Multiple Integrals, A Collection of Solved Problems

Multiple Integrals,

A Collection of Solved Problems

Steven Tan

Multiple Integrals, A Collection of Solved Problems

Copyright © 2024 Steven Tan

ISBN 978-1-300-95955-7

All rights reserved. Apart from any fair dealing for the purposes of research or private study or criticism or review, no part of this publication may be reproduced, stored or transmitted in any form or by any means, without the prior permission in writing of the author.

Title

Copyright

Contents

Preface

1. Integration in Two Variables

1.1 Fundamental Concepts and Theorems

Definition 1.1 Double Integral Over a Rectangle

Theorem 1.1 (Continuous Functions are Integrable)

Theorem 1.2 (Properties of the Double Integral)

Theorem 1.3 (Fubini’s Theorem)

1.2 Problems

2. Double Integrals over Nonrectangular Regions

2.1 Fundamental Concepts and Theorems

Definition 2.1 y-Simple Region and x-Simple Region

Theorem 2.2

Theorem 2.3

Theorem 2.4

Definition 2.5 Average Value of f(x,y) on D

Theorem 2.6 Mean Value Theorem for Double Integrals

2.2 Problems

3. Double Integrals in Polar Coordinates

3.1 Fundamental Concepts and Theorems

Definition 3.1 (Simple Polar Region)

Theorem 3.2 (Double Integral in Polar Coordinates)

Theorem 3.3 (Area of a Polar Region)

3.2 Problems

4. Applications of Double Integrals

4.1 Fundamental Concepts and Theorems

Density and Mass

Moments and Centers of Mass

Moment of Inertia

Probability Density Functions

Monte Carlo Method for Calculating Double Integrals

4.2 Problems

5. Surface Area

5.1 Fundamental Concepts and Theorems

Definition 5.1 Surface area for surface with equation of the form z=f(x,y)

Parametric Representation of Surfaces

Representing Surfaces of Revolution Parametrically

Definition 5.2 Surface area of Parametric Surface

5.2 Problems

6. Triple Integrals

6.1 Fundamental Concepts and Theorems

Theorem 6.1 Fubini’s Theorem for Triple Integrals

Definition 6.2 Simple Solid Region

Theorem 6.3

6.2 Problems

7. Triple Integrals in Cylindrical and Spherical Coordinates

7.1 Fundamental Concepts and Theorems

Theorem 7.1 Triple Integrals in Cylindrical Coordinates

Theorem 7.2 Triple Integrals in Spherical Coordinates

7.2 Problems

8. Change of Variables

8.1 Fundamental Concepts and Theorems

Definition 8.1 Linear Transformation

Theorem 8.2 Change of Variables Formula for Double Integrals

Theorem 8.3 Change of Variables Formula for Triple Integrals

8.2 Problems

9. Applications of Triple Integrals

9.1 Fundamental Concepts and Theorems

Center of Mass of the Solid Region

Centroid of the Solid Region

Moment of Inertia of the Solid Region

Average Value of f on the Solid Region

9.2 Problems

Preface

. . . it’s examples, examples, examples that, for me, all mathematics is based on, and I always look for

them. I look for them first, when I begin to study. I keep looking for them, and I cherish them all.

Paul R. Halmos

This book teaches by solving problems. It is intended as a companion to standard textbooks for calculus students in learning multiple integrals. The first part of each section presents the definitions and theorems (without proofs) necessary for problem solving, and sometimes followed by comments or remarks. These definitions and theorems correspond to those given in most calculus textbooks, where all concepts and theorems are followed by explanations and proofs. The second part contains problems and complete solutions solved in such a simple way that the students find no difficulty to understand.

The book contains over 400 solved problems. They can be used as practicing study guides by students and as supplementary teaching sources by instructors. Since the problems have very detailed solutions, they are helpful for under-prepared students.

1. Integration in Two Variables

1.1 Fundamental Concepts and Theorems

Integrals of functions of several variables are called multiple integrals.

The integral of a function of two variables f(x,y) is called a double integral, denoted

∫ ∫ D f ⁡ ( x ) ⁢ dA

Geometrically, it represents the signed volume of the solid region between the graph of z=f(x,y) and a region D in the x y-plane (Figure 1.1). By convention the volume is positive for regions above the x y-plane and negative for regions below it.

Figure 1.1 The double integral of f(x,y) over the region D equals the volume of the solid region between the                           graph of f(x,y) and the x y-plane over D.

In this section, we consider the simple case where the region is a rectangle denoted by R (Figure 1.2), leaving more general regions for Section 2. Thus, the volume of the solid in Figure 1.2 can be found by computing the volume V under the graph of z=f(x,y) over the rectangle

R={ (x,y)|a≤x≤b,c≤y≤d }

    Figure 1.2 The solid region between the graph of f(x,y) and the rectangle R.

To approximate the volume of the solid in Figure 1.2, we divide R into partitions or sub-rectangles R i ⁢ j as shown in Figure 1.3. Let there be (n×m) sub-rectangles, where

1≤i≤n,     1≤j≤m

Denote the area of the sub-rectangle R i ⁢ j by ΔA i ⁢ j , where ΔA i ⁢ j = Δx i ⁢ Δy j , and the point in R i ⁢ j by ( x i , y j ) . The product f ⁡ ( x i , y j ) ⁢ ΔA i ⁢ j is the volume of a rectangular parallelepiped with base area ΔA i ⁢ j and height f ⁡ ( x i , y j ) . Thus, the sum (known as the Riemann sum)

s n , m = ∑ i = 1 n ⁢ ∑ j = 1 m ⁢ f ⁡ ( x i , y j ) ⁢ ΔA i ⁢ j = ∑ i = 1 n ⁢ ∑ j = 1 m ⁢ f ⁡ ( x i , y j ) ⁢ Δx i ⁢ Δy j

is an approximation to the volume V of the solid.

If we set the sub-rectangles smaller, the error of the approximation becomes smaller. Let || R max || denote the maximum size of the sub-rectangles. As || R max || tends to zero (and thus both n and m tend to infinity), the exact volume of the solid will be

V = lim || R max || → 0 ⁢ ∑ i = 1 n ⁢ ∑ j = 1 m ⁢ f ⁡ ( x i , y j ) ⁢ Δx i ⁢ Δy j

            Figure 1.3 The volume of the box is f ⁡ ( x i , y j ) ⁢ ΔA i ⁢ j , where ΔA i ⁢ j = Δx i ⁢ Δy j .

Definition 1.1 Double Integral Over a Rectangle

The double integral of f(x,y) over a rectangle R in the x y-plane (Figure 1.3) is defined as the limit

(1.1)             ∫ ∫ R f ⁡ ( x ) ⁢ dA = lim || R max || → 0 ⁢ ∑ i = 1 n ⁢ ∑ j = 1 m ⁢ f ⁡ ( x i , y j ) ⁢ ΔA i ⁢ j ,

where ( x i , y j ) denote the point in the sub-rectangle R i ⁢ j .

If this limit exists, we say that f(x,y) is integrable over R.

Remarks:

■ The point ( x i , y j ) is called the sample points for 1≤i≤n and 1≤j≤m. We can choose the upper-left, upper-right, lower-left, lower-right vertices, or the midpoints as sample points. In this book, we only consider sub-rectangles of uniform length, known as regular partitions or uniform partitions. In other words, Δx i = Δx and Δy j = Δy , where

Δx = b - a n ,      Δy = d - c m

■ By Definition 1.1, the volume of the solid in Figure 1.2 is

V = ∫ ∫ R f ⁡ ( x , y ) ⁢ dA = lim || R max || → 0 ⁢ ∑ i = 1 n ⁢ ∑ j = 1 m ⁢ f ⁡ ( x i , y j ) ⁢ Δx i ⁢ Δy j

Theorem 1.1 (Continuous Functions are Integrable)

If a function f is continuous on a rectangle R, then f is integrable over R. That is, ∫ ∫ R f dA exists.

Remarks:

■ If f=c is a constant function, then

∫ ∫ R f ⁢ dA = c · Area ⁡ ( R )

Theorem 1.2 (Properties of the Double Integral)

Suppose that f and g are both integrable on a closed rectangle R. Then the following properties hold:

(i) f+g is also integrable on R and

∫ ∫ R ( f + g ) ⁢ dA = ∫ ∫ R f ⁢ dA = ∫ ∫ R g ⁢ dA

(ii) c f is also integrable on R, where c is any real constant, and

∫ ∫ R c ⁢ f ⁢ dA = c ⁢ ∫ ∫ R f ⁢ dA

(iii) if f(x,y)≤g(x,y) for all (x,y)∈R, then

∫ ∫ R f ⁢ dA ≤ ∫ ∫ R g ⁢ dA

(iv) |f| is also integrable on R and

| ∫ ∫ R f ⁢ dA | ≤ ∫ ∫ R | f | dA

Theorem 1.3 (Fubini’s Theorem)

Let f(x,y) be a continuous function on the rectangle R defined by

R={ (x,y)|a≤x≤b,c≤y≤d }

Then the double integral of f over R exists and

(1.2)             ∫ ∫ R f ⁡ ( x , y ) ⁢ dA = ∫ x = a b [ ∫ y = c d f ⁡ ( x , y ) ⁢ dy ] ⁢ dx = ∫ y = c d [ ∫ x = a b f ⁡ ( x , y ) ⁢ dx ] ⁢ dy

Remarks:

■ The integrals on the right side of Eq, (1.2) are called the iterated integrals.

■  Fubini’s Theorem says that double integrals may be evaluated by iterated integrals.

■ Theorem 1.3 gives us a choice, we can find the value of the double integral by integrating f(x,y) first with respect to y with limits y=c and y=d (treating x as a constant), then the result is integrated with respect to x with limits x=a and x=b. Written explicitly

∫ ∫ R f ⁡ ( x , y ) ⁢ dA = ∫ x = a b [ ∫ y = c d f ⁡ ( x , y ) ⁢ dy ] ⁢ dx

Alternatively, we can integrate the iterated integral first with respect to the x variable and then with respect to the y variable. That is,

∫ ∫ R f ⁡ ( x , y ) ⁢ dA = ∫ y = c d [ ∫ x = a b f ⁡ ( x , y ) ⁢ dx ] ⁢ dy

In other words, the order of integration in the iterated integrals does not matter (although in practice, one order of integration is often easier to use than the other).

■ For convenience we omit the square brackets in the iterated integrals and write Eq. (1.2) as

∫ ∫ R f ⁡ ( x , y ) ⁢ dA = ∫ x = a b ∫ y = c d f ⁡ ( x , y ) ⁢ dy ⁢ dx = ∫ y = c d ∫ x = a b f ⁡ ( x , y ) ⁢ dx ⁢ dy

Observe that the iterated integrals are to be evaluated by working from the inside out. We will sometimes refer to

the first integral as the inner integral and its value as the inner integration.

■ Eq. (1.2) implies that for the rectangle R, the two iterated integrals have the same value. They both represent the signed volume of the solid enclosed between the surface f(x,y) and the region R.  The independence of the order of integration

depends strongly on the fact that the region of integration is rectangular.

It follows that the solid in Figure 1.2 has volume V given by

(1.3)             V = ∫ ∫ R f ⁡ ( x , y ) ⁢ dA = ∫ x = a b ∫ y = c d f ⁡ ( x , y ) ⁢ dy ⁢ dx = ∫ y = c d ∫ x = a b f ⁡ ( x , y ) ⁢ dx ⁢ dy

1.2 Problems

In Problems 1.1 - 1.5, compute the Riemann sum s n , m to estimate the double integral of f(x,y) over R using the choice of sample points indicated.

Problem 1.1 f ⁡ ( x , y ) = x 2 ⁢ y ,   R={ (x,y)|1≤x≤3,1≤y≤2.5 },   n=4, m=3, use the upper-right vertices of the sub-rectangles as sample points.

Solution 1.1:

The rectangle region can be written as R=[1,3]×[1,2.5].

Using the equal length partition, we get the dimensions of the sub-rectangles:

Δx = 3 - 1 4 = 1 2 ,      Δy = 2.5 - 1 3 = 1 2

            Figure 1.4 The upper-right vertices ( x i , y j ) as sample points.

The Riemann sum is s 4 , 3 to estimate the double integral of f ⁡ ( x , y ) = x 2 ⁢ y ,

S 4 , 3 = ∑ i = 1 4 ⁢ ∑ j = 1 3 ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j = 1 4 ⁢ ∑ i = 1 4 ⁢ ∑ j = 1 3 ⁢ f ⁡ ( R i ⁢ j )

Using the upper-right vertices of the sub-rectangles as sample points, we get

S 4 , 3 = 1 4 [ f ⁡ ( 3 2 , 3 2 ) + f ⁡ ( 2 , 3 2 ) + f ⁡ ( 5 2 , 3 2 ) + f ⁡ ( 3 , 3 2 )

+ f ⁡ ( 3 2 , 2 ) + f ⁡ ( 2 , 2 ) + f ⁡ ( 5 2 , 2 ) + f ⁡ ( 3 , 2 )

+ f ⁡ ( 3 2 , 5 2 ) + f ⁡ ( 2 , 5 2 ) + f ⁡ ( 5 2 , 5 2 ) + f ⁡ ( 3 , 5 2 ) ]

= 1 4 ⁢ ( 27 8 + 6 + 75 8 + 27 2 + 9 2 + 8 + 25 2 + 18 + 45 8 + 10 + 125 8 + 45 2 )

S 4 , 3 ≈ 32.25

Thus,

∫ ∫ R f ⁡ ( x , y ) ⁢ dA ≈ 32.25

Problem 1.2 f ⁡ ( x , y ) = x 2 ⁢ y ,   R={ (x,y)|1≤x≤3,1≤y≤2.5 },   n=4, m=3, use the midpoints of the sub-rectangles as sample points.

Solution 1.2:

This problem is similar to Problem 1.1. Here, we use the midpoints of the sub-rectangles as sample points.

From Problem 1.1, we obtain the dimensions of the sub-rectangles:

Δx = 3 - 1 4 = 1 2 ,      Δy = 2.5 - 1 3 = 1 2

            Figure 1.5 The midpoints ( x i , y j ) as sample points.

The Riemann sum to estimate the double integral is

s 4 , 3 = ∑ i = 1 4 ⁢ ∑ j = 1 3 ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j = 1 4 ⁢ ∑ i = 1 4 ⁢ ∑ j = 1 3 ⁢ f ⁡ ( R i ⁢ j )

Using midpoints of the sub-rectangles as sample points yields

s 4 , 3 = 1 4 [ f ⁡ ( 5 4 , 5 4 ) + f ⁡ ( 7 4 , 5 4 ) + f ⁡ ( 9 4 , 5 4 ) + f ⁡ ( 11 4 , 5 4 )

+ f ⁡ ( 5 4 , 7 4 ) + f ⁡ ( 7 4 , 7 4 ) + f ⁡ ( 9 4 , 7 4 ) + f ⁡ ( 11 4 , 7 4 )

+ f ⁡ ( 5 4 , 9 4 ) + f ⁡ ( 7 4 , 9 4 ) + f ⁡ ( 9 4 , 9 4 ) + f ⁡ ( 11 4 , 9 4 ) ]

= 1 4 ⁢ ( 125 64 + 245 64 + 405 64 + 605 64 + 175 64 + 343 64 + 567 64 + 847 64 + 225 64 + 441 64 + 729 64 + 1089 64 )

S 4 , 3 ≈ 22.64

Thus,

∫ ∫ R f ⁡ ( x , y ) ⁢ dA ≈ 22.64

As a check, the exact value of the double integral is

∫ ∫ R f ⁡ ( x , y ) ⁢ dA = ∫ x = 1 2 ∫ y = 1 2.5 x 2 ydx dy = 22.75

Comparing to the result in Problem 1.1, the estimation using midpoints as sample points yields a better approximation.

Problem 1.3 f ⁡ ( x , y ) = x 2 + y 2 ,   R=[0,2]×[0,2],   n=4, m=4, use the midpoints as sample points.

Solution 1.3:

We use uniform length partition to find the length of the sides. Each sub-rectangle has sides of length:

Δx = 2 - 0 4 = 1 2 ,      Δy = 2 - 0 4 = 1 2

And the area is ΔA = Δx ⁢ Δy = 1 4 .

The required Riemann sum is

s 4 , 4 = ∑ i = 1 4 ⁢ ∑ j = 1 4 ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j = 1 4 ⁢ ∑ i = 1 4 ⁢ ∑ j = 1 4 ⁢ f ⁡ ( R i ⁢ j ) ,

where f ⁡ ( x , y ) = x 2 + y 2 .

            Figure 1.6 The midpoints ( x i , y j ) as sample points.

Using the midpoints of the sub-rectangles shown in Figure 1.6, we obtain

s 4 , 4 = 1 4 [ f ⁡ ( 1 4 , 1 4 ) + f ⁡ ( 3 4 , 1 4 ) + f ⁡ ( 5 4 , 1 4 ) + f ⁡ ( 7 4 , 1 4 )

+ f ⁡ ( 1 4 , 3 4 ) + f ⁡ ( 3 4 , 3 4 ) + f ⁡ ( 5 4 , 3 4 ) + f ⁡ ( 7 4 , 3 4 )

+ f ⁡ ( 1 4 , 5 4 ) + f ⁡ ( 3 4 , 5 4 ) + f ⁡ ( 5 4 , 5 4 ) + f ⁡ ( 7 4 , 5 4 )

+ f ⁡ ( 1 4 , 7 4 ) + f ⁡ ( 3 4 , 7 4 ) + f ⁡ ( 5 4 , 7 4 ) + f ⁡ ( 7 4 , 7 4 ) ]

= 1 4 ⁢ ( 1 8 + 5 8 + 13 8 + 25 8 + 5 8 + 9 8 + 17 8 + 29 8 + 13 8 + 17 8 + 25 8 + 37 8 + 25 8 + 29 8 + 37 8 + 49 8 )

s 4 , 4 = 10.5

Therefore,

∫ ∫ R f ⁡ ( x , y ) ⁢ dA ≈ 10.5

Problem 1.4 f(x,y)=x+3y,   R=[1,4]×[1,3],   n=3, m=2, use the sample points as shown in Figure 1.7.

            Figure 1.7 The sample points ( x i , y j ) .

Solution 1.4:

As shown in Figure 1.7, the sample points are

( 3 2 , 3 2 ) , (2,1), ( 7 2 , 3 2 ) , (2,3), ( 5 2 , 5 2 ) , (4,3)

Using the regular partition, we get the dimensions of the sub-rectangles:

Δx = 4 - 1 3 = 1 ,      Δy = 3 - 1 2 = 1

The Riemann sum s 3 , 2 is given by

s 3 , 2 = ∑ i = 1 3 ⁢ ∑ j = 1 2 ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j = ∑ i = 1 3 ⁢ ∑ j = 1 2 ⁢ f ⁡ ( R i ⁢ j )

So,

s 3 , 2 = f ⁡ ( 3 2 , 3 2 ) + f ⁡ ( 2 , 1 ) + f ⁡ ( 7 2 , 3 2 ) + f ⁡ ( 2 , 3 ) + f ⁡ ( 5 2 , 5 2 ) + f ⁡ ( 4 , 3 )

=6+5+8+11+10+13

s 3 , 2 = 53

Problem 1.5 f ⁡ ( x , y ) = x y ,   R=[3,7]×[2,4],   n=4, m=4, use the lower-left vertices as sample points.

Solution 1.5:

We find the dimension of the sub-rectangle.

Δx = 7 - 3 4 = 1 ,      Δy = 4 - 2 4 = 1 2

The area of the sub-rectangle is ΔA = Δx ⁢ Δy = 1 2 .

Thus, the Riemann sum is

s 4 , 4 = ∑ i = 1 4 ⁢ ∑ j = 1 4 ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j = 1 2 ⁢ ∑ i = 1 4 ⁢ ∑ j = 1 4 ⁢ f ⁡ ( R i ⁢ j ) ,

where f ⁡ ( x , y ) = x y .

            Figure 1.8 The lower-left vertices ( x i , y j ) as sample points.

Using the lower-left vertices as sample points, we obtain

s 4 , 4 = 1 2 [ f ⁡ ( 3 , 2 ) + f ⁡ ( 4 , 2 ) + f ⁡ ( 5 , 2 ) + f ⁡ ( 6 , 2 )

+ f ⁡ ( 3 , 5 2 ) + f ⁡ ( 4 , 5 2 ) + f ⁡ ( 5 , 5 2 ) + f ⁡ ( 6 , 5 2 )

+f(3,3)+f(4,3)+f(5,3)+f(6,3)

+ f ⁡ ( 3 , 7 2 ) + f ⁡ ( 4 , 7 2 ) + f ⁡ ( 5 , 7 2 ) + f ⁡ ( 6 , 7 2 ) ]

= 1 2 ⁢ ( 3 2 + 2 + 5 2 + 3 + 6 5 + 8 5 + 2 + 12 5 + 1 + 4 3 + 5 3 + 2 + 6 7 + 8 7 + 10 7 + 12 7 )

s 4 , 4 = 13.67

Thus, the approximation is

∫ ∫ R f ⁡ ( x , y ) ⁢ dA ≈ 13.67

Problem 1.6 Approximate the double integral

∫ ∫ R ( x 2 ⁢ y ) ⁢ dA

over R=[2,4]×[3,4] using Riemann sum s 50 , 50 and the midpoints of each sub-rectangle R i ⁢ j of R as sample points.

Solution 1.6:

We find the dimensions of the sub-rectangle by computing

Δx = 4 - 2 50 = 1 25 ,      Δy = 4 - 3 50 = 1 50

The area of the sub-rectangle is ΔA = Δx ⁢ Δy = 1 1250 .

As required, we have n=50 and m=50. Thus, there are 2500 sub-rectangles in total. The Riemann sum becomes

s 50 , 50 = ∑ i = 1 50 ⁢ ∑ j = 1 50 ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j = 1 1250 ⁢ ∑ i = 1 50 ⁢ ∑ j = 1 50 ⁢ f ⁡ ( x i , y j ) ,

where f ⁡ ( x , y ) = x 2 ⁢ y .

Using the midpoints as sample points, we can express their coordinates as

x i = 2 + ( i - 1 2 ) ⁢ Δx ,      y j = 3 + ( j - 1 2 ) ⁢ Δy ,     for   i,j=1,2,···,50

Therefore,

s 50 , 50 = 1 1250 ⁢ ∑ i = 1 50 ⁢ ∑ j = 1 50 ⁢ f ⁡ ( 2 + ( i - 1 2 ) ⁢ Δx , 3 + ( j - 1 2 ) ⁢ Δy )

= 1 1250 ⁢ ∑ i = 1 50 ⁢ ∑ j = 1 50 [ 2 + 1 25 ⁢ ( i - 1 2 ) ] 2 · [ 3 + 1 50 ⁢ ( j - 1 2 ) ]

We use a computer algebra system to compute this summation and obtain

s 50 , 50 = 65.3324

Thus, our estimation is

∫ ∫ R ( x 2 ⁢ y ) ⁢ dx ⁢ dy ≈ 65.3324

As a check, computing the double integral directly using a computer algebra system yields

∫ y = 3 4 ∫ x = 2 4 ( x 2 ⁢ y ) ⁢ dx ⁢ dy ≈ 65.3333

In Problems 1.7 - 1.8, let s n , n be the Riemann sum for the given double integral using the regular partition and the choice of sample points indicated. Use a computing utility to compute s n , n for n=25,50,100.

Problem 1.7 ∫ x = 0 1 ∫ y = 0 1 e x 3 + y 3 ⁢ dy ⁢ dx

Use the lower-left vertex of each sub-rectangle as sample points.

Solution 1.7:

To estimate the double integral, we use the Riemann sum

s n , n = ∑ i = 1 n ⁢ ∑ j = 1 n ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j ,

where f ⁡ ( x , y ) = ⅇ x 3 + y 3 .

The limits of integration implies that the domain is R=[0,1]×[0,1]. Using the lower-left vertices as sample points, the coordinates are represented by

x i = ( i - 1 ) ⁢ Δx ,      y j = ( j - 1 ) ⁢ Δy ,     for   i,j=1,2,···,n

Case 1. n=25

Using the regular partition, dimensions of the sub-rectangles are uniform:

Δx = 1 - 0 25 = 1 25 ,      Δy = 1 - 0 25 = 1 25 ,      Δx ⁢ Δy = 1 625

Thus,

s 25 , 25 = ∑ i = 1 25 ⁢ ∑ j = 1 25 ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j

= 1 625 ⁢ ∑ i = 1 25 ⁢ ∑ j = 1 25 ⁢ f ⁡ ( x i , y j ) = 1 625 ⁢ ∑ i = 1 25 ⁢ ∑ j = 1 25 ⁢ f ⁡ ( ( i - 1 ) ⁢ Δx , ( j - 1 ) ⁢ Δy )

= 1 625 ⁢ ∑ i = 1 25 ⁢ ∑ j = 1 25 ⁢ e [ ( i - 1 ) / 25 ] 3 + [ ( j - 1 ) / 25 ] 3

Using a computer algebra system, we obtain s 25 , 25 ≃ 1.7125 .

Case 2. n=50

In this case

Δx = 1 - 0 50 = 1 50 ,      Δy = 1 - 0 50 = 1 50 ,      Δx ⁢ Δy = 1 2500

The Riemann sum is

s 50 , 50 = 1 2500 ⁢ ∑ i = 1 50 ⁢ ∑ j = 1 50 ⁢ e [ ( i - 1 ) / 50 ] 3 + [ ( j - 1 ) / 50 ] 3

We obtain

s 50 , 50 ≈ 1.75561

Case 3. n=100

In this case,

Δx = 1 - 0 100 = 1 100 ,      Δy = 1 - 0 100 = 1 100 ,      Δx ⁢ Δy = 1 10000

The Riemann sum is

s 100 , 100 = 1 10000 ⁢ ∑ i = 1 100 ⁢ ∑ j = 1 100 ⁢ e [ ( i - 1 ) / 100 ] 3 + [ ( j - 1 ) / 100 ] 3

s 100 , 100 ≈ 1.7779

As a verification, we compute the integral directly using a computer algebra system which gives

∫ x = 0 1 ∫ y = 0 1 e x 3 + y 3 ⁢ dy ⁢ dx ≈ 1.80071

Clearly, our result yields better approximation when n becomes larger. As n tends to infinity, we will get the exact value of the integral.

Problem 1.8 ∫ x = 0 1 ∫ y = 0 1 sin ⁢ ( x ⁢ y ) ⁢ dy ⁢ dx

Use the midpoint of each sub-rectangle as sample points.

Solution 1.8:

By definition, we have the Riemann sum

s n , n = ∑ i = 1 n ⁢ ∑ j = 1 n ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j ,

where f(x,y)=sin (x y).

Observing the limits of integration, we know that the region is R=[0,1]×[0,1]. The midpoint of each sub-rectangle has coordinates given by

x i = ( i - 1 2 ) ⁢ Δx ,      y j = ( j - 1 2 ) ⁢ Δy ,     for   i,j=1,2,···,n

Case 1. n=25

Each side of the sub-rectangle has length

Δx = 1 - 0 25 = 1 25 ,      Δy = 1 - 0 25 = 1 25 ,      Δx ⁢ Δy = 1 625

Thus,

s 25 , 25 = ∑ i = 1 25 ⁢ ∑ j = 1 25 ⁢ f ⁡ ( R i ⁢ j ) ⁢ Δx i ⁢ Δy j

= 1 625 ⁢ ∑ i = 1 25 ⁢ ∑ j = 1 25 ⁢ f ⁡ ( x i , y j ) = 1 625 ⁢ ∑ i = 1 25 ⁢ ∑ j = 1 25 ⁢ f ⁡ ( 1 25 ⁢ ( i - 1 2 ) , 1 25 ⁢ ( j - 1 2 ) )

= 1 625 ⁢ ∑ i = 1 25 ⁢ ∑ j = 1 25 ⁢ sin ⁢ ( 1 625 ⁢ ( i - 1 2 ) ⁢ ( j - 1 2 ) )

We use a computer algebra system to compute this sum and obtain s 25 , 25 ≃ 0.239828 .

Case 2. n=50

In this case

Δx = 1 - 0 50 = 1 50 ,      Δy = 1 - 0 50 = 1 50 ,      Δx ⁢ Δy = 1 2500

Similar to Part (a), the Riemann sum is

s 50 , 50 = 1 2500 ⁢ ∑ i = 1 50 ⁢ ∑ j = 1 50 ⁢ sin ⁢ ( 1 2500 ⁢ ( i - 1 2 ) ⁢ ( j - 1 2 ) )

A computer algebra system gives

s 50 , 50 ≈ 0.239816

Case 3. n=100

In this case,

Δx = 1 - 0 100 = 1 100 ,      Δy = 1 - 0 100 = 1 100 ,      Δx ⁢ Δy = 1 10000

The Riemann sum becomes

s 100 , 100 = 1 10000 ⁢ ∑ i = 1 100 ⁢ ∑ j = 1 100 ⁢ sin ⁢ ( 1 10000 ⁢ ( i - 1 2 ) ⁢ ( j - 1 2 ) )

s 100 , 100 ≈ 0.23981272720245467

As a check, a direct computation of the integral using a computer algebra system gives

∫ x = 0 1 ∫ y = 0 1 sin ⁢ ( x , y ) ⁢ dy ⁢ dx ≈ 0.23981174200056476

Problem 1.9 Show that if f(x,y)=g(x)h(y), where g and h are continuous functions for all real values of x and y, then

∫ a b ∫ c d f ⁡ ( x , y ) ⁢ dy ⁢ dx = [ ∫ a b g ⁡ ( x ) ⁢ dx ] [ ∫ c d h ⁡ ( y ) ⁢ dy ]

Solution 1.9:

Write

∫ a b ∫ c d f ⁡ ( x , y ) ⁢ dy ⁢ dx = ∫ a b ∫ c d g ⁡ ( x ) ⁢ h ⁡ ( y ) ⁢ dy ⁢ dx

We work the integral from the inside out, so

∫ a b ∫ c d f ⁡ ( x , y ) ⁢ dy ⁢ dx = ∫ a b [ ∫ c d g ⁡ ( x ) ⁢ h ⁡ ( y ) ⁢ dy ] ⁢ dx

Since g(x) is constant relative to y, we can write

∫ a b ∫ c d f ⁡ ( x , y ) ⁢ dy ⁢ dx = ∫ a b g ⁡ ( x ) [ ∫ c d h ⁡ ( y ) ⁢ dy ] ⁢ dx

Also, the integral ∫ c d h ⁡ ( y ) ⁢ dy is constant relative to x, therefore

∫ a b ∫ c d f ⁡ ( x , y ) ⁢ dy ⁢ dx = [ ∫ a b g ⁡ ( x ) ⁢ dx ] [ ∫ c d h ⁡ ( y ) ⁢ dy ]

In Problem 10 - 16, evaluate the iterated integral.

Problem 1.10 ∫ 0 2 ∫ 1 3 x ⁢ ( x + y ) ⁢ dy ⁢ dx

Solution 1.10:

We evaluate the inner integral first, that is, the integration with respect to the y variable, and then with respect to the x variable:

∫ 0 2 ∫ 1 3 x ⁢ ( x + y ) ⁢ dy ⁢ dx = ∫ 0 2 [ ∫ 1 3 ( x 2 + x ⁢ y ) ⁢ dy ] ⁢ dx

= ∫ 0 2 [ ( x 2 ⁢ y + 1 2 ⁢ x ⁢ y 2 ) | 1 3 ] ⁢ dx

= ∫ 0 2 [ ( x 2 ⁢ y ) | 1 3 + ( 1 2 ⁢ x ⁢ y 2 ) | 1 3 ] ⁢ dx

= ∫ 0 2 ( 2 ⁢ x 2 + 4 ⁢ x ) ⁢ dx

= ( 2 3 ⁢ x 3 + 2 ⁢ x 2 ) | 0 2 = 40 3

Problem 1.11 ∫ 0 2 ∫ - 1 2 ( x 2 + y 2 ) ⁢ dx ⁢ dy

Solution 1.11:

We evaluate the inner integral first, that is, the integration with respect to the x variable, and then with respect to the y variable:

Write

∫ 0 2 ∫ - 1 2 ( x 2 + y 2 ) ⁢ dx ⁢ dy = ∫ 0 2 [ ∫ - 1 2 ( x 2 + y 2 ) ⁢ dx ] ⁢ dy

= ∫ 0 2 [ ( 1 3 ⁢ x 3 + x ⁢ y 2 ) | - 1 2 ] ⁢ dy

= ∫ 0 2 [ ( 1 3 ⁢ x 3 ) | - 1 2 + ( x ⁢ y 2 ) | - 1 2 ] ⁢ dy

= ∫ 0 2 ( 3 + 3 ⁢ y 2 ) ⁢ dy

= ( 3 ⁢ y + y 3 ) | 0 2 = 14

Problem 1.12 ∫ 0 1 ∫ 0 1 1 1 - x + y ⁢ dy ⁢ dx

Solution 1.12:

The integral is written such that the integration with respect to y variable is executed first. Thus,

∫ 0 1 ∫ 0 1 1 1 - x + y ⁢ dy ⁢ dx = ∫ 0 1 ( ∫ 0 1 1 1 - x + y ⁢ dy ) ⁢ dx

= ∫ 0 1 [ ln ⁡ ( 1 - x + y ) | 0 1 ] ⁢ dx

= ∫ 0 1 [ ln ⁡ ( 2 - x ) - ln ⁡ ( 1 - x ) ] ⁢ dx

Since

∫ ln ⁡ ( a - x ) ⁢ dx = ( x - a ) ⁢ ln ⁢ ( a - x ) - x + C ,     C is a constant.

Therefore,

∫ 0 1 ∫ 0 1 1 1 - x + y ⁢ dy ⁢ dx = [ ( x - 2 ) ⁢ ln ⁢ ( 2 - x ) - x - ( x - 1 ) ⁢ ln ⁢ ( 1 - x ) + x ] | 0 1

= [ ( x - 2 ) ⁢ ln ⁢ ( 2 - x ) - ( x - 1 ) ⁢ ln ⁢ ( 1 - x ) ] | 0 1

=2ln 2

Problem 1.13 ∫ a b ∫ c d x ⁢ y 2 ⁢ dy ⁢ dx

Solution 1.13:

Notice that in this case, we can separate the double integral and write

∫ a b ∫ c d x ⁢ y 2 ⁢ dy ⁢ dx = ∫ a b x ⁢ dx ⁢ ∫ c d y 2 ⁢ dy

Thus,

∫ a b ∫ c d x ⁢ y 2 ⁢ dy ⁢ dx = [ ( 1 2 ⁢ x 2 ) | a b ] [ ( 1 3 ⁢ y 3 ) | c d ]

∫ a b ∫ c d x ⁢ y 2 ⁢ dy ⁢ dx = 1 6 ⁢ ( b 2 - a 2 ) ⁢ ( d 3 - c 3 )

Problem 1.14 ∫ 0 π ∫ 0 π cos ⁢ ( x - y ) ⁢ dy ⁢ dx

Solution 1.14:

We evaluate the iterated integral by working from the inside out. Thus,

∫ 0 π ∫ 0 π cos ⁢ ( x - y ) ⁢ dy ⁢ dx = ∫ 0 π ∫ 0 π [ cos ⁢ ( x - y ) ⁢ dy ] ⁢ dx

= ∫ 0 π [ - sin ⁢ ( x - y ) | 0 π ] ⁢ dx

= ∫ 0 π [ - sin ⁢ ( x - π ) + sin ⁢ x ] ⁢ dx

= [ cos ⁢ ( x - π ) ] | 0 π - ( cos ⁢ x ) | 0 π = 2 + 2

∫ 0 π ∫ 0 π cos ⁢ ( x - y ) ⁢ dy ⁢ dx = 4

Problem 1.15 ∫ 1 2 ∫ 3 4 1 ( x - y ) 2 ⁢ dx ⁢ dy

Solution 1.15:

By Fubini’s Theorem we may write

∫ 1 2 ∫ 3 4 1 ( x - y ) 2 ⁢ dx ⁢ dy = ∫ 1 2 dy ⁢ ∫ 3 4 1 ( x - y ) 2 ⁢ dx

Written this way, we first compute the definite integral (while y is held constant):

∫ 3 4 1 ( x - y ) 2 ⁢ dx = - ( 1 x - y ) | 3 4 = - 1 4 - y + 1 3 - y

Then,

∫ 1 2 ∫ 3 4 1 ( x - y ) 2 ⁢ dx ⁢ dy = ∫ 1 2 ( - 1 4 - y + 1 3 - y ) ⁢ dy

= [ ln ⁢ ( 4 - y ) - ln ⁢ ( 3 - y ) ] | 1 2

= [ ln ⁢ ( 4 - y ) | 1 2 - [ ln ⁢ ( 3 - y ) ] | 1 2

=ln 2-ln 3+ln 2

∫ 1 2 ∫ 3 4 1 ( x - y ) 2 ⁢ dx ⁢ dy = 2 ⁢ ln ⁢ 2 - ln ⁢ 3 = ln ⁢ 4 3

Problem 1.16 ∫ 0 b ∫ 0 a y ⁢ e x ⁢ y ⁢ dx ⁢ dy ,     for a>0,b>0

Solution 1.16:

Write

∫ 0 b ∫ 0 a y ⁢ e x ⁢ y ⁢ dx ⁢ dy = ∫ 0 b dy ⁢ ∫ 0 a y ⁢ e x ⁢ y ⁢ dx

= ∫ 0 b dy ⁢ ( e x ⁢ y ) | 0 a

= ∫ 0 b ( e a ⁢ y - 1 ) ⁢ dy

= ( 1 a ⁢ e a ⁢ y - y ) | 0 b = ( 1 a ⁢ e a ⁢ b - b ) - 1 a

∫ 0 b ∫ 0 a y ⁢ e x ⁢ y ⁢ dx ⁢ dy = 1 a ⁢ ( e a ⁢ b - 1 ) - b

In Problem 17 - 26, evaluate the double integral by converting it into an iterated integral.

Problem 1.17 ∫ ∫ R ln ⁢ ( x ⁢ y ) x ⁢ dA ,     R={ (x,y)|1≤x≤3,1≤y≤2 }

Solution 1.17:

Write the double integral as an iterated integral:

∫ ∫ R ln ⁢ ( x ⁢ y ) x ⁢ dA = ∫ 1 2 ∫ 1 3 ln ⁢ ( x ⁢ y ) x ⁢ dx ⁢ dy = ∫ 1 2 [ ∫ 1 3 ln ⁢ ( x ⁢ y ) x ⁢ dx ] ⁢ dy

Since

∫ ln ⁢ ( x ⁢ y ) x ⁢ dx = 1 2 [ ln ⁢ ( x ⁢ y ) ] 2 + C ,     where C is a constant

we get

∫ ∫ R ln ⁢ ( x ⁢ y ) x ⁢ dA = ∫ 1 2 ( 1 2 [ ln ⁢ ( x ⁢ y ) ] 2 | 1 3 ) ⁢ dy

= ∫ 1 2 ( 1 2 [ ln ⁢ ( 3 ⁢ y ) ] 2 - 1 2 [ ln ⁢ ( y ) ] 2 ) ⁢ dy

= 1 2 ⁢ ∫ 1 2 [ ln ⁢ ( 3 ⁢ y ) ] 2 ⁢ dx - 1 2 ⁢ ∫ 1 2 [ ln ⁢ ( y ) ] 2 ⁢ dy

Using the fact that

∫ [ ln ⁢ ( a ⁢ y ) ] 2 ⁢ dy = x [ ln ⁢ ( a ⁢ y ) ] 2 - 2 ⁢ y ⁢ ln ⁢ ( a ⁢ y ) + 2 ⁢ y ,

the integral becomes

∫ ∫ R ln ⁢ ( x ⁢ y ) x ⁢ dA = 1 2 [ ( y [ ln ⁢ ( 3 ⁢ y ) ] 2 - 2 ⁢ y ⁢ ln ⁢ ( 3 ⁢ y ) + 2 ⁢ y ] | 1 2

- 1 2 [ y [ ln ⁢ ( y ) ] 2 - 2 ⁢ y ⁢ ln ⁢ ( y ) + 2 ⁢ y ] | 1 2

= 1 2 [ 2 ⁢ ( ln ⁢ 6 ) 2 - 4 ⁢ ln ⁢ 6 + 4 - ( ln ⁢ 3 ) 2 + 2 ⁢ ln ⁢ 3 - 2 ]

- 1 2 [ 2 ⁢ ( ln ⁢ 2 ) 2 - 4 ⁢ ln ⁢ 2 + 4 - 2 ]

= ( ln ⁢ 6 ) 2 - 2 ⁢ ln ⁢ 6 - 1 2 ⁢ ( ln ⁢ 3 ) 2 + ln ⁢ 3 - ( ln ⁢ 2 ) 2 + 2 ⁢ ln ⁢ 2

∫ ∫ R ln ⁢ ( x ⁢ y ) x ⁢ dA = ( ln ⁢ 6 ) ⁢ ( ln ⁢ 6 - 2 ) + ( ln ⁢ 3 ) ⁢ ( - 1 2 ⁢ ln ⁢ 3 + 1 ) + ( ln ⁢ 2 ) ⁢ ( - ln ⁢ 2 + 2 )

Problem 1.18 ∫ ∫ R 1 ( x + 3 ⁢ y ) 3 ⁢ dA ,     R={ (x,y)|2≤x≤3,0≤y≤1 }

Solution 1.18:

By Theorem 1.3, the double integral can be evaluated using an iterated integral:

∫ ∫ R 1 ( x + 3 ⁢ y ) 3 ⁢ dA = ∫ 0 1 ∫ 2 3 1 ( x + 3 ⁢ y ) 3 ⁢ dx ⁢ dy = ∫ 2 3 ∫ 0 1 1 ( x + 3 ⁢ y ) 3 ⁢ dy ⁢ dx

We choose to evaluate the following

∫ 0 1 ∫ 2 3 1 ( x + 3 ⁢ y ) 3 ⁢ dx ⁢ dy = ∫ 0 1 [ ∫ 2 3 1 ( x + 3 ⁢ y ) 3 ⁢ dx ] ⁢ dy

= ∫ 0 1 [ - 1 2 ⁢ 1 ( x + 3 ⁢ y ) 2 ] | 2 3 dy

= - 1 2 ⁢ ∫ 0 1 [ 1 ( 3 + 3 ⁢ y ) 2 - 1 ( 2 + 3 ⁢ y ) 2 ] ⁢ dy

= - 1 2 [ ∫ 0 1 1 ( 3 + 3 ⁢ y ) 2 ⁢ dy - ∫ 0 1 1 ( 2 + 3 ⁢ y ) 2 ⁢ dy ]

= - 1 2 [ - 1 3 ⁢ ( 1 3 + 3 ⁢ y ) | 0 1 + 1 3 ⁢ ( 1 2 + 3 ⁢ y ) | 0 1 ]

= - 1 2 [ - 1 3 ⁢ ( 1 6 - 1 3 ) + 1 3 ⁢ ( 1 5 - 1 2 ) ] = 1 45

Problem 1.19 ∫ ∫ R ( e x + y + ln ⁢ x + y 2 ) ⁢ dA ,     R={ (x,y)|0≤x≤1,1≤y≤2 }

Solution 1.19:

We convert the double integral into an iterated integral. By Theorem 1.3, we have

∫ ∫ R ( e x + y + ln ⁢ x + y 2 ) ⁢ dA = ∫ 1 2 ∫ 0 1 ( e x + y + ln ⁢ x + y 2 ) ⁢ dx ⁢ dy = ∫ 0 1 ∫ 1 2 ( e x + y + ln ⁢ x + y 2 ) ⁢ dy ⁢ dx

Evaluate

∫ ∫ R ( e x + y + ln ⁢ x + y 2 ) ⁢ dA = ∫ 0 1 ∫ 1 2 ( e x + y + ln ⁢ x + y 2 ) ⁢ dy ⁢ dx

= ∫ 0 1 ( e x + y + y ⁢ ln ⁢ x + 1 3 ⁢ y 3 ) | 1 2 dx

= ∫ 0 1 ( e x + 2 + 2 ⁢ ln ⁢ x + 8 3 - e x + 1 - ln ⁢ x - 1 3 ) ⁢ dx

= ∫ 0 1 ( e x + 2 - e x + 1 + ln ⁢ x + 7 3 ) ⁢ dx

Since

∫ ln ⁢ x ⁢ dx = x ⁢ ln ⁢ x - x + C ,          where C is constant

we obtain

∫ ∫ R ( e x + y + ln ⁢ x + y 2 ) ⁢ dA = ( e x + 2 - e x + 1 + x ⁢ ln ⁢ x - x + 7 3 ⁢ x ) | 0 1

= e 3 - e 2 + 4 3 - e 2 + e

∫ ∫ R ( e x + y + ln ⁢ x + y 2 ) ⁢ dA = e 3 - 2 ⁢ e 2 + e + 4 3

Problem 1.20 ∫ ∫ R ln ⁢ x x ⁢ y ⁢ dA ,     R={ (x,y)|2≤x≤e,1≤y≤7 }

Solution 1.20:

We convert the double integral using Theorem 1.3 to obtain the iterated integral.

∫ ∫ R ln ⁢ x x ⁢ y ⁢ dA = ∫ 1 7 ∫ 2 e ln ⁢ x x ⁢ y ⁢ dx ⁢ dy = ∫ 2 e ∫ 1 7 ln ⁢ x x ⁢ y ⁢ dy ⁢ dx

Evaluate

∫ ∫ R ln ⁢ x x ⁢ y ⁢ dA = ∫