This document contains 20 multiple integral exercises with solutions. Some of the exercises involve calculating double integrals over specified regions, while others involve setting up approximations of double integrals using Riemann sums. Exercise 19 involves sketching solid regions in 3D space and Exercise 20 involves sketching surfaces defined by z=f(x,y).

1. January 27, 2005 11:55 L24-CH15 Sheet number 1 Page number 655 black
CHAPTER 15
Multiple Integrals
EXERCISE SET 15.1
1 2 1 3 1 3
1. (x + 3)dy dx = (2x + 6)dx = 7 2. (2x − 4y)dy dx = 4x dx = 16
0 0 0 1 −1 1
4 1 4 0 2 0
1
3. x2 y dx dy = y dy = 2 4. (x2 + y 2 )dx dy = (3 + 3y 2 )dy = 14
2 0 2 3 −2 −1 −2
ln 3 ln 2 ln 3
5. ex+y dy dx = ex dx = 2
0 0 0
2 1 2
1
6. y sin x dy dx = sin x dx = (1 − cos 2)/2
0 0 0 2
0 5 0 6 7 6
7. dx dy = 3 dy = 3 8. dy dx = 10dx = 20
−1 2 −1 4 −3 4
1 1 1
x 1
9. dy dx = 1− dx = 1 − ln 2
0 0 (xy + 1)2 0 x+1
π 2 π
10. x cos xy dy dx = (sin 2x − sin x)dx = −2
π/2 1 π/2
ln 2 1 ln 2
2 1 x
11. xy ey x dy dx = (e − 1)dx = (1 − ln 2)/2
0 0 0 2
4 2 4
1 1 1
12. dy dx = − dx = ln(25/24)
3 1 (x + y)2 3 x+1 x+2
1 2 1
13. 4xy 3 dy dx = 0 dx = 0
−1 −2 −1
1 1
xy 1 √ √
14. dy dx = [x(x2 + 2)1/2 − x(x2 + 1)1/2 ]dx = (3 3 − 4 2 + 1)/3
0 0 x2 + y 2 + 1 0
1 3 1
15. x 1 − x2 dy dx = x(1 − x2 )1/2 dx = 1/3
0 2 0
π/2 π/3 π/2
x π2
16. (x sin y − y sin x)dy dx = − sin x dx = π 2 /144
0 0 0 2 18
17. (a) x∗ = k/2 − 1/4, k = 1, 2, 3, 4; yl = l/2 − 1/4, l = 1, 2, 3, 4,
k
∗
4 4 4 4
f (x, y) dxdy ≈ f (x∗ , yl )∆Akl =
k
∗
[(k/2−1/4)2 +(l/2−1/4)](1/2)2 = 37/4
R k=1 l=1 k=1 l=1
2 2
(b) (x2 + y) dxdy = 28/3; the error is |37/4 − 28/3| = 1/12
0 0
655

2. January 27, 2005 11:55 L24-CH15 Sheet number 2 Page number 656 black
656 Chapter 15
18. (a) x∗ = k/2 − 1/4, k = 1, 2, 3, 4; yl = l/2 − 1/4, l = 1, 2, 3, 4,
k
∗
4 4 4 4
f (x, y) dxdy ≈ f (x∗ , yl )∆Akl =
k
∗
[(k/2 − 1/4) − 2(l/2 − 1/4)](1/2)2 = −4
R k=1 l=1 k=1 l=1
2 2
(b) (x − 2y) dxdy = −4; the error is zero
0 0
19. (a) z (b) z
(1, 0, 4) (0, 0, 5)
(0, 4, 3)
y y
(2, 5, 0) (3, 4, 0)
x x
z z
20. (a) (b) (2, 2, 8)
(0, 0, 2)
y y
(2, 2, 0)
(1, 1, 0)
x x
5 2 5
21. V = (2x + y)dy dx = (2x + 3/2)dx = 19
3 1 3
3 2 3
22. V = (3x3 + 3x2 y)dy dx = (6x3 + 6x2 )dx = 172
1 0 1
2 3 2
23. V = x2 dy dx = 3x2 dx = 8
0 0 0
3 4 3
24. V = 5(1 − x/3)dy dx = 5(4 − 4x/3)dx = 30
0 0 0
1/2 π 1/2 π
25. x cos(xy) cos2 πx dy dx = cos2 πx sin(xy) dx
0 0 0 0
1/2 1/2
1 1
= cos2 πx sin πx dx = − cos3 πx =
0 3π 0 3π

3. January 27, 2005 11:55 L24-CH15 Sheet number 3 Page number 657 black
Exercise Set 15.2 657
5 2 5 3
26. (a) z (b) V = y dy dx + (−2y + 6) dy dx
0 0 0 2
(0, 2, 2)
= 10 + 5 = 15
y
3
5 (5, 3, 0)
x
π/2 1 π/2 x=1 π/2
2 2 2 2
27. fave = y sin xy dx dy = − cos xy dy = (1 − cos y) dy = 1 −
π 0 0 π 0 x=0 π 0 π
1 3 1 3
1 √
28. average = x(x2 + y)1/2 dx dy = [(1 + y)3/2 − y 3/2 ]dy = 2(31 − 9 3)/45
3 0 0 0 9
1 2 1 ◦
1 1 44 14
29. Tave = 10 − 8x2 − 2y 2 dy dx = − 16x2 dx = C
2 0 0 2 0 3 3
b d
1 1
30. fave = k dy dx = (b − a)(d − c)k = k
A(R) a c A(R)
31. 1.381737122 32. 2.230985141
b d b d
33. f (x, y)dA = g(x)h(y)dy dx = g(x) h(y)dy dx
a c a c
R
b d
= g(x)dx h(y)dy
a c
34. The integral of tan x (an odd function) over the interval [−1, 1] is zero.
35. The first integral equals 1/2, the second equals −1/2. No, because the integrand is not continuous.
EXERCISE SET 15.2
1 x 1
1 4
1. xy 2 dy dx = (x − x7 )dx = 1/40
0 x2 0 3
3/2 3−y 3/2
2. y dx dy = (3y − 2y 2 )dy = 7/24
1 y 1
√
3 9−y 2 3
3. y dx dy = y 9 − y 2 dy = 9
0 0 0
1 x 1 x 1
4. x/y dy dx = x1/2 y −1/2 dy dx = 2(x − x3/2 )dx = 13/80
1/4 x2 1/4 x2 1/4

4. January 27, 2005 11:55 L24-CH15 Sheet number 4 Page number 658 black
658 Chapter 15
√ √
2π x3 2π
5. √
sin(y/x)dy dx = √
[−x cos(x2 ) + x]dx = π/2
π 0 π
1 x2 1 π x2 π
1
6. (x2 − y)dy dx = 2x4 dx = 4/5 7. cos(y/x)dy dx = sin x dx = 1
−1 −x2 −1 π/2 0 x π/2
1 x 1 1 x 1
2 2 1 3
8. ex dy dx = xex dx = (e − 1)/2 9. y x2 − y 2 dy dx = x dx = 1/12
0 0 0 0 0 0 3
2 y2 2
2
10. ex/y dx dy = (e − 1)y 2 dy = 7(e − 1)/3
1 0 1
2 x2 4 2
11. (a) f (x, y) dydx (b) √
f (x, y) dxdy
0 0 0 y
√ √
1 x 1 y
12. (a) f (x, y) dydx (b) f (x, y) dxdy
0 x2 0 y2
2 3 4 3 5 3
13. (a) f (x, y) dydx + f (x, y) dydx + f (x, y) dydx
1 −2x+5 2 1 4 2x−7
3 (y+7)/2
(b) f (x, y) dxdy
1 (5−y)/2
√ √
1 1−x2 1 1−y 2
14. (a) √ f (x, y) dydx (b) √ f (x, y) dxdy
−1 − 1−x2 −1 − 1−y 2
2 x2 2
1 5 16
15. (a) xy dy dx = x dx =
0 0 0 2 3
3 (y+7)/2 3
(b) xy dx dy = (3y 2 + 3y)dy = 38
1 −(y−5)/2 1
√
1 x 1
16. (a) (x + y)dy dx = (x3/2 + x/2 − x3 − x4 /2)dx = 3/10
0 x2 0
√ √
1 1−x2 1 1−x2 1
(b) √ x dy dx + √ y dy dx = 2x 1 − x2 dx + 0 = 0
−1 − 1−x2 −1 − 1−x2 −1
8 x 8
17. (a) x2 dy dx = (x3 − 16x)dx = 576
4 16/x 4
4 8 8 8 8
512 4096 8
512 − y 3
(b) x2 dxdy + x2 dx dy = − dy + dy
2 16/y 4 y 4 3 3y 3 4 3
640 1088
= + = 576
3 3
2 y 2
1 4
18. (a) xy 2 dx dy = y dy = 31/10
1 0 1 2
1 2 2 2 1 2
8x − x4
(b) xy 2 dydx + xy 2 dydx = 7x/3 dx + dx = 7/6 + 29/15 = 31/10
0 1 1 x 0 1 3

5. January 27, 2005 11:55 L24-CH15 Sheet number 5 Page number 659 black
Exercise Set 15.2 659
√
1 1−x2 1
19. (a) √ (3x − 2y)dy dx = 6x 1 − x2 dx = 0
−1 − 1−x2 −1
√
1 1−y 2 1
(b) √ (3x − 2y) dxdy = −4y 1 − y 2 dy = 0
−1 − 1−y 2 −1
√
5 25−x2 5
20. (a) y dy dx = (5x − x2 )dx = 125/6
0 5−x 0
√
5 25−y 2 5
(b) y dxdy = y 25 − y 2 − 5 + y dy = 125/6
0 5−y 0
√
4 y 4
1 √
21. x(1 + y 2 )−1/2 dx dy = y(1 + y 2 )−1/2 dy = ( 17 − 1)/2
0 0 0 2
π x π
22. x cos y dy dx = x sin x dx = π
0 0 0
2 6−y 2
1
23. xy dx dy = (36y − 12y 2 + y 3 − y 5 )dy = 50/3
0 y2 0 2
√
π/4 1/ 2 π/4
1
24. x dx dy = cos 2y dy = 1/8
0 sin y 0 4
1 x 1
25. (x − 1)dy dx = (−x4 + x3 + x2 − x)dx = −7/60
0 x3 0
√ √
1/ 2 2x 1 1/x 1/ 2 1
26. 2
x dy dx + √
2
x dy dx = x3 dx + √ (x − x3 )dx = 1/8
0 x 1/ 2 x 0 1/ 2
y
27. (a)
4
3
2
1
x
–2 –1 0.5 1.5
(b) x = (−1.8414, 0.1586), (1.1462, 3.1462)
1.1462 x+2 1.1462
(c) x dA ≈ x dydx = x(x + 2 − ex ) dx ≈ −0.4044
−1.8414 ex −1.8414
R
3.1462 ln y 3.1462
ln2 y (y − 2)2
(d) x dA ≈ x dxdy = − dy ≈ −0.4044
0.1586 y−2 0.1586 2 2
R

6. January 27, 2005 11:55 L24-CH15 Sheet number 6 Page number 660 black
660 Chapter 15
28. (a) y (b) (1, 3), (3, 27)
25
15 R
5
x
1 2 3
3 4x3 −x4 3
224
(c) x dy dx = x[(4x3 − x4 ) − (3 − 4x + 4x2 )] dx =
1 3−4x+4x2 1 15
π/4 cos x π/4 √
29. A = dy dx = (cos x − sin x)dx = 2−1
0 sin x 0
1 −y 2 1
30. A = dx dy = (−y 2 − 3y + 4)dy = 125/6
−4 3y−4 −4
3 9−y 2 3
31. A = dx dy = 8(1 − y 2 /9)dy = 32
−3 1−y 2 /9 −3
1 cosh x 1
32. A = dy dx = (cosh x − sinh x)dx = 1 − e−1
0 sinh x 0
4 6−3x/2 4
33. (3 − 3x/4 − y/2) dy dx = [(3 − 3x/4)(6 − 3x/2) − (6 − 3x/2)2 /4] dx = 12
0 0 0
√
2 4−x2 2
34. 4 − x2 dy dx = (4 − x2 ) dx = 16/3
0 0 0
√
3 9−x2 3
35. V = √ (3 − x)dy dx = (6 9 − x2 − 2x 9 − x2 )dx = 27π
−3 − 9−x2 −3
1 x 1
36. V = (x2 + 3y 2 )dy dx = (2x3 − x4 − x6 )dx = 11/70
0 x2 0
3 2 3
37. V = (9x2 + y 2 )dy dx = (18x2 + 8/3)dx = 170
0 0 0
1 1 1
38. V = (1 − x)dx dy = (1/2 − y 2 + y 4 /2)dy = 8/15
−1 y2 −1
√
3/2 9−4x2 3/2
39. V = √ (y + 3)dy dx = 6 9 − 4x2 dx = 27π/2
−3/2 − 9−4x2 −3/2
3 3 3
40. V = (9 − x2 )dx dy = (18 − 3y 2 + y 6 /81)dy = 216/7
0 y 2 /3 0
√
5 25−x2 5
41. V = 8 25 − x2 dy dx = 8 (25 − x2 )dx = 2000/3
0 0 0

7. January 27, 2005 11:55 L24-CH15 Sheet number 7 Page number 661 black
Exercise Set 15.2 661
√
2 1−(y−1)2 2
1
42. V = 2 (x2 + y 2 )dx dy = 2 [1 − (y − 1)2 ]3/2 + y 2 [1 − (y − 1)2 ]1/2 dy,
0 0 0 3
π/2
1
let y − 1 = sin θ to get V = 2 cos3 θ + (1 + sin θ)2 cos θ cos θ dθ which eventually yields
−π/2 3
V = 3π/2
√
1 1−x2 1
8
43. V = 4 (1 − x2 − y 2 )dy dx = (1 − x2 )3/2 dx = π/2
0 0 3 0
√
2 4−x2 2
1
44. V = (x2 + y 2 )dy dx = x2 4 − x2 + (4 − x2 )3/2 dx = 2π
0 0 0 3
√
2 2 8 x/2 e2 2
45. f (x, y)dx dy 46. f (x, y)dy dx 47. f (x, y)dy dx
0 y2 0 0 1 ln x
√
1 e π/2 sin x 1 x
48. f (x, y)dx dy 49. f (x, y)dy dx 50. f (x, y)dy dx
0 ey 0 0 0 x2
4 y/4 4
1 −y2
e−y dx dy = ye dy = (1 − e−16 )/8
2
51.
0 0 0 4
1 2x 1
52. cos(x2 )dy dx = 2x cos(x2 )dx = sin 1
0 0 0
2 x2 2
3 3
53. ex dy dx = x2 ex dx = (e8 − 1)/3
0 0 0
ln 3 3 ln 3
1 1
54. x dx dy = (9 − e2y )dy = (9 ln 3 − 4)
0 ey 2 0 2
2 y2 2
55. sin(y 3 )dx dy = y 2 sin(y 3 )dy = (1 − cos 8)/3
0 0 0
1 e 1
56. x dy dx = (ex − xex )dx = e/2 − 1
0 ex 0
4 2
57. (a) √
sin πy 3 dy dx; the inner integral is non-elementary.
0 x
2 y2 2 2
1
sin πy 3 dx dy = y 2 sin πy 3 dy = − cos πy 3 =0
0 0 0 3π 0
1 π/2
(b) sec2 (cos x)dx dy ; the inner integral is non-elementary.
0 sin−1 y
π/2 sin x π/2
sec2 (cos x)dy dx = sec2 (cos x) sin x dx = tan 1
0 0 0
√
2 4−x2 2
1
58. V = 4 (x2 + y 2 ) dy dx = 4 x2 4 − x2 + (4 − x2 )3/2 dx (x = 2 sin θ)
0 0 0 3
π/2
64 64 128 64 π 64 π 128 π 1 · 3
= + sin2 θ − sin4 θ dθ = + − = 8π
0 3 3 3 3 2 3 4 3 2 2·4

8. January 27, 2005 11:55 L24-CH15 Sheet number 8 Page number 662 black
662 Chapter 15
59. The region is symmetric with respect to the y-axis, and the integrand is an odd function of x,
hence the answer is zero.
60. This is the volume in the first octant under the surface z = 1 − x2 − y 2 , so 1/8 of the volume of
π
the sphere of radius 1, thus .
6
1 1 1
¯ 1 1 x π
61. Area of triangle is 1/2, so f = 2 dy dx = 2 − dx = − ln 2
0 x 1 + x2 0 1 + x2 1 + x2 2
2
62. Area = (3x − x2 − x) dx = 4/3, so
0
2 3x−x2 2
¯ 3
f= (x2 − xy)dy dx =
3
(−2x3 + 2x4 − x5 /2)dx = −
3 8
=−
2
4 0 x 4 0 4 15 5
1
63. Tave = (5xy + x2 ) dA. The diamond has corners (±2, 0), (0, ±4) and thus has area
A(R)
R
1
A(R) = 4 2(4) = 16m2 . Since 5xy is an odd function of x (as well as y), 5xy dA = 0. Since
2
R
x2 is an even function of both x and y,
2◦
4−2x 2 2
4 1 2 1 1 4 3 1 4
Tave = x2 dA = st x2 dydx = (4 − 2x)x2 dx = x − x = C
16 4 0 0 4 0 4 3 2 0 3
R
x,y>0
64. The area of the lens is πR2 = 4π and the average thickness Tave is
√
2 4−x2 2
4 1 1
Tave = 1 − (x2 + y 2 )/4 dydx = (4 − x2 )3/2 dx (x = 2 cos θ)
4π 0 0 π 0 6
8 π
8 1·3π 1
= sin4 θ dθ = = in
3π 0 3π 2 · 4 2 2
65. y = sin x and y = x/2 intersect at x = 0 and x = a = 1.895494, so
a sin x
V = 1 + x + y dy dx = 0.676089
0 x/2
EXERCISE SET 15.3
π/2 sin θ π/2
1
1. r cos θdr dθ = sin2 θ cos θ dθ = 1/6
0 0 0 2
π 1+cos θ π
1
2. r dr dθ = (1 + cos θ)2 dθ = 3π/4
0 0 0 2
π/2 a sin θ π/2
a3 2
3. r2 dr dθ = sin3 θ dθ = a3
0 0 0 3 9
π/6 cos 3θ π/6
1
4. r dr dθ = cos2 3θ dθ = π/24
0 0 0 2

9. January 27, 2005 11:55 L24-CH15 Sheet number 9 Page number 663 black
Exercise Set 15.3 663
π 1−sin θ π
1
5. r2 cos θ dr dθ = (1 − sin θ)3 cos θ dθ = 0
0 0 0 3
π/2 cos θ π/2
1
6. r3 dr dθ = cos4 θ dθ = 3π/64
0 0 0 4
2π 1−cos θ 2π
1
7. A = r dr dθ = (1 − cos θ)2 dθ = 3π/2
0 0 0 2
π/2 sin 2θ π/2
8. A = 4 r dr dθ = 2 sin2 2θ dθ = π/2
0 0 0
π/2 1 π/2
1
9. A = r dr dθ = (1 − sin2 2θ)dθ = π/16
π/4 sin 2θ π/4 2
π/3 2 π/3 √
10. A = 2 r dr dθ = (4 − sec2 θ)dθ = 4π/3 − 3
0 sec θ 0
5π/6 4 sin θ 3π/2 1
11. A = f (r, θ) r dr dθ 12. A = f (r, θ)r dr dθ
π/6 2 π/2 1+cos θ
π/2 3 π/2 2 sin θ
13. V = 8 r 9 − r2 dr dθ 14. V = 2 r2 dr dθ
0 1 0 0
π/2 cos θ π/2 3
15. V = 2 (1 − r2 )r dr dθ 16. V = 4 dr dθ
0 0 0 1
π/2 3
128 √ π/2
64 √
17. V = 8 r 9 − r2 dr dθ = 2 dθ = 2π
0 1 3 0 3
π/2 2 sin θ π/2
16
18. V = 2 r2 dr dθ = sin3 θ dθ = 32/9
0 0 3 0
π/2 cos θ π/2
1
19. V = 2 (1 − r2 )r dr dθ = (2 cos2 θ − cos4 θ)dθ = 5π/32
0 0 2 0
π/2 3 π/2
20. V = 4 dr dθ = 8 dθ = 4π
0 1 0
π/2 3 sin θ π/2
27
21. V = r2 sin θ drdθ = 9 sin4 θ dθ = π
0 0 0 16
π/2 2 π 2
22. V = 2 4 − r2 r drdθ + 2 4 − r2 r drdθ
0 2 cos θ π/2 0
π/2 π
16 16 32 8
= (1 − cos2 θ)3/2 θ dθ + dθ = + π
0 3 π/2 3 9 3
2π 1 2π
1
e−r r dr dθ = (1 − e−1 ) dθ = (1 − e−1 )π
2
23.
0 0 2 0

10. January 27, 2005 11:55 L24-CH15 Sheet number 10 Page number 664 black
664 Chapter 15
π/2 3 π/2
24. r 9 − r2 dr dθ = 9 dθ = 9π/2
0 0 0
π/4 2 π/4
1 1 π
25. r dr dθ = ln 5 dθ = ln 5
0 0 1 + r2 2 0 8
π/2 2 cos θ π/2
16
26. 2r2 sin θ dr dθ = cos3 θ sin θ dθ = 1/3
π/4 0 3 π/4
π/2 1 π/2
1
27. r3 dr dθ = dθ = π/8
0 0 4 0
2π 2 2π
1
e−r r dr dθ = (1 − e−4 ) dθ = (1 − e−4 )π
2
28.
0 0 2 0
π/2 2 cos θ π/2
8
29. r2 dr dθ = cos3 θ dθ = 16/9
0 0 3 0
π/2 1 π/2
1 π
30. cos(r2 )r dr dθ = sin 1 dθ = sin 1
0 0 2 0 4
π/2 a
r π
31. dr dθ = 1 − 1/ 1 + a2
0 0 (1 + r2 )3/2 2
π/4 sec θ tan θ
1 π/4 √
32. r2 dr dθ = sec3 θ tan3 θ dθ = 2( 2 + 1)/45
0 0 3 0
π/4 2
r π √
33. √ dr dθ = ( 5 − 1)
0 0 1+r 2 4
π/2 5 π/2
1
34. r dr dθ = (25 − 9 csc2 θ)dθ
tan−1 (3/4) 3 csc θ 2 tan−1 (3/4)
25 π 25
= − tan−1 (3/4) − 6 = tan−1 (4/3) − 6
2 2 2
2π a 2π
a2
35. V = hr dr dθ = h dθ = πa2 h
0 0 0 2
π/2 a a
c 2 4c 4 2
36. (a) V = 8 (a − r2 )1/2 r dr dθ = − π(a2 − r2 )3/2 = πa c
0 0 a 3a 0 3
4
(b) V ≈ π(6378.1370)2 6356.5231 ≈ 1,083,168,200,000 km3
3
π/2 a sin θ π/2
c 2 2
37. V = 2 (a − r2 )1/2 r dr dθ = a2 c (1 − cos3 θ)dθ = (3π − 4)a2 c/9
0 0 a 3 0
√
π/4 a 2 cos 2θ π/4
38. A = 4 r dr dθ = 4a2 cos 2θ dθ = 2a2
0 0 0

11. January 27, 2005 11:55 L24-CH15 Sheet number 11 Page number 665 black
Exercise Set 15.4 665
π/4 4 sin θ π/2 4 sin θ
39. A = √ r dr dθ + r dr dθ
π/6 8 cos 2θ π/4 0
π/4 π/2 √
= (8 sin2 θ − 4 cos 2θ)dθ + 8 sin2 θ dθ = 4π/3 + 2 3 − 2
π/6 π/4
φ 2a sin θ φ
1
40. A = r dr dθ = 2a2 sin2 θ dθ = a2 φ − a2 sin 2φ
0 0 0 2
+∞ +∞ +∞ +∞
e−x dx e−y dy = e−x dx e−y dy
2 2 2 2
41. (a) I 2 =
0 0 0 0
+∞ +∞ +∞ +∞
e−x e−y dx dy = e−(x
2 2 2
+y 2 )
= dx dy
0 0 0 0
π/2 +∞
1 π/2
√
e−r r dr dθ =
2
(b) I 2 = dθ = π/4 (c) I= π/2
0 0 2 0
√
42. The two quarter-circles with center at the origin and of radius A and 2A lie inside and outside
of the square with corners (0, 0), (A, 0), (A, A), (0, A), so the following inequalities hold:
√
π/2 A A A π/2 2A
1 1 1
rdr dθ ≤ dx dy ≤ rdr dθ
0 0 (1 + r2 )2 0 0 (1 + x2 + y 2 )2 0 0 (1 + r2 )2
2
πA
The integral on the left can be evaluated as and the integral on the right equals
4(1 + A2 )
2
2πA π
2)
. Since both of these quantities tend to as A → +∞, it follows by sandwiching that
4(1 + 2A 4
+∞ +∞
1 π
dx dy = .
0 0 (1 + x2 + y 2 )2 4
π 1 1
re−r dr dθ = π re−r dr ≈ 1.173108605
4 4
43. (a) 1.173108605 (b)
0 0 0
2π R 2π R R
44. V = D(r)r dr dθ = ke−r r dr dθ = −2πk(1 + r)e−r = 2πk[1 − (R + 1)e−R ]
0 0 0 0 0
tan−1 (2) 2 tan−1 (2) tan−1 (2)
45. r3 cos2 θ dr dθ = 4 cos2 θ dθ = 2 (1 + cos(2θ)) dθ
tan−1 (1/3) 0 tan−1 (1/3) tan−1 (1/3)
√ √
= 2(tan−1 2 − tan−1 (1/3)) + 2/ 5 − 1/ 10
EXERCISE SET 15.4
1. (a) z (b) z (c) z
y
x x
x y y

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666 Chapter 15
z z
2. (a) (b)
x
y
y
x
(c) z
y
x
5 3
3. (a) x = u, y = v, z = + u − 2v (b) x = u, y = v, z = u2
2 2
v 1 2 5
4. (a) x = u, y = v, z = (b) x = u, y = v, z = v −
1 + u2 3 3
5. (a) x = 5 cos u, y = 5 sin u, z = v; 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1
(b) x = 2 cos u, y = v, z = 2 sin u; 0 ≤ u ≤ 2π, 1 ≤ v ≤ 3
6. (a) x = u, y = 1 − u, z = v; −1 ≤ v ≤ 1 (b) x = u, y = 5 + 2v, z = v; 0 ≤ u ≤ 3
7. x = u, y = sin u cos v, z = sin u sin v 8. x = u, y = eu cos v, z = eu sin v
1
10. x = r cos θ, y = r sin θ, z = e−r
2
9. x = r cos θ, y = r sin θ, z =
1 + r2
11. x = r cos θ, y = r sin θ, z = 2r2 cos θ sin θ
12. x = r cos θ, y = r sin θ, z = r2 (cos2 θ − sin2 θ)
√ √
13. x = r cos θ, y = r sin θ, z = 9 − r2 ; r ≤ 5
√
1 1 3
14. x = r cos θ, y = r sin θ, z = r; r ≤ 3 15. x = ρ cos θ, y = ρ sin θ, z = ρ
2 2 2
16. x = 3 cos θ, y = 3 sin θ, z = 3 cot φ 17. z = x − 2y; a plane
18. y = x2 + z 2 , 0 ≤ y ≤ 4; part of a circular paraboloid
19. (x/3)2 + (y/2)2 = 1; 2 ≤ z ≤ 4; part of an elliptic cylinder

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Exercise Set 15.4 667
20. z = x2 + y 2 ; 0 ≤ z ≤ 4; part of a circular paraboloid
21. (x/3)2 + (y/4)2 = z 2 ; 0 ≤ z ≤ 1; part of an elliptic cone
22. x2 + (y/2)2 + (z/3)2 = 1; an ellipsoid
√
23. (a) x = r cos θ, y = r sin θ, z = r, 0 ≤ r ≤ 2; x = u, y = v, z = u2 + v 2 ; 0 ≤ u2 + v 2 ≤ 4
√
24. (a) I: x = r cos θ, y = r sin θ, z = r2 , 0 ≤ r ≤ 2; II: x = u, y = v, z = u2 + v 2 ; u2 + v 2 ≤ 2
25. (a) 0 ≤ u ≤ 3, 0 ≤ v ≤ π (b) 0 ≤ u ≤ 4, −π/2 ≤ v ≤ π/2
26. (a) 0 ≤ u ≤ 6, −π ≤ v ≤ 0 (b) 0 ≤ u ≤ 5, π/2 ≤ v ≤ 3π/2
27. (a) 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π (b) 0 ≤ φ ≤ π, 0 ≤ θ ≤ π
28. (a) π/2 ≤ φ ≤ π, 0 ≤ θ ≤ 2π (b) 0 ≤ θ ≤ π/2, 0 ≤ φ ≤ π/2
29. u = 1, v = 2, ru × rv = −2i − 4j + k; 2x + 4y − z = 5
30. u = 1, v = 2, ru × rv = −4i − 2j + 8k; 2x + y − 4z = −6
31. u = 0, v = 1, ru × rv = 6k; z = 0 32. ru × rv = 2i − j − 3k; 2x − y − 3z = −4
√ √
√ √ 2 π 2
33. ru × rv = ( 2/2)i − ( 2/2)j + (1/2)k; x − y + z=
2 8
34. ru × rv = 2i − ln 2k; 2x − (ln 2)z = 0
35. z = 9 − y 2 , zx = 0, zy = −y/ 9 − y 2 , zx + zy + 1 = 9/(9 − y 2 ),
2 2
2 3 2
3
S= dy dx = 3π dx = 6π
0 −3 9 − y2 0
4 4−x 4
36. z = 8 − 2x − 2y, zx + zy + 1 = 4 + 4 + 1 = 9, S =
2 2
3 dy dx = 3(4 − x)dx = 24
0 0 0
37. z 2 = 4x2 + 4y 2 , 2zzx = 8x so zx = 4x/z, similarly zy = 4y/z thus
1 x √ √ 1 √
zx + zy + 1 = (16x2 + 16y 2 )/z 2 + 1 = 5, S =
2 2
5 dy dx = 5 (x − x2 )dx = 5/6
0 x2 0
38. z 2 = x2 + y 2 , zx = x/z, zy = y/z, zx + zy + 1 = (z 2 + y 2 )/z 2 + 1 = 2,
2 2
√ π/2 2 cos θ √ √ π/2 √
S= 2 dA = 2 2 r dr dθ = 4 2 cos2 θ dθ = 2π
0 0 0
R
39. zx = −2x, zy = −2y, zx + zy + 1 = 4x2 + 4y 2 + 1,
2 2
2π 1
S= 4x2 + 4y 2 + 1 dA = r 4r2 + 1 dr dθ
0 0
R
1 √ 2π √
= (5 5 − 1) dθ = (5 5 − 1)π/6
12 0

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668 Chapter 15
40. zx = 2, zy = 2y, zx + zy + 1 = 5 + 4y 2 ,
2 2
1 y 1 √
S= 5 + 4y 2 dx dy = y 5 + 4y 2 dy = (27 − 5 5)/12
0 0 0
41. ∂r/∂u = cos vi + sin vj + 2uk, ∂r/∂v = −u sin vi + u cos vj,
√ 2π 2 √ √
∂r/∂u × ∂r/∂v = u 4u2 + 1; S = u 4u2 + 1 du dv = (17 17 − 5 5)π/6
0 1
42. ∂r/∂u = cos vi + sin vj + k, ∂r/∂v = −u sin vi + u cos vj,
2v √
√
√ π/2
2 3
∂r/∂u × ∂r/∂v = 2u; S = 2 u du dv = π
0 0 12
43. zx = y, zy = x, zx + zy + 1 = x2 + y 2 + 1,
2 2
π/6 3
1 √ π/6 √
S= x2 + y 2 + 1 dA = r r2 + 1 dr dθ = (10 10 − 1) dθ = (10 10 − 1)π/18
0 0 3 0
R
44. zx = x, zy = y, zx + zy + 1 = x2 + y 2 + 1,
2 2
√
2π 8 2π
26
S= x2 + y2 + 1 dA = r r2 + 1 dr dθ = dθ = 52π/3
0 0 3 0
R
45. On the sphere, zx = −x/z and zy = −y/z so zx + zy + 1 = (x2 + y 2 + z 2 )/z 2 = 16/(16 − x2 − y 2 );
2 2
the planes z = 1 and z = 2 intersect the sphere along the circles x2 + y 2 = 15 and x2 + y 2 = 12;
√
2π 15 2π
4 4r
S= dA = √
√ dr dθ = 4 dθ = 8π
16 − x2 − y 2 0 12 16 − r2 0
R
46. On the sphere, zx = −x/z and zy = −y/z so zx + zy + 1 = (x2 + y 2 + z 2 )/z 2 = 8/(8 − x2 − y 2 );
2 2
2 2
the cone cuts the sphere in the circle x + y = 4;
√
2π 2
2 2r √ 2π √
S= √ dr dθ = (8 − 4 2) dθ = 8(2 − 2)π
0 0 8−r 2
0
47. r(u, v) = a cos u sin vi + a sin u sin vj + a cos vk, ru × rv = a2 sin v,
π 2π π
S= a2 sin v du dv = 2πa2 sin v dv = 4πa2
0 0 0
h 2π
48. r = r cos ui + r sin uj + vk, ru × rv = r; S = r du dv = 2πrh
0 0
h x h y 2 2 h2 x2 + h2 y 2
49. zx = , zy = , zx + zy + 1 = + 1 = (a2 + h2 )/a2 ,
a x2 + y 2 x2 + y 2a a2 (x2 + y 2 )
2π a
√ 2π
a2 + h2 1
S= r dr dθ = a a2 + h2 dθ = πa a2 + h2
0 0 a 2 0