H C A P T E R
5
Multiple Integration
5.1 INTRODUCTION: AREAS AND VOLUMES 1.
2 0 1 3
x2 C y dy dx D
0
2
yD3
x2 y C y2 /2
yD1
dx D
0
2
3x2 C 9/2 - x2 C 1/2
2
dx
D
0
2
2x2 C 4 dx D 2x3 /3 C 4x
0
D 40/3.
2.
0 1 2
y sin x dy dx D
0
y2 sin x 2
yD2
dx D
yD1 0
2 sin x -
1 sin x 2
dx
D 3.
4 -2 0 1
3 2
sin x dx D -
0
3 cos x 2
4 -2
D
0
3 3 C D 3. 2 2
xey dy dx D D
4
yD1
xey
-2 yD0 4
dx D
x e - 1
dx
x2 e - 1 2
/2
-2
D 8 - 2 e - 1 D6 e - 1 .
4.
/2 0 0 1 xD1 /2
e x cos y dx dy D
0
e x cos y
xD0 /2
dy D
0
e - 1 cos y dy
D e - 1 sin y
0
D e - 1.
5.
2 1 0 1
e xCy C x2 C ln y dx dy D
1
2 0 2 1
1
e x e y C x2 C ln y dx dy D
1
2
exey C
x3 C x ln y 3
xD1
dy
xD0 2 1
D
e - 1 ey C
2
1 C ln y dy D 3
e - 1 ey C
y C y ln y - y 3
1 2 C 2 ln 2 - 1 D e 3 - 2e2 C e - C 2 ln 2. D e - 1 e - e C 3 3 6.
9 1 1 e
ln x xy
dx dy D D
1 2 1 2
9 1 9 1 1
e
ln x xy
dx dy 1 2
treat y as a constant--use substitution
9 1
ln x 2y
2 xDe xD1
dy D
1 2y
dy D
1 ln y 4
9
D
1
ln 9 . 4
223
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224
Chapter 5
Multiple Integration
7. (a) Here we are fixing x and finding the area of the slices: A x D
0 2
x2 C y2 C 2 dy D x2 y C
y3 C 2y 3
2
D 2x2 C 20/3.
0
Now we "add up the areas of these slices": V D
2 -1
A x dx D
2 -1
2x2 C 20/3 dx D
2 3 20 x C x 3 3
2 -1
D
40 16 C 3 3
- -
2 20 - D 26. 3 3
(b) Now we fix y and find the area of the slices: A y D D
2 -1
x2 C y2 C 2 dx D - -
x3 C y2 x C 2x 3
2 -1
8 C 2y2 C 4 3
1 - y2 - 2 D 9 C 3y2 . 3
Adding up the area of these slices: V D
0 2
A y dy D
0
2
2
9 C 3y2 dy D 9y C y3
0
D 26.
8. Here we are calculating:
2 1 0 3
x C 3y C 1 dx dy D
1
2
x2 C 3yx C x 2 15 C 9y dy D 2
3
dy D
0 1
2
9 C 9y C 3 dy 2
2 1
D
1
2
15 9 y C y2 2 2
D 15 C 18 - 15/2 C 9/2 D 21. 9. Here we are calculating
2 -1 0 1
2x2 C y4 sin x dx dy D D
2 -1
y4 2 3 x - cos x 3
2 -1
1
dy D
0
2 -1
2y4 2 C 3 -
dy
2y5 2 y C 3 5
D
4 64 C 3 5
-
2 2 66 - D2 C . 3 5 5
10. This is the volume of the "rectangular box" bounded by the plane z D 2, the xy-plane, and the planes x D 1, x D 3, y D 0, and y D 2. Here we could just calculate the volume of this 2 * 2 * 2 box as 8 without integrating--or V D
0 2 1 3
2 dx dy D
0
2
3
2x dy D
1 0
2
2
4 dy D 4y
0
D 8.
11. This is the volume of the region bounded by the paraboloid z D 16 - x2 - z2 , the xy-plane, and the planes x D 1, x D 3, y D -2, and y D 2. The volume is V D
1 3 2 -2
16 - x2 - y2 dy dx D
1
3
16y - x2 y -
y3 3
2 -2
dx D
1
3
64 - 4x2 -
16 3
dx
D 64x -
16 4 3 x - x 3 3
3
D 192 - 36 - 16 - 64 - 4/3 - 16/3 D 248/3.
1
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Section 5.2
Double Integrals
225
12. This is the volume of the region bounded by z D sin x cos y, the xy-plane, and the planes x D 0, x D , y D -/2, and y D /2. The volume is
/2 /2
V D D2
-/2 0 /2 -/2
sin x cos y dx dy D
/2
-/2
- cos x cos y
0
dy
cos y dy D 2 sin y
-/2
D 4.
13. This is the volume of the region bounded by z D 4 - x2 , the xy-plane, and the planes x D -2, x D 2, y D 0, and y D 5. The volume is V D
0 5 2 -2 5 0
4 - x2 dx dy D
0 5
5
2
4x - x3 /3
-2
dy D
0
5
8 - 8/3 - -8 C 8/3
dy
D
32 32 dy D y 3 3
D 160/3.
0
14. This is the volume of the region bounded by z D |x| sin y, the xy-plane, and the planes x D -2, x D 3, y D 0, and y D 1. The volume is V D
3 -2 0 1
|x| sin y dy dx D
3 -2
-
|x| cos y
1
dx D
0
3 -2
2|x| dx.
At this point we use the definition of absolute value to split this into two quantities: V D 15.
5 2
2 - x dx C -2
0
3 0
x2 2 x dx D -
0 -2
C
x2
3
D
0
4 9 13 C D .
-5 -1
5 - |y| dx dy D
5 -5
5 -5
2
5 - |y| x
xD-1 5 -5
dy |y| dy
D
5 - |y| 3 dy D 150 - 3
0 -5 0 -5
D 150 - 3 D 150 C
-y dy - 3
0
5
y dy 75 75 - D 75. 2 2
3 2 y 2
-
3 2 y 2
5
D 150 -
0
The iterated integral gives the volume of the region bounded by the graph of z D 5 - |y|, the xy-plane, and the planes x D -1, x D 2, y D -5, y D 5. (The solid so described is a rectangular prism.)
b d
16. We have V D
a c
f x, y dy dx. Since 0 ... f x, y ... M, the solid bounded by y D f x, y , the xy-plane,
and the planes x D a, x D b, y D c, y D d sits inside the rectangular block of height M and base bounded by x D a, x D b, y D c, y D d. Hence V ... M b - a d - c 5.2 DOUBLE INTEGRALS Note: you may want to discuss Exercise 1 (b) before assigning it, to get your students in the habit of looking critically at problems before working on them.
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226
Chapter 5
Multiple Integration
1. (a) We are computing
2 -2 0 4-x2
x dy dx D
3
2
4-x2
x y
-2 0
3
dx D
2 -2
2
4x3 - x5 dx D x4 - x6 /6
-2
D 0.
(b) The integrand is an odd function depending only on x and the region is symmetric about the y-axis. The a students encounter this situation when they looked at -a x3 dx in first year calculus.
1 x3
2.
0 0
3 dy dx D
0
1
x3 0
3y
dx D
0
1
3x3 dx D
3 4 x 4
1 0
D
3 . The region over which we are integrating is: 4
y 1 0.8 0.6 0.4 0.2 0.2
2 y2
0.4
0.6
2
0.8
1
x
3.
0 0
y dx dy D
0
2
y2
xy
0
dy D
0
2
y3 dy D
y4 4
D 4. The region over which we are integrating is:
0
y 2 1.5 1 0.5 x
1
2 x2 2 0
2
2
3
4
4.
0 0
y dy dx D
y2 2
x2
dx D
0 0
2
x5 x4 dx D 2 10
D
0
32 16 D . The region over which we are integrating is: 10 5
y 4
3
2
1
x 0.5 1 1.5 2
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Section 5.2
Double Integrals
3 -1
227
5.
xy2 1 dx D 2 -1 x -1 2 x The region over which we are integrating is: xy dy dx D
y
3
2xC1
3
2xC1
3 -1
3x3 C 4x2 C x dx D
1 3 4 4 x2 x C x3 C 2 4 3 2
D
152 . 3
6
4
2
x -1
sin x 0 xD xD0 0
1
2
0
3
6.
0
y cos x dy dx D sin3 x 6
0
y2 cos x 2
sin x
dx D
0
1 2
sin2 x cos x dx D using the substitution u D sin x
1 2
u2 du D
D 0. The region over which we are integrating is:
y
1 0.8 0.6 0.4 0.2 x
0.5
1
1.5
2
2.5
3
Note: After you assign Exercises 7 and 8, together you can probe to see whether students see that they are the same. This is a nice set-up for Section 5.3 where they will learn about interchanging the order of integration.
1 1-x2
7.
0
You can also see that the region over which we are integrating is a half-circle of radius 1 so we have found the volume of the cylinder over this region of height 3. This figure is:
- 1-x2
3 dy dx D
0
1
1-x2
3y
- 1-x2
dx D
0
1
6 1 - x2 dx D using the substitution x D sin t D 3/2.
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228
Chapter 5
Multiple Integration
y 1
0.5
x 0.2 0.6 1
-0.5
-1
8. This is the same as Exercise 7 with the limits of integration reversed. The solution is again 3/2.
1 ex -ex
9.
0
y dy dx D
3 0
1
y4 4
y 3 2 1
ex
dx D
-ex 0
1
0 dx D 0. The region over which we are integrating is:
0.2 -1 -2 -3
0.4
0.6
0.8
1
x
10. For each square in the domain we need to estimate the height of the square and multiply it by the length times the width. For our estimate we will choose the value of the height f cij in the lower right corner of the square in row i column j as our height for the square. The heights are then: 4 4 4 4 4 5 5 5 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8 7 9 9 9 8 8 9 10 10 9 8 10 11 10 9 8 9 10 10 9 8 9 9 9 9 8
Each box has a base of area 25 so the sum of the products of 25 times the heights is 92500. Of course, this answer depends on what point in each box we chose for our estimate--your mileage may vary.
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Section 5.2
Double Integrals
229
11. A quick sketch of the region over which we are integrating helps us set up our double integral.
y 2
1.5
1
0.5
x 0.5 1 1.5 2
2 0 0
2-x
1 - xy dy dx D
0
2
y -
xy2 2
2-x
dx D
0 0
2
2 - 3x C 2x2 - x3 /2 dx
2 x4 3 D 2x - x2 C x3 - 2 3 8
2
D 4 - 6 C 16/3 - 2 D 4/3.
0
12. Again a sketch of the region over which we are integrating helps us set up our double integral. The top bounding curve is y D x and the bottom curve is y D 32x3 .
y 0.5 0.4 0.3 0.2 0.1 x -0.05 0.05 0.1 0.15 0.2 0.25
1/4 0
x
32x3
3xy dy dx D
0
1/4
3xy2 2
x
dx D
32x3 1/4 0
1/4
3 2 x - 1536x7 dx 2
D
3 2 x - 192x8 2
D
0
1 3 5 - D . 128 1024 1024
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230
Chapter 5
Multiple Integration
13. We can easily determine the limits of integration from the sketch and/or by solving for where x C y D 2 intersects the parabola y2 - 2y - x D 0.
y 2 1.5 1 0.5 x
-1 -0.5 -1
1
2
3
2
2-y
-1 y2 -2y
x C y dx dy D
2 -1
x2 C xy 2
2-y
dy D
y2 -2y 2
2 -1
-
y2 y4 C y3 - C 2 dy 2 2
y5 y4 y3 D - C - C 2y 10 4 6
D
-1
99 . 20
14. We see from the sketch that we need to divide the integral into two pieces. For 0 ... x ... 1/9 we see that x ... y ... 3 and for 1/9 ... x ... 1 we see that x ... y ... 1/ x.
y 3 2.5 2 1.5 1 0.5 x
0.2
0.4
0.6
0.8
1/ x
1
6 D
3y dA D
0
1/9 x 1/9 0
3
3y dy dx C
3 1 1/9
1 1/9 x
3y dy dx
1/ x
D D
0
3 2 y 2
dx C
x
3 2 y 2
dx
x 1 1/9
1/9
3 27 - x2 dx C 2 2
3 3 - x2 dx 2x 2
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Section 5.2
Double Integrals
231
D
27 1 x - x3 2 2
1/9
C
0
3 1 ln x - x3 2 2
1 1/9
1 3 3 - ln 1/9 D 1 C ln 27. D - 2 2 2 15. From the sketch below we see that this is a fairly straightforward integral.
y 6
4
2
x -2 -1 -2 1 2
6 D
x - 2y dA D D
2
x2 C2
-2 2x2 -2 2 -2
x - 2y dy dx
x2 C2 2x2 -2
xy - y2
dx D
2 -2
3x4 - x3 - 12x2 C 4x dx
2
D 3x5 /5 - x4 /4 - 4x3 C 2x2
-2
D 192/5 - 64 D -128/5
16. From the sketch below we see that this integral needs to be done in two pieces.
y 3 2.5 2 1.5 1 0.5 0.25 0.5 0.75 1 1.25 1.5 x 1.75
6 D
x C y
2
2
dA D
0
1 x 1 0
3x
x C y
2
2
dy dx C
3x 1
3 x
3/x
x2 C y2 dy dx
3/x
D D
0
x2 y C y3 /3
x
dx C
1 1 3
3
x2 y C y3 /3
x
dx
1
32/3 x dx C
3
9/x3 C 3x - 4x3 /3 dx D 8/3 C 10/3 D 6
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232
Chapter 5
Multiple Integration
17. As in the proof of property 1 in the text, we note that the Riemann sum whose limit is
n
6 R 18. g dA D
cf dA is
i,jD1
cf cij 1Aij D c
n
f cij 1Aij ' c
i,jD1
6 R
f dA.
6 R so
6 R
f C [g - f ] dA which, by property 1, equals
6 R
f dA C
6 R
[g - f ] dA. But g - f 0
[g - f ] dA 0 and so g dA f dA. 6 6 6 R R R 19. Define f C D max f , 0 and f - D max -f , 0 . Note that both f C and f - have only non-negative values. Then f D f C - f - and |f | D f C C f - . Since f ; ... |f | D f C C f - we can see that |f | is Riemann integrable. Also we can use property 2 to conclude that f dA D f C - f - dA D f C dA - |f | dA. f - dA
6 R
6 R 6 R
6 R
6 R
...
f C dA C
6 R
f - dA D
6 R
20. (a) Intuitively, the volume of a figure with constant height should be the area of the base times the height. In this case that is just the area of the base. More formally, by Definition 2.3, 6 D 1 dA D
n all 1xi ,1yj '0 i,jD1
lim
1xi 1yj .
We are assuming that D is an elementary region; let's consider the case of a Type 1 region, then we can rewrite the above sum as
n all 1xi '0 iD1 b
lim
ci - ci
1xi D
a
x - x
dx D the area of D.
The proof is not much different for the other elementary regions. a a2 -x2 a 1 dA D dy dx D 2 a2 - x2 dx. We've seen this above in Exercises (b) We integrate -a - a2 -x2 -a 6 D 7 and 8. Let x D a sin t and integrate to get the desired result. 21. Using Exercise 20, the area is 1 dA D
0 1 x2 x3
6 A
1 dy dx D
0 1
1
x2 - x3 dx
D x3 /3 - x4 /4
0
D 1/3 - 1/4 D 1/12.
22. Again using Exercise 20, the area is
6 A
1 dA D
0
5-2
1-2x-x2 2x
1 dy dx D
0 5-2
5-2
1 - 4x - x2 dx
D x - 2x2 - x3 /3
0 a
D 1/3 10 5 - 22 .
a -a
b2 -b2 x2 /a2 b2 -b2 x2 /a2
a
23. We integrate ab.
-a -
dy dx D 2
-a
b2 -
b2 x2 2b dx D 2 a a
a2 - x2 dx
D
b a2 D a
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Section 5.2
Double Integrals
233
24. (a) For x 0 the curve x3 - x lies below the curve y D ax2 between 0 and their positive point of intersection aC a2 C4 /2 ax3 a C a2 C 4 . So the area is given by dy dx. xD 2 x3 -x 0 (b) The graph of area against a is:
Area 2
1.5
1
0.5
a 0.2 0.4 0.6 0.8 1 1.2 1.4
The area is 1 at a L .995. 25. The graphs of y D x2 - 10 and y D 31 - x - 1 2 intersect at x D -4 and x D 5 with the graph of y D x2 - 10 lying below the graph of y D 31 - x - 1 2 on this interval.
5 31- x-1
2
-4 x2 -10
4x C 2y C 25 dy dx D D
5 -4 5 -4
31- x-1
2
4xy C y2 C 25y
dx
x2 -10
-12x3 - 78x2 C 330x C 1800 dx
5
D -3x4 - 26x3 C 165x2 C 1800x
-4
D 11664.
26. (a) This is special case of the region over which we integrated in Exercise 20 (b). The integral is a
2 4-x2 -2 - 4-x2
x2 - y2 C 5 dy dx.
(b) You can use your favorite computer algebra system. Using Mathematica, enter the command: Integrate[Integrate[x2 - y2 C 5, {y, -Sqrt[4 - x2 ], Sqrt[4 - x2 ]}], {x, -2, 2}] or Integrate[x2 - y2 C 5, {x, -2, 2}, {y, -Sqrt[4 - x2 ], Sqrt[4 - x2 ]}] and get the answer 20. 27. By symmetry we see that the volume is four times the volume of the piece over the first quadrant x, y 0 . In this region |x| D x and |y| D y so the volume is
2 2-x 0
4
0
2 - x - y dy dx D 4
0
2
2-x
2y - xy - y2 /2
0 2
dx D 4
0
2
2 - 2x C x2 /2 dx
D 4 2x - x2 C x3 /6
0
D 16/3.
The results demonstrated in Exercises 28 and 29 are arrived at easily but worth seeing. In Exercise 28 we have the dream situation where the double integral of a product can be split into the product of integrals. We quickly see that this only works in a very special case. In Exercise 29 we examine a function where 6 f dy dx exists but
f dA does not. 6 28. (a) The function h x, y D f x g y satisfies the conditions of Theorem 2.6 (Fubini's theorem) on [a, b] * [c, d]. So: 6 R f x g y dA D
a b d
f x g y dy dx.
c
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
234
Chapter 5
Multiple Integration
For emphasis, we rewrite this last integral with parentheses and, since f x does not depend on y, we have:
b a d c d b d
f x g y dy dx D
a
f x
c
g y dy dx.
But
c
g y dy is constant so we can pull it out of this last integral to get the result:
b d d b
f x
a c
g y dy dx D
c
g y dy
a
f x dx .
(b) If D is an elementary region we can perform the first step above, if D is not an elementary region, there's not much we can do. For example, if D is a Type 1 region, D D { x, y | x ... y ... x , a ... x ... b} then
b x b x
6 R
f x g y dA D
a 2 x 2 0
f x g y dy dx D
a
f x
x 2 0
g y dy dx.
1 0
29. (a) If x is rational, then
2 1 0
f x, y dy D
1 dy D 2. If x is irrational, then
f x, y dy D
0 dy C
2 dy D 2.
1 0 0 2
(b) Using our answer from part (a),
f x, y dy dx D
0
1
2 dx D 2.
(c) If cij has a rational x coordinate, then f cij D 1 and so the Riemann sum will converge to the area of the region, which is 2. (d) In this case f cij D 1 for our points in the region [0, 1] * [0, 1] and f cij D 2 for our points in the region [0, 1] * [1, 2]. In short, the Riemann sums will converge to 1 1 C 2 1 D 3. (e) As we saw in parts (c) and (d), the Riemann sum does not have a well defined limit and so f fails to be integrable on R, even though in part (b) we actually computed the iterated integral. 5.3 CHANGING THE ORDER OF INTEGRATION This is a good section in which to encourage students to explore with a computer system. 1. (a)
2 0 2x x2
2x C 1 dy dx D
0
2
2x C 1 2x - x2 dx D
0
2
-2x3 C 3x2 C 2x dx
D -
x4 C x3 C x2 2
2
D 4.
0
(b) The region of integration is bounded above by y D 2x and below by y D x2 :
y 4
3
2
1
x 0.5 1 1.5 2
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 5.3
Changing The Order of Integration
235
(c)
4 0 y
2x C 1 dx dy D
y/2 0
4
y
x C x
2 y/2
dy D
0
4
-
y y2 C C y dy 4 2
D -
y3 y2 2y3/2 C C 12 4 3
4
D 4.
0
Note: In Exercises 29, most students will find the biggest challenge in reversing the order of integration (the topic of this section). You may want to suggest that they reverse the order of integration in all of the exercises, but that they evaluate both iterated integrals only in Exercises 79. 2. The region of integration is:
y 1 0.8 0.6 0.4 0.2 x
0.2
0.4
0.6
0.8
1
1 0 1 0 y 0
x
2 - x - y dy dx D
0
1
2x - 3x2 /2 dx D 1/2
and
1
2 - x - y dx dy D
0
1
3 2 y - 2y C 1 dy D 1/2. 2
3. The region of integration is:
y 4
3
2
1
0.5
1
1.5
2
x
2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
236
Chapter 5
Multiple Integration
2 0 4 0 0 0
4-2x
y dy dx D
0
2
2x2 - 8x C 8 dx D 16/3 and -y2 /2 C 2y dy D 16/3.
2-y/2
y dx dy D
0
4
4. The region of integration is:
y 2
1.5
1
0.5
1
2
3
4
x
2 0 4 0
4-y2 0 0
x dx dy D
0
2
4 - y2 2
2
dy D 128/15 and
4-x
x dy dx D
0
4
x 4 - x dx D 128/15.
5. The region of integration is:
y 8
6
4
2
x 0.5 1 1.5 2 2.5 3
9 0 3 0 0
3 y
x C y dx dy D
0
9
1 -2y3/2 C 5y C 9 dy D 891/20 and 2 x4 /2 C x3 dx D 891/20.
x2
x C y dy dx D
0
9
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Section 5.3
Changing The Order of Integration
237
6. The region of integration is:
y 20
15
10
5
x 0.5 1 1.5 2 2.5 3
3 0 e3 1 1
ex
2 dy dx D
0 3 ln y
3
2ex - 2 dx D 2e3 - 8
and
2 dx dy D
0
3
6 - 2 ln y dy D 2e3 - 8.
7. The region of integration is:
y 1 0.8 0.6 0.4 0.2 x 0.5 1 1.5 2
1 0 1 0 x y 2 1
2y
e x dx dy D
0
1
e 2y - e y dy D xex /2 dy C
1 2
1 2 e - 2e C 1 2 e x - xex /2 dy D
and 1 e2 C - e. 2 2
e x dy dx C
x/2
1 x/2
e x dy dx D
0
1
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238
Chapter 5
Multiple Integration
8. The region of integration is:
y 1 0.8 0.6 0.4 0.2 x 0.25
/2 0 1 0 0 0 cos-1 y cos x
0.5
0.75
1
/2
1.25
1.5
sin x dy dx D
0
cos x sin x dx D 1/2 and
1 0
sin x dx dy D
1 - y dy D 1/2.
9. The region of integration is:
y 2
1.5
1
0.5
x -2
2 0 2 -2 0
-1
1
2 0
2
-
4-y2 4-y2 4-x2
y dx dy D y dy dx D
2y 4 - y2 dy D 16/3 -x2 /2 C 2 dx D 16/3.
and
2 -2
10. The limits of integration describe a region D bounded on the top by the line y D -x and on the bottom by the parabola y D x2 - 2, as shown in the figure.
y 2 1 x
-2
-1.5
-1
-0.5 -1 -2
0.5
1
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Section 5.3
Changing The Order of Integration
239
To reverse the order of integration we must divide D into two regions by the line y D -1. Then the original integral is equivalent to the sum
-1 -2 yC2 - yC2
x - y dx dy C
2
-y
-1 - yC2
x - y dx dy
The first of these integrals is
-1 -2 yC2 - yC2
x - y dx dy D
-1 -2
-2y y C 2 dy
1 0
D
0
1
-2 u - 2
u du D -2
1
u3/2 - 2u1/2 du 2 4 28 - D . 5 3 15
D -2 The second integral is
2 -y
2 5/2 4 u - u3/2 5 3
D -2
0
-1 - yC2
x - y dx dy D
2
2 -1
1 2 1 y - y C 2 C y2 - y y C 2 dy 2 2
2 -1
D D
1 3 1 y - y2 - y 2 4 3 - 4
-1
-
4
y y C 2 dy D
3 - 4
4 1
u - 2
u du
2 5/2 4 u - u3/2 5 3
D
1
64 32 2 4 3 - C C - 4 5 3 5 3
139 D- . 60 28 139 9 - D- . 15 60 20 11. The limits of integration describe a region D bounded on the left by x D y - 4 and on the right by the parabola x D 4y - y2 . Thus the final answer is
y 4 3 2 1 x
-4
-2 -1
2
4
To reverse the order of integration, divide D into two regions by the line x D 0 (the y-axis). The original integral is equivalent to
0 xC4 -5 2- 4-x
y C 1 dy dx C
0 2
4
2C 4-x 2- 4-x
y C 1 dy dx
2
D
0 -5
1 x C 4 2 C
0 4
C x C 4 -
2
1 2 - 4 - x 2
- 2 C
4 - x dx
2
1 2 C 4 - x 2
C 2 C
4 - x -
1 2 - 4 - x 2
- 2 -
4 - x
dx
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240
Chapter 5
Multiple Integration
4 0
D
11 2 1 x C x2 dx C 6 C 3 4 - x C 2 2 -5 625 241 C 32 D . D 12 12
0
6 4 - x dx
12. The limits of integration of the first integral describe the triangular region D1 bounded on top by y D x:
y 1
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1
x
The limits of integration of the second integral describe the triangular region D2 bounded by y D 2 - x:
y 1 0.8 0.6 0.4 0.2 0.25 0.5 0.75 1 1.25 1.5 1.75 2 x
Taken together, we obtain the triangular region D below
y 1 0.8 0.6 0.4 0.2 x 0.25 0.5 0.75 1 1.25 1.5 1.75 2
Reversing the order of integration, we find that the sum of the integrals equals
1 0 y 2-y
sin x dx dy D
0
1
- cos 2 - y C cos y dy
1
D sin 2 - y C sin y
0
D sin 1 C sin 1 - sin 2
D 2 sin 1 - sin 2.
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Section 5.3
Changing The Order of Integration
241
13. The limits of integration of the first integral describe the region D1 bounded on the left by the x-axis, on the right by x D y/3 (or, equivalently, by y D 3x2 ) and on top by x D 8.
y 8
6
4
2
0.25
0.5
0.75
1
1.25
1.5
x
The limits of integration of the second integral describe the region D2 bounded on the bottom by y D 8, on the left by x D y - 8 (which is equivalent to y D x2 C 8), and on the right by x D -y/3.
y 12 10 8 6 4 2 0.25 0.5 0.75 1 1.25 1.5 1.75 2 x
Together, D1 and D2 give the full region D of integration.
y 12 10 8 6 4 2 0.25 0.5 0.75 1 1.25 1.5 1.75 2 x
When we reverse the order of integration, the sum of integrals is equal to
2 0 x2 C8 3x2
y dy dx D
0
2
1 2
2 0
x2 C 8
2
- 9x4 dx
D D
1 2
-8x4 C 16x2 C 64 dx
1 256 128 896 - C C 128 D . 2 5 3 15
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242
Chapter 5
Multiple Integration
14. We reverse the order of integration:
1 0 3 3y
cos x2 dx dy D
0
3 0 3 0
x/3
cos x2 dy dx D
0
3
x/3
y cos x2
0
dx
D 15. We reverse the order of integration:
1 0 y 1
1 3
x cos x2 dx D
sin x2 6
3
D
0
sin 9 . 6
x2 sin xy dx dy D
0
1 0 1 0
x
x2 sin xy dy dx D
0
1
x
-x cos xy
0
dx
1
D 16. We reverse the order of integration:
0 y
x - x cos x2 dx D
1 2 x - sin x2 2
D
0
1 1 - sin 1 . 2
sin x dx dy D x
0 0
x
sin x dy dx D x
0
y sin x x
x
dx D
0 0
sin x dx D - cos x
0
D 2.
17. We reverse the order of integration:
3 0 0 9-x2
xe3y dy dx D 9 - y D
9 0 9 0 0
9-y
xe3y dx dy D 9 - y
9
9 0
x2 e 3y 2 9 - y
9-y
dy
0
e 3y /2 dx D e 3y /6
0
D
e 27 - 1 . 6
18. We reverse the order of integration:
2 0 1 y/2
e -x dy dx D
2
1 0 0 1 0
2x
e -x dy dx D
2
1 0
e -x y
2 2
2x
dx
0
D
2xe-x
2
dx D -e -x
1
D1 -
0
1 . e
Note: It's kind of interesting to see, in Exercises 1921, that order of integration matters to us and to computer algebra systems. 19. (a) After churning for a while the program returned a sum of terms that included Bessel functions, Gamma functions and other non-trivial and non-elightening results. (b) You would use integration by parts twice and then substitute back in to eliminate the integral.
1 2y 0
(c) In a blink of an eye you get
0
y2 cos xy dx dy D 1/4 1 - cos 2 .
20. (a) Again, the program thought for a while and warned that inverse functions were being used and that values could be lost for multivalued inverses. This time, however, it did come up with the correct answer of 1/4 1 - cos 81 .
9 y
(b) The calculation
0 0
x sin y2 dx dy resulted in the same answer, but the solution came much more quickly.
/2 sin x 0
21. (a) The software did nothing more than typeset the integral and leave it unevaluated. (b) This time Mathematica quickly calculated the integral
0
ecos x dy dx D e - 1.
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Section 5.4
Triple Integrals
243
5.4 TRIPLE INTEGRALS In Exercises 13, use Theorem 4.5, Fubini's Theorem, to integrate in the most convenient order. Exercise 4 asks the students to reconsider what happened in Exercise 1. Exercise 3 is a nice opportunity to look back at a result from Section 5.2. 1. If we integrate with respect to x first, the integral simplifies: xyz dV D
1 3 0 3 1 0 2 2 1 -1
l [-1,1] * [0,2] * [1,3]
xyz dx dy dz D
1
3 0
2
x2 yz 2
1
dy dz
-1
D 2. Here order doesn't matter. l * [0,2] * [0,3] [0,1]
0 dy dz D 0.
x2 C y2 C z2 dV D
0
1 0 1 0 0 1 0 0 1 0
2 0 2
3
x2 C y2 C z2 dz dy dx z3 3
3
D D D D
0
x2 z C y 2 z C
dy dx
0
2
3x2 C 3y2 C 9 dy dx
2
3x2 y C y3 C 9y
0
dx
1
6x2 C 26 dx
1
D 2x3 C 26x
0
D 28.
3. You could work this out as Exercise 2, or suggest to your students that they could extend the result they established in Exercise 24 of Section 5.2: 1 xyz
e
l * [1,e] * [1,e] [1,e]
dV D
1 e
1 dx x
3
e 1
1 dy y
e 3
e 1
1 dz z
D
1
1 dx x
D ln x
1
D 13 D 1.
4. This works for the same reason that Exercise 1 simplified. We are integrating an odd function of z on an interval
3
that is symmetric in the z coordinate and so, since 5.
2 -1 1 z2 0 yCz
-3
z dz D 0, the triple integral will also be 0.
3yz2 dx dy dz D D3 D3
2 -1 1 2 -1
z2
yCz
3xyz2
0
dy dz D 3
z2
2 -1 1
z2
y2 z2 C yz3 dy dz
2
y 3 z2 y 2 z3 C 3 2
dz D 3
1 2 -1
-1
z8 z7 z3 z2 C - - 3 2 2 3
dz
z9 z8 z4 z3 C - - 27 16 8 9
D
1539 . 16
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244
Chapter 5
Multiple Integration
3 1 0 z 1 xz 3 1 0 3 1 0 3 1 z z xz
6.
x C 2y C z dy dx dz D D D D
1
xy C y2 C zy
1
dx dz
x2 z C x2 z2 C xz2 - x - z - 1 dx dz x3 z x 3 z2 x 2 z2 x2 C C - - xz - x 3 3 2 2 z5 5z4 3z2 574 C - - z dz D . 3 6 2 9
1 2y yCz z
dz
0
3
7.
0
1
2y 1Cy z
yCz
z dx dz dy D
0
xz
1Cy 1 0 2y 1Cy 1 z
dz dy
1 2y
D D
yz dz dy D
0
yz2 /2
1Cy
dy
y 3y3 5 - y2 - dy D - . 2 2 24 0 8. (a) This is a higher-dimensional analogue of Exercise 20 from Section 5.2. Again the idea would be that if we were in four-dimensional space that a figure of constant height would have volume equal to the volume of the base multiplied by the height. In this case that would be just the volume of the base. Somehow this is a lot less physically appealing or intuitive. By Definition 4.3, l W 1 dA D
n all 1xi ,1yj ,1zk'0 i,j,kD1
lim
1xi 1yj 1zk .
The intuition follows from examining the formula above on the right. This converges to the volume of W . More formally, we are assuming that W is an elementary region; let's consider the case of a Type 1 region, then we can rewrite the sum above as
n all 1xi ,1yj '0 i,jD1
lim
1xi 1yj cij
- cij
D D
n all 1xi '0 iD1 b a ci ci
ci
lim
1xi
ci
y - y
dy
y - y
dy dx D volume of W .
The proof is not much different for the other elementary regions. (b) Work out that the equation of the circle where the two paraboloids intersect is x2 C y2 D 9/2 so 2 2 2 Volume D D D
3/ 2 9/2-x 9-x -y
-3/ 2 - 9/2-x2 3/ 2 9/2-x2 -3/ 2 - 9/2-x2 3/ 2
-3/ 2
x2 Cy2
1 dz dy dx
9 - 2x2 - 2y2 dy dx 9 - x2 dx 2 81 C arcsin 4 2x 3
3/ 2 -3/ 2
12 -
8 2 x 3
D D
2x3 15x 9 - x2 - 2 2 3
81 . 4
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Section 5.4
Triple Integrals
245
9. Of course there are other ways to calculate the volume of the sphere.
a
Volume D D
a2 -x2
-
a2 -x2 -y2 a2 -x2 -y2
a
-a - a2 -x2 a
1 dz dy dx D
2 2
a2 -x2
-a - a2 -x2
2 a2 - x2 - y2 dy dx
a2 -x2
y
-a a
a2
-
x2
-
y2
- a - x
y arcsin 2 - x2 a
a -a
- a2 -x2
dx
D
-a
a2 - x2 dx D a2 x - x3 /3
D
4a3 . 3
10. The students have seen this as the volume of a solid of revolution. We'll orient the cone so that the vertex is down at the origin and the axis is along the z-axis. Then the horizontal cross sections are circles of radius rz/h. This simplifies the following computation:
h rz/h
Volume D
0
rz/h 2 -x2 rz/h 2 -x2
h
-rz/h -
dy dx dz D
0
r2 2 1 z dz D r2 h. h2 3
11.
1 0 2 -2 0 y2
2x - y C z dz dy dx D
0
1
2 -2
2xz - yz C
z2 2
y2
dy dx
0
D
0
1
2 -2
2xy2 - y3 C y4 /2 dy dx
2
D
0
1
y4 y5 2xy3 - C 3 4 10 64 32x C 3 10
1
dx
-2
D
0
1
dx D
0
D 12.
1 1-x2 0 2-x-z -1 - 1-x2
16x2 32 C x 3 5
176 . 15
y dy dz dx D
1
1-x2
D D
-1 - 1-x2 1 1-x2 -1 - 1-x2 1 -1
y2 2
2-x-z
dz dx
0
2 - x - z 2 /2 dz dx
2x3 - 16x2 C 25x C 16 D 1 - x2 12 D
3 9 0 9-y
1 1 - x2 2x2 - 12x C 13 3
9 C arcsin x 4
1 -1
9 . 4
13. Here
symmetric in x (see Exercises 1 and 4).
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-3 x2
8 xyz dz dy dx D 0, because we are integrating an odd function in x over an interval that is
246
Chapter 5
Multiple Integration
14.
3 0 x 3 0
9-y2
z dz dy dx D
0
3 x 3 0 x 3 0
3
z2 2
0
9-y2
dy dx
D D D
0
3
9 - y2 /2 dy dx
3
-y3 C 27y 6
dx
x
3
x3 - 27x C 54 6
dx D
81 . 8
15. Here we are again integrating a polynomial. The only difficulty is in the set up:
1 0 0 2-2x 0 3-3x-3y/2
1 - z2 dz dy dx D
1 . 10
16. Again, the set up and solution are:
2 0 0 4-x2 4 x2 Cy2
3x dz dy dx D
64 . 5
17.
3 0 0 3-x
-
3-x2 /3 3-x2 /3
x C y dz dy dx D
0
3 0 3 0
3-x
2 x C y
3 - x2 /3 dy dx
D
9 - x2
3 - x2 /3 dx 81 3 C arcsin x/3 8
3 0
45x - 2x3 D 3 - x2 /3 8 81 3 D . 16 18.
2
1-x2 /4 1-x2 /4 0
xC2
-2 -
z dz dy dx D D D
2
1-x2 /4 1-x2 /4 2
-2 - 2 -2
x C 2 2 /2 dy dx 1 - x2 /4 dx
2 -2
x C 2
3x3 C 16x2 C 18x - 64 1 - x2 /4 C 5 arcsin x/2 12
D 5. 19.
1
1-y2 /2
2-y2
-1 -
1-y2 /2 4x2 Cy2
dz dx dy D D
1
1-y2 /2 1-y2 /2
-1 - 1 -1
2 - 4x2 - 2y2 dx dy
4 2 2 y - 1 1 - y2 3
dy
D . 2
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Section 5.4
Triple Integrals
247
20.
a a2 -x2 a2 -x2 a
-a - a2 -x2
- a2 -x2
dz dy dx D D D
a2 -x2
-a - a2 -x2 a 2 2 -a
2 a2 - x2 dy dx dx
4 a - x
16a3 . 3
21. The region looks like a wedge of cheese:
The five other forms are:
1 1 0 1-x
-1 y2
f x, y, z dz dx dy D
0
1
x
1-x
- x 0 1 1-x 0 1 1-z 0 1 1-y2
f x, y, z dz dy dx
x
D
0
D
0
- x x - x
f x, y, z dy dz dx
f x, y, z dy dx dz
D D
1-z
f x, y, z dx dz dy
-1 0 y2 1 1-z 1-z 0 - 1-z y2
f x, y, z dx dy dz.
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248
Chapter 5
Multiple Integration
22. The five other forms are:
1 0 0 1 0 x2
f x, y, z dz dx dy D
0
1 0 1 0 0 1 0 0 1 0 0 1 0
1 0 1
x2
f x, y, z dz dy dx
1 1
D D D D 23. The five other forms are:
2 0 0 x 0 y
f x, y, z dx dz dy
z
1
f x, y, z dx dy dz
z 1
x2 0 1 z 0
f x, y, z dy dz dx
1
f x, y, z dy dx dz.
f x, y, z dz dy dx D
0
2 y 2 0 0 2 0 z 2 0 0 2 0 z
2 0 x z 2 z y
y
f x, y, z dz dx dy
x
D D D D
f x, y, z dy dz dx
x
f x, y, z dy dx dz
2
f x, y, z dx dz dy
y 2 y 2
f x, y, z dx dy dz.
24. (a) The solid W is bounded below by the surface z D 5x2 , above by the paraboloid z D 36 - 4x2 - 4y2 , on the left by the xz -plane (i.e., y D 0), and in back by the yz -plane (i.e., x D 0). The solid is shown below.
y 0 1 2 3
30
z
20
10
0 0.5 2 1.5 x
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Section 5.4
Triple Integrals
249
(b) The shadow of the solid in the xy-plane is a quarter of the ellipse 9x2 C 4y2 D 36 (obtained by finding the intersection curve of z D 5x2 and z D 36 - 4x2 - 4y2 .) The shadow looks like:
y 3 2.5 2 1.5 1 0.5 0.5 1 1.5 2 x
Using the shadow region to reverse the order of integration between x and y, we find that the original integral is equivalent to
3
1 3
36-4y2
36-4x2 -4y2
2 dz dx dy.
5x2
0
0
(c) In this case, we need to consider the shadow of W in the xz -plane.
z 35 30 25 20 15 10 5 0.25 0.5 0.75 1 1.25 1.5 1.75 2 x
This region is bounded on the left by x D 0, on the bottom by z D 5x2 , and on top by z D 36 - 4x2 (the 1 36 - 4x2 - z section by y D 0). Now the full solid W is bounded in the y-direction by y D 0 and y D 2 (the latter is just the paraboloid surface). Hence the desired iterated integral is
2 0 36-4x2 5x2 0
1 36-4x2 -z 2
2 dy dz dx.
(d) Here we use the same shadow in the xz -plane as in part (c), only to integrate with respect to x before integrating with respect to z will require dividing the shadow into two regions by the line z D 20. (Equivalently, we are dividing the solid W into two solids by the plane z D 20.) This is why we need a sum of integrals. They are 1 1 1 2 2
20 z/5
2
36-4x -z
2 dy dx dz C
36
2
36-z
2
36-4x -z
2 dy dx dz.
0
0
0
20
0
0
(e) To integrate with respect to x first, we need to divide W in a very different manner. The shadow in the yz plane shows a region bounded on the left by y D 0 and above by z D 36 - 4y2 (the section of the paraboloid by x D 0). However, the curve of intersection of the surfaces z D 5x2 and z D 36 - 4x2 - 4y2 with x 20y2 . It is along this curve that we must divide the yz -shadow and eliminated yields the equation z D 20 - 9 thus the integrals. (Note: This curve is just the shadow of the intersection curve of the two surfaces projected into the yz -plane.)
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250
Chapter 5
Multiple Integration
z 35 30 25 20 15 10 5 0.5 1 1.5 2 2.5 3 y
Thus the desired sum of integrals is 2
3 20-20y /9 0 0 0
z/5
2 dx dz dy C
0
3
36-4y2 20-20y2 /9 0
1 2
36-4y2 -z
2 dx dz dy
25. (a) The solid W is bounded below by the paraboloid z D x2 C 3y2 , above by the surface z D 4 - y2 and in back by the plane y D 0. The solid is shown below.
2 1 x 0 -1 -2 4
3
2
z
1
0 0.25 0.5 y 0.75 1
0
(b) The shadow of W in the xy-plane is half of the region inside the ellipse x2 C 4y2 D 4 (the half with y 0). It may be obtained by finding the intersection curve of z D x2 C 3y2 and z D 4 - y2 and eliminating z. The shadow looks like
y 1 0.8 0.6 0.4 0.2 -2 -1 1 2 x
Using the shadow to reverse the order of integration between x and y, we find that the original integral is equivalent to 2
1 2 1-y2 4-y 0 -2
1-y2
x2 C3y2
x3 C y3 dz dx dy.
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Section 5.4
Triple Integrals
251
(c) We need to consider the shadow of W in the yz -plane.
z 4
3
2
1 y
0.2
0.4
0.6
0.8
1
The region is bounded on the left by y D 0, on the bottom by z D 3y2 (the section by x D 0) and on the top by z D 4 - y2 . The full solid W is bounded in the x-direction by the paraboloid z D x2 C 3y2 , which must be expressed in terms of x as x D ; z - 3y2 . Putting all this information together, we find the desired iterated integral is 2
1 4-y z-3y2 0 3y2 -
z-3y2
x3 C y3 dx dz dy.
(d) Here we use the same shadow in the yz -plane as in part (c), only to integrate with respect to y before integrating with respect to z requires dividing the shadow into two regions by the line z D 3. (Equivalently, we are dividing the solid W by the plane z D 3.) This is why we need a sum of integrals. They are
3 0 0
z/3
-
z-3y2 z-3y2
x C y
3
3
dx dy dz C
3
4 0
4-z
-
z-3y2 z-3y2
x3 C y3 dx dy dz.
(e) To integrate with respect to y first, we need to divide W in a different manner. The shadow in the xz -plane shows a region bounded by z D x2 (the section of the paraboloid by y D 0) and z D 4. However, the curve of intersection of the surfaces z D x2 C 3y2 and z D 4 - y2 with y eliminated yields the equation 2 z D x C 3. It is along this curve that we must divide the xz -shadow and thus the integrals. (Note: This 4 curve is just the shadow of the intersection curve of the two surfaces projected into the xz -plane.)
z 4
3
2
1
x -2 -1 1 2
Thus the desired sum of integrals is
2 -2 x2 x2 /4 C3 0
z-x2 /3
x C y
3
3
dy dz dx C
2 -2
4 x2 /4 C3 0
4-z
x3 C y3 dy dz dx.
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252
Chapter 5
Multiple Integration
5.5 CHANGE OF VARIABLES
1. (a) T u, v D 3 0 0 -1 u v .
(b) In this case we can see by inspection that the transformation stretches by 3 in the horizontal direction and reflects without a stretch in the vertical direction. Therefore the image D D T D where D is the unit square is the rectangle [0, 3] * [-1, 0]. 2. (a) This is similar to the map in Example 4 with a scaling factor of 1/ 2. We can also rewrite 1 2 T u, v D 1 2 1 - 2 1 2
u v
.
This is a rotation matrix (the determinant is 1 so there is no stretching) which rotates the unit square counterclockwise by 45 leaving the vertex at the origin in place. (b) We rewrite 1 1 2 2 u . T u, v D 1 1 v - 2 2 This is a rotation followed by a reflection. You can apply Proposition 5.1 and see where T maps each of the vertices to completely determine the image of the unit square. You will that the vertices (0, 0), (1, 0), (1, see 2, 0 , and 1/ 2, -1/ 2 . 1), and (1, 0) are mapped to 0, 0 , 1/ 2, 1/ 2 , 3. Again, since T has non-zero determinant, we can apply Proposition 5.1 and see where T maps each of the vertices. We conclude that T maps D to the parallelogram whose vertices are: (0, 0), (11, 2), (4, 3), and (15, 5). 4. We are trying to determine the entries a, b, c, and d in the expression: a c b d u v
T u, v D
.
Since T 0, 0 D 0, 0 we know that the motion is not a translation. Also T 0, 5 D 4, 1 so b D 4/5 and d D 1/5. Now T 1, 2 D 1, -1 so a D -3/5 and c D -7/5. We check with the remaining vertex: T -1, 3 D 3, 2 . T u, v D -3/5 -7/5 4/5 1/5 u v .
5. As noted in the text, we have a result for R3 that is analogous to Proposition 5.1, so as in Exercises 3 and 2 (b) we can just compute the images of the vertices of W . We conclude that W maps to the parallelepiped with vertices: (0, 0, 0), (3, 1, 5), -1, -1, 3 , 0, 2, -1 , (2, 0, 8), (3, 3, 4), -1, 1, 2 , and (2, 2, 7). 6. You can see that T u, v D u, uv is not one-one on D by observing that all points of the form (0, v) get mapped to the origin under T. In fact, you can imagine the map by picturing the left vertical side of the unit square being shrunk down to a point at the origin. The image is the triangle:
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Section 5.5
y 1 0.8 0.6 0.4 0.2 x 0.2 0.4 0.6 0.8 1
Change of Variables
253
7. This map should be a happy memory for the students: x, y, z D T , , D sin cos , sin sin , cos is familiar from their work with spherical coordinates. (a) This is the unit ball: D D { x, y, z |x2 C y2 C z2 ... 1}. (b) This is the portion of the unit ball in the first octant: D D { x, y, z |x2 C y2 C z2 ... 1, x, y, z 0}. (c) You can think of this as the region from part (b) with the portion corresponding to 0 ... < 1/2 removed. It is the portion in the first octant of the shell 1/2 unit thick around a sphere of radius 1/2: D D { x, y, z |1/2 ... x2 C y2 C z2 ... 1, x, y, z 0}.
1 y/2 C2
8. (a)
0 y/2
2x - y dx dy D
0
1
x2 - xy
y/2 C2
dy D
0
1
4 dy D 4. A sketch of D is shown below. ' 0, 0 , 2, 0 '
y/2
(b) We again can apply Proposition 5.1 and see that the vertices are mapped: 0, 0 4, 0 , 1/2, 1 ' 0, 1 , and 5/2, 1 ' 4, 1 so D is [0, 4] * [0, 1]. (c) First note that u, v x, y 2 -1 1 D det D 2 so D . 0 1 x, y u, v 2 Then, using the change of variables theorem,
1 0 y/2 C2 y/2
2x - y dx dy D
0
1 0
4
u 1/2 du dv D
0
1
u2 4
4
du D
0 0
1
4 dv D 4.
y 1 0.8 0.6 0.4 0.2 x
0.5
1
1.5
2
2.5
9. First, u, v D det x, y 1 -1 0 2 D 2 so x, y 1 D . u, v 2
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254
Chapter 5
Multiple Integration
2y-x
2
Also we can rewrite x5 2y - x e
2 0 x/2 C1 x/2
D u5 vev , and the transformed region is [0, 2] * [0, 2] so
2 0 0 2
2
x5 2x - y e
2x-y
2
dy dx D
u5 vev 1/2 du dv D
2
16 3
2 0
vev dv D
2
8 4 e - 1 . 3
10. The original region D is sketched below left. The transformation u D x C y and v D x - 2y maps D to the region D sketched below right. x, y / u, v is easily calculated to be -1/3.
y 1 0.8 0.6 0.4 0.2 x 0.2 0.4 0.6 0.8 1
v 1 0.8 0.6 0.4 0.2 u 0.2 0.4 0.6 0.8 1
So, using the change of variables theorem, our integral becomes
1 0 0 u
1 u1/2 3 v1/2
dv du D
0
1
u
2 1/2 1/2 u v 3
du D
0 0
1
2 u2 u du D 3 3
1
D
0
1 . 3
11. Here the problem cries out to you to let u D 2x C y and v D x - y. Once you've made that move you can easily figure that x, y / u, v D -1/3 and that the new region is [1, 4] * [-1, 1]. So the integral is
4 1 1 -1
u2 e v 1/3 dv du D
1 3
4
1
u2 e v
1 -1
du D e - e -1
u3 9
4 1
D 7 e - e -1 .
12. If we sketch the region we get the square:
y 1
0.5
x 0.5 -0.5 1 1.5 2 2.5 3
-1
-1.5
-2
The transformation we use is u D 2x C y - 3 and v D 2y - x C 6 so transformed region is the square:
x, y /
u, v D 1/5 and the
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Section 5.5
v 6
Change of Variables
255
5
4
3
2
1
-3
-2
-1
1
2
u
Our integral is
6 1 2 -3
1 u2 du dv D 5v2 15
6 1
u3 v2
2
dv D
-3
7 3
6 1
6
v-2 dv D
1 7 - 3 v
D
1
35 . 18
Note: In Exercises 1317 the Jacobian for the change of variables is r. Assign Exercise 16 so that your students appreciate the role of the extra r.
1 1-x2 1-x2
13. 14.
0 2 0
-1 - 2 4-x2
3 dy dx D
0
2 0 2 0 3
1
3r dr d D
0 /2
2
3 d D 3. 2
/2
dy dx D
0 2 0
r dr d D
0 2 0 /2 a
2 d D . 243 486 d D . 5 5 re dr d D
/4 r2 /2 -/2 a
3 0
15.
0 a
r4 dr d D
a2 -y2 x2 Cy2
r5 5
d D
0
16.
-a 0 3 x 0
e dy dx
dx dy D
/4
-/2 0
1 r2 e 2
/2
d D
0
-/2
2 1 a2 e - 1 d D ea - 1 /2. 2
17.
2 x2 C y ln 1 C 2 . 0
D
0 0
3 sec
dr d D
0
3 sec d D 3 ln sec C tan |0
/4
D ln 1 C
2 - ln 1 D
18. This is a job for polar coordinates. The given disk has boundary circle with equation x2 C y - 1 polar coordinates this equation becomes r2 cos2 C r sin - 1
2
2
D 1. In
D 1 3 r2 cos2 C r2 sin2 - 2r sin C 1 D 1 3 r2 D 2r sin .
Factoring out r, the boundary circle has equation r D 2 sin . In fact, this circle is completely traced by letting vary from 0 to . Thus the region D inside the disk is given by D D { r, |r ... 2 sin , 0 ... ... }.
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256
Chapter 5
Multiple Integration
Hence 1 6 4 - x2 - y 2 D dA D
0 0 2 sin
1 r dr d 4 - r2
2 sin
D
0
-
1 2 4 - r2 2 cos2 d C
d D -
0 0
4 - 4 sin2 - 2 d
D -2
0
2 d
0
/2
D -2
0
cos d C
/2
- cos d
C 2 C 2
D -2 sin
- sin 0 - sin C sin 2 2
D -4 C 2 D 2 - 4. 19. The region in question looks like
y (-1,1) (1,1)
x
(-1,-1)
(1,-1)
We find
6 D
y2 dA D
6 square
y2 dA -
6 disk
1
y2 dA
1 -1
6 square 6 disk
y2 dA D y2 dA D
1
-1 -1 2 0 0 2 0 1
y2 dx dy D
2y2 dy D
2 0
2 3 y 3
1 -1
D
4 3
r2 sin2 r dr d D
1 2 sin d 4
2
D Thus
1 8
1 - cos 2 d D
1 1 - sin 2 8 2
D
0
. 4
6 D
y2 dA D
4 16 - 3 - D . 3 4 12
20. A sketch of the rose is shown below. One leaf means that 0 ... ... /2. The area of one leaf is
/2 0 0 sin 2
1 r dr d D 2
/2 0
/2
1 sin 2 d D 4 - sin 4 16
2
D
0
. 8
The total area is then four times this, or /2.
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Section 5.5
y 0.75
Change of Variables
257
0.5
0.25
-0.75 -0.5 -0.25 -0.25
0.25
0.5
0.75
x
-0.5
-0.75
21. We sketch the graphs of the cardioid r D 1 - cos and the circle r D 1:
y
1
0.5
-2
-1.5
-1
-0.5
0.5
1
x
-0.5
-1
The two curves intersect when 1 - cos D 1 which is when cos D 0 so the two points of intersection are r, D 1, /2 and 1, 3/2 . The region between the two graphs is where /2 ... ... 3/2. The area is
3/2 /2 1 1-cos 3/2 /2
r dr d D D
cos2 - cos d 2
3/2
1 2 - 8 sin C sin 2 8
D2 C
/2
. 4
22. We want the area "inside" the spiral shown below. The area is
2
2 0 0
3
r dr d D
0
2
3 9 2 d D 3 2 2
D 123 .
0
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258
Chapter 5
Multiple Integration
y 5
-10
-5
5
10
15
x
-5
-10
23. integral The is
/3 0 1
1 r cos r dr d D 2
2
1
sin r
/3
2 0
d D
1 sin 1 2
D
/3
sin 1 . 3
24. Since B is a ball we will use spherical coordinates: dV l x 2 C y 2 C z2 C 3 B D
0 2 0 2 0 0 2 0 0 2
2 sin 2 C 3
d d d
D D
7 -
3 arcsinh 2/ 3 2 d
sin d d
2 7 - 3 arcsinh 2/ 3
D 4 7 - 6 arcsinh 2/ 3 which is the same as the text's solution D 4 7 - 6 ln 2 C 7 C 3 ln 3 . 25. Here we will use cylindrical coordinates: x2 C y2 C 2z2 dV D D D 26. It is natural to use spherical coordinates. dV l x 2 C y 2 C z2 W D
0 2 0 2 0 2 0
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2 -1 0 2 -1 0 2 -1
2 0 2
2
l W
r r2 C 2z2 dr d dz
4z2 C 4 d dz
8z2 C 8 dz D 48.
a
b
sin d d d 1 2 b2 - a2 sin d d
0
D D
b2 - a2 d D 2 b2 - a2 .
Section 5.5
Change of Variables
259
27. Once again we use spherical coordinates. x 2 C y 2 C z2 e x
2 Cy2 Cz2
l W
dV D
0
2 0 2 0 2 0
a 0
b
3 e sin d d d 1 2 [ 1 - a2 e a C b2 - 1 e b ] sin d d
2 2 2 2
2
D D
1 - a2 e a C b 2 - 1 e b
2 2
d
D 2 1 - a2 e a C b2 - 1 e b . 28. We are integrating over a cone with vertex at the origin and base the disk at height 6 with radius 3. We will use cylindrical coordinates. 2 C x2 C y2 dV D
0 3 0 3 0 0 3 0 2 2 6 2r
l W
r 2 C r dz d dr
D D
-2r3 C 2r2 C 12r d dr dr D 63.
2 -2r3 C 2r2 C 12r
You should assign one of Exercises 29 or 30 so that your students see the benefits of using another coordinate system even when it is not explicitly called for. You might want to stress that the symmetries of the problem are what lead you, in this case, to choose cylindrical coordinates. Exercise 31 is fun because students will be tempted to use spherical coordinates--life is much easier if they use cylindrical coordinates. 29. We will use cylindrical coordinates.
1 0 0 1 0 0 1 0 2 2 10-2r 2
l W
dV D D D D
- 10-2r 2
r dz d dr
2r 10 - 2r2 d dr
4r 10 - 2r2 dr
4 5 10 - 8 2 . 3
30. We will again use cylindrical coordinates. dV D
0 2 0 2 0 0 2 0 2 2 0 9-r 2
l W
r dz d dr 9r - r3 d dr
D D
18r - 2r3 dr
D 28.
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260
Chapter 5
Multiple Integration
31. We will again use cylindrical coordinates. 2 C x C y
2 2
l W
dV D
3
5 0 5 3 0
2 0 2
25-z2
2 C r2 r dr d dz d dz
D D
3
5
z4 27z2 725 dz D 656 . 2 - C 4 2 4 5
z4 27z2 725 - C 4 2 4
32. You can draw a million pictures, but the easiest way to visualize this is by taking an apple corer and a potato and cutting in the three orthogonal directions. This will provide you with a model that the students can hold and pass around to aid their discussion. They can easily identify symmetries and cut the model along the coordinate planes to set up the integral. If you do this and look in the first octant, you will see a seam along the line y D x. If we split the integral along this line we will have 1/16 of the desired volume. Using cylindrical coordinates this means that 0 ... ... /4 and the cylinder with axis of symmetry the z-axis gives us that 0 ... r ... a. The hard one to see is z, but because we are only looking at the wedge on one side of D /4 we need only worry about one other cylinder so 0 ... z ... a2 - r2 cos2 . So the volume is /4 a a2 -r 2 cos2 V D 16 r dz dr d D 8a3 2 - 2 .
0 0 0
5.6 APPLICATIONS OF INTEGRATION Exercises 19 concern average value. 1. (a) Let's assume a 30-day month. [f ]avg D D 1 30 1 30
30 0
I x dx D
1 30
30
75 cos
0 30
x C 80 dx 15
1125 x sin C 80x 15
D 2400/30 D 80 cases.
0
(b) Here the 2 cents will be a constant that pulls through the integral so the average holding cost is just 2 cents times the average daily inventory, or $1.60. 2. We will divide the integral by the area: [f ]avg D D 1 2 4 1 82
2 0 0 2 0 4
sin2 x cos2 y dy dx D 1 82
1 82
2 0
sin2 x sin 2y C 2y 4
2
4
dx
0
2 sin2 x dx D
2x - sin 2x 2
D
0
22 1 D . 82 4
3. Again we will divide the integral by the area: [f ]avg D 1 1/2
1 0 1 0 0 1-x
e 2xCy dy dx D 2
0
1
1-x
e 2xCy
0 1
dx D e 2 - 2e C 1.
0
D2
e xC1 - e 2x dx D 2e 1Cx - e 2x
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Section 5.6
Applications of Integration
261
4. Here we are finding the average over a ball of volume 4/3. We'll integrate using cylindrical coordinates because z appears explicitly in the integrand. [f ]avg 1 D 4/3 D 3 4
1 -1 1 -1 0 2 0 1-z2
rez dr d dz D dz D
3 4
1 -1 0
2
ez 1 - z2 2
d dz
e z 1 - z2
3 3 4 D . 4 e e
5. (a) We are told that in the 2 * 2 * 2 cube centered at the origin, T x, y, z D c x2 C y2 C z2 . The average temperature of the cube is [T ]avg D D c 8 c 8
1 1 1
-1 -1 -1 1 -1
x2 C y2 C z2 dx dy dz D c 8 D c. 8
c 8
1
1
-1 -1
2z2 C 2y2 C 2/3 dy dz
4z2 C 8/3 dz D
(b) T x, y, z D c when x2 C y2 C z2 D 1 so the temperature is equal to the average temperature on the surface of the unit sphere. 6. The region looks like
y (-1,1) (1,1)
x
(-1,-1)
(1,-1)
and the area of it is 22 - D 4 - . Hence the average value is 1 1 x2 C y2 dA D x2 C y2 dA - x2 C y2 dA , 4 - 6 4 - 61 62 D D D where D1 denotes the square and D2 the disk. x2 C y2 dA D D x2 C y2 dA D
0 1 1
61 D
-1 -1 1 -1 2 0
x2 C y2 dx dy D
1 -1
1
1 3 x C y2 x 3
1 -1
dy
xD-1
2 C 2y2 dy D 3
1
2 2 y C y3 3 3
2 0
D
4 4 8 C D 3 3 3
62 D
r2 r dr d D
1 d D 4 2
Therefore the average value is 1 4 - 8 16 - 3 3 - 16 - D D L 1.2766. 3 2 24 - 6 6 - 24
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262
Chapter 5
Multiple Integration
7. The volume of W is 8 -
4 24 - 4 xD . Thus the average value is 3 3
3 3 x2 C y2 C z2 dV D x2 C y2 C z2 dV 24 - 4 l 24 - 4 l 1 W W - x2 C y2 C z2 dV l2 W where W1 denotes the cube and W2 the ball. Using Cartesian coordinates to integrate over W1 and spherical coordinates to integrate over W2 , this may be calculated as 3 24 - 4 D
1 1 1
-1 -1 -1
x2 C y2 C z2 dz dy dx -
0
2 0
0
1
4 sin d d d
3 - 30 . 5 - 30
1 y 0 -0.5 -1 1 0.5
0.5 z 0 -0.5 -1 -1 -0.5 0 x 0.5 1
8. We are looking for the average value of the minimum of x and y in the 6 * 6 box. This is 1/36 times the sum of the average value for x in the region where x ... y and the average value for y in the region where y ... x. Because of the symmetry, the average value can be calculated by doubling the result for either region and dividing by 36: [Time]avg D D 2 36
6 0 0 6 x
y dy dx D D 2.
0
1 18
6 0
x2 dx 2
1 x3 18 6
9. This is an extension of Exercise 8. The domain is [0, 6] * [0, 6] * [0, 6]. This time there is six-fold symmetry so we will calculate the average value for z in the region where z ... y ... x and multiply by 6 and then divide by 63 which is the volume of the domain. [Time]avg D D 6 216 1 36
6 0 6 0 0 x 0 6 y
z dz dy dx D
1 36
6 0 0
x
y2 dy dz 2
1 x4 x3 dx D 6 36 24
D 3/2.
0
So with three train lines the average wait is 90 seconds.
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Section 5.6
Applications of Integration
263
Exercises 1024 concern centers of mass. We use the formula:
b
x x dx Center of mass D
a b
x dx
a
and its variants. 10. (a) The curve y D 8 - 2x2 intersects the x-axis at ;2. So
2 -2 0 8-2x2
c dy dx D c
2 -2 2 -2
2
8 - 2x2 dx D c 8x - 2x3 /3 8x - 2x3 dx D c 4x2 - x4 /2
2 -2
-2 2
D 64c/3 D 0 and
2 -2
My D Mx D
2 -2 0 2 -2 0
8-2x2
cx dy dx D c
-2
8-2x2
cy dy dx D c/2
8 - 2x2 2 dx D c
2 5 16 3 x - x C 32x 5 3
D 1024c/15
1024c/15 D 16/5. 64c/3 (b) Again, we see the symmetry with respect to x so x D 0. The following integrals are straightforward so we leave out the details, but So x D 0 and y D
2 8-2x2
yD
-2 0 2 8-2x2 -2 0
3cy2 dy dx D 3cy dy dx
32768c/35 D 32/7. 1024c/5
11. We assume that the plate has uniform density and place it so that the center of the straight border is at the origin and the semicircle is symmetric with respect to the y-axis. Once again this means that x D 0.
a a2 -x2
cy dy dx yD 12. First calculate MD
0 -a 0
a2 c/2
2 2x x2 2x x2 2 0 2x x2
D
2a3 c/3 4a D . 2 c/2 a 3 24 5 28 5 328 35
1 C x C y dy dx D
My D
0
2
[x 1 C x C y ] dy dx D [y 1 C x C y ] dy dx D
Mx D Thus, xD 13. Again we first calculate MD
0 9 0 9 0 0 9 0 0
28/5 D 7/6 24/5
x
and y D
328/35 D 41/21. 24/5
xy dy dx D
243 2 6561 8 1458 7
x
My D Mx D
x2 y dy dx D
x
xy2 dy dx D
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264
Chapter 5
Multiple Integration
so xD
6561/8 D 27/4 and 243/2
yD
1458/7 D 12/7. 243/2
14. We'll take to be 1. A look at the figure below tells us again that x D 0. We'll use polar integrals to calculate y.
y
-1
-0.5
0.5
1
x
-0.5
-1
-1.5
-2
We first calculate M D -5/4 D -5/6. 3/2 15. We first calculate so y D
0
2 0
1-sin
r dr d D
3 and Mx D y dA D 2 6 D
2 0 0
1-sin
r2 sin dr d D
-5 , 4
/3
MD
0 0
4 cos
r dr d D
/3 0 /3 0
3 C
4 3 and
My D Mx D
6 D 6 D
x dA D y dA D
4 cos
7 8 r2 cos dr d D C 3 3 r2 sin dr d D 5,
4 cos 0
0
7 3 C 8 15 and y D . so x D 3 3 C 4 3 3 C 4 16. The region is a slice of pie:
y 2
1.5
1
0.5
x 0.5 1 1.5 2 2.5 3
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Section 5.6
/4
Applications of Integration
265
Total mass M D D
6 D
dA D
0 0
3
/4
4 - r r dr d D
0
2r2 -
1 3 r 3
3 rD0
/4
18 - 9 d D
0 /4
9 4
/4
My D
6 D 63 D 4 2
x dA D
0 0
3
4r2 - r3 cos dr d D
0
36 -
81 cos d 4
Mx D
6 D 63 2 - 2 D 8 My
y dA D
/4 0 0
3
4r2 - r3 sin drd D
0
/4
36 -
81 sin d 4
Thus
63 7 2 4 D D xD M 2 4 2 9 63 2 - 2 4 7 2 - 2 Mx yD D D M 8 9 2
y
17.
1
0.5
0.5 -0.5
1
1.5
2
x
-1
Total mass M D D
6 D
0
dA D
0
2 0
1Ccos
r2 dr d
2
1 1 C cos 3 d 3 1 5 1 C 3 cos C 3 cos2 C cos3 d D 3 3
2 0 0 1Ccos
D
0
2
My D
6 D 7 D 4
x dA D
r3 cos dr d D
0
2
1 1 C cos 4
4
cos d
after some effort!
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266
Chapter 5
Multiple Integration
Mx D D
6 D
y dA D
0
2 0 4
1Ccos
r3 sin dr d 1 4
2 2
2 0
1 1 C cos 4
sin d D -
u4 du D 0
(It's also possible to see this from symmetry.) Thus xD 7 3 21 D , 4 5 20 y D 0.
18. Because the volume of the tetrahedron is 1, we can find the centroid by calculating: xD
0 1 0 1 0 0 1 0 0 2-2x 0 2-2x 0 3-3y/2-3x 2-2x 0 3-3y/2-3x 3-3y/2-3x
x dz dy dx D y dz dy dx D z dz dy dx D
1 4 1 2 3 . 4 and,
yD zD 19. (a) First calculate:
MD Myz D Mxz D Mxy D
2
1
3
-1 -1 3y2 2 1 3
dz dy dx D 12 x dz dy dx D 6 y dz dy dx D 0 z dz dy dx D 108 . 5
-1 -1 3y2 2 1 3
-1 -1 3y2 2 1 3
-1 -1 3y2
This means that x, y, z D 1/2, 0, 9/5 . (b) Next we calculate the center of mass with the given density function. MD Myz D Mxz D Mxy D
2 1 3 3y2 3
-1 -1 2 1
z C x2 dz dy dx D
168 5 129 5
-1 -1 3y2 2 1 3
x z C x2 dz dy dx D
-1 -1 3y2 2 1 3
y z C x2 dz dy dx D 0 z z C x2 dz dy dx D 2376 . 35
-1 -1 3y2
This means that x, y, z D 43/56, 0, 99/49 . Note that in Exercises 2022, the symmetry with respect to the z-axis implies that x D 0 and y D 0. We only explicitly set up all of the integrals in the solution of Exercise 20.
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Section 5.6
Applications of Integration
267
20. First calculate: MD
0 3 0 3 0 0 3 0 0 3 0 0 2
18-r 2
Myz D Mxz D Mxy D This means that x, y, z D 0, 0,
r 2 /3 2 18-r 2 r 2 /3 2 18-r 2 r 2 /3 2 18-r 2 r 2 /3
63 r dz d dr D 36 2 - 2 r cos dz d dr D 0 r sin dz d dr D 0 rz dz d dr D 189 . 4
21 . 2 8 2 - 7 21. As noted above, x D 0 and y D 0. First calculate: MD
0 5/2 0 5/2 0 0 2 2 9-r 2 3r 2 -16 9-r 2 3r 2 -16
rdz d dr D
625 8 10625 96
Mxy D This means that x, y, z D 0, 0, -
rz dz d dr D -
17 . 12 22. As noted above, x D 0 and y D 0. First calculate:
MD
0
5 0 5 0
2
25-r 2
r dz d dr D
Mxy D
0
2
2r
250 - 100 5 3 125 4
25-r 2
rz dz d dr D
2r
This means that x, y, z D 0, 0,
15 8 5 - 2 5
. 1 8 4 3 a3 a D . 3 6
23. By symmetry x D y D z. z is easiest to find. Volume of W is
x 0.25 0 0.5 0.75 1 1
0.75 z 0.5 0.25 0 0 0.25 0.5 y 0.75 1
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268
Chapter 5
Multiple Integration
/2 0 0 /2 0 /2 0 a
Thus zD D 6 6 z dV D 3 a l a3 W
/2 0 0 /2
cos 2 sin d d d 1 d 2
6 a3 3a . D 8
a4 3a cos sin d d D 4 2
24. If we put the bottom of the cylinder in the xy-plane, then x, y, z D h - z 2 .
z
y x
Therefore the total mass is MD D
0
l W
2
dV D
0 a 0 a 0
2 0
a 0
h
h - z 2 r dz dr d
h 3 zD0
1 - h - z 3
r dr d
D
0
2
h3 a2 h3 r dr d D . 3 3
x D y D 0 by symmetry, so we compute Mxy D D
0
l W
2
z dV D
0 a 0 0 h
2 0
a 0
h
z h - z 2 r dz dr d
2 0 0 a
h2 z - 2hz2 C z3 r dz dr d D
r
1 4 h4 a2 h dr d D . 12 12
Thus zD
h4 a2 h 3 D . 12 h3 a2 4
25. (a) By symmetry we can see that the moment of inertia about each of the coordinate axes is the same. Ix D Iy D Iz D
0 1 0 1-x 0 1-x-y
y2 C z2 dz dy dx D
1 . 30
(b) Again, by symmetry we see that the radius of gyration about each of the coordinate axes is the same. We calculate 1 1-x 1-x-y 1 MD dz dy dx D . 6 0 0 0
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Section 5.6
Applications of Integration
269
1/30 D 1/ 5. 1/6 26. By symmetry we can see that the moment of inertia about each of the coordinate axes is the same Then rx D ry D rz D Ix D Iy D Iz D
0 2 0 2 0 2
y 2 C z2
x C y C z C 1 dz dy dx D 96.
Again, by symmetry we see that the radius of gyration about each of the coordinate axes is the same. We calculate MD
0 2 0 2 0 2
x C y C z C 1 dz dy dx D 32.
96 D 3. 32 27. (a) The problem cries out to be solved using cylindrical coordinates. For Iz this means that the x2 C y2 in the integrand is r2 so Then rx D ry D rz D Iz D
0 3 0 3 0 0 2 2 9 r2 9 r2
2zr3 dz d dr D
6561 4
and so
MD
2zr dz d dr D 486,
3 3 rz D . 2 2 (b) This time Iz D
0 3 0 3 0 0 2 2
9 r2 9 r2
r4 dz d dr D r2 dz d dr D
8748 35 324 , 5
and so
MD
3 3 rz D . 7
28. Although it may be tempting to move to spherical coordinates, it is nice to have a z-coordinate so we will stay with cylindrical coordinates. (a)
a
Iz D
0 0 a 0 0
2
a2 -r 2
MD rz D a (b)
a
- a2 -r 2 2 a2 -r 2 - a2 -r 2
r3 c dz d dr D rc dz d dr D
8ca5 15
and
4ca3 3
so
2 . 5
a2 -r 2
Iz D
0 0 a 0 0
2
MD rz D a
- a2 -r 2 2 a2 -r 2 - a2 -r 2
r3 r2 C z2 dz d dr D r r2 C z2 dz d dr D
8a7 21
and
4a5 5
so
10 . 21
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270
Chapter 5
Multiple Integration
(c)
a
Iz D
0 a 0
2
a2 -r 2
MD
0 0
- a2 -r 2 2 a2 -r 2 - a2 -r 2
r5 dz d dr D r3 dz d dr D
32a7 105 8a5 15
and
so
2a rz D . 7 29. Ix D D D y2 dA D 9 -
1 3
6 D
1 -1
-1 x2 C2 3
y2 x2 C 1 dy dx x2 C 1 dx D
1 -1
1 2 x C 2 3
1 19 C 7x2 - 18x4 - 7x6 - x8 dx 3
1 14 36 2 1496 38 C - - 2 - D 3 3 5 9 135
y 3 2.5 2 1.5 1 0.5 x -1 -0.5 0.5 1
30. Iz D D
0
6 * [0,1] [0,2]
2 0 1
x2 C y2 dA D
0
2 0
1
x2 C y 2
2 0
1 C y dy dx 1 1 1 C x2 C dx 3 2 4
x2 C y2 C x2 y C y3 dy dx D
x2 C
2 4 1 31 8 C C D D C 3 3 3 2 6 31. The only adjustment in the formula for Ix is because we are using the square of the distance from the line y D 3 and not the formula given in text which squares the distance from the x-axis. This is a straightforward application of formula (8). IyD3 D
2 4-x2 4-x2 -2 -
x2 3 - y
2
dy dx D
116 . 3
What follows is preliminary work for Exercises 3234. You should probably assign all three together. We will be calculating Gm x, y, z dV . V 0, 0, r D - 2 C y2 C z - r 2 l x W
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Section 5.6
Applications of Integration
271
In this special case, W is the shell bound by spheres centered at the origin of radii a and b where a < b. The volume of W is therefore 4 b3 - a3 /3. The density is assumed to be constant and so the density is mass divided by volume, so M 3M D D . [4 b3 - a3 /3] 4 b3 - a3 So V 0, 0, r D -Gm D- D- D- D- dV l x2 C y 2 C z - r W
2
3GmM dV 4 b3 - a3 l x2 C y2 C z - r W 3GmM 4 b3 - a3 3GmM 4 b3 - a3 3GmM 4 b3 - a3
2 0 2 0 2 0 a a b a b b 0
2
2 sin 2 C r2 - 2r cos
d d d
r r
2 C r2 - 2r cos
D0
d d
| C r| - | - r|
d d.
Our final note before proceeding to Exercises 3234 is that r | C r| - | - r| D 2 22 /r if r, and if < r.
32. See preliminary work above. When r b, then in the range a ... ... b, we have that ... r, so V 0, 0, r D - D- D- D- 3GmM 4 b3 - a3 3GmM 4 b3 - a3 3GmM 4 b3 - a3 GmM 2r
2 0 2 0 2 0 2 0 a b
22 d d r
b
23 3r
d
a
2 b 3 - a3 d 3r GmM . r
d D -
33. See preliminary work above. When r ... a, then in the range a ... ... b, we have that r, so V 0, 0, r D - D- D 3GmM 4 b3 - a3 3GmM 4 b3 - a3
2 0 2 0 a b
2 d d b2 - a2 d
-3GmM b2 - a2 . 2 b 3 - a3
What is striking about this result is that V (0, 0, r) is independent of r. Therefore, since F D -V , we see that there is no gravitational force.
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272
Chapter 5
Multiple Integration
34. (b) Students might consider the connection before they explicitly find it in part (a). If a < r < b, then we have a combination of the two cases dealt with in Exercises 32 and 33. For a ... ... r, we are in a case similar to Exercise 32, and for r ... ... b we are in a case similar to Exercise 33. (a) We must break the integral at D r: V 0, 0, r D - D- D- D- 3GmM 4 b3 - a3 3GmM 4 b3 - a3 3GmM 2 b 3 - a3 GmM 2r b3 - a3
2 0 2 0 a r
22 3GmM d d - r 4 b3 - a3
2 0 r 2 0
b
2 d d b2 - r2 d
2 r 3 - a3 3GmM d - 3r 4 b3 - a3 - 3GmM 2 b 3 - a3 b2 - r 2
2 r 3 - a3 3r
3b2 r - 2a3 - r3 .
5.7 TRUE/FALSE EXERCISES FOR CHAPTER 5 False. (Not all rectangles must have sides parallel to the coordinate axes.) True. True. True. False. (Let f x, y D x, for example.) True. False. (The integral on the right isn't even a number!) True. True. False. (It's a type 1 region.) True. True. False. (The value of the integral is 3.) True. (Use symmetry.) True. False. (The y part of the integrand gives a nonzero value.) True. (The inner integral with respect to z is zero because of symmetry.) False. (The triple integral of y is zero because of symmetry, but not the triple integral of x.) True. False. (The area is 30 square units.) False. (The integrals are opposites of one another.) True. False. (A factor of r should appear in the integrand.) True. False. (A factor of is missing in the integrand.) True. True. 1 28. False. (The centroid is at 0, 0, h .) 4 29. True. 30. True.
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1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27.
Section 5.8
Miscellaneous Exercises for Chapter 5
273
5.8 MISCELLANEOUS EXERCISES FOR CHAPTER 5 1. First let's split the integrand: l B z3 C 2 dV D l B z3 dV C l B 2 dV D l B z3 dV C 2 l B dV .
Here B is the ball of radius 3 centered at the origin. The integral of an odd function of z over a region which is symmetric with respect to z is 0. The other integral is twice the volume of a sphere of radius 3 so l B z3 C 2 dV D 2 4 3 3 D 72. 3
2. As in Exercise 1 we see that our integrand is the sum of odd functions in x and y and a constant which we are integrating over a region which is symmetric with respect to x and y. Our answer will be -3 times the volume of the hemisphere of radius 2. In symbols, V D l W x3 C y - 3 dV D -3 l W dV D -3 1 2 4 3 2 D -16. 3
3. (a) We'll use the bounds given for z in both integrals and just reverse the order of integration for x and y. We are integrating over the ellipse:
y 3
2
1
x -2 -1.5 -1 -0.5 -1 0.5 1 1.5 2
-2
-3
l W
3 dV D D
3
3-y2 /3
9-x2
3 dz dx dy
-3 0 3 9-3x2 0 - 9-3x2 2x2 Cy2 9-x2 2x2 Cy2
and
3 dz dy dx.
(b) Using Mathematica, the result was 81 3 /4 in either order. 4. First follow the hint (noting that x and y are just dummy variables) and write
x y
F x, y D
a
g x , y dx
where g x , y D
c
f x , y dy .
By the fundamental theorem of calculus, F D x x Also, again by the fundamental theorem,
2F y x
g x , y dx D g x, y .
a
y x
D
y
[g x, y ] D
y
f x, y dy D f x, y .
c
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274
Chapter 5
Multiple Integration
x a b y y x
By Fubini's theorem, f x , y dy dx D
c a
f x , y dx dy .
As above, write F x, y D
c
y
h x, y dy
where h x, y
D
a
x
f x , y dx .
Proceeding as above we see that F D h x, y y 5. I think the given form is the easiest to integrate:
2 0 0 1 0 9-r 2
and
2F
x y
D f x, y .
r dz dr d D
0
2 0
1
r 9 - r2 dr d
16 2 D d 9 - 3 0 16 2 D 2 9 - . 3
2
(a) In Cartesian coordinates, z doesn't really change and for the outer two limits, we are integrating over a unit circle so our answer is
1 1-x2 9-x2 -y2 -1 - 1-x2
dz dy dx.
0
(b) The solid is the intersection of the top half of a sphere of radius 3 centered at the origin and a cylinder of radius 1 with axis of symmetry the z-axis. In spherical coordinates this means that we have to split the integral into two pieces: one that corresponds to the spherical cap and one that corresponds to the straight sides. The "cone" of intersection is when D sin-1 1/3. For the integral that corresponds to the "straight sides", 0 ... r ... 1. In spherical coordinates that is 0 ... sin ... 1 or 0 ... ... csc . The integrals are, therefore,
2 0 0 sin-1 1/3 0 3
2 sin d d d C
0
2
/2 sin-1 1/3 0
csc
2 sin d d d.
6. (a) This solid is similar to that in Exercise 5. It is the intersection of a cylinder over the circle of radius 2 with center 0, 2 (i.e., x2 D 4y - y2 ) and the plane x D 0 with caps on either end that are portions of the sphere of radius 4 centered at the origin z D 16 - x2 - y2 . ; (b) In cylindrical coordinates, - 16 - r2 ... z ... 16 - r2 and we are above the first quadrant so 0 ... ... /2. Since x2 C y2 D 4y, r2 D 4r sin so in the first quadrant, r D 4 sin . The volume is therefore
/2
V D
0 0
4 sin
16-r 2
/2
- 16-r 2 /2
r dz dr d D
0 0
4 sin
2r 16 - r2 dr d
D
128 3
1 - cos3 d D
0
64 3 - 4 . 9
Exercises 7 and 8 are a good lesson in the advantage of choosing the right coordinate system in which to work. This simple problem in Cartesian coordinates is a pain using either cylindrical or spherical coordinates. 7. Orient the cube so that a vertex is at the origin and the edges that meet at that vertex lie along the x-, y- and z-axes so that the cube is in the first octant. We'll double the volume of half of the cube. In this case 0 ... z ... a,
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Section 5.8
Miscellaneous Exercises for Chapter 5
275
0 ... ... /4 and the only difficulty is with r. The radius varies from 0 to the line x D a. In cylindrical coordinates x D r cos so r D a sec and our limits for r are 0 ... r ... a sec . The volume is V D2
0 a 0 /4 0 a sec
r dr d dz D
0
a 0
/4
a2 sec2 d dz D
0
a
a2 dz D a3 .
The above calculation wouldn't change much if you followed the hint in the text and placed the center of the cube at the origin. In this case you would have 1/8 of the figure in the first octant and you would be calculating the volume of a cube with sides a/2. 8. We again orient the cube so that a vertex is at the origin and the edges that meet at that vertex lie along the x-, yand z-axes so that the cube is in the first octant. As in Exercise 7 we will double the volume of half of the cube corresponding to 0 ... ... /4. We will have to split the integral into two pieces: the piece in which is bounded by the top of the cube (z D a or D a sec ) and the piece in which is bounded by the side of the cube (x D a or D a csc sec ). The boundary value of depends on . Set a D cos equal to a D sin cos and solve to obtain D cot-1 cos . So the volume is
/4
V D2
0 0
cot-1 cos 0
a sec
/4
/2 cot-1 cos 0
2 sin d d d C 2
0
a csc sec
2 sin d d d
D2
a3 6
C
a3 3
D a3 .
Exercises 917 are examples where a change of variables helps. Exercise 14 depends on Exercise 11 and together they are much less difficult than they may first appear. 9. Here we will let u D x - 2y and v D x C y. We calculate u, v D x, y 1 1 -2 1 D3 so x, y D 1/3. u, v
The three boundary lines x C y D 1, x D 0, and y D 0 correspond to v D 1, 2v D -u, and u D v. We have all of the pieces to assemble our integral: cos x - 2y x C y dA D
0 1 v -2v 1 0
6 D
u 1 cos 3 v
du dv D
0
1
v
u v sin 3 v
du
-2v 1
D
v sin 1 - sin -2 3
dv D
v2 sin 1 C sin 2 6
D
0
1 sin 1 C sin 2 . 6
10. Let u D y3 and v D x C 2y. Then 0 ... u ... 216, 0 ... v ... 1, and u, v D x, y Then
6 0 1-2y -2y
0 1
3y2 2
D -3y2 D -3u2/3
so
x, y 1 D- . u, v 3u2/3
y3 x C 2y 2 e
xC2y
3
dx dy D
0
216 0 216 0 0 216 0
1
u 2 v3 v e 3u2/3 1 1/3 2 v3 u v e 3
1
dv du
D D D
0
1
dv du
1 1/3 v3 u e 9
du
vD0
216
1 1/3 u e - 1 9
du
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276
Chapter 5
Multiple Integration
216
D
1 9
3 4/3 u 4
e - 1
0
D 108 e - 1 .
a b
1-x2 /a2
11. (a) As we've seen before, we can write the integral as
(b) When we scale by letting x D ax and y D by, the ellipse is transformed into the unit circle E in the xyplane. To rewrite the integral we also quickly calculate that x, y / x, y D ab. The transformed integral
1 1-x2
-a -b
dy dx.
1-x2 /a2
is
(c) Because we are integrating over a unit circle, we transform to polar coordinates (o.k., really we do it because the text tells us to):
1 1-x2
-1 - 1-x2
ab dy dx.
-1 - 1-x2
ab dy dx D
0
2 0
1
abr dr d D
0
2
1 1 ab d D ab 2 D ab. 2 2
12. (a) With u D 2x - y, v D x C y, we see u C v D 3x so x D
u C v 2v - u which implies y D . Substituting 3 3 these expressions into the equation for the ellipse, we obtain 13 u C v 3
2
C 14
u C v 3
2v - u 3
C 10
2v - u 3
2
D 9.
Expanding and simplifying, we find u2 C v2 D 1. 9 x, y du dv where E denotes the corresponding ellipse in the u, v 6 6 6 E E E x, y 1/3 1/3 1 D det uv -plane given above. Now D - so 2/3 -1/3 u, v 3 1 1 1 Area D du dv D area of E D 3 1 D using the result of part (c) of Exercise 11. 3 3 6 3 E u C v v - u 13. With u D x - y, v D x C y we find that x D ,yD . Substituting these expressions into the 2 2 equation for E, we find (b) Area D 1 dA D 1 dx dy D 5 which simplifies to u2 C v2 D 1. 4 The area of ellipse E in the uv -plane is 2. The area of the original ellipse E is 1 dA D D dx dy D x, y u, v du dv D det 1/2 -1/2 1/2 1/2 du dv u C v 2
2
C 6
u C v 2
v - u 2
C 5
v - u 2
2
D 4,
6 E
6 E
6 E
6 E
1 1 1 du dv D area of E D 2 D . 2 2 6 2 E
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Section 5.8
Miscellaneous Exercises for Chapter 5
277
14. We follow the steps in Exercise 11, inserting the same letters for ease in locating the corresponding parts. (a) First we write the integral in Cartesian coordinates as
a b
1-x2 /a2 1-x2 /a2
c
1-x2 /a2 -y2 /b2
-a -b
-c
dz dy dx.
1-x2 /a2 -y2 /b2
(b) We now scale the variables using x D ax, y D by, and z D cz. Note that the ellipsoid E is transformed into the unit sphere E and that x, y, z / x, y, z D abc. The transformed integral is:
1 1-x2
-
1-x2 -y2
-1 - 1-x2
abc dz dy dx.
1-x2 -y2
(c) Because we are integrating over a unit sphere, we will transform to spherical coordinates:
1 1-x2
-
1-x2 -y2 1-x2 -y2
-1 - 1-x2
abc dz dy dx D
0
2 0 2 0 0 2 0
1 0 1
abc 2 sin d d d
D D
- cos abc 2
0
d d D
0
2 0
1
2abc 2 d d
4 2 abc d D abc. 3 3
15. If you didn't first sketch the region you may be tempted to use the numerator and denominator of the integrand as your new variables. The diamond-like shape is bounded on two sides by the hyperbolas x2 - y2 D 1 and x2 - y2 D 4 and on the other two sides by the ellipses x2 /4 C y2 D 1 and x2 /4 C y2 D 4.
2 1.75 1.5 1.25 1 0.75 0.5 0.25 0.5 1 1.5 2 2.5 3 3.5 4
We, therefore, make the change of variables u D x2 - y2 and v D x2 /4 C y2 . Then u, v D x, y The integral greatly simplifies: xy dA D 2 - x2 6y D
4 1 1 4
2x x/2
-2y 2y
D 5xy
so
x, y D 1/ 5xy . u, v
xy 1 -u 5xy
du dv D -
1 5
4 1 1
4
1 du dv u
D-
1 5
4 1
3 ln 4 dv D - ln 4. 5
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278
Chapter 5
Multiple Integration
16. The region D is bounded on the left and right by x2 - y2 D 1 and x2 - y2 D 9 and on the bottom and top by xy D 1 and xy D 4. It looks
y 3 2.5 2 1.5 1 0.5 0.5 1 1.5 2 2.5 3 3.5 4 x
This suggests we try the change of variables u D x2 - y2 , v D xy. Then u, v D det x, y so that 2x y -2y x D 2 x2 C y 2
x, y 1 D . u, v 2 x2 C y 2
Moreover, the region D* in the uv -plane corresponding to D is
v 4
1 u
1
9
Thus, using the change of variables theorem, we have 6 D x2 C y2 ex
2 -y2
dA D D
1 u e du dv D 6 2 D
4 1
4 1 1
9
1 u e du dv 2
3 9 1 9 e - e dV D e - e . 2 2
17. The region D is bounded on the top and bottom by y D 1 and y D 2 and on the left and right by xy D 1 and xy D 4; the region looks like the following figure.
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Section 5.8
y 3 2.5 2 1.5 1 0.5 1 2 3
Miscellaneous Exercises for Chapter 5
279
4
x
With this in mind, we try the change of variables u D xy, v D y. Then u, v D det x, y y 0 x 1 D y D v.
Moreover, the region D in the uv -plane corresponding to D is the rectangle [1, 4] * [1, 2]:
v 2
1
u 1 4
The change of variables theorem tells us that 6 D 1 1 dA D v du dv D C 1 6 u2 C 1 D D
1 4 4 1 4 1 1 2
x2 y 2
v dv du u2 C 1
3/2 3 du D tan-1 u u2 C 1 2
3 D tan-1 4 - . 2 4 18. (a) Follow the same steps as in defining the double and triple integrals. Define a partition of B D [a, b] * [c, d] * [p, q] * [r, s] of order n to be four collections of partition points that break up B into a union of n4 subboxes. See Definition 4.1 and add that r D w0 < w1 < < wn D s and 1wl D wl - wl-1 . Define a Riemann sum. For a function f defined on B, partition B as above and let cijkl be any point in the subbox Bijkl D [xi-1 , xi ] * [yj-1 , yj ] * [zk-1 , zk ] * [wl-1 , wl ]. The Riemann sum of f on B corresponding to the partition is SD
n i,j,k,lD1
f cijkl 1xi 1yj 1zk 1wl D
n
f cijkl 1Vijkl .
i,j,k,lD1
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280
Chapter 5
Multiple Integration
Define the quadruple integral of f on B, written o B to be o B f x, y, z, w dV D lim f x, y, z, w dV D o B f x, y, z, w dx dy dz dw
n all 1xi ,1yj ,1zk ,1wl '0 i,j,k,lD1
f cijkl 1xi 1yj 1zk 1wl .
We extend the definition to compact non-box regions W by defining the function f ext which is f everywhere in W and is 0 everywhere else. Then if B is a box containing W we can define o W f dV D o B f ext dV .
As in the cases of the double and triple integrals, Fubini's theorem allows us to evaluate the integral as an iterated integral. (b) We calculate: o W x C 2y C 3z - 4w dV D
0 2 3 -1 0 2 0 3 -1 0 2 0 3 -1 2 0 4 4 2 -2
x C 2y C 3z - 4w dw dz dy dx
D4 D4 D4
x C 2y C 3z dz dy dx
4x C 8y C 24 dy dx
2 0
16x C 32 C 96 dx D 64
x C 8 dx
D 64 2 C 16 D 1152. 19. (a) We are just generalizing what we've done to set up the area of a circle or the volume of a sphere (more recently see Exercises 11 and 14 from this section). Here our integral is:
a a2 -x2 a2 -x2 -y2 a2 -x2 -y2 -z2 -a - a2 -x2 -
a2 -x2 -y2
-
dw dz dy dx.
a2 -x2 -y2 -z2
(b) You should get 2 a2 /4. (c) For n D 5 you should get 82 a5 /15, and for n D 6 you should get 3 a6 /6. If you include the cases for n D 2 and n D 3 you may begin to see a pattern for the even exponents. If n is even, the volume of the n-sphere of radius a is n/2 an / n/2 !. Fitting in the odd terms looks really hard and the pattern shouldn't occur to any of your students. In fact, the general formula depends on the Gamma function which is beyond what we would expect the students to know at this point. For kicks, the volume of the n-sphere of radius a is n/2 an . 0 n/2 C 1 Note that the volume of an n-sphere of radius a decreases to 0 as n increases. 20. (a) To obtain the mass we compute the following integral (which is straightforward so the details are omitted): MD
0 2 0 3 4
.122 2 sin d d d D 74.976 L 235.5440508g.
(b) Because the shell is sealed, the volume is V D 4/3 43 D 256/3 cm3 , so the mass of that volume of water is greater, and so the shell would float.
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Section 5.8
Miscellaneous Exercises for Chapter 5
281
(c) If the core of the shell fills with water then the volume that the shell has to displace is V D 4/3 43 - 33 D 4/3 37 . The water for that volume would have mass of about 155 grams so the shell would sink. 21. When you average the height of the hemisphere of radius a, first you integrate
2 0 0 a
zr dr d D
0
2 0
a
2 r a2 - r2 dr d D a3 . 3
For the average height, we divide this by the area of the region over which we are integrating: 2/3 a3 2 D a. 2 a 3 We now solve to see which values of r correspond to this height: 2/3 a D a2 - r2 when 4/9 a2 D a2 - r2 , which is when r D 5a/3. Therefore, the pole can be installed at most 5a/3 from the center of the floor of the dome. 22. (a) By the fundamental theorem of calculus d dy
y
G y dy D G y
c
so
d dy
y c a
b
b
fy x, y dx dy D
a
fy x, y dx.
(b) On the other hand, by Fubini's theorem, d dy
y c a b
fy x, y dx dy D D
d dy d dy
b a b c
y
fy x, y dy dx f x, y - f x, c
a
dx D
d dy
b
f x, y dx.
a
Combine parts (a) and (b) to obtain the desired results. 23. (a) I , D D4 (b)
, ' 0,0 1- 1- 1- 1 2 1 - - dx dy dx D xy x - 1 - -
1 -
lim
I , D 4 1 1 D 4
,
24. For 0 <
<
1 1 1 , 0 < < we consider I , D dA where D 2 2 x C y 6 , D I , D D
1- 1- 1-
D [ , 1 - ] * [, 1 - ]. Then
1-
1 dy dx D x C y
1-
ln x C y
yD
dx
ln x C 1 - - ln x C
dx.
Using integration by parts, we find that
ln u du D u ln u - u C C so that
I , D [ x C 1 - ln x C 1 - - x C 1 - - x C ln x C C x C |1- xD D 2 - C - ln 2 - C - C ln - - 2 - C 1 - ln C . - - 1 - C 1 - C C ln 1 - C 1 - C C 1 -
C -
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282
Chapter 5
Multiple Integration
, ' 0C ,0C
To evaluate
lim
I , we first note that, by l'H^ pital's rule, o lim u ln u D lim 1/u ln u D lim D - lim u D 0. C -1/u2 u'0 u'0C 1/u
u'0C
u'0C
Thus that
25. For 0 <
, ' 0C ,0C < 1, 2
C ln C ' 0 as , ' 0C , 0C . The other terms in the expression have evident limits so I , D 2 ln 2 - 2 - ln 1 C 1 - ln 1 C 1 C 0 - 0 D 2 ln 2. lim 0 < <
1 , 2
let D
,
D [ , 1 - ] * [, 1 - ] and consider
1- 1-
I , D
x dA D 6 , y D
1- 1 2
x dx dy y ln 1 - - ln .
D Note that lim
- y
dy D
1 - 2
, ' 0,0
I , D -q since
1 2
-
and ln 1 - remain finite, but ln ' -q. Thus the improper
integral does not converge. In Exercises 26 and 27 the students will need integration by parts and l'H^ pital's rule. o 26. 6 D ln x2 C y2 dA D lim D lim
'0 6 D 2
ln x2 C y2 dA D lim r2 r2 ln r - 2 4
2 1 - ln 4 2 1
2
1
'0 0
r ln r dr d
2
'0 0
d D lim C
2
'0 0
-
2 1 - ln 4 2
C
2
4
d
D lim 2 -
'0
4
D -/2.
27. Define B D { x, y, z | ... x2 C y2 C z2 ... 1}. Then l B ln x2 C y2 C z2 dV D lim
'0 l B 2
ln x2 C y2 C z2 dV
1 0
D lim D lim D lim
'0 0 2 1
ln 2 sin d d d 22 ln d d
1
'0 0 2
'0 0
2/3 3 ln - 23 /9 2 2 3 - ln 9 3
a
d D -4/9.
D lim 2 -
'0
C
2 3 9
28. (a)
a b 1 a
I a, b D
1
1 dy dx D x2 y 3
-
1
1 2y2 x2
b
dx
yD1
D
1
1 1 - 2 2b2
1 dx D x2
1 1 - 2 2b2
1 -
1 a
(b) As a, b ' q, I a, b '
1 . 2
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Section 5.8
Miscellaneous Exercises for Chapter 5
283
29. Let Da,b D [1, a] * [1, b] and consider, for p, q Z 1, 1 dA D 6 a,b xp yq D
b b 1 a 1 a
I a, b D D
1 dx dy xp y q dy D 1 1 - p 1 bq-1 1 ap-1
b
1
1 1 - p yq xp-1 1
- 1
1
xD1
1 dy yq
D
1 ap-1
1 - p 1 - q
- 1
- 1 .
If p > 1, q > 1 then as a, b ' q, I a, b ' If p < 1, then 1/ap-1 ' q. Similarly if q < D ln a If p D 1, q Z 1, then
b a 1 1 1 xyq dx dy
1 1-p 1-q 1, 1/bp-1 1 bq-1
D
1 p-1 q-1
, so the integral converges in this case.
' q.
1 1-q
- 1 . This becomes infinite as a, b ' q. Similarly, if
q D 1, p Z 1, I a, b becomes infinite as a, b ' q. If p D q D 1, then I a, b D ln a ln b ' q as a, b ' q. To summarize: the integral converges if and only if p > 1 and q > 1--in which case the value of the integral is 1/ p - 1 q - 1 . 30. (a) We use polar coordinates to make the evaluation. Let
2 0 0 a
I a D
6a D
1 C x2 C y 2 1 2 - 1 C r2 1 . 1 C a2
-2
dA D
a
1 C r2
-2
r dr d
D 2
-1
D -
rD0
1 - 1 1 C a2
D 1 -
lima'q I a D . Thus the integral converges. (b) Let I a D 6a D 1 C x2 C y 2
p
dA D
0
2 0
a
1 C r2
p
r dr d. If p Z -1, then
I a D
p C 1
1 C a2
pC1
- 1 .
is finite (and equals 0) just in case p C 1 < 0 or p < -1. In such case, the integral 2 a r dr d D ln 1 C a2 ' q converges and its value is - . If p D -1, then I -1 D 2 p C 1 0 0 1 C r as a ' q. So the integral converges if and only if p < -1.
pC1
Now lima'q 1 C a2
31. Consider 1 1 C x 2 C y 2 C z2
2 0 0 2 0 0 a 3/2 0 a
I a D D
la B
a 0
3/2
dV D
2 sin d d d 1 C 2 3/2 2 1 C 2
3/2
2 1 C 2
sin d d d D
0
4
d
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284
Chapter 5
Multiple Integration
Now let D tan u so d D sec2 u du. Then I a D 4
0 tan-1 a
tan2 u sec2 u du D 4 sec3 u 1 - cos u cos2 u du D 4
0
tan-1 a 0
sin2 u du cos u
D 4
0
tan-1 a
tan-1 a
sec u - cos u du a2 C 1 C a - a . C 1
tan-1 a
D 4 ln | sec u C tan u| - sin u
0
D 4 ln
a2
Since lim I a D q the integral does not converge.
a'q
32. Consider I a D D
0
la B
a
e-
x2 Cy2 Cz2
dV D
0
2 0
0
a
e- 2 sin d d d
a 0
2 0 0
e- 2 sin d d d D 4
2 e- d
Now use integration by parts twice: First let u D 2 and dv D e- d. Then I a D 4 -2 e-
a a
C 2
0 0
e- d D -4a2 e-a C 8
0
a
e- d.
Now let u D and dv D e- d so that I a D -4a2 e-a C 8 -e- |a C 0
a 0
e- d
D -4a2 e-a - 8ae-a - 8e-a C 8.
a'q
lim I a D 8, so the integral converges and has value 8.
The importance of Exercise 33 can not be overemphasized. The students have come from a course where they learned one technique of integration after another. They also learned some numerical methods (at least a brief introduction to Riemann sums, the trapezoid rule and Simpson's rule). In a way Exercise 33 is the payoff--it is a chance to mention: Until now they couldn't calculate analysis) is pretty cool.
b
2
q -x2 dx. -q e
The fact that you need the tools of multivariable calculus (or complex
They still can't calculate a e-x dx. There is a need for numerical methods to calculate a function as common as the bell curve (with a constant that stretches in the vertical direction and another constant that stretches in the horizontal direction, this is the normal curve). Many will encounter this function in a course on statistics and use the tables; they should know that this is because we can't find the definite integral over a general finite interval. The technique is pretty and unexpected and is one of the tricks that they should see some time in their mathematical training. The problem is surprisingly straightforward once someone shows you the trick.
1
33. (a)
-1
e-x dx is finite since e-x is bounded on [-1, 1]. Since 0 ... e-x ... 1/x2 on both [1, q
2 2 2
and
q
-q, -1] and the improper integrals
-1 -q
1/x2 dx and
1 q -q
-1 -q
2
q
1/x2 dx converge, we see that
1 -1 1
e-x dx and
2
e-x dx both converge. Hence
2
e-x dx D
2
-1 -q
e-x dx C
e-x dx C
2
q 1
e-x dx converges.
2
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Section 5.8
Miscellaneous Exercises for Chapter 5
285
(b) We have
q
I2 D D
e-x dx
2
q -q
-q q q
2
e-x dx D
2
q -q
e-x dx
2
q -q
e-y dy
2
-q -q
e-x e-y dx dy D
2
6
e-x
2 -y2
dA.
R2
(c) We'll use polar coordinates. e-x
2 -y2
6a D
dA D
0
2 0
a
e-r r dr d D -
2 2
1 2
2 0
e-a - 1 d
2
D 1 - e-a
(d) Note that, as a ' q, the disk Da fills out more and more of R2 . Thus lim dA D I 2 . 62 R (e) Putting everything together: I 2 D lim 1 - e-a
a'q
2
a'q 6 Da
e-x
2 -y2
dA D
e-x
2 -y2
D .
2 q Thus I D -q e-x dx D . 34. (a) First, clearly f x 0. And second, by symmetry,
q -q
e-2|x| dx D 2
0
q
q
e-2x dx D -e-2x
0
D 1.
(b) We will reduce the calculations in Egbert's problem by recentering. P 250 ... x ... 350 D D
0 350 250 50
1 -|x-300| e dx D 2 2
50 0
350 300
1 - e 2
x-300
dx
e-x dx D -e-x
D 1 - e-50 .
35. (a) Since
2x C y 0 on [0, 5] * [0, 4], f x, y 0 for all (x, y). Now 140 f x, y D
0 4 0 5
62 R
2x C y 1 dx dy D 140 140
4 0
25 C 5y dy D
1 5 25y C y2 140 2
4 0
1 D 100 C 40 D 1. 140 (b) Since f x, y D 0 for x < 0 or y < 0, Prob x ... 1, y ... 1 D Prob x, y [0, 1] * [0, 1] D
0 1 0 1
2x C y dx dy 140
D
1 140
1 0
1 C y dy D
1 1 3 1 C D L 0.0107. 140 2 280
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286
Chapter 5
Multiple Integration
36. (a) Since ye-x-y 0 for y 0 (note the exponential term is strictly positive), we have that f x, y 0 for all (x, y). Now we check 62 R ye-x-y dA D
0 q q 0 q
ye-y e-x dx dy
D
0
ye-y dy
q
D -ye-y - e-y
0
D 1.
(b) Prob x C y ... 2 D Prob x, y D where D is the triangular region bounded by x D 0, y D 0 and x C y D 2.
y 2
1.5
1
0.5
0.5
1
1.5
2
x
Thus the desired probability is
2 0 0 2-x
ye-y e-x dy dx D
0
2
-ye-y - e-y |2-x e-x dx 0
2 0
D
0
2
x - 2 ex-2 - ex-2 C 1 e-x dx D
x - 2 e-2 - e-2 C e-x dx
D 1 - 5e-2 . 37. First, we know that C 0. Second,
q q
1D
-q -q q
Ce-a|x|-b|y| dx dy D 4
0 q 0
q 0
q
Ce-ax-by dx dy
q 0
D 4C
0
e-ax dx
1 e-by dy D 4C - e-ax a
1 - e-by b
q
D
0
4C . ab
So C D ab/4. 38. Note that if C 0, then f x, y 0 for all x since a and b are nonnegative. Thus we calculate 62 R f x, y dA D
0 1 0 1
C ax C by dx dy D C
0
1
1 a C by dy 2
DC For this to equal 1, we must have C D
1 1 a C b a C b DC . 2 2 2
2 . a C b
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Section 5.8
Miscellaneous Exercises for Chapter 5
287
39. (a) If C 0, then f x, y 0 for all (x, y). Thus we calculate
b a b
62 R
f x, y dA D
0 0
C xy dx dy D C
0
a2 y dy 2
a2 b 2 DC . 4 For this to equal 1 we must have C D 4 a2 b 2 .
(b) Prob bx - ay 0 is the probability that a point (x, y) falls below the line y D
b x. Since f is zero a outside the rectangle [0, a] * [0, b], we see that the desired probability is the same as the probability Prob x, y D where D is the triangular region shown below.
y b
a
x
This last probability is calculated as
a b/a x 0 a
6 D
f dA D
0
4 xy dy dx a2 b 2 b x a
2 a
D
0
4 1 x a2 b 2 2
a
dx D
0
2 3 x dx a4
D
1 4 x 2a4
D
0
1 . 2
40. We are integrating over the triangle where 0 ... x C y ... 60. The integral is fairly straightforward so the details are omitted: 1 250
60 0 0 60-x
e-x/50 e-y/5 dy dx D -
1 50
60 0
e-x/50 [e-12Cx/5 - 1] dx
10 -6/5 1 D1 - C e-12 L .665340. e 9 9 41. We use polar coordinates:
1/2
1/4 -x2 1/4 -x2
-1/2 -
1 -x2 -y2 dy dx D e D
2 0 2 0 0
1/2
1 -r2 re 1 2
dr d
1
1 - e-1/4
d
D 1 - e-1/4 L .22199. 42. The joint density function of the components is F x, y D f x f y D 0
1 2000
2
e-
xCy /2000
if x 0, y 0 otherwise.
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288
Chapter 5
Multiple Integration
2000 0 0 2000
So we want Prob x ... 2000, y ... 2000 D D D 1 2000 1 2000 6 W
1 e-x/2000 e-y/2000 dx dy 20002
2000 0
2000 2 0
-2000e-x/2000
e-y/2000 dy
2
2000 0
1 -
1 1 -y/2000 e dy D 1 - e e
.
43. Formula (8) in Section 5.6 is I D at the origin, then
d 2 x, y, z dV . If we choose the coordinates so that the center of mass is
x x, y, z dV D 0 and y x, y, z dV D 0. We can also choose the coordinates so l l W W that L is the z-axis. L is a line parallel to the z-axis distance h away, so L is the line corresponding to x D a and y D b where a2 C b2 D h2 . Then IL D and IL D so IL - IL D l W h2 x, y, z dV D h2 l W x, y, z dV D h2 M. l W x2 C y2 C h2 - 2ax - 2by x, y, z dV D l W x2 C y2 C h2 x, y, z dV l W x2 C y2 x, y, z dV
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