CSM_Chapters14.pdf
September 14, 2017 | Author: Clau Amaiia | Category: N/A
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Chapter 14
Multiple Integrals 14.1
The Double Integral
1. With f (x, y) = x + 3y + 1 and ∆Ak = 1, Z Z (x + 3y + 1)dA ≈ f (1/2, 1/2) + f (3/2, 1/2) + f (5/2, 1/2) + f (1/2, 3/2) R
+ f (3/2, 3/2) + f (5/2, 3/2) + f (1/2, 5/2) + f (3/2, 5/2) = 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 52. 2. With f (x, y) = 2x + 4y and ∆Ak = 1/4, Z Z (2x + 4y)dA ≈ R
1 [2(3/2) + 4(1/2) + 2(2) + 4(1/2) + 2(5/2) + 4(1/2) + 2(1/2) + 4(1) 4 + 2(3/2) + 4(1) + 2(1) + 4(1) + 2(1/2) + 4(3/2) + 2(1) + 4(3/2) + 2(3/2)
+ 4(3/2) + 2(1) + 4(2) + 2(1/2) + 4(2) + 2(1/2) + 4(5/2)] 1 = (3 + 2 + 4 + 2 + 5 + 2 + 4 + 4 + 3 + 4 + 2 + 4 + 1 + 6 + 2 + 6 + 3 4 93 + 6 + 2 + 8 + 1 + 8 + 1 + 10) = . 4 3. (a) With f (x, y) = x + y, and ∆Ak = 1, Z Z (x + y) dA ≈ (−3/2 + 1/2) + (−1/2 + 1/2) + (1/2 + 1/2) + (3/2 + 1/2) R
+ (−3/2 + 3/2) + (−1/2 + 3/2) + (1/2 + 3/2) + (3/2 + 3/2) 1 16 = (−2 + 0 + 2 + 4 + 0 + 2 + 4 + 6) = = 8. 2 2 (b) With f (x, y) + y + 4 and ∆Ak = 1,
139
140
CHAPTER 14. MULTIPLE INTEGRALS Z Z (x + y) dA ≈ (−2 + 1) + (−1 + 1) + (0 + 1) + (1 + 1) + (−2 + 2) + (−1 + 2) + (0 + 2) + (1 + 2) R
= 8. 4. With Z Z f (x, y) = xy and ∆Ak = 1/4, 1 xydA ≈ [0(1/2) + (1/2)(1/2) + (−1/2)(1) + (0)(1) + (1/2)(1) 4 R + (1)(1) + (−1/2)(3/2) + (0)(3/2) + (1/2)(3/2)
y
4
+ (1)(3/2) + (−1/2)(2) + (0)(2) + (1/2)(2) + (1)(2) 1
+ (−1)(5/2) + (−1/2)(5/2) + (0)(5/2) + (1/2)(5/2)
x
+ (1)(5/2) + (3/2)(5/2) + (−1)(3) + (−1/2)(3) + (0)(3) + (1/2)(3) + (1)(3) + (3/2)(3) + (−1)(7/2) + (−1/2)(7/2) + (0)(7/2) + (1/2)(7/2) + (1)(7/2) + (3/2)(7/2)] = 73/16 5.
RR
6.
RR
7.
RR
8.
RR
R R
R
R
10dA = 10
RR
10dA = 10
RR
10dA = 10
RR
10dA = 10
RR
R
dA = 10(6) = 60
R
dA = 10(12) = 120
1 2 dA = 10 π(2) = 10π R 4
1 dA = 10 (5) R 2
5 125 = 2 2
9. No, since x + 5y is negative at (3, −1) which is in R. 10. Yes, since x2 + y 2 is nonnegative on R. 11.
RR
12.
RR
13.
RR
14.
RR
15.
RR
R
10dA = 10
RR
dA = 10(8) = 80
R
−5xdA = −5
R
(2x + 4y)dA = 2
R
(2x + 4y)dA =
R
(3x + 7y + 1)dA = 3
Z Z
R
R
RR R
xdA + 4
xdA − RR R
RR R
RR R
RR
RR R
Z Z
ydA +
R1
f (x, y)dA + RR R1
R
RR R2
f (x, y)dA +
dA = 3(3) + 7(7) + 8 = 66 Z Z
y dA − 4 R
RR
RR
Z Z
2
(2 + y) dA =
f (x, y)dA = f (x, y)dA = −5. R2 R
ydA = 2(3) + 4(7) = 34
ydA = 3 − 7 = −4
xdA + 7
R
f (x, y)dA =
18. Since RR
RR
2
y dA − R
RR
xdA = −5(3) = −15
R
Z Z
2
16. 17.
R
RR
Z Z
dA − 4 R
y 2 dA
ydA − R
= −4(8) − 4(7) = −60
R
f (x, y)dA = 4 + 14 = 18 RR R2
f (x, y)dA, 25 = 30 +
RR R2
f (x, y)dA and
14.2. ITERATED INTEGRALS
14.2 1.
R
Iterated Integrals dy = y + c1 (x)
2. RBy holding y fixed, (1 − 2y)dy = x − 2yx + c2 (y) 3. By 2 3 Z holding y fixed, x √ x √ 2 (6x y − 3x y)dx = 6 y−3 y + c2 (y) 3 2 3 √ = 2x3 y − x2 y + c2 (y) 2 4. By 2 Z holding x fixed, y 3/2 y √ (6x2 y − 3x y)dy = 6x2 − 3x + c1 (x) 2 (3/2) = 3x2 y 2 − 2xy 3/2 + c1 (x) 5. By holding x fixed, R 1 ln |y + 1| dy = + c1 (x) x(y + 1) x 6. By 2 Z holding x fixed, x − 5xy 4 + c2 (y) (1 + 10x − 5y 4 )dx = x + 10 2 = x + 5x2 − 5xy 4 + c2 (y) 7. By Z holding y fixed, sin 4x (12y cos 4x − 3 sin y)dx = 12y − 3x sin y + c2 (y) 4 = 3y sin 4x − 3x sin y + c2 (y) 8. By Z holding x fixed, tan 3xy + c1 (x) sec2 3xydy = 3x 9. By Z holding y fixed, p y √ dy = y 2x + 3y + c2 (y) 2x + 3y 10. By Z holding x fixed, 1 (2x + 5y)7 (2x + 5y)6 dy = + c1 (x) 5 7 (2x + 5y)7 = + c1 (x) 35 Z 3 3 11. (6xy − 5ey ) dx = (3x2 y − 5xey ) −1 = (27y − 15ey ) − (3y + 5ey ) = 24y − 20ey −1
141
142 12.
CHAPTER 14. MULTIPLE INTEGRALS Z
2
tan xy dy = 1
13.
Z
3x
1
Z 14.
15.
y3
√
Z
y
3x 2 x3 exy = x2 exy 1 = x2 (e3x − ex ) y 3 (8x3 y − 4xy 2 ) dx = (2x4 y − 2x2 y 2 ) √y = (2y 13 − 2y 8 ) − (2y 3 − 2y 3 ) = 2y 13 − 2y 8
2x
2x xy x x x 2 2 dy = ln(x + y ) = [ln(x2 + 4x2 ) − ln x2 ] = ln 5 2 2 x +y 2 2 2 0
0
Z
x
e2y/x dy =
16. x3
Z
2 1 1 ln | sec xy| = ln | sec 2x − sec x| x x 1
sec y
17. tan y
3 2 x x 2y/x x x e 3 = (e2y/x − e2x /x ) = (e2 − e2x ) 2 2 2 x
sec y (2x + cos y) dx = (x2 + x cos y) tan y = sec2 y + sec y cos y − tan2 y − tan y cos y = sec2 y + 1 − tan2 y − sin y = 2 − sin y
Z 18.
1
√
y ln x dx
Integration by parts
y
= y(x ln x − Z
1 x)|√
y
π/2
cos x sin3 ydy = cos x cos y
19.
√ √ √ √ = y(0 − 1) − y( y ln y − y) = −y − y y
x
1 ln y − 1 2
π/2 − sin2 x 2 − sin2 y 2 2 = 0 − cos x − − 3 3 x 3 3
cos2 (1 − cos2 x) 2 cos2 x cos2 x sin3 x 2 cos2 x + = + 3 3 3 3 cos2 x cos4 x 2 cos2 x 1 = − + = cos2 x − cos4 x 3 3 3 3 1 1 Z 1 sin xy cos xy xy sin xy cos xy x 2 = 20. y cos xydx = y + + 2y 2 1/2 2 2 1/2 1/2 x x sin 2 cos 2 sin x cos x − sin x2 cos x2 sin x cos x x x x = + − + = + 2 2 2 4 2 4 =
Z
2
Z
x2
Z
2
(8x − 10y + 2) dy dx =
21. 1
−x
x2 (8xy − 5y + 2y) 2
1 Z 2
=
dx
−x
[(8x3 − 5x4 + 2x2 ) − (−8x2 − 5x2 − 2x)] dx
1
Z = 1
2
2 (8x3 − 5x4 + 15x2 + 2x) dx = (2x4 − x5 + 5x3 + x2 ) 1
= 44 − 7 = 37
14.2. ITERATED INTEGRALS Z
1
−1
(x + y)2 dx dy =
0
√
Z
y 1 (x + y)3 dy −1 0 −1 3 0 1 Z 1 Z 1 1 1 1 7 4 3 3 3 = [(y + y) − (0 + y) ] dy = 7y dy = y =0 3 −1 3 −1 3 4 −1
y
Z
22.
143 1
Z
Z
Z √2−y2
2
23. −
0
√
y
(x + y)2 dx
√
Z
2
(2x − y) dx dy = 2−y 2
0
√
Z
2
= 0
π/4
cos x
Z
24. 0
Z
0
π
3y
Z
25. 0
Z
y
2
√
Z
2−y
h i p p (2 − y 2 − y 2 − y 2 ) − (2 − y 2 + y 2 − y 2 ) dy √2 p 2 2 2 4√ (−2 2 − y 2 ) dy = (2 − y 2 )3−2 = (0) − 23/2 = − 2 3 3 3 3 0
cos x Z π/4 (1 + 4y tan2 x) dy dx = = (y + 2y 2 tan2 x) (cos x + 2 cos2 x tan2 x) dx 0 0 0 π/4 Z π/4 1 = (cos x + 2 sin2 x) dx = sin x + x − sin 2x 2 0 0 √ √ 2 2+π−2 2 π 1 = + − = 2 4 2 4 π/4
Z
3y Z 1 π 1 dy sin(2x + y) cos(2x + y) dx dy (sin 7y − sin 3y) dy 2 0 0 2 y π 1 1 1 − cos 7y + cos 3y = 2 7 3 0 1 1 1 1 1 4 − (−1) + (−1) − − + =− = 2 7 3 7 3 21 Z
x 2
26.
2y sin πx 1
2
1
√2−y2 (x2 − xy) √ 2
=
Z
Z
dy =
−
√
Z 0
π
2
Z
"Z
dy dx =
0
√
#
x 2
2y sin πx 1
dy
Z
2
dx =
0
√x Z dx = y sin πx 2
1
2
0
2
x sin πx2 dx
1
2 1 1 1 1 2 =− cos πx = − (cos 4π − cos π) = − (1 − (−1)) = − 2π 2π 2π π 1 Z
ln 3
Z
27. 1
x
6ex+2y dy dx =
0
Z
ln 3
1
x Z ln 3 ln 3 x+2y e (3e3x − 3ex ) dx = (e3x − 3ex ) 1 dx = 1 0
= (27 − 9) − (e3 − 3e) = 18 − e3 + 3e 28.
R 1 R 2y 0
0
e−y
2
dx dy =
R1 0
R1 2 2y 2 2 1 xe−y dy = 0 2ye−y dy = −e−y = −e−1 − (−1) = 1 − e−1 0
0
144
CHAPTER 14. MULTIPLE INTEGRALS Z
3
2x+1
Z
29. x+1
0
2x+1 Z 3 √ √ dx = 2 ( x + 1 − 1) dx 2 y − x
3
Z
1 √ dy dx = y−x
0
0
x+1
3 16 2 2 10 =2 (x + 1)3/2 = 2 −3 − = 3 3 3 3 0
30.
Z
1
y
Z
2
2 3/2
x(y − x ) Z
9
0 x
Z
1 dy dx = x+ y 2
31. 0
1
Z
1/2
y
Z
32. 0
0
Z
e
y
Z
1
1
4
√
Z
−x
2ye
Z
1
6
Z √25−y2 /2
35. 0
0
Z
2
Z √20−y2
36. y2
0
Z
π
37. π/2
1
x Z 9 9 1 π π π −1 y = tan dx = ln |x| = ln 9 x x 0 4x 4 4 1 1 1/2
Z
1 √ dx dy = 1 − x2
x
34. 1
9
Z
y 1 Z 1 2 1 1 1 1 6 2 5/2 5 − (y − x ) = − (−y ) dy = − y = 5 5 30 30 0 0 0
−1
sin 0
y Z 1/2 sin−1 y dy x dy = 0
√x Z 4 dy dx = y e dx = (xe−x − e−x ) dx 1 1 1 4 = (−xe−x − e−x + e−x ) 1 = −4e−4 + e−1 Z
4
2 −x
1 p (25 − y 2 ) − x2
0
x
Integration by parts
! √25−y2 /2 x dx dy = dy sin−1 p 2 25 − y 0 0 Z 6 Z 6 π −1 1 = sin dy = dy = π 2 0 0 6 Z
√
20−y 2
6
2 1 1 4 2 3/2 y dx dy = dy = (y 20 − − y ) dy = − (20 − y ) − y 3 4 0 0 y2 0 √ √ 1 1 40 5 − 76 = − (64) − 4 − − (40 5) − 0 = 3 3 3 Z
2
xy
0 e sin y dx dy = e sin y cos y π/2
Z
Integration by parts
0
√ √ 1/2 p 1 π π + 6 3 − 12 3 = y sin−1 y + 1 − y 2 = −1= + 2 6 2 12 0 y Z e Z e y y ln y dy Integration by parts dx dy = y ln x dy = x 1 1 1 e 1 2 1 2 1 2 1 2 1 1 = y ln y − y = e − e − − = (e2 + 1) 2 4 2 4 4 4 1
33.
Z
dx dy =
0
0
1
Z
Z
Z
2
p
π
x
= (− cos y +
Z
3
π
dy =
cos y cos y π e )|π/2
y2
(sin y − ecos y sin y) d
π/2
= (1 + e−1 ) − (0 + 1) = e−1
14.2. ITERATED INTEGRALS Z
1
145
y 1/3
Z
38.
1
Z
2
6x ln(y + 1) dx dy = 0
0
y1/3 Z dy = 2 2x ln(y + 1)
1
3
0
y ln(y + 1) dy
0
0
Integration by parts 1 1 2 2 = y ln(y + 1) − y + y − ln(y + 1) 2 0 1 1 = (ln 2 + 1 − ln 2) − (0 − 0 + 0 − ln 1) = 2 2
Z
2π
x
Z
Z
2π
(cos x − sin y) dy dx =
39.
π
0
π
x Z (y cos x + cos y) dx =
2π
(x cos x + cos x − 1) dx
π
0
Integration by parts 2π
= (cos x + x sin x + sin x − x)|π = (1 − 2π) − (−1 − π) = 2 − π
Z
3
1/x
Z
40. 1
0
1 dy dx = x+1
3
Z 1
1/x Z 3 Z 3 1 y 1 1 dx = − dx dx = x + 1 0 x x+1 1 1 x(x + 1) 3
= [ln x − ln(x + 1)]|1 = (ln 3 − ln 4) − (0 − ln 2) = ln 3/2
Z
5π/12
√
Z
2 sin 2θ
41. π/12
Z
π/3
1
Z
√2 sin 2θ 5π/12 Z 5π/12 1 2 1 1 r dr dθ = r dθ = sin 2θ − dθ = − (cos 2θ + θ) 2 2 2 π/12 π/12 π/12 1 " ! !# √ √ √ 1 3 5π 3 5π 3 π =− − + − + = − 2 2 12 2 12 2 6 Z
1+cos θ
42.
π/3
Z r dr dθ =
0
3 cos θ
5π/12
0
1+cos θ 1 2 r dθ 2 3 cos θ
π/3 1 1 (1 + 2 cos θ − cos2 θ) dθ = (θ + 2 sin θ − 4θ − 2 sin 2θ) 2 0 2 0 √ √ 1 π = (−π + 3 − 3) = − 2 2 Z
=
π/3
146
CHAPTER 14. MULTIPLE INTEGRALS
43. 44.
y=2x+1
y
y
x=My x=-My
x
x
45.
46.
y
y=x2+1
y
x=M16-y2
x
x
y=-x2
√x Z 4 x2 1 2 1 2 47. x ydydx = x y dx = x x− dx 4 0 x/2 0 2 0 2 x/2 4 Z 4 1 4 1 1 2 1 4 x − x dx = x − x5 = 2 8 8 40 0 0 128 32 = 32 − = 5 5 Z 2 Z 2y Z 2 Z 2 2y 1 1 x2 ydxdy = x3 y dy = y(8y 3 − y 6 )dy 0 y2 0 3 0 3 y2 2 Z 2 8 4 1 7 8 5 1 8 = y − y dy = y − y 3 3 15 24 0 0 256 32 32 − = = 3 Z 5Z Z Z √ 15 Z
4
√
Z
x
Z
2
4
4
x
2
2y
x2 ydydx =
Therefore 0
x/2
x2 ydxdy
0
y2
y
y=Mx y=1/2x
x
y
x=y2 x=2y
x
14.2. ITERATED INTEGRALS
Z
1
√
Z
48.
1−x2
2xdydx =
√
− 1−x2
0
√1−x2 dx 2xy √ 2
1
Z Z
0 1
=
147
y
− 1−x
2x
p
1 − x2 + 2
y=M1-x2
p 1 − x2 dx
0 1
Z =
4x
p
1−
x2 dx
0
Z
1
4 = 3 Z √1−y2 Z 2xdxdy =
−1
√ 1
−1
0
x2
1 4 2 3/2 = − (1 − x ) 3 0
1−y 2
Z
y=-M1-x2
1
dy =
x
y
(1 − y 2 )dy
−1
0
1 1 = y − y 3 3 −1 1 4 1 − −1 + = = 1− 3 3 3 Z Z √ 2 2
Therefore,
Z
1
√
Z
√ − 1−x2
0
49.
1−x
x=0
x=M1-y2 x
1−y
1
2xdydx =
2xdxdy −1
0
3 Z 2 2 3x2 dx = x3 −1 = 8 − (−1) = 9 x dy dx = x y dx = −1 −1 0 −1 0 2 Z 3Z 2 Z 3 Z 3 Z 3 8 1 1 3 3 2 x dy = + dy = 3 dy = 3y|0 = 9 x dx dy = 3 3 0 0 0 −1 0 3 −1 2
Z
Z
2
Z
3
2
4
Z
50.
2
Z
2
Z
2
(2x + 4y) dx dy = −2
−2 Z 2
2
= −2
Z
4
Z
2
Z
4
(2x + 4y) dy dx = 2
−2
2 Z 4
= 2
Z
3Z
51. 1
0
π
4 Z (x + 4xy) dy =
2
2
2
[(16 + 16y) − (4 + 8y)] dy
−2
2 (12 + 8y) dy = (12y + 4y 2 ) −2 = (24 + 16) − (−24 + 16) = 48
2 (2xy + 2y ) 2
−2
Z
4
[(4x + 8) − (−4 + 8)] dx
dx = 2
4 8x dx = 4x2 2 = 64 − 16 = 48
π Z 3 2 3π 2 3 2 2 x y − 4 cos y dx = x − 4 − (4) dx (3x y − 4 sin y) dy dx = 2 2 1 1 0 2 3 Z 3 2 3π 2 π 3 = x − 8 dx = x − 8x 2 2 1 1 2 2 27π π = − 24 − − 8 = 13π 2 − 16 2 2 2
Z
3
148
CHAPTER 14. MULTIPLE INTEGRALS Z
π
3
Z
3 (x3 y − 4x sin y) 1 dy =
0
1
0
π
Z
(3x2 y − 4 sin y) dy dx =
π
Z = 0
π
Z
[(27y − 12 sin y) − (y − 4 sin y)]dy 0
π (26y − 8 sin y)dy = (13y 2 + 8 cos y) 0
= (13π 2 − 8) − (8) = 13π 2 − 16 2 Z 1 4 x2 52. dy = 8y ln 3 − 2 dxdy = 8y ln |x + 1| − 2 dy y + 1 0 y +1 0 0 0 0 1 = (4y 2 ln 3 − 4 tan−1 y) 0 = 4 ln 3 − π 1 Z 2 Z 2Z 1 Z 2 8y 4 4y 2 2x π − 2 dydx = − 2x tan−1 y = − dx x+1 y +1 x+1 x+1 2 0 0 0 0 0 π 2 = 4 ln |x + 1| − x2 = 4 ln 3 − π 4 0 Rβ Rβ 53. We use the fact that α kF (t)dt = k α F (t)dt. Then "Z # "Z # "Z # Z Z Z Z
1
2
Z
8y 2x − 2 x+1 y +1
d
Z
b
Z
∞
Z
54.
∞
a
b
g(y)
−(2x2 +3y 2 )
Z
a ∞
∞
Z
d
f (x)dx
g(y)dy
a
c
2 2 xe−2x ye−3y dxdy 0 0 Z Z ∞ ∞ −2x2 −3y 2 = xe dx · ye dy
dxdy =
0
b
f (x)dx dy =
c
xye 0
d
f (x)g(y)dxdy = c
1
0
0
=
a
Z lim
a→∞
2 xe−2x dx ·
0
Z lim
b→∞
!
b
−3y 2
ye
dy
0
! ! 2 2 a e−3y e−2x = lim − · lim − a→∞ b→∞ 4 6 0 #! " #! " 2 2 1 −e−3b 1 −e−2a + · lim + = lim a→∞ b→∞ 4 4 6 6 1 1 1 = · = 4 6 24
14.3
Evaluation of Double Integrals
Z Z
x3 y 2 dA =
1. R
Z 0
1
Z
x
x3 y 2 dydx =
0
1 1 7 1 = x = 21 0 21
Z 0
1
x Z 1 1 6 1 3 3 x y dx = x dx 3 3 0 0
y
x
1
x
14.3. EVALUATION OF DOUBLE INTEGRALS Z Z
2
Z
2.
Z
4−x
149 2
Z (x + 1)dydx =
(x + 1)dA = R
x
0
0
4−x dx (xy + y)
y
x
2
Z
[(4x − x2 + 4 − x) − (x2 + x)]dx
=
4-x
0 2
Z
(2x − 2x2 + 4)dx
= 0
=
x
2 20 2 x − x3 + 4x = 3 3 0 2
Z Z
Z
3.
1
x2
Z
(2x + 4y + 1)dA = R
x
(2x + 4y + 1)dydx Z
y
x3
0 1
x2 (2xy + 2y + y) dx 3
x2
2
= 0 Z 1
=
x3
x
x
[(2x3 + 2x4 + x2 ) − (2x4 + 2x6 + x3 )]dx
0
Z
1 3
2
6
(x + x − 2x )dx =
= 0
1 1 4 1 3 2 7 x + x − x 4 3 7 0
1 1 2 25 + − = 4 3 7 84 x Z Z Z 1Z x Z 1 4. xey dA = xey dydx = xey dx =
R
0
0
0
y
0
1
Z
(xex − x)dx
=
Integration by parts
x
0
1 1 1 2 1 x x = xe − e − x = e − e − − (−1) = 2 2 2 0
Z Z
Z
5.
2
Z
8
2xydA = R
2
Z 2xydydx =
x3
0
Z = 0
2
(64x − x7 )dx =
0
8 xy dx 3 2
1
x
y
x
2 1 32x2 − x8 = 96 8 0
x3
x
150
CHAPTER 14. MULTIPLE INTEGRALS
Z Z 6. R
x √ dA = y
1
Z
3−x2
Z
xy
Z
1
dydx =
x2 +1
−1
Z
−1/2
−1
3−x2 √ 2x y dx 2
y
x +1
x2+1
3
1
(x
=2
p
3 − x2 − x
p
x2 + 1)dx
−1 3-x2
1 1 1 = 2[− (3 − x2 )3/2 − (x2 + 1)3/2 ] 3 3 −1 2 3/2 3/2 3/2 3/2 = − [(2 + 2 ) − (2 + 2 )] = 0 3 Z Z 7. R
1
Z
y dA = 1 + xy
1
Z 0
Z0
y dxdy = 1 + xy
Z 0
1
x
1 ln(1 + xy) dy
y
0
1
1
1 ln(1 + y)dy = [(1 + y) ln(1 + y) − (1 + y)]|0
= 0
= (2 ln 2 − 2) − (−1) = 2 ln 2 − 1
1 2
y Z 2 πx y πx dxdy = − cos dy y x y 0 R 1 0 1 Z 2 y y = − cos πy + dy Integration by parts π π 1 2 y 1 y 2 = − 2 sin πy − 3 cos πy + π π 2π 1 1 2 1 1 3π 2 − 4 = − 3+ − + = 3 π π π 2π 2π 3 x Z Z p Z √3 Z x p Z √3 p 9. x2 + 1dA = x2 + 1dydx = y x2 + 1 dx R 0 0 −x Z Z
8.
sin
πx dA = y
Z
2
Z
y2
Z
3
(x 0
Z =
10.
π/4
Z
p
3
2x 0
=
√
y2
x
y y=x
−x
=
Z
y
sin
√
Z Z
x
x2 + 1 + x
p
p
x2 + 1)dx x
√3 2 2 3/2 x2 + 1dx = (x + 1) 3
y=-x
0
2 3/2 14 (4 − 13/2 ) = 3 3 1
Z
π/4
1 1 2 x 2
1 xdA = xdxdy = dy = 2 R 0 tan y 0 tan y π/4 Z π4 1 1 2 (2 − sec y)dy = (2y − tan y) = 2 0 2 0 π 1 1 π = −1 = − 2 2 4 2
Z 0
y
π4
(1 − tan2 y)dy π/4
x=tany
1
x
14.3. EVALUATION OF DOUBLE INTEGRALS Z Z
Z
11.
4
Z
2
2
Z
4
(x + y)dxdy
(x + y)dxdy +
(x + y)dA =
2
0
0
0
R
Z
151
2 4 Z 2 1 2 1 2 = x + xy dy + x + xy dy 2 2 0 0 0 2 Z 4 Z 2 4 2 = (2 + 2y)dy + [(8 + 4y) − (2 + 2y)]dy = (2y + y 2 ) 0 + (6y + y 2 ) 0 Z
4
0
0
= 24 + 16 = 40 Z Z
Z
12.
4
Z
R
4
Z
3
Z
(x + y)dxdy −
(x + y)dA = 0
0
3
(x + y)dxdy 1
1
4 3 Z 3 1 2 1 2 = x + xy dy − x + xy dy 2 2 0 1 0 1 Z 4 Z 3 9 1 = (8 + 4y)dy − + 3y − + y dy 2 2 0 1 2 4 2 3 = (8y + 2y ) − (4y + y ) = 64 − (21 − 5) = 48 Z
4
0
Z
3Z
2x−x
2
13. A = −x
y
(2x − x2 + x)dx
dydx = 0
1
3
Z 0
=
2x-x2
3 9 3 2 1 3 x − x = 2 3 2 0
x
-x
14. Using symmetry, Z
1
Z
y
2−y 2
A=2
Z dxdy = 2
y2
0
0
1
y2
1 8 2 3 2 2 (2−y −y )dy = 2 2y − y = . 3 3 0
2-y2 x
Z
4
Z
ex
15. A =
Z dydx =
1 4
ln x
1
4
4
y
(ex − ln x)dx = (ex − x ln x + x)|1
ex
= (e − 4 ln 4 + 4) − (e + 1) = e4 − e − 4 ln 4 + 3
lnx
10 4
x
152
CHAPTER 14. MULTIPLE INTEGRALS 4
Z
√ (2− x)2
0 4
Z
4−x
Z
16. A = =
4
Z
[4 − x − (2 −
dydx =
√
y
x)2 ]dx
0
√ (4 x − 2x)dx =
0
4 8 3/2 16 x − x2 = 3 3 0
4-x
(2-Mx)2 x
Z
1
Z
−2x+3
x3
−2
=
Z
1
dydx =
17. A =
(−2x + 3 − x3 )dx
−2
y
-2x+3
1 1 4 7 63 −x + 3x − x = − (−14) = 4 4 4 −2 2
x3
x
18. Expressing y = −x2 + 3x and y = −2x + 4 as functions of 3 1√ 1 y, we have x = − 9 − 4y and x = 2 − y. 2 2 2 Z 2 Z 2−y/2 Z 2 y 3 1p − A= dxdy = 2 − − 9 − 4y dy √ 2 2 2 0 3/2− 9−4y/2 0 2 1 1 1 27 13 1 y − y 2 − (9 − 4y)3/2 = − − − = = 2 4 12 12 12 6 0
19. The correct integral is ( c). Z 2 Z √4−y2 Z V =2 (4 − y)dxdy = 2 −2
0
p = 2 2y 4 − y 2 + 8 sin−1
20. The correct integral is (b).
y
2-y/2
3/2-1/2 M9-4y
√ 4−y2 Z 2 p (4 − y)x dy = 2 (4 − y) 4 − y 2 dy −2 −2 0 2 y 1 + (4 − y 2 )3/2 = 2(4π − (−4π)] = 16π 2 3 −2 2
x
14.3. EVALUATION OF DOUBLE INTEGRALS Z
r
Z √r2 −y2
V =8 0
(r2 − y 2 )1/2 dxdy = 8
0
0
r
2 2 =8 (r − y )dy = 8 ry − 0 3 r3 2r 3 =8 r − =8 = 3 3 Z
Z
r
153
√r2 −y2 (r2 − y 2 )1/2 x dy 0
r y 3 0 3
16 3 r 3
21. Setting z = 0 we have y = 6 − 2x. Z Z 3 Z 6−2x (6 − 2x − y)dydx = V =
6−2x 1 2 6y − 2xy − y dx 2 0 0 0 0 Z 3 Z 3 1 6(6 − 2x) − 2x(6 − 2x) − (6 − 2x)2 dx = = (18 − 12x + 2x2 )dx 2 0 0 3 2 = 18x − 6x2 + x3 = 18 3 3
y
6-2x
0
22. Setting z = 0 we have y ± 2. R3R2 V = (4 − y 2 )dydx = 0 0 R 3 16 dx = 16 0 3
x
y
R3 0
2 1 4y − y 3 dx 3
=
0
x
1 1 23. Solving for z, we have x = 2 − x + y. Setting z = 0, we 2 2 see that this surface (plane) intersects the xy-plane in the line y = x − 4. Since z(0, 0) = 2 > 0, the surface lies above the xy-plane over the quarter-circular region. Z 2 Z √4−x2 1 1 V = 2 − x + y dydx 2 2 0 0 √4−x2 Z 2 1 1 = 2y − xy + y 2 dx 2 4 0 0 Z 2 p 1 p 1 = 2 4 − x2 − x 4 − x2 + 1 − x2 dx 2 4 0 p 2 1 1 3 −1 x 2 3/2 2 = x 4 − x + 4 sin + (4 − x ) + x − x 2 6 12 0 2 4 = 2π + 2 − − = 2π 3 3
y
M4-x2 2
2
x
154
CHAPTER 14. MULTIPLE INTEGRALS y
24. Setting z = 0 we have y = 3. Using symmetry, 3 Z √3 Z √3 Z 3 1 2 (3 − y)dydx = 2 (3y − y ) dx V =2 2 2 x2 0 0 x √ √3 Z 3 9 1 4 9 1 5 2 3 =2 ( − 3x + x )dx = 2 x−x + x 2 2 2 10 0 0 √ √ √ √ 9 24 3 9 3−3 3+ 3 = =2 . 2 10 5
3 x2
x
25. Note that z = 1 + x2 + y 2 is always positive. Then 3−3x Z 1 Z 3−3x Z 1 1 3 2 2 2 (1 + x + y )dydx = dx V = y+x y+ y 3 0 0 0 0 Z 1 [(3 − 3x) + x2 (3 − 3x) + 9(1 − x)3 ]dx =
y
3-3x
0
Z =
1
(12 − 30x + 30x2 − 12x3 )dx
0 x
1 = (12x − 15x2 + 10x3 − 3x4 ) 0 = 4.
26. In the first octant, z = x + y is nonnegative. Then √9−x2 Z 3 Z √9−x2 Z 3 1 V = (x + y)dydx = xy + y 2 dx 2 0 0 0 0 Z 3 p 9 1 2 2 = x 9 − x + − x dx 2 2 0 3 9 1 1 = − (9 − x2 )3/2 + x − x3 3 2 6 0 27 9 = − − (−9) = 18. 2 2
y
3
M9-x2
3
x
y 27. In the first octant z = 6/y is positive. Then R6R5 6 R 6 6x 5 R 6 dy 6 dy = 30 V = 1 0 dxdy = 1 = 30 ln y|1 = 30 ln 6. 6 1 y y y 0
1
5
x
14.3. EVALUATION OF DOUBLE INTEGRALS
155
28. Setting z = 0, we have x2 /4 + y 2 /16 = 1. Using symmetry, Z
2
y
√
Z
2 4−x2
1 (4 − x2 − y 2 )dydx 4 0 0 2√4−x2 Z 2 1 (4y − x2 y − y 3 ) =4 dx 12 0 0 Z 2 p p 2 =4 [8 4 − x2 − 2x2 4 − x2 − (4 − x2 )3/2 ]dx 3 0
V =4
4 2M4-x2
Trig substitution p p x 1 x = 4 4x 4 − x2 + 16 sin−1 − x(2x2 − 4) 4 − x2 − 4 sin−1 2 4 2 i 2 p 1 x + x(2x2 − 20) 4 − x2 −4 sin 2 0 12 16π 4π 4π =4 − − − (0) = 16π. 2 2 2 29. Note that z = 4−y 2 is positive for |y| ≤ 1. Using symmetry, √2x−x2 Z 2 Z √2x−x2 Z 2 1 dx V =2 (4 − y 2 )dydx = 2 (4y − y 3 ) 3 0 0 0 0 p Z p 1 2 2 2 2 = 2 0 4 2x − x − (2x − x ) 2x − x dx 3 Z 2 p p 1 =2 (4 1 − (x − 1)2 − [1 − (x − 1)2 ] 1 − (x − 1)2 )dx 3 0
2
x
y 1
M2x-x2
u = x − 1, du = dx Z 1 p Z 1 p p 1 1 p 11 =2 1 − u2 + u2 1 − u2 du [4 1 − u2 − (1 − u2 ) 1 − u2 ]du = 2 3 3 3 −1 −1 Trig substitution 1 p 11 p 11 1 1 −1 2 2 2 =2 u 1−u + sin u + x(2x − 1) 1 − u + sin u 6 6 24 24 −1 1 π 11 π 1 π 15 11 π =2 + − − − = . 6 2 24 2 6 2 24 2 4
x
156
CHAPTER 14. MULTIPLE INTEGRALS
30. From z = 1 − x2 and z = 1 − y 2 we have 1 − x2 = 1 − y 2 or y = x (in the first octant). Thus, the surfaces intersect in the plane y = x. Using symmetry, 1 Z 1 Z 1Z 1 1 3 2 y − y dx 1 − y dydx = 2 V =2 3 0 x 0 x Z 1 2 1 =2 − x + x3 dx 3 3 0 1 1 1 2 1 4 2 x− x + x = . =2 3 2 12 2
z
1
1
y
y=x
1
0
x
31. From z = 4−x−2y and z = x+y, we have 4−x−2y = x+y 3 or x = 2 − y. 2 Z Z 4/3
z 4
2−3y/2
[4 − x − 2y) − (x + y)]dxdy
V = 0
0
2−3y/2 = 4x − x − 3xy dy 0 0 2 # Z 4/3 " 3 3 3 = 4(2 − y) − 2 − y − 3 2 − y y dy 2 2 2 0 Z 4/3 9 = 4 − 6y + y 2 dy 4 0 4/3 16 3 = = 4y − 3y 2 + y 3 4 9 0 Z
4/3
2
2
y
x=2-3y/2
x
32. Using symmetry, √9−x2 Z 3 Z √9−x2 Z 3 1 3 2 2 2 V =4 (9 − x − y )dydx = 4 [(9 − x )y − y ] 3 0 0 0 0 Z 8 3 2 3/2 0 (9 − x ) dx Trig substitution = 3 3 p 8 x 243 8 243 π 81π x = [− (2x2 − 45) 9 − x2 + sin−1 ] = ( )= . 3 8 8 3 3 8 2 2
z 9
y
0
x
33. From z = x2 and z = −x + 2 we have x2 = −x + 2 or x = 1 (in the first octant). Then 1 Z 5Z 1 Z 5 1 2 1 3 2 V = (−x + 2 − x )dxdy = (− x + 2x − x ) dy 2 3 0 0 0 0 Z 5 7 35 = dy = . 6 0 6
y=M
9-x2
z
y x
14.3. EVALUATION OF DOUBLE INTEGRALS
157
34. From 2z = 4 − x2 − y 2 and z = 2 − y we have z 4 − x2 − y 2 = 4 − 2y or x2 + (y − 1)2 = 1. We find the volume in√the first octant and use symmetry. Z 2 Z 1−(y−1)2 1 2 1 2 2 − x − y − (2 − y) dxdy V =2 2 2 0 0 √ 1−(y−1)2 Z 2 1 1 dy =2 − x3 − xy 2 + xy y 6 2 0 0 x Z 2 p p 3/2 1 2 1 − 1 − (y − 1)2 =2 − y 1 − (y − 1)2 + y 1 − (y − 1)2 dy 6 2 0 Z 2 p 1 1 2 3/2 2 2 − [1 − (y − 1) ] + (2y − y ) 1 − (y − 1) dy =2 6 2 0 Z 2 1 1 =2 − [1 − (y − 1)2 ]3/2 + [1 − (y − 1)2 ]3/2 dy 6 2 0 Z 2 2 [1 − (y − 1)2 ]3/2 dy Trig substitution = 3 0 2 p 2 3 π 3 π 2 3 π y−1 −1 2 2 = [2(y − 1) − 5] 1 − (y − 1) + sin (y − 1) = − − = − 3 8 8 3 82 8 2 4 0 √ 35. Solving x = y 2 for y, we obtain y = x. Thus, y
2Z
Z
y
2
4Z
Z f (x, y)dxdy =
0
0
2
√
0
f (x, y)dydx. x
x=y2
36. Solving x = Thus, Z Z √
p
25−y 2
5
5
Z f (x, y)dxdy =
−5
x
√ 25 − y 2 for y, we obtain y = ± 25 − x2 .
0
0
√
Z
y x=M25-y2
25−x2
√ − 25−x2
f (x, y)dydx. x
x
37. Solving y = ex for x, we obtain x = ln y. Thus, Z
3
Z
ex
Z
e3
Z
3
f (x, y)dydx = 0
1
f (x, y)dxdy. 1
y
y=ex
ln y
3
x
158
CHAPTER 14. MULTIPLE INTEGRALS
38. Solving x = 3 − y and x = y/2 for y, we obtain y = 3 − x and y = 2x. Thus, Z 3 Z 3−x Z 1 Z 2x Z 2 Z 3−y f (x, y)dydx. f (x, y)dydx+ f (x, y)dxdy = y/2
0
0
0
y
x=y/2
0
1
x=3-y
x
√ 39. Solving y = 3 x and y = 2 − x for x, we obtain x = y 3 and x = 2 − y. Thus, 1
Z 0
√ 3
Z
x
2
Z f (x, y)dydx+
0
2−x
Z
Z
1
3 y=M x
2−y
Z
f (x, y)dxdy.
f (x, y)dydx = 0
1
y
y3
0
y=2-x
x
√ √ 40. Solving x = y and x = 2 − y for y, we obtain y = x2 and y = 2 − x2 . Thus, 1
Z
√
Z
y
Z
2
√
Z
2−y
f (x, y)dxdy+
1
Z
2−x2
f (x, y)dxdy =
0
0
Z
0
1
y
f (x, y)dydx. x2
0
x=My 1 x=M2-y
Z
1
Z
1
41.
x 0
x
2
p
y 1 3p x 1 + y 4 dy 1 + y 4 dydx = y 0 0 3 0 0 1 Z 1 p 1 1 1 (1 + y 4 )3/2 = y 3 1 + y 4 dy = 1 3 0 3 6 0 1 √ = (2 2 − 1) 18 Z
1
Z
y
Z p x 1 + y 4 dxdy =
x
1
2
y=x
x
Z
1
Z
2
42.
e 0
−y/x
Z
2
Z
dxdy =
2y
x/2
e 0
Z = 0
−y/x
2
Z
−xe
dydx =
0 2
(−xe−1/2 + x)dx =
x/2
−y/x
0
Z
0
2
y x=2y
(1 − e−1/2 )xdx
0
2 1 −1/2 2 )x = 2(1 − e−1/2 ) = (1 − e 2 0
x
14.3. EVALUATION OF DOUBLE INTEGRALS
Z
2
√x cos x3/2 dxdy = cos x3/2 dydx = y cos x3/2 dx 0 y2 0 0 0 4 Z 4 √ 2 2 x cos x3/2 dx = sin x3/2 = sin 8 = 3 3 0 0 4
Z
43. 0
Z
Z
1−x2
√ − 1−x2
−1
4
Z
√
1
44.
Z
√ x
Z
Z p x 1 − x2 − y 2 dydx =
1
1
Z
1
0
−
x=y2
x
y y=M1-x2
x
1−y
y=-M1-x2
1
Z
(0 − 0)dy = 0 −1
1 dydx = 1 + y4
x
y
p 2 2 √ 2 x 1 − x − y dxdy −1 − 1−y √ 1−y2 Z 1 1 = [− (1 − x2 − y 2 )3/2 ] √ dy 3 2 1 =− 3
45.
4
Z √1−y2
−1
Z
159
Z
y x dy 4 0 1+y 0 0 1 π y 1 −1 2 dy = tan y = 4 1+y 2 8 0
1
Z
0
Z
1
= 0
y
1 dxdy = 1 + y4
Z
1
y y=x
x
Z
4
Z
46. 0
2
√
p
y
x2 x3 + 1dxdy = y x3 + 1 dx 0 0 0 0 2 Z 2 p 2 = x2 x3 + 1dx = (x3 + 1)3/2 9 Z
2
Z
x2
Z p x3 + 1dydx =
0
2
p
y
x=My
0
2 52 = (93/2 − 13/2 ) = 9 9
47. fave
1 = A
Z
1 A
Z
=
d
c
c
b
Z a
d
2
1 xydxdy = A
d
Z
2
(b − a )y 1 dy = 2 A
c
b x2 y dy 2 a
d (b − a2 )y 2 4 c 2
1 (b2 − a2 )(d2 − c2 ) = A 4 But A = (b − a)(d − c), so fave =
(b2 − a2 )(d2 − c2 ) (b + a)(d + c) = 4(b − a)(d − c) 4
x
160
CHAPTER 14. MULTIPLE INTEGRALS √
48. fave
1 = A
Z
3
√ − 3
Z √9−3y2 −
√
9 − x2 − 3y 2 dxdy
9−3y 2
√9−3y2 1 x = dy − 3y 2 x √ 9x − − √ A − 3 3 2 Z
√ 3
3
−
√
9−3y
3
p p 1 (9 − 3y 2 )3/2 2 2 2 = 9 9 − 3y − − 3y 9 − 3y A −√3 3 p p (9 − 3y 2 )3/2 − −9 9 − 3y 2 + + 3y 2 9 − 3y 2 dy 3 Z √3 p 1 9 − 3y 2 9 − 3y 2 2 2 2 = 9 − 3y 9 − − 3y + 9 − − 3y dy A −√3 3 3 Z √3 p 1 9 − 3y 2 (12 − 4y 2 )dy = A −√3 √ " √ ! √ !# 3 √ √ p 1 12 9 4 81 3y p 3y 3y 3y √ = 9 − 3y 2 + sin−1 − √ (6y 2 − 9) 9 − 3y 2 + sin−1 √ A 2 2 3 3 8 9 3 3 3 − 3 √ 27π 3 = 2A 49. Let S be the solid with base R and height described by the function f (x, y). The volume of S is equal to the volume of the solid with base R and constant height fave . Z Z Z dZ b Z d 1 50. (a) cos 2π(x + y)dA = [sin 2π(b + y) − sin 2π(a + b)]dy cos 2π(x + y)dxdy = 2π c R c a Z d 1 [(sin 2πb cos 2πy + cos 2πb sin 2πy) − (sin 2πa cos 2πy + cos 2πa sin 2πy)] dy = 2π c Z d 1 = (S1 cos 2πy + C1 sin 2πy)dy 2π c 1 1 [S1 (sin 2πd − sin 2πc) − C1 (cos 2πd − cos 2πc)] = (S1 S2 − C1 C2 ) = 2 2 4π 4π Z Z Z dZ b Z d 1 sin 2π(x + y)dA = sin 2π(x + y)dxdy − [cos 2π(b + y) − cos 2π(a + y)]dy 2π R c a c Z d 1 =− [(cos 2πb cos 2πy − sin 2πb sin 2πy) − (cos 2πa cos 2πy − sin 2πa sin 2πy)]dy 2π c Z d 1 (C1 cos 2πy − S1 sin 2πy)dy =− 2π c 1 1 = − 2 [C1 (sin 2πd − sin 2πc) + S1 (cos 2πd − cos 2πc)] = − 2 (C1 S2 + S1 C2 ) 4π 4π (b) If b − a = n is an integer, then b = a + n and Z
sin 2πb = sin 2π(a + n) = sin 2πa cos 2πn + cos 2πa sin 2πn = sin 2πa cos 2πb = cos 2π(a + n) = cos 2πa cos 2πn − sin 2πa sin 2πn = cos 2πa.
14.4. CENTER OF MASS AND MOMENTS
161
RR R In this case, S1 = 0 and C1 = 0, so cos 2π(x+y)dA = 0 and sinR sin 2π(x+y)dA = R 0. Similarly, is d − c is an integer, the double integrals are zero. (c) If both integrals are 0, then 0 = (S1 S2 − C1 C2 )2 + (C1 S2 + S1 C2 )2 = S12 S22 + C12 C22 + C12 S22 + S12 C22 = (S12 + C12 )(S22 + C22 ). Thus, either S12 + C12 = 0, in which case S1 = C1 = 0, or S22 + C22 = 0, in which case S2 = C2 = 0. Suppose S1 = C1 = 0, and b − a = k or b = a + k. We want to show that k is an integer. Consider S1 = sin 2πb − sin 2πa = sin 2π(a + k) − sin 2πa = sin 2πa cos 2πk + cos 2πa sin 2πk − sin 2πa C1 = cos 2πb − cos 2πa = cos 2π(a + k) − cos 2πa = cos 2πa cos 2πk − sin 2πa sin 2πk − cos 2πa S1 − C1 = (sin 2πa − cos 2πa) cos 2πk + (sin 2πa + cos 2πa) sin 2πk − (sin 2πa − cos 2πa) = (sin 2πa − cos 2πa)(cos 2πk − 1) + (sin 2πa + cos 2πa) sin 2πk. Since a is arbitrary we must have cos 2πk − 1 = 0 and sin 2πk = 0, which implies k is an integer. Similarly, if S2 = C2 = 0, d − c must be an integer. RR RR 51. By Problem 50 (a) we have cos 2π(x + y)dA = sin 2π(x + y)dA = 0 for k = Rk Rk 1, 2, · · · , n. Then Z Z Z Z Z Z cos 2π(x + y)dA = cos 2π(x + y)dA + · · · + cos 2π(x + y)dA = 0 + · · · + 0 = 0 R
R1
and Z Z
Rn
Z Z
Z Z sin 2π(x + y)dA + · · · +
sin 2π(x + y)dA = R
R1
sin 2π(x + y)dA = 0 + · · · + 0 = 0. Rn
Therefore by Problem 45 (c), at least one of the two sides of R must have integer length.
14.4
Center of Mass and Moments 3
Z
4
Z
1. m =
4 Z 3 1 2 x y dy = 8ydy 2 0 0
3
xydxdy = 0
=
Z
0
0
3 4y 2 0 = 36 Z 3Z 4 2
y
3
My =
x=4
Z
x ydxdy =
0
Z = 0
0 3
0
3
4 1 3 x y dy 3 0
3 64 32 2 ydy = y = 96 3 3 0
x
162
CHAPTER 14. MULTIPLE INTEGRALS Z
3
Z
4
4 1 2 2 x y 2 0
3
Z
2
xy dxdy =
Mx = 0
0
Z = 0
3
0
3 8 8y 2 dy = y 3 = 72 3 0
x = My /m = 96/36 = 8/3; y = Mx /m = 72/36 = 2. The center of mass is (8/3, 2). Z 2 Z 4−2x Z 2 Z 2 2 2 4−2x y x dydx = 2. m = x y0 dx = x2 (4 − 2x)dx 0
0
0
0
2 4 3 1 4 32 8 (4 − 22 − 2x3 )dx = = x − x = −8= 4 3 2 3 3 0 4−2x 0 Z 2 Z 3 Z 2 Z 4−2x x3 (4 − 2x)dx dx = x3 y x3 dydx = My = y=4-2x 0 0 0 0 0 2 Z 2 2 (4x3 − 2x4 )dx = x4 − x5 = 5 0 0 64 16 = 16 − = 2 5 5 4−2x Z 2 Z 2 Z 4−2x Z 2 1 1 2 2 x y Mx = dx = x2 (4 − 2x)2 dx x2 ydydx = 2 0 0 0 0 2 0 2 Z Z 2 1 5 1 2 4 3 4 2 3 4 2 3 4 x −x + x = (16x − 16x + 4x )dx = 2 (4x − 4x x )dx = 2 2 0 3 5 0 0 32 32 32 − 16 + = =2 15 15 15 32/15 x = My /m = 16/5 = 6/5; y = Mx /m = = 4/5. 8/3 The center of mass is (6/5, 4/5). Z
2
3. Since both the region and ρ are symmetric with respect to the line x = 3, x = 3. 6−y Z 3 Z 6−y Z 3 m= 2ydxdy = 2xy 0 y 0 y Z 3 Z 3 = 2y(6 − y − y)dy = (12y − 4y 2 )dy 0
0
y
0 4
3 x=y
3
6−y Z 2xy 2 dxdy = y
x=6-y
x
3
2y 2 (6 − y − y)dy =
0
3 = (4y − y ) 0 = 27 y = Mx /m = 27/18 = 3/2. The center of mass is (3, 3/2). 3
y
0
3 4 = 6y 2 − y 3 = 18 3 0 Z Z 3 Z 6−y Mx = 2y 2 dxdy =
x
Z 0
3
(12y 2 − 4y 3 )dy
14.4. CENTER OF MASS AND MOMENTS
163 y
4. Since both the region and ρ are symmetric with respect to the y-axis, x = 0. Using symmetry,
3
y x=y 1 3 2 (x + y )dxdy = m= x + xy dy 3 x 0 0 0 0 3 Z 3 Z 3 1 1 3 4 y 3 dy = y 4 = 27 = y + y 3 dy = 3 3 3 0 0 0 y Z 3 Z 3 Z Z 3Z y 1 3 1 4 4 3 4 (x2 y + y 3 )dxdy = y dy x y + xy 3 dy = y + y 4 dy = Mx = 3 3 3 0 0 0 0 0 0 3 4 5 324 = y = 15 0 5 324/5 = 12/5. The center of mass is (0, 12/5). y = Mx /m = 27 Z
3
Z
y
2
2
3
Z
x2 y 1 2 5. m = (x + y)dydx = xy + y dx 2 0 0 0 0 1 Z 1 1 7 1 1 1 4 = x3 + x4 dx = x + x5 = 2 4 10 20 0 0 x2 Z 1 Z x2 Z 1 1 2 2 2 My = (x + xy)dydx = x y + xy dx 2 0 0 0 0 1 Z 1 1 17 1 1 = x4 + x5 dx = x5 + x6 = 2 5 12 60 0 0 x2 Z 1 Z 1 Z x2 Z 1 1 5 1 6 1 2 1 3 2 Mx = xy + y = x + x dx (xy + y )dydx = 2 3 2 3 0 0 0 0 0 1 11 1 1 6 x + x7 = = 12 21 84 0 17/60 11/84 x = My = m = = 17/21; y + Mx /m = = 55/147. 7/20 7/20 The center of mass is (17/21, 55/147). Z
1
Z
x2
Z
1
√x 1 2 6. m = (y + 5)dydx = ( y + 5y) dx 0 0 0 2 0 4 Z 4 √ 1 1 2 10 3/2 92 = x + 5 x dx = x + x = 3 2 4 3 0 0 √x Z 4 Z √x Z 4 1 2 My = (xy + 5x)dydx = xy + 5xy dx 2 0 0 0 0 4 1 3 224 = x + 2x5/2 = 6 3 0 Z
4
√
Z
x
Z
y=x2
1
x
y
4
2
y=Mx
4
x
164
CHAPTER 14. MULTIPLE INTEGRALS Z
4
√
Z
Mx = 0
x
(y 2 + 5y)dydx =
4
Z 0
0
√x Z 4 1 3 5 2 1 3/2 5 y + y dx = x + x dx 3 2 3 2 0 0
4 2 5/2 5 2 364 = x + x = 15 4 15 0 224/3 364/15 = 56/23; y = Mx /m = = 91/115. x = My /m = 92/3 92/3 The center of mass is (56/23, 91/115).
y
7. The density is ρ = ky. Since both the region and ρ are symmetric with respect to the y-axis, x = 0. Using symmetry, 1−x2 Z 1 Z 1−x2 Z 1 Z 1 2 m=2 kydydx = 2k dx = k 01 (1 − x2 )2 dx y 2 0 0 0 0 1 Z 1 2 1 5 3 2 4 (1 − 2x + x )dx = k(x − x + x ) =k 3 5 0 0 2 1 8 =k 1− + = k 3 5 15 1−x2 Z 1 Z 1−x2 Z 1 Z 1 1 3 2 2 Mx = 2 ky dydx = 2k y (1 − x2 )3 dx dx = k 3 0 0 0 0 3 0 1 Z 1 2 3 1 2 (1 − 3x2 + 3x4 − x6 )dx = k x − x3 + x5 − x7 = k 3 0 3 5 7 0 2 3 1 32 = k 1−1+ − = k 3 5 7 105 32k/105 y = Mx /m = = 4/7. The center of mass is (0,4/7). 8k/15 8. The density is ρ = kx. Z π Z sin x Z m= kxdydx = 0
0
1
y=1-x2
1
x
y π
0
sin x Z dx = kxy 0
y=sin x
π
kx sin xdx
1
0
Integration by parts π
= k(sin x − x cos x)|0 = kπ Z Z π Z sin x 2 My = kx dydx =
sin x Z π kx y dx = kx2 sin xdx Integration by parts 0 0 0 0 0 π = k(−x2 cos x + 2 cos x + 2x sin x) 0 = k[(π 2 − 2) − 2] = k(π 2 − 4) sin x Z π Z sin x Z π 1 2 Mx = kxydydx = kxy dx 0 0 0 2 0 Z π Z π 1 1 = kx sin2 xdx = kx(1 − cos 2x)dx 4 0 2 Z π Z π 0 1 = k xdx − x cos 2xdx Integration by parts 4 0 0 π
2
π
x
14.4. CENTER OF MASS AND MOMENTS
165
π π 1 1 1 1 1 2 1 = k[ x − (cos 2x + 2x sin 2x) ] = k( π 2 ) = kπ 2 4 2 0 4 4 2 8 0 k(π 2 − 4) kπ 2 /8 x = My /m = = π − 4/π; y = Mx /m = = π/8. kπ kπ The center of mass is (π − 4/π, π/8). 1
Z
ex
Z
Z
3
9. m =
ex Z 1 1 4 1 4x y dx = e dx 4 0 0 4
1
y dydx = 0
0
0
1 1 4 1 4x = = e (e − 1) 16 0 16 Z 1 Z ex Z 3 My = xy dydx = 0
0
0
1
ex 1 4 xy dx 4 0
1
1
1
Z
y=ex
y
x
1 4x = xe dx Integration by parts 4 0 1 1 3 4 1 4x 1 1 1 1 4x 4 xe − e = = 4 16 e + 16 = 64 (3e + 1) 4 4 16 0 x 1 R 1 1 5x R 1 R ex 4 R 1 1 5 e 1 5 1 5x = Mx = 0 0 y dydx = 0 y dx = 0 e dx = e (e − 1) 5 0 5 25 0 25 (3e4 + 1)/64 (e5 − 1)/25 3e4 + 1 16(e5 − 1) x = My /m = = ; y = Mx /m = 4 = 4 4 (e − 1)/16 4(e − 1) (e − 1)/16 25(e4 − 1) 5 4 3e + 1 16(e − 1) , ≈ (0.77, 1.76). The center of mass is ( 4 4(e − 1) 25(e4 − 1) 10. Since both the region and ρ are symmetric with respect to the y-axis, x = 0. √Using symmetry, 9−x2 Z 3 Z 3 Z √9−x2 x2 dydx = 2 m=2 x2 y dx 0
0
Z =2
3
x2
0
p
9 − x2 dx
3
y=M9-x2
0
Trig substitution 3
0
3 81 π x x 81π 81 = (2x2 − 9) 9 − x2 + sin−1 · = . =2 8 8 3 0 √ 4 2 8 9−x2 Z 3 Z √9−x2 Z 3 Z 3 1 2 2 2 dydx = x2 (9 − x2 )dx Mx = 2 x ydydx = 2 x y 0 0 0 2 0 0 3 1 162 = (3x2 − x5 = 5 0 5 162/5 y = Mx /m = = 16/5π. The center of mass is (0, 16/5π). 81π/8
y
p
x
166
CHAPTER 14. MULTIPLE INTEGRALS
y−y2 Z 1 dy = (y − y 2 )2 y 2 dy 11. Ix = 2xy dxdy = x y 0 0 0 0 0 1 Z 1 1 5 1 6 1 7 1 = (y 4 − 2y 5 + y 6 )dy = y − y + y = 5 3 7 105 0 0 Z
1
Z
y−y 2
1
Z
2
y
2 2
1 x=y-y2
x
Z
1
√
Z
√
x
12. Ix =
1
Z
2 2
x y dydx = x2
0
0
x Z 1 2 3 1 1 7/2 (x − x8 )dx x y dx = 3 3 2 0 x
y y=x2 y=Mx
1
1 1 1 2 9/2 1 9 = ( x − x ) = 3 9 9 27 0
1
13. Using symmetry, Z π/2 Z cos x Z 2 Ix = 2 ky dydx = 2k
y
cos x Z π/2 2 1 3 y dx = k cos3 xdx 3 3 0 0 0 0 0 π/2 Z π/2 2 4 1 2 = k = k. cos x(1 − sin2 x)dx = k(sin x − sin3 x) 3 0 3 3 9 0 π/2
√4−x2 Z 1 2 1 3 4 dx = y dydx = y 14. Ix = (4 − x2 )2 dx 4 0 0 0 4 0 0 2 Z 8 1 1 2 1 16x − x3 + x5 = (16 − 8x2 + x4 )dx = 4 0 4 3 5 0 1 64 32 2 1 64 = 32 − + =8 1− + = 4 3 5 3 5 15 √ √ y Z 4Z y Z 4 Z Z 1 3 1 4 3/2 1 4 5/2 15. Iy = x2 ydxdy = x y dy = y ydy = y dy 3 0 3 0 0 0 0 3 0 4 2 7/2 256 1 2 7/2 = y (4 ) = = 3 7 21 21 0 Z
2
√
Z
4−x2
Z
x
1
y=cos x
π/2
2
y
2
y=M4-x2
2
Z
1
√
Z
16. Iy = 0
x2
x
x4 dydx =
Z 0
1
x4 y
1 2 11/2 1 7 3 =( x − x ) = 11 7 77 0
Z dx =
x2
x
y 4
x=My
2
√x
x
1
x
y
(x9/2 − x6 )dx
y=x2
0
y=Mx
1
1
x
14.4. CENTER OF MASS AND MOMENTS
1
Z
Z
17. Iy =
3
(4x3 + 3x2 y)dxdy =
y
0
Z
1
0
167
3 (x4 + x3 y) dy y
y
x=y
1
1
Z
3
(81 + 27y − 2y 4 )dy
=
x
0
=
81y +
1 27 2 2 5 941 y − y = 2 5 10 0 y
18. The density is ρ = ky. Using symmetry, 1−x2 Z 1 Z 1−x2 Z 1 Z 1 1 2 2 2 Iy = 2 kx ydydx = 2 dx = k x2 (1 − x2 )2 dx kx y 0 0 0 0 2 0 1 Z 1 8k 1 2 1 =k (x2 − 2x4 + x6 )dx = k( x3 − x5 + x7 ) = . 3 5 7 105 0 0 19. Using symmetry, √a2 −y2 Z a Z a Z √a2 −y2 Z a 1 2 x dy = m=2 xdxdy = 2 (a2 − y 2 )dy 2 0 0 0 0 0 a 2 1 = (a2 y − y 3 ) = a3 . 3 3 0 √a2 −y2 Z a Z a Z √a2 y2 Z 1 4 1 a 2 3 x dxdy = 2 dy = Iy = 2 x (a − y 2 )2 dy 4 2 0 0 0 0 0 a Z 4 5 1 a 4 2 1 1 4 2 3 5 2 2 4 = a (a − 2a y + y )dy = (a y − a y + y ) = 2 0 2 3 5 15 0 s r r Iy 4a5 /15 2 Rg = = = a 3 m 2a /3 5
1
y=1-x2 x
1
y a
y=Ma2-x2
a
x
Ra R a R a−x R a a−x y kdydx = 0 ky 0 dx = k 0 (a − x)dx = 0 0 a 1 1 a k(ax − x2 ) = ka2 y=a-x 2 2 0 a−x Z a Z a−x Z a Z a 1 3 1 ky 2 dydx = ky (a − x)3 dx Ix = dx = k 3 3 a x 0 0 0 0 0 a Z a 1 1 3 1 1 = k (a3 − 3a2 x − x3 )dx = k(a3 x − a2 x2 + ax3 − x4 ) = ka4 3 0 s 3 2 4 12 0 r r Ix ka4 /12 1 Rg = = = a m ka2 /2 6
20. m =
21. (a) Using symmetry,
168
CHAPTER 14. MULTIPLE INTEGRALS √ b a2 −x2 /a
Z 4b3 a 2 (a − x2 )3/2 dx x = a sin θ, dx = a cos θdθ Ix = 4 0 y dydx = 3 3a 0 0 Z π/2 Z π/2 4 1 4 = ab3 cos4 θdθ = ab3 (1 + cos 2θ)2 dθ 3 3 4 0 0 π/2 Z 1 3 π/2 1 1 1 3 3 1 1 = ab (1 + cos 2θ + + cos 4θ)dθ = ab ( θ + sin 2θ + sin 4θ) 3 2 2 3 2 2 8 0 0 Z
a
Z
2
ab3 π . 4 (b) Using symmetry, Z a Z b√a2 −x2 /a Z 4b a 2 p 2 x2 dydx = x a − x2 dx x = a sin θ, dx = a cos θdθ Iy = 4 a 0 0 0 Z π/2 Z π/2 1 sin2 θ cos2 θdθ = 4a3 b = 4a3 b (1 − cos2 2θ)dθ 4 0 0 π/2 Z π/2 a3 bπ 1 1 1 3 3 1 = =a b . (1 − − cos 4θ)dθ = a b( θ − sin 4θ) 2 2 2 8 4 0 0 p 1p 3 1 (c) Using m = πab, Rg = Ix /m = ab π/πab = b. 2 2 p 1p 3 1 (d) Rg = Iy /m = a bπ/πab = a 2 2 =
22. The equation of the ellipse is 9x2 /a2 + 4y 2 /b2 = 1 and the equation of the parabola is y = ±(9bx2 /8a2 − b/2). Letting Ie and Ip represent the moments of inertia of the ellipse and parabola, respectively, about the x-axis, we have Z 0 Z b√a2 −9x2 2a Z 0 b3 a a 2 Ie = 2 y dydx = (a2 − 9x2 )3/2 dx x = sin θ, dx = cos θdθ 3 12a 3 3 −a/3 0 −a/3 Z 0 b3 a4 b3 a 3π ab3 π = cos4 θdθ = = 3 12a 3 −π/3 36 16 192 and Z 3 Z 2a/3 Z b/2−9bx2 /8a2 9b 2 2a/3 b − 2 x2 dx Ip = 2 y 2 dydx = 3 0 2 8a 0 0 3 3 Z 2a/3 3 Z 2a/3 2b b 243 4 729 6 9 2 27 2 = dx = x − x dx 1 − 2x 1 − 2x + 3 8 0 4a 12 0 4a 16a4 64a6 2a/3 b3 9 3 243 5 729 7 8ab3 b3 32a = x − 2x + x − x = = . 12 4a 80a4 64a6 12 105 315 0 ab3 π 8ab3 Then Ix = Ie + Ip = + . 192 315
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES 1 1 4 23. From Problem 20, m = ka2 and Ix = ka . 2 Z a−x12 Z Z a Z a−x a dx = k kx2 y kx2 dydx = Iy = 0
0
0
0
169
y a
a
x2 (a − x)dx
y=a-x
0
4 1 3 1 4 1 4 =k ax − x = ka 3 4 12 0 1 4 1 1 I0 = Ix + Iy = ka + ka4 = ka4 12 12 6
a
x
1 3 1 3 24. From Problem 12, Ix = , and from Problem 16, Iy = . Thus, I0 = Ix + Iy = + = 27 77 27 77 158 . 2079 25. The density is ρ = k/(x2 + y 2 ). Using symmetry, Z √2 Z 6−y2 Z √2 6−y2 k 2 2 I0 = 2 (x + y ) 2 dxdy = 2 kx dy 2 2 x + y 2 y +2 0 0
y 2 x=6-y2
y +2
√2 2 = 2k (6 − y 2 − y 2 − 2)dy = 2k 4y − y 3 3 0 0 √ 8√ 16 2 = 2k k. 2 = 3 3 4 Z 3Z 4 Z 3 1 3 26. I0 = k(x2 + y 2 )dxdy = k x + xy 2 dy 3 0 y 0 y Z 3 1 64 =k + 4y 2 − y 3 − y 3 dy 3 3 0 3 64 4 3 1 4 y + y − y = 73k =k 3 3 3 0 Z
6
x=y2+2
√ 2
x
y 3
x=y
4
x
1 1 27. From Problem 20, m = ka2 , and from Problem 21, I0 = ka4 . 2 6 s r 4 p ka /6 1 Then Rg = I0 /m = = a. ka2 /2 3 28. Since the plate is homogeneous, the density is ρ = m/lw. Using symmetry, w/2 Z l/2 Z w/2 Z m 2 4m l/2 1 3 2 2 I0 = 4 (x + y )dydx = x y+ y dx lw lw 0 3 0 0 0 l/2 Z 4m l/2 w 2 w3 4m w 3 w3 4m wl3 lw3 l2 + w 2 x + dx = x + x = + =m . = lw 0 2 24 lw 6 24 lw 48 48 12 0
14.5
Double Integrals in Polar Coordinates
170
CHAPTER 14. MULTIPLE INTEGRALS
1. Using symmetry, Z Z π/2 Z 3+3 sin θ rdrdθ = 2 A=2 −π/2
Z
−π/2
0
π/2
9(1 + sin θ)2 dθ = 9
=
6
π/2
Z
−π/2
3+3 sin θ 1 2 dθ r 2 0
π/x
(1 + 2 sin θ + sin2 θ)dθ
1 = 9 θ − 2 cos θ + θ − 2 3π 3 π =9 − = − 22 2 2
π
3. Solving r = 2 sin θ and r = 1, we obtain sin θ = 1/2 or θ = π/6. Using symmetry, Z π/6 Z 2 sin θ Z π/2 Z 1 A=2 rdrdθ + 2 rdrdθ 0
π/6
5. Using symmetry, Z π/6 Z 5 cos 3θ Z V =2 4rdrdθ = 4
= 100
polar axis
0
2
2
0
2 sin θ 1 Z π/2 Z π/6 Z π/2 π/6 1 2 1 2 dθ + 2 =2 r r dθ = 4 sin2 θdθ + dθ 2 π/6 2 0 0 π/6 0 0 √ π π π √3 π 4π − 3 3 π/6 − = − + = = (2θ − sin 2θ)|0 + 2 6 3 2 3 6 8 sin 4θ Z Z π/4 Z 8 sin 4θ Z π/4 1 π/4 1 2 dθ = r 4. A = 64 sin2 4θdθ rdrdθ = 2 2 0 0 0 0 0 π/4 1 1 = 4π = 32 θ− sin 8θ 2 16 0 Z
3
27π 2
2+cos θ Z π 1 2 dθ = (2 + cos θ)2 dθ r 0 0 0 2 0 0 π Z π 1 1 2 = (4 + 4 cos θ + cos θ)dθ = 4θ + 4 sin θ + θ + cos 2θ 2 4 0 0 π 1 1 9π = 4π + + − = . 2 4 4 2
0
polar axis
π/2 1 sin 2θ 4 −π/2
2. Using symmetry, Z Z π Z 2+cos θ rdrdθ = 2 A=2
0
3
−π/2
π/6
0
π/6 1 1 25π θ+ sin 6θ = 2 12 3 0
5 cos 3θ Z π/6 r dθ = 4 25 cos2 3θdθ 0 2
0
1 polar axis
polar axis
polar axis
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES 2π
Z
2
Z
0 2π
0
1 =− 3
Z 0
2π
Z
3
Z
=−
1 3
Z 8. V =
2
1 2π
Z
= 0
(73/2 − 153/2 )dθ =
Z π
5
√
r2 rdrdθ
0 2π
2π
0
0
0
Z
0
2 1 − (9 − r2 )3/2 dθ 3 0
√ 1 2π(27 − 5 5) 3/2 3/2 (5 − 27)dθ = (27 − 5 )2π = 3 3
Z p 16 − r2 rdrdθ =
7. V = 0
2π
Z p 9 − r2 rdrdθ =
6. V =
Z = 0
2π
polar axis
3 1 2 3/2 − (16 − r ) dθ 3 1
√ √ 1 2π(15 15 − 7 7) (153/2 − 73/2 )2π = 3 3
polar axis
5 1 3 r dθ 3 0
125 250π dθ = 3 3
polar axis
1+cos θ 1 3 9. V = r sin θ dθ (r sin θ)rdrdθ = 3 0 0 0 0 π/2 Z 1 π/2 1 1 4 3 = − (1 + cos θ) (1 + cos θ) sin θdθ = 3 0 3 4 0 1 5 4 = − (1 − 2 ) = 12 4 Z
171
π/2
Z
1+cos θ
Z
10. Using symmetry, Z π/2 Z cos θ Z 2 V =2 (2 + r )rdrdθ =
π/2
cos θ 1 r2 + r4 dθ 4 0 0 0 0 Z π/2 Z π/2 1 1 1 + cos 2θ 2 2 4 2 =2 cos θ + cos θ dθ = 2 ) dθ cos θ + ( 4 4 2 0 0 Z π/2 1 1 1 = 2 cos2 θ + + cos 2θ + cos2 2θ dθ 8 4 8 0 π/2 1 1 1 1 1 19π = θ + sin 2θ + θ + sin 2θ + θ + sin 4θ = . 2 8 8 16 64 32 0 π/2
1
2
polar axis
1
polar axis
172
CHAPTER 14. MULTIPLE INTEGRALS
R π/2 1 2 3 R dθ = 1 k π/2 8dθ = 2kπ r 0 1 0 2 1 2 0 3 Z π/2 Z 3 Z π/2 Z π/2 Z 3 1 3 2 r cos θdrdθ = k kxrdrdθ = k My = r cos θ dθ 3 polar 3 1 1 0 0 0 1 axis π/2 Z π/2 1 26 26 = k = 26 cos θdθ = k sin θ k 3 0 3 3 0 26k/3 13 x = My /m = = . 2kπ 3π Since the region and density function are symmetric about the ray θ = π/4, y = x = 13/3π and the center of mass is (13/3π, 13/3π).
11. m =
R π/2 R 3
krdrdθ = k
12. The interior of the upper-half circle is traced from θ = 0 to π/2. The density is kr. Since both the region and the density are symmetric about the polar axis, y = 0. Z
π/2
Z
cos θ
Z
2
m=
kr drdθ = k 0
0
0
π/2
1
cos θ Z k π/2 1 3 r dθ = cos3 θdθ 3 3 0
polar axis
0
π/2 2k k 2 1 2 = + cos θ sin θ = 3 3 3 9 0 Z π/2 Z cos θ Z My = k (r cos θ)(r)(rdrdθ) = k
1
0
0
0
π/2
Z
cos θ 3
Z
r cos θdrdθ = k 0
0
π/2
cos θ 1 4 dθ r cos θ 4 0
π/2 k 2k 2 1 k 3 5 5 = = sin θ − sin θ + sin θ cos θdθ = 4 0 4 3 5 15 0 2k/15 Thus, x = = 3/5 and the center of mass is (3/5, 0). 2k/9 Z
π/2
13. In polar coordinates the line x = 3 becomes r cos θ √ = 3 or r = 3 sec θ. The angle of inclination of the line y = 3x is π/3. 3 sec θ Z π/3 Z Z π/3 1 4 3 sec θ 2 m= 0 r rdrdθ = r dθ 4 0 0 0 Z Z 81 π/3 81 π/3 4 sec θdθ = (1 + tan2 θ) sec2 θdθ = 4 0 4 0 π/3 √ 81 1 81 √ 81 √ = (tan θ + tan3 θ) = ( 3 + 3) = 3 4 3 4 2 0
3
polar axis
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES π/3
Z
3 sec θ
Z
xr2 rdrdθ =
My = π/3
= 0
π/3
Z
3 sec θ
r4 cos θdrdθ
0
0
0
0
Z
Z
173
3 sec θ Z 1 5 243 π/3 5 dθ = sec θ cos θdθ r cos θ 5 5 0 0
Z 243 π/3 4 243 √ 486 √ = sec θdθ = 3 (2 3) = 5 0 5 5 Z Z π/3 Z 3 sec θ Z π/3 Z 3 sec θ 4 2 r sin θdθ = yr rdrdθ = Mx =
3 sec θ 1 5 r sin θ 5 0 0 0 0 0 0 Z Z Z π/3 243 π/3 5 243 243 = sec θ sin θdθ = 0π/3 tan θ sec4 θdθ = tan θ(1 + tan2 θ) sec2 θdθ 5 0 5 5 0 π/3 Z 243 3 9 243 1 243 π/3 1 729 = (tan θ + tan3 θ) sec2 θdθ = = tan2 θ + tan4 θ ( + )= 5 0 5 2 4 5 2 4 4 0 √ 486 3/5 729/4 √ x = My /m = = 12/5; y = Mx /m = √ = 81 3/2 81 3/2 √ √ 3 3/2. The center of mass is (12/5, 3 3/2). π/3
14. Since both the region and the density are symmetric about the x-axis, y = 0. Using symmetry, 4 cos 2θ Z π/4 Z 4 cos 2θ Z π/4 1 2 m=2 r dθ krdrdθ = 2k 2 0 0 0 0 π/4 Z π/4 1 1 = 16k = 2kπ cos2 2θdθ = 16k( θ + sin 4θ) 2 8 0 0 Z π/4 Z 4 cos 2θ Z π/4 Z 4 cos 2θ Z My = 2 kxrdrdθ = 2k r2 cos θdrdθ = 2k 0
0
128 k 3
Z
128 = k 3
Z
=
0
π/4
cos3 2θ cos θdθ =
0
128 k 3
0
Z
0
π/4
(1 − 2 sin2 θ)3 cos θdθ
0
π/4
(1 − 6 sin2 θ + 12 sin4 θ − 8 sin6 θ) cos θdθ
0
π/4 128 12 8 k(sin θ − 2 sin3 θ + sin5 θ − sin7 θ) 3 5 7 0 √ √ √ √ ! √ 128 2 2 3 2 2 1024 = k − + − = 2k 3 2 2 10 14 105 √ √ 1024 2/105 512 2 = . x = My /m = 2kπ √ 105π The center of mass is (512 2/105π, 0) or approximately (2.20, 0). =
polar axis
π/4
4 cos 2θ 1 3 r cos θ dθ 3 0
174
CHAPTER 14. MULTIPLE INTEGRALS
2
15. The density is ρ = k/r. Z π/2 Z 2+2 cos θ Z π/2 Z 2+2 cos θ 4 polar k axis drdθ rdrdθ = k m= r 2 0 2 0 Z π/2 π/2 2 cos θdθ = 2k(sin θ)|0 = 2k =k 2+2 cos θ Z 0π/2 Z 2+2 cos θ Z Z π/2 Z 2+2 cos θ k 1 x rdrdθ = k My = r cos θdrdθ = k 0π/2 r2 cos θdθ r 2 2 2 0 2 0 Z π/2 Z π/2 1 2 (8 cos θ + 4 cos θ) cos θdθ = 2k (2 cos2 θ + cos θ − sin2 θ cos θ)dθ = k 2 0 0 π/2 π 2 1 3π + 4 1 = 2k( + ) = k = 2k(θ + sin 2θ + sin θ − sin3 θ) 2 3 2 3 3 0 2+2 cos θ Z π/2 Z 2+2 cos θ Z Z 2+2 cos θ Z π/2 k 1 2 Mx = y rdrdθ = k r sin θdrdθ = k r sin θdθ r 2 0 2 π/2 2 0 2 π/2 Z π/2 1 1 4 = k (8 cos θ + 4 cos2 θ) sin θdθ = k −4 cos2 θ − cos3 θ 2 0 2 3 0 1 4 8 = k − −4 − = k 2 3 3 3π + 4 4 (3π + 4)k/3 8k/3 = ; y = Mx /m = = x = My /m = 2k 6 2k 3 The center of mass is ((3π + 4)/6, 4/3).
Z
π
Z
16. m =
2+2 cos θ
Z
π
krdrdθ = k 0
Z = 2k
0 π
0
2+2 cos θ Z π 1 2 r dθ = 2k (1 + cos θ)2 dθ 2 0 0
2
4 polar axis
(1 + 2 cos θ + cos2 θ)dθ
0
π 1 1 = 2k θ + 2 sin θ + θ + sin 2θ = 3πk 2 4 0 2+2 cos θ Z π Z 2+2 cos θ Z π Z 2+2 cos θ Z π 1 3 My = kxrdrdθ = k r2 cos θdrdθ = k r cos θdθ 0 0 0 0 0 3 0 Z π Z π 8 8 = k (1 + cos θ)3 cos θdθ = k (cos θ + 3 cos2 θ + 3 cos3 θ + cos4 θ)dθ 3 0 3 0 π 8 3 3 3 1 1 = k sin θ + θ + sin 2θ + (3 sin θ − sin3 θ) + θ + sin 2θ + sin 4θ 3 2 4 8 4 32 0 8 15 = k π = 5πk 3 8
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES π
Z
2+2 cos θ
Z
π
Z
Z
kyrdrdθ = k
Mx = 0
0
0
175
2+2 cos θ
r2 sin θdrdθ = k
0
π
Z 0
2+2 cos θ 1 3 sin θdθ r 3 0
Z π Z π 8 8 = k (1 + cos θ)3 sin θdθ = k (1 + 3 cos θ + 3 cos2 θ + cos3 θ) sin θdθ 3 0 3 0 π 8 1 8 1 15 32 = k − cos θ − 32 cos2 θ − cos3 θ − cos4 θ = k − − = k 3 4 3 4 4 3 0 5πk 32k/3 = 5/3; y = Mx /m = = 32/9π. The center of mass is (5/3, 32/9π). x = My /m = 3πk 3πk
2π
Z
a
Z
y 2 krdrdθ = k
17. Ix = 0
0
ka4 = 4
Z
0
Z
2π
= 0
ka4 sin2 θdθ = 4
Z
a
r3 sin2 θdrdθ = k
0
Z
2π
0
a 1 4 2 r sin θ dθ 4 0
a polar axis
2π kπa4 1 1 θ − sin 2θ = 2 4 4 0
Z 2π Z a r3 1 y rdrdθ = sin2 θdrdθ 4 1 + r4 0 0 0 1+r a 2π 1 1 1 1 ln(1 + r4 ) sin2 θdθ = ln(1 + a4 ) θ − sin 2θ 4 4 2 4
Z
18. Ix =
2π
0
2π
0
2π
Z
Z
a
2
0
a polar axis
0
π = ln(1 + a4 ) 4
19. Solving a = 2a cos θ, cos θ = 1/2 or θ = π/3. The density is k/r3 . Using symmetry, Z π/3 Z 2a cos θ Z π/3 Z 2a cos θ 2 k cos2 θdrdθ Iy = 2 x 3 rdrdθ = 2k r 0 a 0 a Z π/3 = 2k (2a cos3 θ − a cos2 θ)dθ
a
2a
polar axis
0
π/3 2 1 1 3 = 2ak 2 sin θ − sin θ − θ − sin 2θ 3 2 4 0 √ √ ! √ √ 3 π 3 5ak 3 akπ = 2ak 3− − − = − 4 6 8 4 3
20. Solving 1 = 2 sin 2θ, we obtain sin 2θ = 1/2 or θ = π/12 and θ = 5π/12. 1
polar axis
176
CHAPTER 14. MULTIPLE INTEGRALS Z
5π/12
Z
2 sin 2θ
x2 sec2 θrdrdθ =
Iy =
5π/12
π/12
1
π/12
Z
Z
2 sin 2θ
r3 drdθ
1
2 sin 2θ Z 5π12 5π/12 1 4 = π/12 r dθ = 4 sin4 2θdθ 4 1 π/12 5π/12 3 1 1 =2 θ − sin 4θ + sin 8θ 4 4 32 π/12 " √ √ √ ! √ √ !# 5π 3 3 π 3 3 8π + 7 3 =2 − = + − − + 16 8 64 16 8 64 16 Z
21. From Problem 17, Ix = kπa4 /4. By symmetry, Iy = Ix . Thus I0 = kπa4 /2. a polar axis
π
Z
θ
Z
22. The density is ρ = kr. I0 = 0
=
1 k 5
Z
0 π
0
π
θ 1 5 r dθ 5 0
r (kr)rdrdθ = k 0 π kπ 6 1 1 θ6 = θ5 dθ = k 5 6 30 0
Z 3 Z 1/r k r2 rdθdr = k f 2 dθdr r 1 0 1 0 3 Z 3 1 2 1 =k dr = k r = 4k r2 r 2 1 1 Z
3
Z
2
1
polar axis
1/r
Z
23. The density is ρ = k/r. I0 =
2a cos θ Z π 1 4 4 2 dθ = 4ka cos4 θdθ 24. I0 = r krdrdθ = k r 4 0 0 0 0 π0 4 3 3π 1 1 3kπa = 4ka4 θ + sin 2θ + sin 4θ = 4ka4 = 8 4 32 8 2 0 Z
Z
3
√
Z
25. −3
π
Z
2a cos θ
Z
9−x2
p
x2 + y 2 dydx =
π
Z
3
polar axis
π
Z
03 |r|rdrdθ
2a
polar axis r=3
0
0
Z = 0
π
3 Z 1 3 r dθ = 9 3 0
0
π
dθ = 9π
3 polar axis
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES √
Z
Z √1−y2
2/2
26. 0
y2 p
y
Z
x2 + y 2
π/4
Z
π/4
Z
1
dxdy = 0
0
Z
177
r2 sin2 θ rdrdθ |r|
r=1
1
r2 sin2 θdrdθ
=
1
0
0
polar axis
1 Z 1 π/4 2 1 3 2 sin θdθ r sin θ dθ = = 3 3 0 0 0 π/4 1 1 π−2 1 = ( θ − sin 2θ) = 3 2 4 24 0 π/4
Z
Z
1
Z √1−y2
27. 0
ex
2
+y
2
28.
√
x
Z
1 2
er rdrdθ = (e − 1)dθ =
0
0
29. 0
1
Z
√
Z √
4−x2
1−x2
π/2
1 1 r2 e dθ 2
r=1
0
π(e − 1) 4
1
x
(sin r2 )rdrdθ
r=Mπ
√x 1 2 = − cos r dθ 2 0 0 Z 1 π =− (−1 − 1)dθ = π 2 0 x2 dydx = 2 x + y2
Z 1
Z
2
Z
π
√ 4−x2
0 π/2 Z 2
= 0
Z
1 π/2
Z
= 0
x2 2
2
r cos θ rdrdθ r2
2
r cos2 θdrdθ
1 π/2
Mπ
x2 dydx + y2
2 Z 1 2 3 π/2 = r cos2 θdθ = cos2 θdθ 2 2 0 0 1 π/2 3 1 1 3π = θ + sin 2θ = 2 2 4 8 0 Z
polar axis
0
0
Z
Z
√
π
Z
2
sin(x + y )dydx =
√ − x
Z 0
π/2
Z
π−x2 2
2
0
0
=
√
1
Z
dxdy =
0
Z
π/2
Z
polar axis
r=2 1 2 polar axis
178
CHAPTER 14. MULTIPLE INTEGRALS
Z
1
Z √2y−y2
r=csc θ 2
2
(1 − x − y )dxdy
30. 0
r=2 sin θ
0
Z
π/4
Z
2 sin θ
(1 − r2 )rdrdθ +
= 0
0
Z
π/2
π/4
Z
csc θ
(1 − r2 )rdrdθ
polar axis
1
0
2 sin θ csc θ Z π/2 1 2 1 4 = dθ + dθ r − r 2 4 0 π/4 0 0 Z π/4 Z π/2 1 1 = (2 sin2 θ − 4 sin4 θ)dθ + csc2 θ − csc4 θ dθ 2 4 0 π/4 π/2 1 3 1 1 1 1 = θ − sin 2θ − θ − sin 2θ + sin 4θ + − cot θ − (− cot θ − cot3 θ) 2 2 8 2 4 3 π/4 π 1 1 1 16 − 3π = − + + 0− − + = 8 2 4 12 24 Z
Z
5
√
Z
π/4
1 2 1 4 r − r 2 4
25−x2
31.
Z
π
Z
−5
5
5
(4x + 3y)dydx =
(4r cos θ + 3r sin θ)rdrdθ
0
0
Z
r=5
0 π
Z
5
= 0
5
(4r2 cos θ + 3r2 sin θ)drdθ
polar axis
0
5 4 3 3 r cos θ + r sin θ dθ = 3 0 0 π Z π 500 500 = cos θ + 125 sin θ dθ = sin θ − 125 cos θ = 250 3 3 0 0 π
Z
Z
1
Z √1−y2
32. 0
0
Z
2
∞
33. I =
e 0
Z =
−(x2 +y 2 )
Z
π/2
∞
Z
e 0
t→∞
π/2
dxdy =
0
lim
0
1
Z π/2 Z 1 1 1 p rdrdθ dxdy = 1 + r 1 + x2 + y 2 0 0 1 Z π/2 Z 1 Z π/2 1 = (1 − )drdθ = [r − ln(1 + r)] dθ 1+r 0 0 0 0 Z π/2 π (1 − ln 2)dθ = (1 − ln 2) = 2 0
∞
Z
2 1 1 − e−t + 2 2
0
Z dθ = 0
π/2
−r 2
Z rdrdθ = 0
π/2
r=1
1
t 1 −r2 lim − e dθ t→∞ 2
√ 1 π π dθ = ; I = 2 4 2
0
polar axis
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES Z Z
π/2
Z
34.
Z
2
π/2
Z
Z
2
r2 (cos θ + sin θ)drdθ
(r cos θ + r sin θ)rdrdθ =
(x + y)dA = Z = 0
π/2
2 sin θ
0
2 sin θ
0
R
179
2 1 3 r (cos θ + sin θ) 3
dθ =
2 sin θ
8 3
Z
π/2
(cos θ + sin θ − sin3 θ cos θ − sin4 θ)dθ
0
π/2 8 1 1 3 3 = ] sin θ − cos θ − sin4 θ + sin3 θ cos θ − θ + sin 2θ 3 4 4 8 16 0 8 1 3π 28 − 3π = 1− − − (−1) = 3 4 16 6 35. The volume of the cylindrical portion of the tank is Vc = π(4.2)2 19.3 ≈ 1069.56m3 . We take the equation of the ellipsoid to be x2 5.15 p x2 + = 1 or z = ± (4.2)2 − x2 − y 2 . 2 2 (4.2) (5.15) 4.2 The volume of the ellipsoid is Z Z Z Z p 5.15 10.3 2π 4.2 2 2 2 Ve = 2 [(4.2)2 − r2 ]1/2 rdrdθ (4.2) − x − y dxdy = 4.2 4.2 0 R 0 " 4.2 # Z Z 10.3 1 2π 10.3 2π 1 2 2 2 3/2 dθ = = [(4.2) − r ] − (4.2)3 dθ 4.2 0 2 3 4.2 3 0 0 2π 10.3 (4.2)3 ≈ 380.53. 3 4.2 The volume of the tank is approximately 1069.56 + 380.53 = 1450.09m3 . =
36. (a) With b > 2 we have Z Z Z Z 1 2π R r IdA = dr u = r + c, du = dr b 2 0 C 0 (r + c) 2 R+c Z R+c Z R+c u−c r −b r1−b 1−b −b = πa du = πa −c (u − cu )du = πa ub 2−b 1 − b c c c 2−b c (R + c)2−b c2 − b c(R + c)1−b = πa − − πa − b−2 b−1 b−2 b−1 πa 1 c = − πa − . (b − 1)(b − 2)cb−2 (b − 2)(R + c)b−2 (b − 1)(R + c)b−1 Z Z πa I(r)dA = (b) lim R→∞ (b − 1)(b − 2)cb−2 C (c) Identifying a = 68.585, b = 2.351, and c = 0.248 in part b we find that the total number of infections in the plane is approximately 741.25. Z Z
Z
37. (a) P =
2π
Z
D(r)dA = C
R
D0 e 0
0
−r/d
Z rdrdθ = 2πD0
R
re−r/d dr
0
R = 2πD0 (−dre−r/d − d2 er/d ) = 2πdD0 [d − (R + d)e−R/d ] 0
180
CHAPTER 14. MULTIPLE INTEGRALS (b) Using Z Z
Z
2π
Z
rD(r)dA = C
R
rD0 e−r/d rdrdθ = 2πD0
0
0
Z
R
r2 e−r/d dr
0
R = 2πD0 (−2d3 e−r/d − 2d2 re−r/d − dr2 e−r/d ) 0 i h 2 2 2 −R/d = 2πdD0 2d − (R + 2dR + 2d )e we have RR rD(r)dA 2d2 − (R2 + 2dR + 2d2 )e−R/d R RC = D(r)dA d − (R + d)e−R/d C (c) Letting R −→ ∞ in the result of parts (a) and (b) we find that the total population is 2πd2 D0 and the average commute for the total population is 2d2 /d = 2d. 38. In the first case, let the circle centered at (D/2, 0) be described by the equation r = D cos θ for −π/2 ≤ θ ≤ π/2 and assume that the snow is plowed to the origin. Then Z Z
Z
π/2
D cos θ
Z
r2 drdθ =
rdA = −π/2
R
0
D3 3
Z
π/2
(1 − sin2 θ) cos θdθ
0
π/2 4D3 2D 1 sin θ − sin3 θ = . = 3 3 9 0 3
In the second case, let the circle centered at the origin be described by the equation r = D/2 for 0 ≤ θ ≤ 2π, and assume the snow is plowed to the origin. Then Z Z
Z
2π
Z
rdA = R
0
0
D/2
D/2 2π 3 πD3 r drdθ = = r . 3 12 0 2
3
4D /9 16 = ≈ 1.698, which means that plowing snow to one πD3 /12 3π point on the perimeter is approximately 69.8% more costly than plowing to the center. The ratio of these integrals is
14.6
Surface Area
1. Letting z = 0, we have 2x + 3y = 12. Using f (x, y) = z = 1 3 1 3 29 3− x− y we have fx = − , fy = − , 1+fx2 +fy2 = . 2 4 2 4 16 Then √ Z 6 Z 6 Z 4−2x/3 p 29 2 A= 29/16dydx = (4 − x)dx 4 3 0 0 0 6 √ √ √ 29 1 29 = (4x − x2 ) = (24 − 12) = 3 29. 4 3 4 0
y 4 y=4-2x/3
6 x
14.6. SURFACE AREA
181
2. We see from the graph in Problem 1 that the plane is entirely above the region bounded by r = sin 2θ in the 3 1 first octant. Using f (x, y) = z = 3 − x − y we have 2 4 1 3 29 2 2 fx = − , fy = − , 1 + fx + fy = . Then 2 4 16 sin 2θ √ Z Z π/2 Z sin 2θ p 29 π/2 1 2 dθ A= 29/16rdrdθ = r 4 2 0 0 0 0 π/2 √ Z π/2 √ √ 29 29 1 1 29π 2 = sin 2θdθ = = ( θ − sin 4θ) . 8 8 2 8 32 0
r=sin 2 θ
1
polar axis
0
√
3. Using f (x, y) = z = 16 − x2 we see that for 0 ≤ x ≤ 2 and 0 ≤ y ≤ 5, z.0. Thus, the surface is entirely above the x , fy = 0, 1 + fx2 + fy2 = region. Now fx = − √ 16 − x2 x2 16 1+ = and 2 16 − x 16 − x2 2 Z 5Z 2 Z 5 Z 5 4 π 10π −1 x √ A= dxdy = 4 sin dy = 4 dy = . 2 4 6 3 16 − x 0 0 0 0 0
y 5
x=2
2
4. The region in the xy-plane beneath the surface is bounded by the graph of x2 + y 2 = 2. Using f (x, y) = z = x2 + y 2 we have fx = 2x, fy = 2y, 1 + fx2 + fy2 = 1 + 4(x2 + y 2 ). Then, √
2π
Z
Z
2
p
A= 0
1 = 12
0 2π
Z 0
1+
4r2 rdrdθ
2π
Z = 0
√2 1 (1 + 4r2 )3/2 dθ 12
2π
Z
2
A= 0
=
1 12
Z 0
r=M2
M2 polar axis
0
13π . (27 − 1)dθ = 3
5. Letting z = 0 we have x2 + y 2 = 4. Using f (x, y) = z = 4 − (x2 + y 2 ) we have fx = −2x, fy = −2y, 1 + fx2 + fy2 = 1 + 4(x2 + y 2 ). Then Z
x
p
1+
4r2 rdrdθ
Z = 0
0 2π
(173/2 − 1)dθ =
2π
2 1 2 3/2 (1 + 4r ) dθ 3 0
π (173/2 − 1). 6
r=2
2 polar axis
182
CHAPTER 14. MULTIPLE INTEGRALS
6. The surfaces x2 + y 2 + z 2 = 2 and z 2 = x2 + y 2 intersect on the cylinder 2x2 + 2y 2 = 2 or x2 + y 2 = 1. There are portions of the sphere within the cone p both above and below the xy-plane. Using f (x, y) = 2 − x2 − y 2 we have x y fx = − p , fy = − p , 1 + fx2 + fy2 = 2 2 2 2 2−x −y 2−x −y 2 . Then 2 − x2 − y 2 "Z # √ 1 2π Z 1 √ Z 2π p 2 2 √ A=2 rdrdθ = 2 2 − 2 − r dθ 2 2−r 0 0 0 0 Z 2π √ √ √ √ =2 2 ( 2 − 1)dθ = 4π 2( 2 − 1).
r=1
1 polar axis
0
p
7. Using f (x, y) = z = 25 − x2 − y 2 we have x y fx = − p , fy = − p , 2 2 25 − x − y 25 − x2 − y 2 25 . Then 1 + fx2 + fy2 = 25 − x2 − y 2 Z 5 Z √25−y2 /2 5 p A= dxdy 25 − x2 − y 2 0 0 √25−y2 /2 Z 5 Z 5 π 25π x −1 =5 dy = 5 dy = . sin p 2 6 6 25 − y 0 0
y 5
x=M25-y2/2
3
x
0
8. In the first octant, the graph of z = x2 −y 2 intersects the xyplane in the line y = x. The surface is in the first octant for x > y. Using f (x, y) = z = x2 − y 2 we have fx = 2x, fy = −2y, 1 + fx2 + fy2 = 1 + 4x2 + 4y 2 . Then Z
π/4
Z
2
A= 0
1 = 12
0
Z 0
π/4
Z p 2 1 + 4r rdrdθ = 0
π/4
2 1 (1 + 4r2 )3/2 dθ 12
r=2 2 polar axis
0
π (173/2 − 1)dθ = (173/2 − 1). 48
9. There are portions of the sphere within the cylinder both above and below the xy-plane. Using f (x, y) = z = p x a2 − x2 − y 2 we have fx = − p , fy = 2 1 − x2 − y 2 a2 y , 1 + fx2 + fy2 = 2 −p . Then, using a − x2 − y 2 a2 − x2 − y 2 symmetry,
r=a sin θ
a
polar axis
14.6. SURFACE AREA
183
a sin θ # Z π/2 p a 2 2 √ dθ − a −r rdrdθ = 4a a2 − r2 0 0 0 0 Z π/2 Z π/2 p 2 2 = 4a (a − a 1 − sin θ)dθ = 4a (1 − cos θ)dθ
" Z A=2 2
π/2
a sin θ
Z
0
0
π/2 π = 4a (θ − sin θ) 0 = 4a2 ( − 1) = 2a2 (π − 2). 2 2
10. There are portions of the cone within the cylinder both above and 1p 2 below the xy-plane. Using f (x, y) = x + y 2 , we have fx = 2 y 5 x p , fy = p , 1 + fx2 + fy2 = . Then, using 2 2 2 2 4 2 x +y 2 x +y symmetry, 2 cos θ " Z # r π/2 Z 2 cos θ √ Z π/2 1 2 5 dθ A=2 2 rdrdθ = 2 5 r 4 2 0 0 0 0 π/2 √ Z π/2 √ √ 1 1 2 =4 5 = 5π. cos θdθ = 4 5 θ + sin 2θ 2 4 0 0 11. There are portions of the surface in each octant with areas equal p to the area of the portion in the first octant. Using f (x, y) = z = a2 − y 2 y a2 . Then we have fx = 0, fy = p , 1 + fx2 + fy2 = 2 a − y2 a2 − y 2 Z a Z √a2 −y2 A=8 0
= 8a 0
a
2
polar axis
y x=Ma2-y2
a
x
a
dxdy a2 − y 2 √a2 −y2 Z a x p dy = 8a dy = 8a2 . a2 − y 2 0
0
Z
r=2 cos θ
p
0
12. √ From Example 1, the area of the portion of the hemisphere √with x2 + y 2 = b2 is 2πa(a − a2 − b2 ). Thus, the area of the sphere is A = 2 lim 2πa(a − a2 − b2 ) = 2(2πa2 ) = 4πa. b→a
13. The projection of the surface onto √ the xy-plane is shown in the graph. Using f (x, z) = y = a2 − x2 − z 2 we have x z fx = − √ , fz = − √ , 1 + fx2 + 2 2 2 2 a −x −z a − x2 − z 2 a2 fz2 = 2 . Then a − x2 − z 2 √ a2 −C12 Z 2π Z √a2 −c21 Z 2π p a A= dθ rdrdθ = a − a2 − r2 √ √ 2 2 √ 2 2 2 a − r2 a −c2
0
Z
2π
(c2 − c1 )dθ = 2πa(c2 − c1 ).
=a 0
0
a −c2
z
r=Ma2-c12 r=Ma2-c22
x
184
CHAPTER 14. MULTIPLE INTEGRALS
14. The surface area of the cylinder x2 + z 2 = a2 from y = c1 to y = c2 is the area of a cylinder of radius a and height c2 − c1 . This is 2πa(c2 − c1 ). 15. The equations of the spheres are x2 + y 2 + z 2 = a2 and x2 + y 2 + (z + a)2 = 1. Subtracting these equations, we obtain(z − a)2 − z 2 = 1 − a2 or −2az + a2 = 1 − a2 . Thus, the spheres intersect on the plane z = a − 1/2a. The region of integration is x2 + y 2 + (a − 1/2a)2 = a2 or r2 = 1 − 1/4a2 . The area is Z √1−1/4a2
√1−1/4a2 (a − r ) rdrdθ = 2πa[−(a − r ) ] A=a 0 0 0 " 1/2 ! 2 #1/2 1 1 = π. = 2πa a − a− = 2πa a − a2 − 1 − 2 4a 2a Z
16.
2π
2 −1/r
2
2
2 1/2
(a) Both states span 7 degrees of longitude and 4 degrees of latitude, but Colorado is larger because it lies to the south of Wyoming. Lines of longitude converge as they go north, so the east-west dimensions of Wyoming are shorter than those of Colorado. p (b) We use the function f (x, y) = R2 − x2 − y 2 to describe the northern hemisphere, where R ≈ 3960 miles is the radius of the Earth. We need to compute the surface area over a polar rectangle P of the form θ1 ≤ θ ≤ θ2 , R cos φ2 ≤ r ≤ R cos φ1 . We have −y −x fx = p and fy = p 2 2 2 2 2 2 R −x −y s R −x −y q x2 + y 2 so that 1 + fx2 + fy2 = 1+ 2 = R − x2 − y 2 R √ . 2 R − r2 Thus
A=
Z Z q
1+
fx2
P
= (θ2 − θ1 )R
+
fy2 dA
Z
θ2
R cos φ1
= θ1
p
Z
R cos φ2
√
φ2 φ1
R R θ2
θ1
R rdrdθ − r2
R2
R cos φ2 R2 − r 2 = (θ2 − θ1 )R2 (sin φ2 − sin φ1 ). R cos φ1
The ratio of Wyoming to Colorado is then sin 45◦ − sin 41◦ ≈ 0.941. Thus Wyoming is about 6% sin 41◦ − sin 37◦ smaller than Colorado. (c) 97,914/104,247 ≈ 0.939, which is close to the theoretical value of 0.941. (Our formula for the area says that the area of Colorado is approximately 103,924 square miles, while the area of Wyoming is approximately 97,801 square miles.
14.7. THE TRIPLE INTEGRAL
14.7
185
The Triple Integral
1 R4R2 1 2 1. 2 −2 −1 (x + y + z)dxdydz = 2 −2 ( x + xy + xz) dydz 2 −1 2 4 R4 R4R2 R4 2 = 2 −2 (2y + 2z)dydz = 2 (y + 2yz) dz = 2 8zdz = 4z 2 2 = 48 R4R2 R1
−2
Z
3
x
Z
xy
Z
2. 2
1
1
xy Z Z x 3 (24x2 y 2 − 48xy)dydx 24xyz dydx = 1 24xydzdydx = 1 1 1 2 x Z 3 Z 3 (8x5 − 24x3 − 8x2 + 24x)dx (8x2 y 3 − 24xy 2 ) dx = = 3
Z
x
Z
1
=
Z
6
6−x
Z
3 4 6 14 8 3 1552 4 2 x − 6x − x + 12x = 522 − = 3 3 3 3 1
6−x 1 dx (6z − xz − z 2 ) 2 0 0 0 0 Z 6 Z 6 1 1 2 = 6(6 − x) − x(6 − x) − (6 − x) dx = (18 − 6x + x2 )dx 2 2 0 0 6 1 = 18x − 3x2 + x3 = 36 6
6−x−z
Z
3.
6
Z
6−x
Z
0
0
Z
6
(6 − x − z)dzdx =
dydzdx = 0
1
1
0
Z
1
1−x
Z
Z
√
4.
0 0
Z
0
π/2
y2
Z
y
Z
5. 0
0
0
y
√
Z 1 Z 1−x y 4x z dzdydx = x z dydx = x2 y 2 dydx 0 0 0 0 0 1−x Z Z Z 1 1 1 2 1 2 3 1 1 2 dx = x y x (1 − x)3 dx = (x − 3x3 + 3x4 − x5 )dx = 3 3 3 0 0 0 0 1 1 1 1 3 3 4 3 5 1 6 = x − x + x − x = 3 3 4 5 6 180 0 1
Z
2 3
Z
1−x
2 4
y 2 Z π/2 Z y2 Z π/2 x x x 2 cos dzdxdy = y cos dxdy = y sin dy y y y 0 0 0 0 Z π/2 = y 2 sin ydy Integration by parts 0
π/2 = (−y 2 cos y + 2 cos y + 2y sin y) 0 = π − 2 Z 6. 0
√ 2
Z
2
√
√
2
Z
ex
Z
2
Z
xdzdxdy = y
0
√
0
√
=
2
√ x2
Z
xe dxdy = y
0
2
2 1 x2 e (e4 − ey )dy 2 √ y
√ √ √ √ 2 1 1 1 (ye4 − ey ) = [(e4 2 − e 2 ) − (−1)] = (1 + e4 2 − e 2 ) 2 2 2 0
186
CHAPTER 14. MULTIPLE INTEGRALS
2−x2 y2 Z 1Z 1 2 2 (xye2−x −y − xy)dxdy xye dxdy = xye dzdxdy = 7. 0 0 0 0 0 0 0 0 1 Z 1 Z 1 1 2−x2 −y2 1 2 1 1−y2 1 1 2−y2 = − ye − ye − x y dy = − y + ye dy 2 2 2 2 2 0 0 0 1 1 1 1 1 1 2 1 1−y2 1 2 1 2−y2 1 2 1 e = − y − e = 4 − 4 − 4 e) − ( 4 e − 4 e = 4 e − 2 e 4 4 4 0 R 4 R 1/2 R 4 R 1/2 R x2 R 4 R 1/2 −1 y x2 1 p 8. 0 0 dydxdz = sin dxdz = 0 0 sin−1 xdxdz Integration by parts 0 0 0 2 2 x 0 x −y ! √ 1/2 √ √ R R4 1π 3 π 4 −1 2 = 0 x sin x + 1 − x dz = 0 + − 1 dz = + 2 3 − 4 26 2 3 0 Z 5Z 3 Z Z Z Z 5 Z 3 Z y+2 z 2xdydz zdxdydz = 9. zdV = Z
1
Z
1
Z
2−x2 −y 2
1
0
D
Z = 0
5
1
Z
z
Z
1
z
5
5 4zdz = 2z 2 0 = 50
0
1
5
1
0
y
3 Z 2yz dz =
3 y
x=y
3 x=y+2 x
10. Using Z Z Zsymmetry, Z (x2 + y 2 )dV = 2 D
2
z
4−y
Z
2
4
2
(x + y )dzdydx x2
0
Z
2
Z
=2 Z
4
Z
0 2
Z
0
4−y dydx (x + y )z 2
4
2
2
x 4
0
2
2
2
3
(4x − x y + 4y − y )dydx
=2
4
2 x
y
y=x2
x2
0
4 1 2 2 4 3 1 4 =2 (4x y − x y + y − y ) dx 2 3 4 0 x2 2 8 23, 552 64 4 5 1 = 2( x3 + x − x5 − x7 + x9 ) = . 3 3 5 42 36 315 0 Z
2
2
11. The other five integrals are R 4 R 2−x/2 R 4 R 4 R z R (z−x)/2 f (x, y, z)dydxdz, f (x, y, z)dzdydx, 0 0 0 0 0 x+2y R 4 R z/2 R z−2y R 4 R 4 R (z−x)/2 f (x, y, z)dydzdx, 0 0 f (x, y, z)dxdydz, 0 R02 Rx4 R0 z−2y f (x, y, z)dxdzdy. 0 2y 0
z
4 z=2y z=x
2 x+2y=4
4 x
y
14.7. THE TRIPLE INTEGRAL
187
12. The √ other five integrals are R 3 R 36−4y2 /3 R 3 f (x, y, z)dzdxdy, R03 R02 R √36−9x2 /21 f (x, y, z)dydxdz, 1 0 0 R 3 R 3 R √36−4y2 /3 f (x, y, z)dxdydz, 1 0 0 R 3 R 3 R √36−4y2 /3 f (x, y, z)dxdzdy, R02 R13 R0√36−9x2 /2 f (x, y, z)dydzdx. 0 1 0
z
3
y=M36-9x2 /2
3 2
R2R8 R4 R 8 R 4 R y1/3 13. (a) V = 0 x3 0 dzdydx (b) V = 0 0 0 dxdz R4R2R8 (c) V = 0 0 x2 dydxdz √ 14. Solving z = x and x + z = 2, we obtain x = 1, z = 1. R 3 R 1 R 2−z R 1 R 2−z R 3 (a) V = 0 0 z2 dxdzdy (b) V = 0 z2 dydxdz 0 R 3 R 1 R √x R 3 R 2 R 2−x (c)V = 0 0 0 dzdxdy + 0 1 0 dzdxdy 15.
x=M
x
16.
z
y
36-4y2 /3
z
5 3
x=2-2z/3
4
2
y
3
x
y
x=M9-y2
3 x
The region in the first octant is shown. 17.
18.
z
z
6
4 y=-M1-x2
2 y=M1-x2
2
x
y
2
2
y=M
4-x2
x
y
188
CHAPTER 14. MULTIPLE INTEGRALS
19.
20.
z
z
1
2
3
y
3 2
1
y
x
x
√ 21. Solving x = y 2 and 4 − x = y 2 , we obtain x = 2, y = ± 2. Using symmetry, √
3
Z
Z
2
4−y 2
Z
V =2
Z
√
3
Z
2
dxdydz = 2 0
y2
0
(4 − 2y 2 )dydz
0
0
√2 Z 3 √ √ 8 2 2 3 dz = 16 2. 4y − y dz = 2 3 3 0 0
3
Z
z
5
=2 0
4
x=y2
y
x=4-y2
5 x
2
Z
√
Z
4−x2
Z
x+y
Z
2
Z
√ 4−x2
dzdydx =
22. V = 0
0
0
0
0
x+y dydx z
z
0
2
√4−x2 1 2 = (x + y)dydx = dx xy + y 2 0 0 0 0 2 Z 2 p 1 1 1 = x 4 − x2 + (4 − x2 ) dx = − (4 − x2 )3/2 + 2x − x3 2 3 6 0 0 4 8 16 = 4− − − = 3 3 3 Z
2Z
√
4−x2
Z
2
23. Adding the two equations, we obtain 2y = 8. Thus, the paraboloids intersect in the plane y = 4. Their intersection is a circle of √ radius 2. Using symmetry, Z 2 Z 4−x2 Z 8−x2 −z2 V =4 dydzdx Z
2
√
Z
=4 0
Z
x2 +z 2
0
0
0 2
4−x2
(8 − 2x2 − 2x2 )dzdx = 4
0
2
x
2 z=M4-x2
2 x
4−x2 2 2(4 − x2 )z − z 3 dx 3 0
4 (4 − x2 )3/2 dx Trig substitution =4 0 3 2 p 16 h x x i = − (2x2 − 20) 4 − x2 + 6 sin−1 = 16π. 3 8 2 0
y
y=M4-x2
z
√
Z
2
2
8 y
14.7. THE TRIPLE INTEGRAL
189
24. Solving x = 2, y = x, and z = x2 + y 2 , we obtain the point (2,2,8). Z 2Z x Z 2 Z x Z x2 +y2 (x2 + y 2 )dydx dzdydx = V = 0
0 2
Z = 0
0
z
(2,2,8)
0
0
x 2 Z 2 4 3 16 1 4 1 3 2 (x y + y ) dx = x dx = x = . 3 3 3 3 0 0 0
2 x
25. We are given ρ(x, y, z) = kz.
2
y
y=x
y1/3 Z 8Z 4 y 1/3 zdzdy dzdy = k xz kzdxdzdy = k m= 0 0 0 0 0 0 0 0 4 8 Z 8 Z 8 1 1/3 2 3 =k y 1/3 dy = 8k y z dy = 8k y 4/3 = 96k 4 0 2 0 0 0 y1/3 Z 8Z 4 Z 8Z 4 Z 8 Z 4 Z y1/3 2 2 Mxy = kz dxdzdy = k xz dzdy = k y 1/3 z 2 dzdy Z
8
Z
4
0
y 1/3
0
0
8
Z
4
Z
0
0
0
0
0
4 8 Z 8 3 4/3 64 64 1 1/3 3 1/3 y z dy = k k y y dy = =k = 256k 3 3 4 0 0 3 0 0 1/3 y Z 8Z 4 Z 8Z 4 Z 8 Z 4 Z y1/3 = kyzdxdzdy = k xyz dzdy = i y 4/3 zdzdy Z
Mxz
Z
8
0
0
0
0
0
0
0
0
4 8 Z 8 3072 3 7/3 1 4/3 2 y z dy = 8k y y 4/3 dy = 8k =k = 7 k 2 7 0 0 0 0 y1/3 Z 8 Z 4 Z y1/3 Z 8Z 4 Z 8Z 4 1 1 2 Myz = x z dzdy = k kxzdxdzdy = k y 2/3 zdzdy 2 0 0 0 0 0 0 0 2 0 4 8 Z 8 Z 8 384 1 2/3 2 3 5/3 1 2/3 = y z dy = 4k y k y dy = 4k = k 2 0 2 5 5 0 0 0 384k/5 3072k/7 256k = 4/5; y = Mxz /m = = 32/7; z = Mxy /m = = 8/3 x = Myz /m = 96k 96k 96k The center of mass is (4/5, 32/7, 8/3). Z
8
26. We use the form of the integral in Problem 14(b) of this section. Without loss of generality, we take 1. Z Z 1ρZ=2−z Z 1 Z 2−z Z 1 3 m= dydxdz = 3dxdz = 3 (2 − z − z 2 )dz 0
z2
0
0
z2
1 1 2 1 3 7 = 3(2z − z − z ) = 2 3 2 0 Z 1 Z 2−z Z 3 Z Mxy = zdydxdz =
0
3 Z 1 Z 2−z yz dxdz = 3zdxdz 0 z2 0 0 z2 0 z2 0 1 Z 1 2−z Z 1 1 3 1 4 5 2 3 2 =3 xz dz = 3 (2z − z − z )dz = 3 z − z − z = 3 4 4 0 0 z2 0 1
Z
2−z
190
CHAPTER 14. MULTIPLE INTEGRALS 3 Z Z 1 2 9 1 2−z dxdz ydydxdz = = y dxdz = 2 0 2 0 z2 z2 z2 0 0 0 1 Z 9 1 9 1 21 1 = (2 − z − z 2 )dz = (2z − z 2 − z 3 ) = 2 0 2 2 3 4 0 Z 1 Z 2−z Z 1 Z 2−z 3 Z 1 Z 2−z Z 3 3xdxdz xy dxdz = xdydxdz = = 1
Z
Mxz
Myz
Z
2−z
z2
0
3
Z
1
Z
0
2−z
Z
z2
0
0
0
z2
2−z 1 Z 1 1 2 3 1 3 16 1 1 =3 dz = (4 − 4z + z 2 − z 4 )dz = (4z − 2z 2 + z 3 − z 5 ) = x 2 2 2 3 5 5 2 0 0 z 0 16/5 21/4 5/4 x = Myz /m = = 32/35, y = Mxz /m = = 3/2, z = Mxy /m = = 5/14. 7/2 7/2 7/2 The centroid is (32/35, 3/2, 5/14). Z
27. The density is ρ(x, y, z) = ky. Since both the region and the density function are symmetric with respect to the xyand yz-planes, x = z = 0. Using symmetry, √ Z 3 Z 2 Z √4−x2 Z 3 Z 2 4−k2 m=4 kydzdxdy = 4k dxdy yz 0
0
0
3
Z
Z
= 4k 0
2
0
Z p 2 y 4 − x dxdy = 4k 0
0 3
Z
πydy = 4πk
= 4k
3
0
0
z z=M4-x2
2 3
2
y
x
0
2 xp −1 x 2 dy 4 − x + 2 sin y 2 2 0
3 1 2 y = 18πk 2 0
√4−x2 Z 3Z 2 p 2 2 Mxz = 4 dxdy = 4k y 2 4 − x2 dxdy ky dzdxdy = 4k y z 0 0 0 0 0 0 0 0 3 Z 3 p Z 3 2 1 3 −1 x 2 x 2 2 = 4k dy = 4k y = 36πk. y 4 − x + 2 sin πy dy = 4πk 2 2 0 3 0 0 0 36πk y = Mxz /m = = 2. The center of mass is (0,2,0). 18πk Z
3
Z
2
√
Z
4−x2
28. The density is ρ(x, y, z) = kz. Z 1 Z x Z y+2 Z m= kzdzdydx = k 0
=
1 k 2
1 = k 2 =
1 k 6
x2 0 Z 1Z x
1
0 Z 1 0
Z
2
z 1
Z
0
x
x2
y+2 1 2 z dydx 2
(y + 2)2 dydx x 1 3 (y + 2) dx 3 x2
[(x + 2)3 − (x2 + 2)3 ]dx =
y=x
1
1 k 6
Z
y=x2 x
1
[(x + 2)3 − (x6 + 6x4 + 12x2 + 8)]dx
0
1 1 1 1 7 6 5 407 4 3 = k (x + 2) − x − x − 4x − 8x = k 6 4 7 5 840 0
2
0
x2
0
Z
3
Z
y
14.7. THE TRIPLE INTEGRAL 1
Z
x
Z
Z
y+2 2
0
=
Mxz
Myz
Z
1
Z
x
kz dzdydx = k
Mxy = =
191
1 k 3
x2 Z 1
1 k 12
0
Z
0
0
x2
y+2 Z 1Z x 1 3 1 dydx = k (y + 2)3 dydx z 3 0 3 0 x2
x Z 1 1 1 (y + 2)4 dx = k [(x + 2)4 − (x2 + 2)4 ]dx 4 12 2 0 x
1
[(x + 2)4 − (x8 + 8x6 + 24x4 + 32x2 + 16)]dx
0
1 1 1493 1 1 9 8 7 24 32 3 5 = k (x + 2) − x − x − − x − 16x = k 12 5 9 7 5 3 1890 0 Z 1Z x Z 1 Z x Z y+2 Z 1Z x y+2 1 2 1 kyzdzdydx = k dydx = k y(y + 2)2 dydx = yz 2 2 x2 2 x2 0 0 x 0 0 0 x Z 1Z x Z 1 1 1 1 4 4 3 3 2 2 (y + 4y + 4y)dydx = k = k y + y + 2y dx 2 0 x2 2 0 4 3 x2 Z 1 1 1 8 4 6 4 = k − x − x − 74x4 + x3 + 2x2 dx 2 0 4 3 3 1 1 68 1 4 7 1 2 = k − x9 − x7 − x5 + x4 + x3 = k 2 36 21 20 3 3 315 0 y+2 Z 1 Z x Z y+2 Z 1Z x Z 1Z x 1 1 2 dydx = k = xz x(y + 2)2 dydx kxzdzdydx = k 2 0 x2 0 x2 0 0 x2 2 0 x Z 1 Z 1 1 1 1 x(y + 2)3 dx = k [x(x + 2)3 − x(x2 + 2)3 ]dx = k 2 0 3 6 2 0 x Z 1 1 = k [x4 + 6x3 + 12x2 + 8x − x(x2 + 2)3 ]dx 6 0 1 21 1 2 1 1 5 3 4 3 2 k = k x + x + 4x + 4x − (x + 2) = 6 5 2 8 80 0
21k/80 68k/315 x = Myz /m = = 441/814, y = Mxz /m = = 544/1221, 407k/840 407k/840 1493k/1890 z = Mxy /m = = 5972/3663. The center of mass is (441/814,544/1221,5972/3663). 407k/840 29. m =
Z
1
−1
√
Z
1−x2
√ − 1−x2
Z
8−y
z
(x + y + 4)dzdydx 2+2y
8
y=-M1-x2 2 x
1 y=M1-x2
y
192
CHAPTER 14. MULTIPLE INTEGRALS
30. Both the region and the density function are symmetric with respect to the xy- and√ yz-planes. Thus, Z 2 Z 1+z2 Z √1+z2 −y2 z 2 dxdydz. m=4 −1
0
z 2
y=M1+x2 2
0
y
2 x
31. We are given ρ(x, y, z) = kz. Z Z 8 Z 4 Z y1/3 2 2 kz(x + z )dxdzdy = k Iy =
8
32. We are given ρ(x, y, z) = k. Z 1 Z 2−z Z 3 Z Ix = k(y 2 + z 2 )dydxdz = k
Z
y1/3 1 3 3 ( x z + xz ) dzdy 3 0 0 0 0 0 0 4 Z 8Z 4 Z 8 1 1 2 1 1/3 4 1/3 3 =k yz + y z dzdy = k yz + y z dy 3 6 4 0 0 0 0 8 Z 8 8 2560 4 =k y + 64y 1/3 dy = k y 2 + 48y 4/3 = k 3 3 3 0 0r √ p 4 5 2560k/3 = . From Problem 25, m = 96k. Thus, Rg = Iy /m = 96k 3
z2
0
0
0
1
Z
4
2−z
z2
3 Z 1 Z 2−z 1 3 (9 + 3z 2 )dxdz y + yz 2 dxdz = k 3 2 0 z 0
2−z Z 1 dz = k (18 − 9z − 3z 2 − 3z 3 − 3z 4 )dz =k (9x + 3xz 2 ) 2 0 0 z 1 223 9 2 3 3 3 4 5 = k 18z − z − z − z − z = k 2 4 5 20 0 Z 1 Z 2−z Z 3 Z 1 Z 2−z Z 1 m= kdydxdz = k 3dxdz = 3k (2 − z − z 2 )dz Z
1
z2
0
0
0
z2
0
1 7 1 2 1 3 = 3k 2z − z − z = k 3 2 0 s2 r r Ix 223k/20 223 Rg = = = m 7k/2 70 Z 1 Z 1−x Z 1−x−y 33. Iz = k (x2 + y 2 )dzdydx
0
Z
0 1
Z
z
0
1
1−x 2
2
(x + y )(1 − x − y)dydx
=k 0
Z
0 1Z
=k 0
0
1
1−x
(x2 − X 3 − x2 y + y 2 − xy 2 − y 3 )dydx
1
y
y=1-x
x 1−x 1 1 1 dx =k (x2 − x3 )y − x2 y 2 + (1 − x)y 3 − y 4 2 3 4 0 0 1 Z 1 1 1 1 1 1 1 1 k =k [ x2 − x3 + x4 + (1 − x)4 ]dx = k x6 − x4 + x5 − (1 − x)5 = 2 12 6 4 10 60 30 0 2 0
Z
1
14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS 34. We are given ρ(x, y, z) = kx. Z Z 1 Z 2 Z 4−z 2 2 kx(x + z )dydxdz = k Iy = z
0
0
Z
1
0
z
1
Z 0
2
4−z 3 2 (x + xz )y dxdz z
2
Z
3
193
y=z 1
y=4-z 4
2
y
x
2
(x + xz )(4 − 2z)dxdz =k 0 0 2 Z 1 1 4 1 2 2 =k ( x + x z )(4 − 2z) dz 2 0 4 0 Z 1 Z 1 2 =k (4 + 2z )(4 − 2z)dz = 4k (4 − 2z + 2z 2 − z 3 )dz 0
0
1 2 3 1 4 41 2 = 4k 4z − z + z − z = k 3 4 3 0 p 35. We are given ρ(x, y, z) = k x2 + y 2 + z 2 . Both the region and the integrand are symmetric with respect to the yzand xz-planes. Z 5 Z √25−x2 Z 5 p Iz = 4 k(x2 + y 2 ) x2 + y 2 + z 2 dzdydx √ 0
z 5
x2 +y 2
0
5 y
5 x
36. We are given ρ(x, y, z) = kz. Both the region and the integrand are symmetric with respect to the xz- and xy-planes. Z 1 Z √1−x2 Z √1−z2 Iy = 4 kx(x2 + z 2 )dydzdx −1
0
0
y=M25-x2
z 1
z=M1-x2
1 x
14.8
Triple Integrals in Other Coordinate Systems
√ √ √ √ 1. x = 10 cos 3π/4 = −5 2; y = 10 sin 3π/4 = 5 2; (−5 2, 5 2, 5) √ √ 2. x = 2 cos 5π/6 = − 3; y = 2 sin 5π/6 = 1; (− 3, 1, −3) √ √ √ √ 3. x = 3 cos π/3 = 3/2; y = 3 sin π/3 = 3/2; ( 3/2, 3/2, −4) √ √ √ √ 4. x = 4 cos 7π/4 = 2 2; y = 4 sin 7π/4 = −2 2; (2 2, −2 2, 0) 5. x = 5 cos π/2 = 0; y = 5 sin π/2 = 5; (0, 5, 1) √ √ 6. x = 10 cos 5π/3 = 5; y = 10 sin 5π/3 = −5 3; (5, −5 3, 2)
y=M1-x2
1
y
194
CHAPTER 14. MULTIPLE INTEGRALS
√ 7. With x = 1 and y = −1 we have r2 = 2 and tan θ = −1. The point is ( 2, −π/4, −9) √ √ 8. With x = 2 3 and y = 2 we have r2 = 16 and tan θ = 1/ 3. The point is (4, π/6, 17). √ √ √ √ 9. With x = − 2 and y = 6 we have r2 = 8 and tan θ = − 3. The point is (2 2, 2π/3, 2). √ 10. With x = 1 and y = 2 we have r2 = 5 and tan θ = 2. The point is ( 5, tan−1 2, 7). 11. With x = 0 and y = −4 we have r2 = 16 and tan θ undefined. The point is (4, −π/2, 0). √ √ √ 12. With x = 7 and y = − 7 we have r2 = 14 and tan θ = −1. The point is ( 14, −π/4, 3). 13. r2 + z 2 = 25 14. r cos θ + r sin θ − z = 1 15. r2 − z 2 = 1 16. r2 cos2 θ + z 2 = 16 17. z = x2 + y 2 18. z = 2y 19. r cos θ = 5, z = 5 √ √ √ 20. tan θ = 1/ 3, y/z = 1/ 3, z = 3y, x > 0 21. The equations are r2 = 4, r+ z 2 = 16, and z = 0. Z 2π Z 2 Z √16−r2 Z 2π Z 2 p V = rdzdrdθ = r 16 − r2 drθ 0
Z
0 2π
= 0
0
0
2 Z 1 2 3/2 − (16 − r ) dθ = 3
z=M16-r2
4
0
2π
0
0
z
√ √ 2π (64 − 24 3)dθ = (64 − 24 3) 3 2 2
y
r=2
x
22. The equation is z = 10 − r2 . Z 2π Z 3 Z 10−r2 Z V = rdzdrdθ = 0
Z = 0
0 2π
1
z
2π
0
3 Z 9 2 1 4 r − r dθ = 2 4 0
0
2π
Z
10
3
r(9 − r2 )drdθ z=10-r2
0
81 81π dθ = . 4 2 3 3
r=3 x
y
14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS 23. The equations are z = r2 , r = 5, and z = 0. Z Z 2π Z 5 Z 2π Z 5 Z r2 r3 drdθ = rdzdrdθ = V = 0
0 2π
Z = 0
0
0
0
0
195 z
2π
5 1 4 r dθ 4 0
z=r2
625 625π dθ = 4 2 5
y
r=5
5 x
z 24. Substituting the first equation into the second, we see y=r2 2 that the surfaces intersect in the plane y = 4. Using polar coordinates in the xz-plane, the equations of the surfaces become y = r2 and y = 21 r2 + 2. Z 2π Z 2 Z r2 /2+2 Z 2π Z 2 2 2 r rdydrdθ = r + 2 − r2 drdθ x V = y=r2/2+2 2 0 0 r2 0 0 2 Z 2π Z 2π Z 2π Z 2 1 1 2dθ = 4π r2 − r4 dθ = = 2r − r3 drdθ = 2 8 0 0 0 0 0
√ 25. The equation is √ z = a2 − r2 . By symmetry, x = y = 0. Z 2π Z a Z a2 −r2 Z 2π Z a p m= rdzdrdθ = r a2 − r2 drdθ 0 0 0 0 0 a Z 2π Z 2π 1 3 2 1 = a dθ = πa3 − (a2 − r2 )3/2 dθ = 3 3 3 0 0 0 √a2 −r2 Z 2π Z a Z √a2 −r2 Z 2π Z a 1 2 drdθ Mxy = zrdzdrdθ = rz 0 0 0 0 0 2 0 Z Z 1 2π a = r(a2 − r2 )drdθ 2 0 0 a Z Z 1 2π 1 4 1 1 2π 1 2 2 1 4 a r − r dθ = a dθ = πa4 = 2 2 4 2 4 4 0
4
y
z
a
z=Ma2-r2
a
a
y
x
0
0
πa4 /4 z = Mxy /m = = 3a/8. The centroid is (0, 0, 3a/8). 2πa3 /3 z
26. We use polar coordinates in the yz-plane. The density is ρ(x, y, z) = kz. By symmetry, y = z = 0. Z
2π
4
Z
Z
m=
5
Z
2π
Z
kxrdxdrdθ = k 0
k = 2
0
Z
2π
Z
Z
0 2π
0
25k = 2
0 4
0
25k 25rdrdθ = 2
8dθ = 200kπ 0
Z 0
2π
0
4
5 1 2 rz drdθ 2 0
4 1 2 r dθ 2 0
4
r=4
4 5 x
y
196
CHAPTER 14. MULTIPLE INTEGRALS 2π
Z
4
Z
5
Z
Z
2
2π
Z
kx rdxdrdθ = k
Myz = 0
0
0
0
0
4
5 Z 2π Z 4 1 3 1 125rdrdθ rx drdθ = k 3 3 0 0 0
4 Z 2π 1 125 2 1 2000 = k 1000dθ = r dθ = k kπ 3 2 3 3 0 0 0 2000kπ/3 x = Myz /m = = 10/3, The center of mass of the given solid is (10/3, 0, 0). 200kπ 2π
Z
√ 27. The equation is z√= 9 − r2 and the density is ρ = k/r2 . When x = 2,√r = √ 5. Z 2π Z 5 Z 9−r2 Iz = r2 (k/r2 )rdzdrdθ(k/r2 ) 0
0
Z
5
2π
0 Z √
5
√9−r2 rz drdθ
=k 0
Z
2
3 x
√5 1 − (9 − r2 )3/2 − r2 dθ 3 0
2π
=k 0 2π
4 8 dθ = πk 3 3
=k 0
28. The equation Z 2π Z 1 is Z z1 = r and the density is ρ = kr. Ix = (y 2 + z 2 )(kr)rdzdrdθ 0
0
Z
2π
0
1
0
Z
1
1
Z
z=r
(r4 sin2 θ + r2 z 2 )dzdrdθ
r
1 1 2 3 2 4 =k (r sin θ)z + r z drdθ 3 0 0 r Z 2π Z 1 1 1 =k r4 sin2 θ + r2 − r5 sin2 θ − r4 drdθ 3 3 1 Z 2π0 0 1 5 2 1 3 1 6 2 1 6 =k r sin θ + r − r sin θ − r dθ 5 9 6 18 0 0 Z 2π 1 1 2 =k sin θ + dθ 30 18 0 2π 1 1 1 13 =k θ− sin 2θ + θ = πk 60 120 18 90 2π
z
r
Z
=k Z
y
0
0
Z
3
p r 9 − r2 − 2r drdθ
=k Z
z=M9-r2
3
2 √
2π
Z
z
1
1
1
y
x
0
√ 29. (a) x = (2/3) sin(π/2) sin(π/2) cos(π/6) = 3/3; y = (2/3) sin(π/2) sin(π/6) = 1/3; √ √ (b) With x = 3/3 and y = 1/3 we have r2 = 4/9 and tan θ = 3/3. The point is (2/3, π/6, 0).
14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS
197
√ √ 30. (a) x = 5 sin(5π/4) cos(2π/3) 2/4; y = 5√sin(5ı/4)√sin(2π/3) = −5 6/4; √ =5 √ z = 5 cos(5π/4) = −5 2/2; (5 2/4, −5 6/4, −5 2/2) √ √ √ (b) With x = 5 2/4√and y = −5√ 6/4 we have r2 = 25/2 and tan θ = − 3. The point is (5/ 2, 2π/3, −5 2/2). √ 31. (a) x = 8 sin(π/4) cos(3π/4) = −4; y = 8 sin(π/4) sin(3π/4) = 4; z = 8 cos(π/4) = 4 2; √ (−4, 4, 4 2) √ √ (b) With x = −4 and y = 4 we have r2 = 32 and tan θ = −1. The point is (4 2, 3π/4, 4 2). √ 32. (a) x = (1/3) sin(5π/3) cos(π/6) = −1/4;√y = (1/3) sin(5π/3) sin(π/6) = − 3/12; z = (1/3) cos(5π/3) = 1/6; (−1/4, − 3/12, 1/6) √ √ 2 (b) With x = −1/4 and √ y = − 3/12 we have r = 1/12 and tan θ = 3/3. The point is (1/2 3, π/6, 1/6). √ √ 33. (a) x = √ 4 sin(3π/4) √ cos 0 = 2 2; y = 4 sin(3π/4) sin 0 = 0; z = 4 cos(3π/4) = −2 2; (2 2, 0, −2 2) √ √ √ (b) With x = 2 2 and y = 0 we have r2 = 8 and tan θ = 0. The point is (2 2, 0, −2 2). √ 34. (a) x = 1 sin(11π/6) cos π = 1/2; y = 1 sin(11π/6) sin π = 0; z = 1 cos(11π/6) = 3/2; √ (1/2, 0, ( 3/2) √ (b) With x = 1/2 and y = 0 we have r2 = 1/4 and tan θ = 0. The point is (1/2, 0 3/2). 2 35. With √ x = −5, y = −5, and z = 0, we have ρ = 50, tan θ = 1, and cos φ = 0. The point is (5 2, π/2, 5π/4). √ √ √ 36. With z = 1, y√= − 3, and z = 1, we have ρ2 = 5, tan θ = − 3, and cos φ = 1/ 5. √ The point is ( 5, cos−1 1/ 5, −π/3). √ √ √ 2 37. With x = 3/2, √ y = 1/2, and z = 1, we have ρ = 2, tan θ = 1/ 3, and cos φ = 1/ 2. The point is ( 2, π/4, π/6). √ 38. With x = − 3/2, y = 0, and z = −1/2, we have ρ2 = 1, tan θ = 0, and cos φ = −1/2. The point is (1, 2π/3, 0). √ √ 39. With x = 3, y = −3, and z = 3 2, we have ρ2 = 36, tan θ = −1, and cos φ = − 2/2. The point is (6, π/4, −π/4) √ √ 40. With x = 1, y √ = 1, and z = − 6, we have ρ2 = 8, tan θ = 1, and cos φ = − 3/2 The point is (2 2, 5π/6, π/4).
41. ρ = 8 42. ρ2 = 4ρ cos φ; ρ = 4 cos φ √ 43. 4z 2 = 3x2 + 3y 2 + 3z 2 ; 4ρ2 cos2 φ = 3ρ2 ; cos φ = ± 3/2; φ = π/6, 5π/6 44. −x2 − y − z 2 = 1 − 2z 2 ; −ρ2 = 1 − 2ρ2 cos2 φ; ρ2 (2 cos2 φ − 1) = 1 45. x2 + y 2 + z 2 = 100
198
CHAPTER 14. MULTIPLE INTEGRALS
46. cos φ = 1/2; ρ2 cos2 φ = ρ2 /4; 4z 2 = x2 + y 2 + z 2 ; z 2 + y 2 = 3z 2 47. ρ cos φ = 2; z = 2 48. ρ(1 − cos2 φ) = cos φ; ρ2 − ρ2 cos2 φ = ρ cos φ; x2 + y 2 + z 2 − z 2 = z; z = x2 + y 2 z
49. The
equations are φ = π/4 and ρ = 3. 3 Z 2π Z π/4 Z 2π Z π/4 Z 3 1 3 2 ρ sin φdrhodφdθ = ρ sin φ dφdθ V = 3 0 0 0 0 0 0 π/4 Z 2π Z 2π Z π/4 dθ −9 cos φ 9 sin φdφdθ = = 0 0 0 0 ! Z 2π √ √ 2 = −9 − 1 dθ = 9π(2 − 2) 2 0
50. The equations are ρ = 2, θ = π/4, and θ = π/3. 2 Z π/3 Z π/2 Z 2 Z π/3 Z π/2 1 3 2 ρ sin φdρdφdθ = ρ sin φ dφdθ 3 π/4 0 0 π/4 0 0 Z π/3 Z π/2 8 sin φdφdθ = 3 π/4 0 π/2 Z 8 π/3 dθ − cos φ = 3 π/4 0 Z 8 π/3 2π = (0 + 1)dθ = 3 π/4 9
0
0 π/2
= 0
0 π/2
8 = 3
Z
8 3
Z
=
0
z
2 ρ=2
2
0
y
2
x
z ρ=2secφ
2
2 sec φ 1 2 ρ sin φ dφdθ 3 0
Z
π/6
sec3 φ sin φdφdθ =
0 π/2
y
x
0 π/6
Z
2
2
51. Using Problem 43, the equations are φ = π/6, θ = π/2, and ρ cos φ = 2. Z π/2 Z π/6 Z 2 sec φ V = ρ2 sin φdρdφdθ Z
ρ=3
3
8 3
Z 0
π/2
Z
π/6
0
π/6 Z 1 4 π/2 1 2 2 tan φ dθ = dθ = π 2 3 0 3 9 0
sec2 φ tan φdφdθ
1
x
1
y
14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS
199 z
52. The equations are ρ = 1 and φ = π/4. We find the volume above the xy-plane and double. 1 1 ρ2 sin φdρdφdθ = 2 ρ3 sin φ dφdθ V =2 3 π/4 0 0 π/4 0 0 π/2 Z 2π Z π/2 Z 2π Z 2π √ 2 2 2 2 sin φdφdθ = = − cos φ = dθ 3 0 3 3 2 π/4 0 0 π/4 √ 2π 2 = 3 2π
Z
π/2
Z
Z
1
2π
Z
1
ρ=1
π/2
Z
1
1 x
z 53. By symmetry, x = y = 0. The equations are φ = π/4 and 2 ρ = 2 cos φ. 2 cos φ Z 2π Z π/4 Z 2 cos φ Z 2π Z π/4 1 3 dφdθ m= ρ2 sin φdρdφdθ = ρ sin φ 3 0 0 0 0 0 0 π/4 Z Z Z 8 2π π/4 8 2π 1 3 4 dθ = sin φ cos φdφdθ = − cos φ 1 3 0 3 4 0 0 0 Z 2π x 2 1 =− − 1 dθ = π 3 0 4 Z 2π Z π/4 Z 2 cos φ Z 2π Z π/4 Z 2 cos φ 2 Mxy = zρ sin φdρdφdθ ρ3 sin φ cos φdρdφdθ 0
0
0
0
0
y
ρ=2cosφ
1
y
0
2 cos φ Z 2π Z π/4 1 4 = ρ sin φ cos φ dφdθ = 4 cos5 φ sin φdφdθ 4 0 0 0 0 0 π/4 Z 2π Z 2 2π 1 7 1 6 =− − 1 dθ = π =4 − cos φ 6 3 8 6 0 0 0 7π/6 z = Mxy /m = = 7/6. The centroid is (0, 0, 7/6). π 2π
Z
π/4
Z
z
54. We are given density= kz. By symmetry, x = y = 0. The equation is ρ = 1. 2π
Z
π/2
Z
Z
1
kzρ2 sin φdρdφdθ = k
m= 0
0 2π
Z
0 π/2
Z
=k 0
= =
0 2π
1 k 4
Z
1 k 4
Z
Z 0
2π
Z 0
1 1 4 ρ sin φ cos φ dφdθ 4 0
Z
π/2
sin φ cos φdφdθ 0
0
0 2π
π/2 Z 2π 1 1 kπ sin2 φ dθ = k dθ = 2 8 4 0 0
π/2
Z
1
ρ=1
1
ρ3 sin φ cos φdρdφdθ
0 1 x
1
y
200
CHAPTER 14. MULTIPLE INTEGRALS 2π
Z
π/2
Z
Z
Mxy =
kz 2 ρ2 sin φdρdφdθ = k
Z 0
0
0
0
1
2π
Z
π/2
0
Z
1
ρ3 cos2 φ sin φdρdφdθ
0
1 Z 2π Z π/2 1 5 1 2 =k cos2 φ sin φdφdθ ρ cos φ sin φ dφdθ = k 5 5 0 0 0 0 0 π/2 Z 2π Z 2π 1 1 2 1 − cos3 φ =− k (0 − 1)dθ = = k kπ 5 0 3 15 15 0 0 2kπ/15 z = Mxy /m = = 8/15. The center of mass is (0, 0, 8/15). kπ/4 2π
Z
π/2
Z
55. We are given density= k/ρ. Z 2π Z cos−1 4/5 Z 5 k 2 m= ρ sin φdρdφdθ 4 sec φ ρ 0 0 5 Z 2π Z cos−1 4/5 1 2 dφdθ ρ sin φ =k 2 0 0
z 5
ρ=4sec φ
4 sec φ
Z
2π
Z
3
cos−1 4/5
1 = k (25 sin φ − 16 tan φ sec φ)dφdθ 2 0 0 cos−1 4/5 Z 2π 1 k dθ (−25 cos φ − 16 sec φ) 2 0 0 Z 2π 1 = k [−25(4/5) − 16(5/4) − (−25 − 16)]dθ 2 0 Z 2π 1 dθ = kπ = k 2 0 Z 2π Z π Z a 56. Iz = (x2 + y 2 )(kρ)ρ2 sin φdρdφdθ 0
0
Z
2π
π
x
z a
Z
=k
a 2
2
2
2
2
14.9
0
ρ=a
3
sin π cos θ + ρ sin φ sin θ)ρ sin φdρdφdθ a Z 2π Z π Z a Z 2π Z π 1 6 3 3 5 ρ sin φ dφdθ =k ρ sin φdρdφdθ = k 0 0 0 0 0 6 0 a Z 2π Z π Z 2π Z π 1 6 1 3 3 2 = ka sin φdφdθ = ka (1 − cos φ) sin φdφdθ x 6 6 0 0 0 0 π Z 2π Z 2π 1 1 1 4 4π 6 = ka3 − cos φ + cos3 φ dθ = ka3 dθ = ka 6 3 6 3 9 0 0 0 0
y
3
0
Z
ρ=5
0
a
y
Change of Variables in Multiple Integrals
1. T : (0, 0) −→ (0, 0); (0, 2) −→ (−2, 8); (4, 0) −→ (16, 20); (4, 2) −→ (14, 28) 2. Writing x2 = v − u and y = v + u and solving for u and v, we obtain u = (y − √ x2 )/2 and 2 −1 v = (x +y)/2. Then the images under T are (1, 1) −→ (0, 1); (1, 3) −→ (1, 2); ( 2, 2) −→
14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
201
(0, 2). 3. The uv-corner points (0, 0), (2, 0), (2, 2) correspond to xy-points (0, 0), (4, 2), (6, −4). v = 0 : x = 2u, y = u =⇒ y = x/2 y u = 2 : x = 4 + v, y = 2 − 3v =⇒ y = 2 − 3(x − 4) = −3 + 14 v = u : x = 3u, y = −2u =⇒ y = −2x/3 v=u
y y=x/2
y=14-3x x
u=2 y=-2x/3
S v=0
4. Solving for x and y we see that the transformation is x = 2u/3 + v/3, y = −u/3 + v/3 > The uv-corner points (−1, 1), (4, 1), (4, 5), (−1, 5) correspond u=-1 to the xy-points (−1/3, 2/3), (3, −1), (13/3, 1/3), (1, 2). v = 1 : x + 2y = 1; v = 5 : x + 2y = 5; u = −1 : x − y = −1; u=4:x−y =4
y
y
v=5 x+2y=5
x-y=-1
u=4
S v=1
x
x-y=4
x+2y=1
x
5. The uv-corner points (0.0), (1, 0), (1, 2), (0, 2) correspond to the u=0 xy-points (0, 0), (1, 0), (−3, 2), (−4, 0). v = 0 : x = u2 , y = 0 =⇒ y = 0 and 0 ≤ x ≤ 1 u = 1 : x = 1 − v 2 , y = v =⇒ x = 1 − y 2 v = 2 : x = u2 − 4, y = 2u =⇒ x = y 2 /4 − 4 u = 0 : x = −v 2 , y = 0 =⇒ y = 0 and −4 ≤ x ≤ 0
6. The uv-corner points (1, 1), (2, 1), (2, 2), (1, 2) correspond to the xy-points (1, 1), (2, 1), (4, 4), (2, 4). v = 1 : x = u, y = 1 =⇒ y = 1, 1 ≤ x≤2 u = 2 : x = 2v, y = v 2 =⇒ y = x2 /4 v = 2 : x = 2u, y = 4 =⇒ y = 4, 2 ≤ x≤4 u = 1 : x = v, y = v 2 =⇒ y = x2
x
y
y v=2 S u=1 v=0
x=y2/4-4
x=1-y2
x
y=0
x
y y
y=4 y=x2 y=x2/4
v=2 u=1
S u=2 v=1
y=1 x
x
202
CHAPTER 14. MULTIPLE INTEGRALS
∂(x, y) = 7. ∂(u, v) ∂(x, y) 8. = ∂(u, v) ∂(u, v) = 9. ∂(x, y) ∂(u, v) 10. = ∂(x, y)
−ve−u veu
e−u = −2v eu
e3u cos v = −3e6u −e3u sin v 3y 2 y ∂(x, y) 1 1 −2y/x3 1/x2 = − = −3( 2 )2 = −3u2 ; =− 2 = 2 2 −y /x 2y/x x4 x ∂(u, v) −3u2 3u 2(y 2 − x2 ) −4xy 2 2 2 2 2 2 4 x +y ) (x + y ) = 4xy 2(y 2 − x2 ) (x2 + y 2 )2 (x2 + y 2 )2 (x2 + y 2 )2 From u = 2x/(x2 + y 2 ) and v = −2y(x2 + y 2 ) we obtain u2 = v 2 = 4/(x2 + y 2 ). Then x2 + y 2 = 4/(u2 + v 2 ) and ∂(x, y)/∂(u, v) = (x2 + y 2 )2 /4 = 4/(u2 + v 2 )2 .
11.
3e3u sin v 3e3u cos v
(a) The uv-corner points (0, 0), (1, 0), (1, 1), (0, 1) correspond to the xy-points (0, 0), (1, 0), (0, 1), (0, 0). v = 0 : x = u, y = 0 =⇒ y = 0, 0 ≤ x ≤ 1 u=0 u = 1 : x = 1 − v, y = v =⇒ y = 1 − x v = 1 : x = 0, y = u =⇒ x = 0, 0 ≤ y ≤ 1 u = 0 : x = 0, y = 0
y
y v=1
y=1-x S
u=1
v=0
x=0 y=0
x
x
(b) Since the segment u = 0, 0 ≤ v ≤ 1 in the uv-plane maps to the origin in the xy-plane, the transformation is not one-to-one.
∂(x, y) 1 − v 12. = ∂(u, v) −u
v = u. The transformation is 0 when u is 0, for 0 ≤ v ≤ 1. u
y 13. R1 : x + y = −1 =⇒ v = −1 R4 R2 : x − 2y = 6 =⇒ u = 6 R3 R R3 : x + y = 3 =⇒ v = 3 R4 : x − 2y R1 = −6 =⇒ u = −6 ∂(u, v) 1 −2 ∂(x, y) 1 R2 = = 3 =⇒ = ∂(x, y) 1 1 ∂(u, v) 3 RR RR 1 1 R3 R6 vdudv = (x + y)dA = v( )dA0 = R S 3 −1 −6 3 3 R3 1 1 (12) −1 vdv = 4( )v 2 = 16 3 2
−1
y
S
x
x
14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS 14. R1 : y = −3x + 3 =⇒ v = 3 y R2 : y = x − π =⇒ u = π R3 : y = −3x + 6 =⇒ v = 6 R4 R4 : y = x =⇒ u = 0 R3 ∂(x, y) ∂(u, v) 1 −1 1 = 4 =⇒ = ∂(x, y) 3 1 ∂(u, v) 4 x R1 R Z Z cos 1 (x − y) Z Z cos u/2 1 R2 2 dA = ( )dA0 3x + y v 4 R S π Z Z Z 1 6 π cos u/2 1 6 2 sin u/2 = dudv = dv 4 3 0 v 4 3 v 0 6 Z 1 6 dv 1 1 = = ln v = ln 2 2 3 v 2 2 3
15. R1 : y = x2 =⇒ u = 1 y R2 : x = y 2 =⇒ v = 1 1 R4 R3 : y = x2 =⇒ u = 2 2 R R3 R1 1 R4 : x = y 2 =⇒ v = 2 R2 2 ∂(u, v) 2x/y −x2 /y 2 =3 = x 2y/x ∂(x, y) −y 2 /x2 ∂(x, y) 1 =⇒ = ∂(u, v) 3 2 RR R R y2 1 1 R2R2 1 1 2 1 R2 0 dA = v( )dA = vdv = v = vdudv = S R x 1 1 1 3 3 3 6 1 2
203 y S
x
y
S
x
x2 + y 2 = 2y =⇒ v = 1 y x2 + y 2 = 2x =⇒ u = 1 y x2 + y 2 = 6y =⇒ v = 1/3 x2 + y 2 = 4x =⇒ u = 1/2 R4 2(y 2 − x2 ) −4xy S R ∂(u, v) (x2 + y 2 )2 (x2 + y 2 )2 R1 = R3 2 2 −4xy 2(x − y ) ∂(x, y) R2 (x2 + y 2 )2 (x2 + y 2 )2 x x −4 = 2 (x + y 2 )2 2 Using u + v 2 = 4/(x2 + y 2 ) we see that ∂(x, y)/∂(u, v) = −4/(u2 + v 2 )2 . RR RR 4 −4 1 R1 R1 115 (u2 + v 2 )dudv = (x2 + y 2 )−3 dA = ( )−3 | 2 |dA0 = R S u2 + v 2 (u + v 2 )2 16 1/3 1/2 5184
16. R1 : R2 : R3 : R4 :
204
CHAPTER 14. MULTIPLE INTEGRALS
y
y
17. R1 : 2xy = c =⇒ v = c R2 : x2 − y 2 = b =⇒ u = b d d S R3 : 2xy = d =⇒ v = d R4 : x2 − y 2 = a =⇒ u = a c R3 c R R4 ∂(u, v) 2x −2y R2 2 2 = 4(x + y ) = R1 ∂(x, y) 2y 2x a b b a x ∂(x, y) 1 =⇒ = ∂(u, v) 4(x2 + y 2 ) RR R R 1 1 RdRb 1 (x2 + y 2 )dA = (x2 + y 2 ) dA0 = dudv = (b − a)(d − c) R S 4(x2 + y 2 ) 4 c a 4
x
y
18. R1 : xy = −2 =⇒ v = −2 R2 : x2 − y 2 = 9 =⇒ u = 9 R3 : xy = 2 =⇒ v = 2 R4 : x2 − y 2 = 1 =⇒ u = 1 ∂(u, v) 2x −2y = 2(x2 + y 2 ) = x ∂(x, y) y ∂x, y) 1 =⇒ = 2 ∂(u, v) 2(x + y 2 ) Z Z Z Z (x2 + y 2 ) sin xydA = (x2 + y 2 ) sin v( R
1 = 2
Z
S 2
R3 R4
R
R2 4
2
y
x
S
R1
x
1 1 )dA0 = 2(x2 + y 2 ) 2
Z
2
Z
9
sin vdudv −2
1
8 sin vdv = 0 −2
19. R1 : y = x2 =⇒ v + u = v − u =⇒ u = 0 R2 : y = 4 − x2 =⇒ v + u = 4 − (v − u) =⇒ v + u = 4 − v + u =⇒ v = 2 R3 : x = 1 =⇒ v − u = 1 =⇒ v = 1 + u 1 1 √ ∂(x, y) − √ 1 = 2 v − u 2 v − u = − √ ∂(u, v) v−u 1 1
y
y
R2 R3
R
2 R1
x
S
x
14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS Z Z R
205
Z Z √
v − u 1 0 dA −√ 2v v − u S Z Z Z 1 1 2 1 1 1 = [ln 2 − ln(1 + u)]du dvdu = 2 0 1+u v 2 0 1 1 1 1 1 = ln 2 − [(1 + u) ln(1 + u) − (1 + u)] = ln 2 − [2 ln 2 − 2 − (0 − 1)] 2 2 2 2 0 1 1 = − ln 2 2 2
x dA = y + x2
20. Solving x = 2u − 4v, y = 3u + v for y 2 1 x + y, u and v we obtain u = 14 7 3 1 y v = − x + y. The xy-corner points (−4, 1), 14 7 (0, 0), (2, 3) correspond to the uv-points R3 S R2 R (0, 1), (0, 0), (1, 0). R1 x ∂(x, y) 2 −4 = 14 = x 3 1 ∂(u, v) 1−u Z Z Z Z Z 1 Z 1−u Z 1 1 2 0 ydA = (3u + v)(14)dA = 14 du (3u + v)dvdu = 14 (3uv + v ) 2 R S 0 0 0 0 1 Z 1 28 1 5 2 35 3 2 = 14 ( + 2u − u )du = (7u + 14u − u ) = 2 2 3 3 0 0
21. R1 : R2 : R3 : R4 :
y y y y
= 1/x =⇒ u = 1 = x =⇒ v = 1 = 4/x =⇒ u = 4 = 4x =⇒ v = 4 ∂(u, v) y x 2y = = =⇒ −y/x2 1/x ∂(x, y) x Z Z Z Z 1 4 y dA = u2 v 2 ( )dudv = 2v R S 4 21 2 315 = v = 4 4 1
y
y
R4
R3
R1
S
R2
∂(x, y) x = x ∂(u, v) 2y 4 Z 4 Z 4 Z 1 1 1 3 1 4 2 u vdudv = u v dv = 63vdv 2 1 2 1 3 6 1 1
x
22. Under the transformation u = y + z, v = −y + z, w = x − y the parallelepiped D is mapped
206
CHAPTER 14. MULTIPLE INTEGRALS to the parallelepiped E : 1 ≤ u ≤ 3, −1 ≤ v ≤ 1, 0 ≤ w ≤ 3. 0 1 1 ∂(x, y, z) ∂(u, v, w) 1 = 0 −1 1 = 2 =⇒ = ∂(x, y, z) ∂(u, v, w) 2 1 −1 0 Z Z Z Z Z Z Z 1 3 1 3 1 0 (2u + 2v + 2w)dudvdw (4z + 2x − 2y)dV = (2u + 2v + 2w) dV = 2 2 0 −1 1 D E 3 Z Z Z 1 3 1 2 1 3 = (u + 2uv + 2uw) dvdw = int1−1 (8 + 4v + 4w)dvdw 2 0 −1 2 0 1 1 Z 3 Z 3 3 (8 + 4w)dw = (8w + 2w2 ) 0 = 42 (4v + v 2 + 2vw) dw = = 0
0
−1
23. We let u = y − x and V=y+x. R1 : y = 0 =⇒ u = −x, v = x =⇒ v = −u R2 : x + y = 1 =⇒ v = 1 R3 : x = 0 =⇒ u = y, v = y, =⇒ v = u ∂(u, v) −1 1 = −2 = ∂(x, y) 1 1 ∂(x, y) 1 =⇒ =− ∂(u, v) 2 Z Z Z Z 1 (y−x)/(y+x) e dA = eu/v − dA0 2 R S Z 1Z v 1 = eu/v dudv = 2 0 −v Z 1 1 v(e − e−1 )dv = = 2 0
y
S
R3
R2 R x
R1
x
v ve dv 0 −v 1 1 1 −1 1 2 (e − e ) v = (e − e−1 ) 2 2 0 4 1 2
Z
1
u/v
24. We let u = y − x and v = y. R1 : y = 0 =⇒ v = 0, u = −x =⇒ v = 0, 0 ≤ u ≤ 2 R2 : x = 0 =⇒ v = u R3 R3 : y = x + 2 =⇒ u = 2 ∂(x, y) ∂(u, v) −1 1 R = = −1 =⇒ ∂(u, v) = −1 0 1 ∂(x, y) Z Z Z Z 2 2 2 R1 ey −2xy+x dA = eu |−1| dA0 R
y 1
y
y
R2
S
x
S
Z
2
Z
= 0
0
u
2
eu dvdu =
Z 0
2
2
ueu du =
2 1 u2 1 e = (e4 − 1) 2 2 0
x
14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
25. Noting that R2, R3, and R4 have equations y + 2x = 8, y − 2x = 0, and y + 2x = 2, we let u = y/x and v = y + 2x. R1 : y = 0 =⇒ u = 0, v = 2x =⇒ u = 0, 2 ≤ v ≤ 8 y R2 : y + 2x = 8 =⇒ v = 8 R3 : y − 2x = 0 =⇒ u = 2 R3 R4 : y + 2x = 2 =⇒ v = 2 ∂(u, v) −y/x2 1/x y + 2x = = R 2 1 ∂(x, y) x2 R4 ∂(x, y) x2 =⇒ = ∂(u, v) y + 2x RR RR x2 0 (6x + 3y)dA = 3 S (y + 2x) − dA = R y + 2x RR 2 0 3 S x dA From y = ux we see that v = ux + 2x and x = v/(u + 2). Then 8 RR 2 0 R2R8 2 R2 v 3 2 3 S x dA = 3 0 2 v (u+2) dvdu = 0 du = (u + 2)2 2 2 R2 du 504 504 0 =− = 126. 2 (u + 2) u + 2 0
207
y
S R2
R1
x
x
y
y 26. We let u = x + y and v = x − y. R1 : x + y = 1 =⇒ u = 1 R4 R3 R2 : x − y = 1 =⇒ v = 1 R S R3 : x + y = 3 =⇒ u = 3 R2 R1 R4 : x − y = −1 =⇒ v = −1 x ∂(x, y) ∂(u, v) 1 1 1 = −2 =⇒ = =− 1 −1 ∂(x, 2 Z Z y) Z Z ∂(u, v) 1 (x + y)4 ex−y dA = u4 ev − dA0 2 R S 1 Z 3Z 1 Z 1 1 3 4 v 4 v = u e dvdu = u e du 2 1 −1 2 1 −1 3 −1 −1 Z 3 e−e 242(e − e−1 121 e−e u4 du = u5 = = (e − e−1 ) = 2 10 10 5 1 1
5 ∂(x, y) 27. The image of the ellipse is the unit circle x + y = 1. From = 0 ∂(u, v) 2
2
x
0 = 15 we 3
208
CHAPTER 14. MULTIPLE INTEGRALS obtain Z Z
y2 x2 ( + )dA = 9 R 25
Z Z
2π
Z
0
2
Z
(u + v )15dA = 15 15 4
0
0
s
= sin ω cos θ ∂(x, y, z) 28. = sin ω sin θ ∂(ρ, ω, θ) cos ω
2
2π
Z
15 r rdrdθ = 4 2
Z 0
2π
1 4
r dθ 0
15π . 2
dθ = 0
ρ cos ω cos θ ρ cos ω sin θ −ρ sin ω
1
−ρ sin ω sin θ ρ sin ω cos θ 0
= cos ω(ρ2 sin ω cos ω cos2 θ + ρ2 sin ω cos ω sin2 θ) + ρ sin ω(ρ sin2 ω cos2 θ + ρ sin2 ω sin2 θ) = ρ2 sin ω cos2 ω(cos2 θ + sin2 θ) + ρ2 sin3 ω(cos2 θ + sin2 θ) = ρ2 sin ω(cos2 ω + sin2 ω) = ρ2 sin ω 29. The image of the ellipsoid x2 /a2 + y 2 /b2 + z 2 /c2 = 1 under the transformation u = x/a, v = 4 y, w = z/c, is the unit sphere u2 + v 2 + w2 = 1. The volume of this sphere is π. Now 3 ¯ a 0 0 ∂(x, y, z) = 0 b 0 = abc ∂(u, v, w) 0 0 c and
Z Z Z
Z Z Z dV =
EabcdV 0 = abc
Z Z Z
D
dV 0 = abc E
4 π 3
=
4 πabc. 3
30. Let u = xy and v = xy 1.4 . Then xy 1.4 = c =⇒ v = c; xy = b =⇒ u = b; xy 1.4 = d =⇒ v = d; xy = a =⇒ u = a. ∂(u, v) y x = 0.4xy 1.4 = 0.4v =⇒ ∂(x, y) = 5 = ∂(x, y) y 1.4 1.4xy 0.4 ∂(u, v) 2v Z Z Z Z Z dZ b Z d 5 5 5 dv 5 dA = dA0 = dudv = (b − a) = (b − a)(ln d − ln c) 2v 2v 2 v 2 R S c c a
Chapter 14 in Review A. True/False 1. True; use ex
2
−y
2
= ex e−y and Problem 53 in Section 14.2
2. True 3. True 4. False; consider f (x, y) = x. 5. False; both the density function and the lamina must be symmetric about an axis. 6. True; the equation of the plane is θ =
π 4,
θ=
5π 4
CHAPTER 14 IN REVIEW
209
B. Fill in the Blanks Z
5
1. y 2 +1
5y 8y − x 3
5 dx = (8xy 3 − 5y ln x) y2 +1 = 40y 3 − 5y ln 5 − [8(y 2 + 1)y 3 − 5y ln(y 2 + 1)] = −8y 5 + 32y 3 + 5y ln
2. 16 6.
Z
a
Z b√1−x2 /a2
7.
3. square Z √ 2 2
√
4. II
5. f (x, 4) − f (x, 2)
1−x /a −y 2 /b2
c
√
−b 1−x2 /a2 −c R 4 R √x f (x, y)dydx 0 x/2 −a
y2 + 1 y
ρ(x, y, z)dzdydx 1−x2 /a2 −y 2 /b2 y y=Mx y=x/2
1 1
4 x √ 8. x = 6 sin(5π/3) cos(5π/6) = 9/2; y = 6 sin(5π/3) sin(5π/6) = −3 3/2; z = 6 cos(5π/3) = 3 √ The point is (9/2, −3 3/2, 3). √ √ √ √ 9. r = 2 sin(π/4) = 2; θ = 2π/3; z = 2 cos(π/4) = 2; ( 2, 2π/3, 2) Z 4 Z √4−y y 10. f (x, y)dxdy 4 √ y=4-x 2
− 4−y
0
1
11. z = r2 ; ρ = cot φ csc φ 12. circle
C. Exercises Z 1. Holding x fixed,
12x2 e−4xy = −5xy + y + c1 (x) −4x = −3xe−4xy − 5xy + y + c1 (x)
(12x2 e−4xy − 5x + 1)dy =
2. Holding y fixed, R 1 ln |3xy + 4| dx = + c2 (y) 4 + 3xy 3y Z y y y 2 sin xydx = −y cos xy|y3 = y(cos y 4 − cos y 2 ) 3. y3
Z
ex
4. 1/x
ex x x dy = − = x2 − x/ex y2 y 1/x
x
210
CHAPTER 14. MULTIPLE INTEGRALS Z
2
2x
Z
y−x
ye
5.
Integration by parts
(2xex − ex + e−x )dx
Integration by parts
y−x
dydx =
0
0
2x ) dx
2
Z
−e
(ye 0 Z 2
=
y−x
0
0
2 = (2xex − 2ex − ex − e−x ) 0 = e2 − e−2 + 4 Z
4
4 Z 4 4 4 y x 1 −1 x 2 = − dx = tan − ln(16 + x ) 2 2 2 16 + x 16 + x 4 2 0 0 16 + x x 0 π 1 1 π 1 π 1 1 = − ln 32 − 0 − ln 16 = + (ln 16 − ln 32) = + ln 4 2 2 4 2 4 2 2
4
Z
1 dydx = 16 + x2
6. x
0
Z
1
√
Z
x
7. x
0
sin y dydx = y
Z
4
1
Z
Z
y
y2
0
sin y dxdy = y
Z 0
1
y sin y x dy y y2
y
y=x
1
y=Mx
1
Z
(sin y − y sin y)dy
=
Integration by parts
0
1 x
1
= (− cos y − s ∈ y + y cos y)|0 = (− cos 1 − sin 1 + cos 1) − (−1) = 1 − sin 1 8.
Z
e2
1/x
Z
Z ln xdydx =
e
Z
0
5
e
π/2
Z
Z
cos θ
0
0
0
1/x e2 Z e2 1 1 2 1 2 ln xdx = (ln x) = (2x2 − 12 ) = y ln x dx = x 2 2 3 e e 0
cos θ Z 5 Z π/2 cos3 θdθdz 3r drdθdz = dθdz = r 0 0 0 0 0 π/2 Z 5 Z π/2 Z 5 1 dz = (1 − sin2 θ) cos θdθdz = sin θ − sin3 θ 3 0 0 0 0 Z 5 = 2/3dz = 10/3 2
9.
e2
Z
5
Z
π/2
3
0
Z
π/2
Z
sin x
Z
10. π/4
0
0
ln x
ln x Z π/2 Z sin x y y e dydxdz = e dxdz = (x − 1)dxdz π/4 0 π/4 0 0 sin z Z π/2 Z π/2 1 2 1 = = x − x sin2 z − sin z dz 2 2 π/4 π/4 0 √ ! π/2 1 1 π π 1 2 = z − sin 2z + cos z = − − + 4 8 8 16 8 2 π/4 √ π+2−8 2 = 16 Z
π/2
Z
sin x
CHAPTER 14 IN REVIEW
211
11. Using polar coorindates, r=8
Z Z
Z
2π
8
Z
5rdrdθ = 5
5dA = R
0
0
0
Z Z 12. Using symmetry,
Z
π
Z
8 Z 1 2 r dθ = 5 2
2π
32dθ = 320π.
8 polar axis
0
0
1+cos θ
Z
π
1+cos θ 1 2 dθ r 2 0
r=1+cos θ
rdrdθ = 2 dA = 2 0 0 R 0 Z π (1 + 2 cos θ + cos2 θ)dθ = 0 π 1 1 = θ + 2 sin θ + θ + sin 2θ = 3π/2. 2 4 0
Z Z
Z
13.
2π
Z
1
Z
y 2 +1
(2x + y)dA =
1
Z
0
Z =
2y
y2 +1 dy (x + xy)
0
x=2y
y
2
(2x + y)dxdy =
R
2 polar axis
x=y2+1
2y
1
[(y 2 + 1)2 + (y 2 + 1)y − (4y 2 + 2y 2 )]dy
x
0
= (y 4 + y 3 − 4y 2 + y + 1)dy =
1 37 1 5 1 4 4 3 1 2 y + y − y + y + y = 5 4 3 2 60 0
14. Substracting z = 6 − x − y from z = x + y, we obtain x + y = 3. Z Z Z Z 3 Z 3−x Z 6−x−y Z 3 Z 3−x 6−x−y xdV = xdzdydx = dydx xz R 0 0 x+y 0 0 x+y Z 3 Z 3−x = (6x − 2x2 − 2xy)dydx 0
Z
0 3
3−x dx (6xy − 2x2 y − xy 2 )
0
Z
3
(9x − 6x2 + x3 )dx
0
=
z=6-x-y
3 y
0
[6x(3 − x) − 2x2 (3 − x) − x(3 − x)2 ]dx
= =
6
0 3
= Z
z
3 9 2 1 27 x − 2x3 + x4 = 2 4 4 0
2 x
212
CHAPTER 14. MULTIPLE INTEGRALS
√ 15. The circle x2 + y 2 = 1 intersects y = x at x√= 1/ 2. The circle x2 + y 2 = 9 intersects y = x at x = 3/ 2. Z Z Z 1/√2 Z √9−x2 1 1 dA = dydx √ 2 2 2 2 R x +y 0 1−x2 x + y Z 3/√2 Z √9−x2 1 + dydx √ 2 x + y2 1/ 2 x
y y=M9-x2
y=x
x
y
16. The circles are r = 1 and r = 3; the line is θ = π/4. 3 Z π/2 Z 3 Z π/2 Z Z 1 1 ln r dθ dA = rdrdθ = 2 2 2 x + y r 1 π/4 π/4 R 1 Z π/2 π = ln 3dθ = ln 3 4 π/4
r=3
r=1
x
17. y
y=x2
4
2
x
y=-x2
z
18. The region is symmetric with respect to the xz- and yzplanes and is shown in the first octant.
z=x2+y2 1
1
y
1
x
Z
1
Z
19.
√ 3
y
Z
2
1
Z
x
cos x dxdy = 0
y
2
Z
cos x dydx = x3
0
Z = 0
0
1
y
x y cos x2 x3
1
(x cos x2 − x3 cos x2 )dx
y=x
dx
1
y=x3 1
x
CHAPTER 14 IN REVIEW
213
1 Z 1 1 x2 (x cos x2 )dx sin x2 = 2 0 0
=
Integration by parts 1 1 2 1 1 2 2 x sin x + cos x = sin 1 − 2 2 2 0 1 1 1 1 = sin 1 − sin 1 + cos 1 − 2 2 2 2 1 − cos 1 = 2
z
20. The six Z Z 2 Z 4−2x
forms
of
the
integral
are:
0 4
0 2−y/2
Z
y+z=8
8−2x−y
F (x, y, z)dzdydx; Z
8
2x+z=8
4 8−2x−y
Z
F (x, y, z)dzdxdy; Z
0 8
Z
0 8−z
2x+y=4
4 4−y/2−z/2
Z
F (x, y, z)dxdydz; 4
0 4
Z
8−y
Z
y
0 4−y/2−z/2
Z
x
F (x, y, z)dxdzdy; 0
4 8
Z
0 4−z/2
Z
Z
8−2x−z
F (x, y, z)dydxdz; 4
0
Z
0Z
0 8−2x Z
8−2x−z
F (x, y, z)dydzdx. 0
Z
2
4
Z
1
0
√
Z
x−x2
2
Z
Z
1
(4z − 1)
(4z + 1)dydxdz =
21. 0
1/2
0
0
Z
2
Z
=
(4z + 1) 0
1/2 2
x−2 dxdz
1/2
s
1
p
2 1 1 − x− dxdz 4 2
Trig substitution
x − 1/2 p 1 x − 1/2 = (4z + 1) x − x2 + sin−1 2 8 1/2 0 Z 2 2 π π 5π = (4z + 1) − 0 dz = (2z 2 + z) = 16 16 8 0 0 Z
1
dz
1/2
22. The region is the portion of the sphere of radius 1 centered at the origin in the first octant and the octant below that. Using spherical coordinates, we have
214
CHAPTER 14. MULTIPLE INTEGRALS Z
1
√
Z
0
Z √1−x2 −y2
1−x2
−
0
√
(x2 + y 2 + z 2 )4 dzdydx =
Z
1−x2 −y 2
π/2
Z
0
π
1
Z
ρ8 ρ2 sin φdρdφdθ
0
0
1 Z π/2 Z π 1 1 1 = ρ 1 sin φ dφdθ = sin φdφdθ 11 11 0 0 0 0 0 π Z π/2 Z π/2 1 1 π = − cos φ dθ = 2dθ = 11 0 11 0 11 Z
π/2
π
Z
0
23. fx = y; fy = z, 1 + fx2 + x2y = 11 + x2 + y 2 . Using cylindrical coordinates, 2π
Z
1
Z
A= 0
Z p 2 1 + r rdrdθ =
0
1 Z 1 2π 3/2 1 2π √ 2 3/2 (1 + r ) dθ = (2 − 1)dθ = (2 2 − 1). 3 3 0 3 0
2π
0
3 Z √3 Z √3 2 3 2 6 2 2 2 6y − y dx = (18 − 6) − 6x − x dx 24. V = 6 − y dydx = 2 3 9 9 0 0 0 x2 x √3 Z √3 √ √ 2 6 2 7 6√ 48 √ 2 3 = 12x − 6x + x dx = 12x − 2x + x = 12 3 − 6 3 + 3= 3 9 63 7 7 0 0 √
Z
3
Z
3
Z
1
Z
2x
25. (a) V = 0
1
Z p 2 1 − x dydx =
y
p
1−
0
x
2x Z dx =
x2
x
1
p x 1 − x2 dx
0
1 1 1 = − (1 − x2 )3/2 = 3 3 0 1
Z
Z
y
(b) V = 0
p
2
Z
1 − x2 dxdy +
y/2
1
Z
1
p
1 − x2 dxdy
y/2
y
26. We are given ρ = k(x2 + y 2 ). Z 1Z x Z m= k(x2 + y 2 )dydx = k x3
0
Z =k 0
=k
1
1
0
x2 1 3 x y + y dx 3 x3
1
2
1 1 x4 + x− x5 − x9 dx 3 3
1 1 5 1 1 1 k x + x7 − x6 − x10 = 5 21 6 30 21 0
y=x2 y=x3 1
x
CHAPTER 14 IN REVIEW 1
Z
Z
215
x 3
Z
2
1
k(x + xy )dydx = k
My = x3
0
0
x2 Z 1 1 3 1 1 x5 + x7 − x6 − x10 dx x y + xy dx = k 3 3 3 0 x3 3
1 1 6 65k 1 8 1 7 1 11 =k x + x − x − x = 1848 6 24 7 33 0 x2 Z 1 Z 1Z x 1 2 2 1 4 2 3 k(x y + y )dydx = k Mx = x y + y dx 2 4 0 x3 0 x3 Z 1 1 6 1 8 1 8 1 12 =k x + x − x − x dx 2 4 2 4 0 1 20k 1 9 1 13 1 7 x − x − x = =k 14 36 52 819
0
65k/1848 20k/819 x = My /m = = 65/88; y = Mx /m = = 20/39 k/21 k/21 The center of mass is (65/88, 20/39). x2 1 2 3 27. Iy = k(x + x y )dydx = k x y + x y dx 3 0 x3 0 x3 1 Z 1 41 1 1 1 1 1 1 =k x7 + x9 − x8 − x12 = k x6 + x8 − x7 − x11 dx = k 3 3 7 27 8 36 1512 0 0 Z
1
Z
x2
4
Z
2 2
1
4
28. (a) Using symmetry, Z a Z √a2 −x2 Z √a2 −x2 −y2 Z V =8 dzdydx = 8 0
0
0
0
a
Z
√ a2 −x2
p
a2 − x2 − y 2 dydx
0
Trig substitution √a2 −x2 Z a Z a p 2 2 a − x y π a2 − x2 y −1 sin √ dx = 8 dx =8 a2 − x2 − y 2 + 2 2 2 a2 − x2 0 0 2 0 a 4 1 3 2 = 2π a x − x = πa3 3 3 0 (b) Using symmetry, Z 2π Z a Z V =2
√
a2 −r 2
Z 2π Z a p rdzdrdθ = 2 r a2 − r2 drdθ 0 0 0 0 0 a Z Z 2π 2 2π 3 4 1 =2 − (a2 − r2 )3/2 dθ = a dθ = πa3 3 3 3 0 0 0 a Z 2π Z π Z a Z 2π Z π 1 3 2 (c) V = ρ sin φdρdφdθ = ρ sin φ dφdθ 3 0 0 0 0 0 0 π Z 2π Z π Z 2π Z 1 1 2π 3 4 1 = a3 sin φdφdθ = −a3 cos φ dθ = 2a dθ = πa3 3 0 3 0 3 3 0 0 0
29. We use spherical coordinates.
216
CHAPTER 14. MULTIPLE INTEGRALS 2π
Z
π/4
Z
3 sec φ
Z
ρ2 sin φdρdφdθ
V = tan−1
0 2π
Z
Z
1/2
π/4
= tan−1 1/2
0
=
1 3
2π
Z
2π
=9 0
3 sec φ 1 3 dφdθ ρ sin φ 3 0
π/4
27 sec3 φ sin φdφdθ = 9
tan−1
0
Z
Z
0
tan
2π
Z
0
1/2
π/4 1 2 tan φ 2 −1
Z
dθ = 1/2
9 2
Z 0
2π
π/4
tan φ sec2 φdφdθ
tan−1
1/2
1 1− dθ = 8π 9
2 1 2 ρ sin φdρdφdθ = 30. V = ρ sin φ dφdθ 3 0 0 1 0 0 1 π/6 Z 2π Z π/6 Z 2π Z π/6 Z 1 7 8 7 2π sin φ − sin φ dφdθ = dθ − cos φ sin φdφdθ = = 3 3 3 0 3 0 0 0 0 0 " √ # √ ! Z √ 7 2π 7 7π 3 3 = − − (−1) dθ = 1− 2π = (2 − 3) 3 0 2 3 2 3 Z
2π
Z
π/6
Z
2
Z
2
2π
Z
π/6
y
y
2
31. x = 0 =⇒ u = 0, v = −y =⇒ u = 0, −1 ≤ v ≤ 0 x = 1 =⇒ u = 2y, v = 1 − y 2 = 1 − u2 /4 x = 1 =⇒ u = 2y, v = 1 − y 2 = 1 − u2 /4 y = 0 =⇒ u = 0, v = x2 =⇒ u = 0, 0 ≤ v ≤ 1 y = 1 =⇒ u = 2x, v = x2 − 1 = u2 /4 − 1 ∂(u, v) 2y 2x = −4(x2 + y 2 ) = ∂(x, y) 2x −2y
1
1 R
S 1
1
x
x
-1
∂(x, y) 1 =− ∂(u, v) 4(x2 + y 2 ) Z Z Z Z p √ (x2 + y 2 ) 3 x2 + y 2 dA = (x2 + y 2 ) 3 v − =⇒
R
Z 2 Z 1−u2 /4 0 1 dA = 1 v 1/3 dvdu 4(x2 + y 2 ) 4 0 u2 /4−1 S 1−u2 /4 Z Z 2h i 3 1 2 3 4/3 v du = (1 − u2 /4)4/3 − (u2 /4 − 1)4/3 du = 4 0 4 16 0 u2 /4−1 Z 2h i 3 (1 − u2 /4)4/3 − (1 − u2 /4)4/3 du = 0 = 16 0 y
32. y = x =⇒ u + uv = v + uv2 =⇒ v = u 2 x = 2 =⇒ u + uv = 2 =⇒ v = (2 − u)/u y = 0 =⇒ v = 0 or u = −1 ∂(x, y) 1 − w u we take v = 0 = =1+u+v v 1+u ∂(u, v)
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2
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CHAPTER 14 IN REVIEW
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Using x = u + uv and y = v + uv we find (x − y)2 = (u + uv − v − uv)2 = (u − v)2 = u2 − 2uv + v 2 x + y = u + uv + v + uv = u + v + 2uv 2
(x + y) + 2(x + y) + 1 = u2 + 2uv + v 2 + 2(u + v) + 1 = (u + v)2 + 2(u + v) + 1 = (u + v + 1)2 . Then Z Z R
Z 1 Z 2/(1+v) 1 0 p dA = dudv (u + v + 1)dA = (x − y)2 + 2(x + y) + 1 S u+v+1 v 0 1 Z 1 2 1 2 1 = − v dv = 2 ln(1 + v) − v = 2 ln 2 − . 1 + v 2 2 0 0 1
Z Z
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