Solução_Calculo_Stewart_6e

F.
254 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
35. In spherical coordinates z = x 2 + y 2 becomes cos φ =sinφ or φ = π 4 .Then
V = 2π
0
M xy = 2π
0
π/4
1
0 0 ρ2 sin φdρdφdθ = 2π
dθ π/4
sin φdφ 1
0 0
π/4
0
Hence (x, y, z) =
1
0 ρ3 sin φ cos φdρdφdθ =2π − 1 cos 2φ π/4
4 0
0, 0,
3
8 2 − √ 2 .
0 ρ2 dρ = 1 π 2 − √ 2 ,
3
1
4 =
π
and by symmetry Myz = Mxz =0.
8
37. In cylindrical coordinates the paraboloid is given by z = r 2 and the plane by z =2r sin θ and they intersect in the circle
r =2sinθ. Then zdV = π 2sinθ
2r sin θ
rz dz dr dθ = 5π E 0 0 r 2 6
[using a CAS].
39. The region E of integration is the region above the cone z = x 2 + y 2 and below the sphere x 2 + y 2 + z 2 =2in the first
octant. Because E is in the first octant we have 0 ≤ θ ≤ π 2 . The cone has equation φ = π 4 (as in Example 4), so 0 ≤ φ ≤ π 4 ,
and 0 ≤ ρ ≤ √ 2. So the integral becomes
π/4
0
π/2
0
√ 2
0
(ρ sin φ cos θ)(ρ sin φ sin θ) ρ 2 sin φdρdθdφ
= π/4
sin 3 φdφ π/2
sin θ cos θdθ √ 2
ρ 4 dρ =
0 0 0
= 1
3 cos3 φ − cos φ π/4
· 1 · 1
0 2 5
√
2
5
=
√2
41. In cylindrical coordinates, the equation of the cylinder is r =3, 0 ≤ z ≤ 10.
The hemisphere is the upper part of the sphere radius 3, center (0, 0, 10), equation
π/4
1 − cos 2 φ
sin φdφ
1
2 sin2 θ π/2
0
−
√
2
− 1
− 1 · 2√ 2
= 4√ 2−5
12 2 3 5 15
r 2 +(z − 10) 2 =3 2 , z ≥ 10.InMaple,wecanusethecoords=cylindrical option
in a regular plot3d command. In Mathematica, we can use ParametricPlot3D.
0
1
5 ρ5√ 2
0
43. If E is the solid enclosed by the surface ρ =1+ 1 sin 6θ sin 5φ, it can be described in spherical coordinates as
5
E = (ρ, θ, φ) | 0 ≤ ρ ≤ 1+ 1 5 sin 6θ sin 5φ, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π . Its volume is given by
V (E) = E dV = π
0
2π
1+(sin6θ sin 5φ)/5
ρ 2 sin φdρdθdφ= 136π [using a CAS].
0 0 99
45. (a) From the diagram, z = r cot φ 0 to z = √ a 2 − r 2 , r =0
to r = a sin φ 0 (or use a 2 − r 2 = r 2 cot 2 φ 0 ). Thus
V = 2π
0
a sin φ0
0
√ a 2 −r 2
r cot φ 0
rdzdrdθ
=2π a sin φ 0
√
0 r a2 − r 2 − r 2 cot φ 0 dr
a sin φ0
−(a 2 − r 2 ) 3/2 − r 3 cot φ 0
= 2π 3
= 2π 3
− a 2 − a 2 sin 2 φ 0
3/2
− a 3 sin 3 φ 0 cot φ 0 + a 3
= 2 3 πa3 1 − cos 3 φ 0 +sin 2 φ 0 cos φ 0
=
2
3 πa3 (1 − cos φ 0 )
0