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620
CHAPTER 15
Multiple Integrals
EXERCISE SET 15.
0
0
(x + 3)dy dx =
0
(2x + 6)dx = 7 2.
1
− 1
(2x − 4 y)dy dx =
1
4 x dx = 16
2
0
x^2 y dx dy =
2
y dy = 2 4.
− 2
− 1
(x^2 + y^2 )dx dy =
− 2
(3 + 3y^2 )dy = 14
∫ (^) ln 3
0
∫ (^) ln 2
0
ex+y^ dy dx =
∫ (^) ln 3
0
exdx = 2
0
0
y sin x dy dx =
0
sin x dx = (1 − cos 2)/ 2
− 1
2
dx dy =
− 1
3 dy = 3 8.
4
− 3
dy dx =
4
10 dx = 20
0
0
x (xy + 1)^2 dy dx^ =
0
x + 1
dx = 1 − ln 2
∫ (^) π
π/ 2
1
x cos xy dy dx =
∫ (^) π
π/ 2
(sin 2x − sin x)dx = − 2
∫ (^) ln 2
0
0
xy ey (^2) x dy dx =
∫ (^) ln 2
0
(ex^ − 1)dx = (1 − ln 2)/ 2
3
1
(x + y)^2 dy dx =
3
x + 1
x + 2
dx = ln(25/24)
− 1
− 2
4 xy^3 dy dx =
− 1
0 dx = 0
0
0
√^ xy x^2 + y^2 + 1
dy dx =
0
[x(x^2 + 2)^1 /^2 − x(x^2 + 1)^1 /^2 ]dx = (
0
2
x
1 − x^2 dy dx =
0
x(1 − x^2 )^1 /^2 dx = 1/ 3
∫ (^) π/ 2
0
∫ (^) π/ 3
0
(x sin y − y sin x)dy dx =
∫ (^) π/ 2
0
x 2 − π
2 18 sin x
dx = π^2 / 144
- (a) x∗ k = k/ 2 − 1 / 4 , k = 1, 2 , 3 , 4; y∗ l = l/ 2 − 1 / 4 , l = 1, 2 , 3 , 4 , ∫ ∫
R
f (x, y) dxdy ≈
∑^4
k=
∑^4
l=
f (x∗ k, y∗ l )∆Akl =
∑^4
k=
∑^4
l=
(k/ 2 − 1 /4)^2 +(l/ 2 − 1 /4)^2 = 37/ 4
(b)
0
0
(x^2 + y) dxdy = 28/3; the error is | 37 / 4 − 28 / 3 | = 1/ 12
Exercise Set 15.1 621
- (a) x∗ k = k/ 2 − 1 / 4 , k = 1, 2 , 3 , 4; y l∗ = l/ 2 − 1 / 4 , l = 1, 2 , 3 , 4 , ∫ ∫
R
f (x, y) dxdy ≈
∑^4
k=
∑^4
l=
f (x∗ k, y∗ l )∆Akl =
∑^4
k=
∑^4
l=
(k/ 2 − 1 /4) − 2(l/ 2 − 1 /4)^2 = − 4
(b)
0
0
(x − 2 y) dxdy = −4; the error is zero
19. V =
3
1
(2x + y)dy dx =
3
(2x + 3/2)dx = 19
20. V =
1
0
(3x^3 + 3x^2 y)dy dx =
1
(6x^3 + 6x^2 )dx = 172
21. V =
0
0
x^2 dy dx =
0
3 x^2 dx = 8
22. V =
0
0
5(1 − x/3)dy dx =
0
5(4 − 4 x/3)dx = 30
- (a) (1, 0, 4)
(2, 5, 0) x
y
z (^) (b) (0, 0, 5)
(3, 4, 0)
(0, 4, 3)
z
x
y
- (a)
(1, 1, 0)
(0, 0, 2)
x
y
z (^) (b)
(2, 2, 0)
(2, 2, 8)
x
y
z
0
∫ (^) π
0
x cos(xy) cos^2 πx dy dx =
0
cos^2 πx sin(xy)
]π 0 dx
0
cos^2 πx sin πx dx = −
3 π cos^3 πx
] 1 / 2
0
3 π
Exercise Set 15.2 623
∫ √ 2 π √π
∫ (^) x^3
0
sin(y/x)dy dx =
∫ √ 2 π √π^ [−x^ cos(x
(^2) ) + x]dx = π/ 2
− 1
∫ (^) x^2
−x^2
(x^2 − y)dy dx =
− 1
2 x^4 dx = 4/ 5 7.
∫ (^) π
π/ 2
∫ (^) x^2
0
x cos(y/x)dy dx =
∫ (^) π
π/ 2
sin x dx = 1
0
∫ (^) x
0
ex 2 dy dx =
0
xex 2 dx = (e − 1)/ 2 9.
0
∫ (^) x
0
y
x^2 − y^2 dy dx =
0
x^3 dx = 1/ 12
1
∫ (^) y 2
0
ex/y 2 dx dy =
1
(e − 1)y^2 dy = 7(e − 1)/ 3
- (a)
0
∫ (^) x^2
0
xy dy dx =
0
x^5 dx =
(b)
1
∫ (^) (y+7)/ 2
−(y−5)/ 2
xy dx dy =
1
(3y^2 + 3y)dy = 38
- (a)
0
∫ √x
x^2
(x + y)dy dx =
0
(x^3 /^2 + x/ 2 − x^3 − x^4 /2)dx = 3/ 10
(b)
− 1
∫ √ 1 −x 2
−√ 1 −x^2
x dy dx +
− 1
∫ √ 1 −x 2
−√ 1 −x^2
y dy dx =
− 1
2 x
1 − x^2 dx + 0 = 0
- (a)
4
∫ (^) x
16 /x
x^2 dy dx =
4
(x^3 − 16 x)dx = 576
(b)
2
16 /y
x^2 dxdy +
4
y
x^2 dx dy =
4
[
3 y^3
]
dy +
4
512 − y^3 3 dy
- (a)
1
∫ (^) y
0
xy^2 dx dy =
1
y^4 dy = 31/ 10
(b)
0
1
xy^2 dydx +
1
x
xy^2 dydx =
0
7 x/ 3 dx +
1
8 x − x^4 3 dx = 7/6 + 29/15 = 31/ 10
- (a)
− 1
∫ √ 1 −x 2
−√ 1 −x^2
(3x − 2 y)dy dx =
− 1
6 x
1 − x^2 dx = 0
(b)
− 1
∫ √ 1 −y 2
−
1 −y^2
(3x − 2 y) dxdy =
− 1
− 4 y
1 − y^2 dy = 0
- (a)
0
∫ √ 25 −x 2
5 −x
y dy dx =
0
(5x − x^2 )dx = 125/ 6
(b)
0
∫ √ 25 −y 2
5 −y
y dxdy =
0
y
25 − y^2 − 5 + y
dy = 125/ 6
0
∫ √y
0
x(1 + y^2 )−^1 /^2 dx dy =
0
y(1 + y^2 )−^1 /^2 dy = (
624 Chapter 15
∫ (^) π
0
∫ (^) x
0
x cos y dy dx =
∫ (^) π
0
x sin x dx = π
0
∫ (^6) −y
y^2
xy dx dy =
0
(36y − 12 y^2 + y^3 − y^5 )dy = 50/ 3
∫ (^) π/ 4
0
sin y
x dx dy =
∫ (^) π/ 4
0
cos 2y dy = 1/ 8
0
∫ (^) x
x^3
(x − 1)dy dx =
0
(−x^4 + x^3 + x^2 − x)dx = − 7 / 60
0
∫ (^2) x
x
x^2 dy dx +
1 /√ 2
∫ (^1) /x
x
x^2 dy dx =
0
x^3 dx +
1 /√ 2
(x − x^3 )dx = 1/ 8
- (a)
-2 -1.5 -1-0.5 0.5 1 1.
x
y
1
2
3
4
(b) x = (− 1. 8414 , 0 .1586), (1. 1462 , 3 .1462)
(c)
R
x dA ≈
− 1. 8414
∫ (^) x+
ex
x dydx =
− 1. 8414
x(x + 2 − ex) dx ≈ − 0. 4044
(d)
R
x dA ≈
- 1586
∫ (^) ln y
y− 2
x dxdy =
- 1586
[
ln^2 y 2 − (y^ −^ 2)
2 2
]
dy ≈ − 0. 4044
- (a)
1 2 3
5
15
25
x
y
R
(b) (1, 3), (3, 27)
(c)
1
∫ (^4) x (^3) −x 4
3 − 4 x+4x^2
x dy dx =
1
x[(4x^3 − x^4 ) − (3 − 4 x + 4x^2 )] dx =
25. A =
∫ (^) π/ 4
0
∫ (^) cos x
sin x
dy dx =
∫ (^) π/ 4
0
(cos x − sin x)dx =
26. A =
− 4
∫ (^) −y^2
3 y− 4
dx dy =
− 4
(−y^2 − 3 y + 4 )dy = 125/ 6
626 Chapter 15
0
∫ (^) e
ey
f (x, y)dx dy 45.
∫ (^) π/ 2
0
∫ (^) sin x
0
f (x, y)dy dx 46.
0
∫ √x
x^2
f (x, y)dy dx
0
∫ (^) y/ 4
0
e−y 2 dx dy =
0
ye−y 2 dy = (1 − e−^16 )/ 8
0
∫ (^2) x
0
cos(x^2 )dy dx =
0
2 x cos(x^2 )dx = sin 1
0
∫ (^) x^2
0
ex 3 dy dx =
0
x^2 ex 3 dx = (e^8 − 1)/ 3
∫ (^) ln 3
0
ey
x dx dy =^1 2
∫ (^) ln 3
0
(9 − e^2 y^ )dy =^1 2 (9 ln 3 − 4)
0
∫ (^) y 2
0
sin(y^3 )dx dy =
0
y^2 sin(y^3 )dy = (1 − cos 8)/ 3
0
∫ (^) e
ex
x dy dx =
0
(ex − xex)dx = e/ 2 − 1
- (a)
0
√x^ sin^ πy
(^3) dy dx; the inner integral is non-elementary.
∫ (^2)
0
∫ (^) y 2
0
sin
πy^3
dx dy =
0
y^2 sin
πy^3
dy = −
3 π cos
πy^3
) ]^2
0
(b)
0
∫ (^) π/ 2
sin−^1 y
sec^2 (cos x)dx dy ; the inner integral is non-elementary. ∫ (^) π/ 2
0
∫ (^) sin x
0
sec^2 (cos x)dy dx =
∫ (^) π/ 2
0
sec^2 (cos x) sin x dx = tan 1
54. V = 4
0
∫ √ 4 −x 2
0
(x^2 + y^2 ) dy dx = 4
0
x^2
4 − x^2 +
(4 − x^2 )^3 /^2
dx (x = 2 sin θ)
∫ (^) π/ 2
0
sin^2 θ −
sin^4 θ
dθ =
π 2
π 4
π 2
= 8π
- The region is symmetric with respect to the y-axis, and the integrand is an odd function of x, hence the answer is zero.
- This is the volume in the first octant under the surface z =
1 − x^2 − y^2 , so 1/8 of the volume of the sphere of radius 1, thus π
Area of triangle is 1/2, so f¯ = 2
0
x
1 + x^2 dy dx = 2
0
[
1 + x^2
x 1 + x^2
]
dx = π 2 − ln 2
- Area =
0
(3x − x^2 − x) dx = 4/3, so
f¯ =^3 4
0
∫ (^3) x−x 2
x
(x^2 − xy)dy dx =
0
(− 2 x^3 + 2x^4 − x^5 /2)dx = −
Exercise Set 15.3 627
- Tave = 1 A(R)
R
(5xy + x^2 ) dA. The diamond has corners (± 2 , 0), (0, ±4) and thus has area
A(R) = 4
2(4) = 16m^2. Since 5xy is an odd function of x (as well as y),
R
5 xy dA = 0. Since
x^2 is an even function of both x and y,
Tave = 4 16
x,y>^ R 0
x^2 dA =^1 4
0
∫ (^4) − 2 x
0
x^2 dydx =^1 4
0
(4 − 2 x)x^2 dx =^1 4
x^3 − 1 2 x^4
) ] 2
0
=^2
◦ C
- The area of the lens is πR^2 = 4π and the average thickness Tave is
Tave =
4 π
0
∫ √ 4 −x 2
0
1 − (x^2 + y^2 )/ 4
dydx =
π
0
(4 − x^2 )^3 /^2 dx (x = 2 cos θ)
= 8 3 π
∫ (^) π
0
sin^4 θ dθ = 8 3 π
π 2
=^1
in
- y = sin x and y = x/2 intersect at x = 0 and x = a = 1.895494, so
V =
∫ (^) a
0
∫ (^) sin x
x/ 2
1 + x + y dy dx = 0. 676089
EXERCISE SET 15.
∫ (^) π/ 2
0
∫ (^) sin θ
0
r cos θdr dθ =
∫ (^) π/ 2
0
sin^2 θ cos θ dθ = 1/ 6
∫ (^) π
0
∫ (^) 1+cos θ
0
r dr dθ =
∫ (^) π
0
(1 + cos θ)^2 dθ = 3π/ 4
∫ (^) π/ 2
0
∫ (^) a sin θ
0
r^2 dr dθ =
∫ (^) π/ 2
0
a^3 3 sin^3 θ dθ =
a^3
∫ (^) π/ 6
0
∫ (^) cos 3θ
0
r dr dθ =
∫ (^) π/ 6
0
2 cos
(^2 3) θ dθ = π/ 24
∫ (^) π
0
∫ (^1) −sin θ
0
r^2 cos θ dr dθ =
∫ (^) π
0
(1 − sin θ)^3 cos θ dθ = 0
∫ (^) π/ 2
0
∫ (^) cos θ
0
r^3 dr dθ =
∫ (^) π/ 2
0
cos^4 θ dθ = 3π/ 64
7. A =
∫ (^2) π
0
∫ (^1) −cos θ
0
r dr dθ =
∫ (^2) π
0
(1 − cos θ)^2 dθ = 3π/ 2
8. A = 4
∫ (^) π/ 2
0
∫ (^) sin 2θ
0
r dr dθ = 2
∫ (^) π/ 2
0
sin^2 2 θ dθ = π/ 2
9. A =
∫ (^) π/ 2
π/ 4
sin 2θ
r dr dθ =
∫ (^) π/ 2
π/ 4
(1 − sin^2 2 θ)dθ = π/ 16
Exercise Set 15.3 629
∫ (^) π/ 2
0
0
cos(r^2 )r dr dθ =
sin 1
∫ (^) π/ 2
0
dθ = π 4 sin 1
∫ (^) π/ 2
0
∫ (^) a
0
r (1 + r^2 )^3 /^2 dr dθ = π 2
1 + a^2
∫ (^) π/ 4
0
∫ (^) sec θ tan θ
0
r^2 dr dθ =^1 3
∫ (^) π/ 4
0
sec^3 θ tan^3 θ dθ = 2(
∫ (^) π/ 4
0
0
√^ r 1 + r^2
dr dθ = π 4
∫ (^) π/ 2
tan−^1 (3/4)
3 csc θ
r dr dθ =^1 2
∫ (^) π/ 2
tan−^1 (3/4)
(25 − 9 csc^2 θ)dθ
[ (^) π 2 − tan−^1 (3/4)
]
tan−^1 (4/3) − 6
31. V =
∫ (^2) π
0
∫ (^) a
0
hr dr dθ =
∫ (^2) π
0
h a
2 2 dθ = πa^2 h
- (a) V = 8
∫ (^) π/ 2
0
∫ (^) a
0
c a (a^2 − r^2 )^1 /^2 r dr dθ = − 4 c 3 a π(a^2 − r^2 )^3 /^2
]a
0
πa^2 c
(b) V ≈
π(6378.1370)^26356. 5231 ≈ 1 , 083 , 168 , 200 ,000 km^3
33. V = 2
∫ (^) π/ 2
0
∫ (^) a sin θ
0
c a (a^2 − r^2 )^1 /^2 r dr dθ =
a^2 c
∫ (^) π/ 2
0
(1 − cos^3 θ)dθ = (3π − 4)a^2 c/ 9
34. A = 4
∫ (^) π/ 4
0
∫ (^) a√2 cos 2θ
0
r dr dθ = 4a^2
∫ (^) π/ 4
0
cos 2θ dθ = 2a^2
35. A =
∫ (^) π/ 4
π/ 6
∫ (^) 4 sin θ √8 cos 2θ^ r dr dθ^ +
∫ (^) π/ 2
π/ 4
∫ (^) 4 sin θ
0
r dr dθ
∫ (^) π/ 4
π/ 6
(8 sin^2 θ − 4cos 2 θ)dθ +
∫ (^) π/ 2
π/ 4
8 sin^2 θ dθ = 4π/3 + 2
36. A =
∫ (^) φ
0
∫ (^2) a sin θ
0
r dr dθ = 2a^2
∫ (^) φ
0
sin^2 θ dθ = a^2 φ −
2 a
(^2) sin 2φ
- (a) I^2 =
[∫ +∞
0
e−x 2 dx
] [∫ +∞
0
e−y 2 dy
]
0
[∫ +∞
0
e−x 2 dx
]
e−y 2 dy
0
0
e−x 2 e−y 2 dx dy =
0
0
e−(x (^2) +y (^2) ) dx dy
(b) I^2 =
∫ (^) π/ 2
0
0
e−r 2 r dr dθ =^1 2
∫ (^) π/ 2
0
dθ = π/ 4 (c) I =
π/ 2
- (a) 1. 173108605 (b)
∫ (^) π
0
0
re−r 4 dr dθ = π
0
re−r 4 dr ≈ 1. 173108605
630 Chapter 15
39. V =
∫ (^2) π
0
∫ R
0
D(r)r dr dθ =
∫ (^2) π
0
∫ R
0
ke−r^ r dr dθ = − 2 πk(1 + r)e−r
]R
0
= 2πk[1 − (R + 1)e−R]
∫ (^) tan−^1 (2)
tan−^1 (1/3)
0
r^3 cos^2 θ dr dθ = 4
∫ (^) tan−^1 (2)
tan−^1 (1/3)
cos^2 θ dθ =^1 5
- 2[tan−^1 (2) − tan−^1 (1/3)] =^1 5
- π 2
EXERCISE SET 15.
- (a) z
x (^) y
(b)
x
y
z
(c)
x y
z
- (a)
y
x
z (^) (b) z
y
x
(c) z
x y
632 Chapter 15
- ru × rv = (
2 /2)i − (
2 /2)j + (1/2)k; x − y +
2 z^ =^
π
- ru × rv = 2i − ln 2k; 2x − (ln 2)z = 0
- z =
9 − y^2 , zx = 0, zy = −y/
9 − y^2 , z x^2 + z^2 y + 1 = 9/(9 − y^2 ),
S =
0
− 3
√^3
9 − y^2
dy dx =
0
3 π dx = 6π
- z = 8 − 2 x − 2 y, z^2 x + z^2 y + 1 = 4 + 4 + 1 = 9, S =
0
∫ (^4) −x
0
3 dy dx =
0
3(4 − x)dx = 24
- z^2 = 4x^2 + 4y^2 , 2zzx = 8x so zx = 4x/z, similarly zy = 4y/z thus
z x^2 + z^2 y + 1 = (16x^2 + 16y^2 )/z^2 + 1 = 5, S =
0
∫ (^) x
x^2
5 dy dx =
0
(x − x^2 )dx =
- z^2 = x^2 + y^2 , zx = x/z, zy = y/z, z^2 x + z^2 y + 1 = (z^2 + y^2 )/z^2 + 1 = 2,
S =
R
2 dA = 2
∫ (^) π/ 2
0
∫ (^) 2 cos θ
0
2 r dr dθ = 4
∫ (^) π/ 2
0
cos^2 θ dθ =
2 π
- zx = − 2 x, zy = − 2 y, z x^2 + z y^2 + 1 = 4x^2 + 4y^2 + 1,
S =
R
4 x^2 + 4y^2 + 1 dA =
∫ (^2) π
0
0
r
4 r^2 + 1 dr dθ
∫ (^2) π
0
dθ = (
5 − 1)π/ 6
- zx = 2, zy = 2y, z x^2 + z y^2 + 1 = 5 + 4y^2 ,
S =
0
∫ (^) y
0
5 + 4y^2 dx dy =
0
y
5 + 4y^2 dy = (27 − 5
- ∂r/∂u = cos vi + sin vj + 2uk, ∂r/∂v = −u sin vi + u cos vj,
‖∂r/∂u × ∂r/∂v‖ = u
4 u^2 + 1; S =
∫ (^2) π
0
1
u
4 u^2 + 1 du dv = (
5)π/ 6
- ∂r/∂u = cos vi + sin vj + k, ∂r/∂v = −u sin vi + u cos vj,
‖∂r/∂u × ∂r/∂v‖ =
2 u; S =
∫ (^) π/ 2
0
∫ (^2) v
0
2 u du dv =
π^3
- zx = y, zy = x, z^2 x + z^2 y + 1 = x^2 + y^2 + 1,
S =
R
x^2 + y^2 + 1 dA =
∫ (^) π/ 6
0
0
r
r^2 + 1 dr dθ =
∫ (^) π/ 6
0
dθ = (
10 − 1)π/ 18
Exercise Set 15.4 633
- zx = x, zy = y, z^2 x + z^2 y + 1 = x^2 + y^2 + 1,
S =
R
x^2 + y^2 + 1 dA =
∫ (^2) π
0
0
r
r^2 + 1 dr dθ =
∫ (^2) π
0
dθ = 52π/ 3
- On the sphere, zx = −x/z and zy = −y/z so z^2 x + z y^2 + 1 = (x^2 + y^2 + z^2 )/z^2 = 16/(16 − x^2 − y^2 ); the planes z = 1 and z = 2 intersect the sphere along the circles x^2 + y^2 = 15 and x^2 + y^2 = 12;
S =
R
√^4
16 − x^2 − y^2
dA =
∫ (^2) π
0
√ 12 √^4 r 16 − r^2
dr dθ = 4
∫ (^2) π
0
dθ = 8π
- On the sphere, zx = −x/z and zy = −y/z so z^2 x + z^2 y + 1 = (x^2 + y^2 + z^2 )/z^2 = 8/(8 − x^2 − y^2 ); the cone cuts the sphere in the circle x^2 + y^2 = 4 ;
S =
∫ (^2) π
0
0
√^2 r 8 − r^2
dr dθ = (8 − 4
∫ (^2) π
0
dθ = 8(2 −
2)π
- r(u, v) = a cos u sin vi + a sin u sin vj + a cos vk, ‖ru × rv ‖ = a^2 sin v,
S =
∫ (^) π
0
∫ (^2) π
0
a^2 sin v du dv = 2πa^2
∫ (^) π
0
sin v dv = 4πa^2
- r = r cos ui + r sin uj + vk, ‖ru × rv ‖ = r; S =
∫ (^) h
0
∫ (^2) π
0
r du dv = 2πrh
- zx = h a √ x x^2 + y^2
, zy = h a √ y x^2 + y^2
, z^2 x + z^2 y + 1 = h^2 x^2 + h^2 y^2 a^2 (x^2 + y^2 ) + 1 = (a
(^2) + h (^2) )/a (^2) ,
S =
∫ (^2) π
0
∫ (^) a
0
a^2 + h^2 a r dr dθ^ =
2 a
a^2 + h^2
∫ (^2) π
0
dθ = πa
a^2 + h^2
- Revolving a point (a 0 , 0 , b 0 ) of the xz-plane around the z-axis generates a circle, an equation of which is r = a 0 cos ui + a 0 sin uj + b 0 k, 0 ≤ u ≤ 2 π. A point on the circle (x − a)^2 + z^2 = b^2 which generates the torus can be written r = (a + b cos v)i + b sin vk, 0 ≤ v ≤ 2 π. Set a 0 = a + b cos v and b 0 = a + b sin v and use the first result: any point on the torus can thus be written in the form r = (a + b cos v) cos ui + (a + b cos v) sin uj + b sin vk, which yields the result.
- ∂r/∂u = −(a + b cos v) sin ui + (a + b cos v) cos uj, ∂r/∂v = −b sin v cos ui − b sin v sin uj + b cos vk, ‖∂r/∂u × ∂r/∂v‖ = b(a + b cos v);
S =
∫ (^2) π
0
∫ (^2) π
0
b(a + b cos v)du dv = 4π^2 ab
- ‖ru × rv ‖ =
u^2 + 1; S =
∫ (^4) π
0
0
u^2 + 1 du dv = 4π
0
u^2 + 1 du = 174. 7199011
- z = −1 when v ≈ 0. 27955 , z = 1 when v ≈ 2. 86204 , ‖ru × rv ‖ = | cos v|;
S =
∫ (^2) π
0
- 27955
| cos v| dv du ≈ 9. 099
Exercise Set 15.5 635
∫ (^) π
0
0
∫ (^) π/ 6
0
xy sin yz dz dy dx =
∫ (^) π
0
0
x[1 − cos(πy/6)]dy dx =
∫ (^) π
0
(1 − 3 /π)x dx = π(π − 3)/ 2
− 1
∫ (^1) −x^2
0
∫ (^) y
0
y dz dy dx =
− 1
∫ (^1) −x^2
0
y^2 dy dx =
− 1
3 (1^ −^ x
(^2) ) (^3) dx = 32/ 105
0
∫ (^) x
0
∫ (^2) −x 2
0
xyz dz dy dx =
0
∫ (^) x
0
xy(2 − x^2 )^2 dy dx =
0
x^3 (2 − x^2 )^2 dx = 1/ 6
∫ (^) π/ 2
π/ 6
∫ (^) π/ 2
y
∫ (^) xy
0
cos(z/y)dz dx dy =
∫ (^) π/ 2
π/ 6
∫ (^) π/ 2
y
y sin x dx dy =
∫ (^) π/ 2
π/ 6
y cos y dy = (5π − 6
0
1
− 2
x + z^2 y dzdydx ≈ 9. 425
0
∫ √ 1 −x 2
0
∫ √ 1 −x (^2) −y 2
0
e−x (^2) −y (^2) −z 2 dz dy dx ≈ 2. 381
15. V =
0
∫ (^) (4−x)/ 2
0
∫ (^) (12− 3 x− 6 y)/ 4
0
dz dy dx =
0
∫ (^) (4−x)/ 2
0
(12 − 3 x − 6 y)dy dx
0
(4 − x)^2 dx = 4
16. V =
0
∫ (^1) −x
0
∫ √y
0
dz dy dx =
0
∫ (^1) −x
0
y dy dx =
0
(1 − x)^3 /^2 dx = 4/ 15
17. V = 2
0
x^2
∫ (^4) −y
0
dz dy dx = 2
0
x^2
(4 − y)dy dx = 2
0
8 − 4 x^2 +
2 x
4
dx = 256/ 15
18. V =
0
∫ (^) y
0
∫ √ 1 −y 2
0
dz dx dy =
0
∫ (^) y
0
1 − y^2 dx dy =
0
y
1 − y^2 dy = 1/ 3
- The projection of the curve of intersection onto the xy-plane is x^2 + y^2 = 1,
V = 4
0
∫ √ 1 −x 2
0
∫ (^4) − 3 y^2
4 x^2 +y^2
dz dy dx
- The projection of the curve of intersection onto the xy-plane is 2x^2 + y^2 = 4 ,
V = 4
0
∫ √ 4 − 2 x 2
0
∫ (^8) −x^2 −y^2
3 x^2 +y^2
dz dy dx
21. V = 2
− 3
∫ √ 9 −x (^2) / 3
0
∫ (^) x+
0
dz dy dx 22. V = 8
0
∫ √ 1 −x 2
0
∫ √ 1 −x 2
0
dz dy dx
636 Chapter 15
- (a)
(0, 0, 1)
(0, –1, 0) (1, 0, 0)
z
y
x
(b) (0, 9, 9)
(3, 9, 0)
z
x y
(c)
(0, 0, 1)
(1, 2, 0) x
y
z
- (a)
(3, 9, 0)
(0, 0, 2)
x
y
z (^) (b) (0, 0, 2)
(0, 2, 0)
(2, 0, 0) x
y
z
(c)
(2, 2, 0)
(0, 0, 4)
x
y
z
25. V =
0
∫ (^1) −x
0
∫ (^1) −x−y
0
dz dy dx = 1/ 6 , fave = 6
0
∫ (^1) −x
0
∫ (^1) −x−y
0
(x + y + z) dz dy dx =
- The integrand is an odd function of each of x, y, and z, so the answer is zero.
- The volume V = 3 π √ 2
, and thus
rave =
3 π
G
x^2 + y^2 + z^2 dV =
3 π
− 1 /√ 2
∫ √ 1 − 2 x 2
−√ 1 − 2 x^2
∫ (^6) − 7 x (^2) −y 2
5 x^2 +5y^2
x^2 + y^2 + z^2 dzdydx ≈ 3. 291
638 Chapter 15
EXERCISE SET 15.
- Let a be the unknown coordinate of the fulcrum; then the total moment about the fulcrum is 5(0 − a) + 10(5 − a) + 20(10 − a) = 0 for equilibrium, so 250 − 35 a = 0, a = 50/7. The fulcrum should be placed 50/7 units to the right of m 1.
- At equilibrium, 10(0 − 4) + 3(2 − 4 ) + 4 (3− 4 ) +m(6 − 4 ) = 0, m = 25
- A = 1, x =
0
0
x dy dx =
2 , y^ =
0
0
y dy dx =
- A = 2, x =
G
x dy dx, and the region of integration is symmetric with respect to the x-axes
and the integrand is an odd function of x, so x = 0. Likewise, y = 0.
5. A = 1/2,
R
x dA =
0
∫ (^) x
0
x dy dx = 1/3,
R
y dA =
0
∫ (^) x
0
y dy dx = 1/6;
centroid (2/ 3 , 1 /3)
6. A =
0
∫ (^) x 2
0
dy dx = 1/3,
R
x dA =
0
∫ (^) x 2
0
x dy dx = 1/4,
∫ ∫
R
y dA =
0
∫ (^) x^2
0
y dy dx = 1/10; centroid (3/ 4 , 3 /10)
7. A =
0
∫ (^2) −x^2
x
dy dx = 7/6,
R
x dA =
0
∫ (^2) −x^2
x
x dy dx = 5/12,
∫ ∫
R
y dA =
0
∫ (^2) −x^2
x
y dy dx = 19/15; centroid (5/ 14 , 38 /35)
8. A =
π 4 ,
R
x dA =
0
∫ √ 1 −x 2
0
x dy dx =
3 ,^ x^ =^
3 π , y^ =^
3 π by symmetry
- x = 0 from the symmetry of the region,
A =
π(b^2 − a^2 ),
R
y dA =
∫ (^) π
0
∫ (^) b
a
r^2 sin θ dr dθ =
(b^3 − a^3 ); centroid x = 0, y = 4(b^3 − a^3 ) 3 π(b^2 − a^2 )
- y = 0 from the symmetry of the region, A = πa^2 /2, ∫ ∫
R
x dA =
∫ (^) π/ 2
−π/ 2
∫ (^) a
0
r^2 cos θ dr dθ = 2a^3 /3; centroid
4 a 3 π ,^0
11. M =
R
δ(x, y)dA =
0
0
|x + y − 1 | dxdy
0
[∫ (^1) −x
0
(1 − x − y) dy +
1 −x
(x + y − 1) dy
]
dx =
Exercise Set 15.6 639
x = 3
0
0
xδ(x, y) dy dx = 3
0
[∫ (^1) −x
0
x(1 − x − y) dy +
1 −x
x(x + y − 1) dy
]
dx =^1 2 By symmetry, y =
2 as well; center of gravity (1/^2 ,^1 /2)
- x =
M
G
xδ(x, y) dA, and the integrand is an odd function of x while the region is symmetric
with respect to the y-axis, thus x = 0; likewise y = 0.
13. M =
0
∫ √x
0
(x + y)dy dx = 13/20, Mx =
0
∫ √x
0
(x + y)y dy dx = 3/10,
My =
0
∫ √x
0
(x + y)x dy dx = 19/42, x = My /M = 190/273, y = Mx/M = 6/13;
the mass is 13/20 and the center of gravity is at (190/ 273 , 6 /13).
14. M =
∫ (^) π
0
∫ (^) sin x
0
y dy dx = π/4, x = π/2 from the symmetry of the density and the region,
Mx =
∫ (^) π
0
∫ (^) sin x
0
y^2 dy dx = 4/9, y = Mx/M =
9 π ; mass π/4, center of gravity
π 2
9 π
15. M =
∫ (^) π/ 2
0
∫ (^) a
0
r^3 sin θ cos θ dr dθ = a^4 /8, x = y from the symmetry of the density and the
region, My =
∫ (^) π/ 2
0
∫ (^) a
0
r^4 sin θ cos^2 θ dr dθ = a^5 /15, x = 8a/15; mass a^4 /8, center of gravity (8a/ 15 , 8 a/15).
16. M =
∫ (^) π
0
0
r^3 dr dθ = π/4, x = 0 from the symmetry of density and region,
Mx =
∫ (^) π
0
0
r^4 sin θ dr dθ = 2/5, y =
5 π ; mass π/4, center of gravity
5 π
- V = 1, x =
0
0
0
x dz dy dx =
, similarly y = z =
; centroid
- symmetry,
G
z dz dy dx =
0
∫ (^2) π
0
0
rz dr dθ dz = 2π, centroid = (0, 0 , 1)
- x = y = z from the symmetry of the region, V = 1/6,
x =
V
0
∫ (^1) −x
0
∫ (^1) −x−y
0
x dz dy dx = (6)(1/24 ) = 1/4; centroid (1/ 4 , 1 / 4 , 1 /4)
- The solid is described by − 1 ≤ y ≤ 1 , 0 ≤ z ≤ 1 − y^2 , 0 ≤ x ≤ 1 − z;
V =
− 1
∫ (^1) −y 2
0
∫ (^1) −z
0
dx dz dy =
, x =
V
− 1
∫ (^1) −y 2
0
∫ (^1) −z
0
x dx dz dy =
, y = 0 by symmetry,
z =
V
− 1
∫ (^1) −y^2
0
∫ (^1) −z
0
z dx dz dy =
; the centroid is
