can be thought of as the image of E∗,

E = {(x, y, z) = (x(u, v, w), y(u, v, w), z(u, v, w)) : (u, v, w) ∈ E∗},

where each (x, y, z) ∈ E is the image of a unique (u, v, w) ∈ E∗,

4. x(u, v, w), y(u, v, w), z(u, v, w) have continuous partial derivatives,

5. the determinant of the matrix of partial derivatives of x(u, v, w), y(u, v, w), z(u, v, w) isdenoted by

∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣xu xv xwyu yv ywzu zv zw

∣∣∣∣∣∣and is called the Jacobian of the transformation.

Then

Change of Variables Theorem. Under the conditions above,∫∫Ef(x, y, z) dV =

∫∫E∗f(x(u, v, w), y(u, v, w))

∣∣∣∣ ∂(x, y, z)

∂(u, v, w)

∣∣∣∣ dV ∗,where dV = dx dy dz and dV ∗ = du dv dw are volume elements.

Notes:

1. Substitutions are made for each component in the triple integral of f over E, including

dV =

∣∣∣∣ ∂(x, y, z)

∂(u, v, w)

∣∣∣∣ dV ∗.Thus, the absolute value of the Jacobian can be interpreted as a stretching or shrinkingfactor relating the volume elements in the two coordinate systems.

2. The parametric description of E must be a one-to-one transformation from E∗ to E.

The theorem remains true, however, as long as each point in the interior of E is theimage of a unique point in the interior of E∗.

3. Substitutions in triple integrals are made to simplify f , to simplify E, or both.

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Page 58: Contents · 3.2 Riemann Sums, Double Integrals and Iterated Integrals Double integrals over rectangles. Let fbe a function of 2 variables whose domain contains

Volumes of ellipsoids. A solid ellipsoid centered at the origin can be described as follows:

E = {(x, y, z) : (x/a)2 + (y/b)2 + (z/c)2 ≤ 1},

where a, b, c are positive constants. Alternatively, we can write

E ={

(x, y, z) = (au, bv, cw) : u2 + v2 + w2 ≤ 1}.

Exercise. Use an appropriate change of variables tofind the volume of the solid

4x2 + 9y2 + 36z2 ≤ 36.

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