Solução_Calculo_Stewart_6e

F.
25.
E x2 dV = π
π
0 0
SECTION 16.8TX.10
TRIPLE INTEGRALS IN SPHERICAL COORDINATES ET SECTION 15.8 ¤ 253
4
3 (ρ sin φ cos θ)2 ρ 2 sin φdρdφdθ = π
0 cos2 θdθ π
0 sin3 φdφ 4
= 1
2 θ + 1 4 sin 2θ π
0
−
1
(2 + 3 sin2 φ)cosφ π 1 ρ5 4
=
π 2 + 2
0 5 3 2 3 3
3 ρ4 dρ
1
5 (45 − 3 5 )= 1562
15 π
27. The solid region is given by E = (ρ, θ, φ) | 0 ≤ ρ ≤ a, 0 ≤ θ ≤ 2π, π 6 ≤ φ ≤ π 3
and its volume is
V = E dV = π/3
π/6
=[− cos φ] π/3
1
π/6 [θ]2π 0
ρ3 a
= 3 0
2π
a
0 0 ρ2 sin φdρdθdφ= π/3
sin φdφ 2π
dθ a
π/6 0
− 1 2 + √ 3
2
(2π) 1
3 a3 = √ 3−1
3
πa 3
0 ρ2 dρ
29. (a) Since ρ =4cosφ implies ρ 2 =4ρ cos φ, the equation is that of a sphere of radius 2 with center at (0, 0, 2). Thus
V = 2π
0
π/3
4cosφ
ρ 2 sin φdρdφdθ = 2π
0 0 0
π/3
1 ρ3 ρ=4 cos φ
0 3 ρ=0
= 2π
0 −
16
3 cos4 φ φ=π/3
dθ = 2π
− 16 1
− 1 dθ =5θ
φ=0
0 3 16
(b) By the symmetry of the problem M yz = M xz =0.Then
M xy = 2π
0
Hence (x, y, z) =(0, 0, 2.1).
2π
0
sin φdφdθ = 2π
0
=10π
π/3
64
0 3
π/3
4cosφ
ρ 3 cos φ sin φdρdφdθ = 2π π/3
cos φ sin φ 64 cos 4 φ dφ dθ
0 0 0 0
= 2π
64 − 1 0 6 cos6 φ φ=π/3
dθ = 2π 21
dθ =21π
φ=0
0 2
31. By the symmetry of the region, M xy =0and M yz =0. Assuming constant density K,
and
m = KV = K π
π
E 0 0
= Kπ − cos φ π 1 ρ3 4
=2Kπ · 37 = 74
0 3 3 3
M xz = yKdV = K π
π
E 0 0
= K − cos θ π
0
Myz
Thus the centroid is (x, y, z) =
m , M xz
m , M xy
m
4
3
ρ2 sin φdρdφdθ = K π
0 dθ π
0 sin φdφ 4
3 ρ2 dρ
cos3 φ sin φdφdθ
πK 3
4 (ρ sin φ sin θ) 3 ρ2 sin φdρdφdθ = K π
sin θdθ π
0 0 sin2 φdφ 4
1 φ − 1 sin 2φ π 1 ρ4 4
= K(2)
π 1 175
(256 − 81) = πK
2 4 0 4 3 2 4 4
= 0, 175πK/4
74πK/3 , 0 = 0, 525 , 0 .
296
3 ρ3 dρ
33. (a) The density function is ρ(x, y, z) =K, a constant, and by the symmetry of the problem M xz = M yz =0.Then
M xy = 2π
0
π/2
a
0 0 Kρ3 sin φ cos φdρdφdθ = 1 2 πKa4 π/2
sin φ cos φdφ= 1 0 8 πKa4 .ButthemassisK(volume of
the hemisphere) = 2 3 πKa3 , so the centroid is 0, 0, 3 8 a .
(b) Place the center of the base at (0, 0, 0); the density function is ρ(x, y, z) =K. By symmetry, the moments of inertia about
any two such diameters will be equal, so we just need to find I x:
I x = 2π
0
= K 2π
0
π/2
a
0 0 (Kρ2 sin φ) ρ 2 (sin 2 φ sin 2 θ +cos 2 φ) dρ dφ dθ
π/2
0
(sin 3 φ sin 2 θ +sinφ cos 2 φ) 1
a5 dφ dθ
5
= 1 5 Ka5 2π
0 sin 2 θ − cos φ + 1 3 cos3 φ + − 1 3 cos3 φ φ=π/2
dθ = 1 φ=0 5 Ka5 2π
0
= 1
5 Ka5 2 1 θ − 1 sin 2θ + 1 θ 2π
= 1 3 2 4 3 0 5 Ka5 2
(π − 0) + 1 (2π − 0) = 4
3 3 15 Ka5 π
2
3 sin2 θ + 1 3 dθ