Solução_Calculo_Stewart_6e

F.
SECTION 16.7 TX.10 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES ET SECTION 15.7 ¤ 249
15. The region of integration is given in cylindrical coordinates by
E = {(r, θ, z) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 4, r ≤ z ≤ 4}. This represents the solid
region bounded below by the cone z = r and above by the horizontal plane z =4.
4
2π
4 rdzdθdr = 4
2π
z=4
0 0 r 0 0 rz dθ dr = 4
2π
r(4 − r) dθ dr
z=r 0 0
= 4
0 (4r − r2 ) dr 2π
0
dθ = 2r 2 − 1 3 r3 4
0
= 32 − 64
3
(2π) =
64π
3
17. In cylindrical coordinates, E is given by {(r, θ, z) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 4, −5 ≤ z ≤ 4}. So
x2 + y
E
2 dV = 2π
4
4
√
r2 rdzdrdθ= 2π
dθ 4
0 0 −5 0 0 r2 dr 4
= θ 2π 1 r3 4 4
0 3 0 z =(2π)
64
−5 3 (9) = 384π
−5 dz
θ
2π
19. In cylindrical coordinates E is bounded by the paraboloid z =1+r 2 , the cylinder r 2 =5or r = √ 5,andthexy-plane,
so E is given by (r, θ, z) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ √ 5, 0 ≤ z ≤ 1+r 2 .Thus
E ez dV = 2π
0
√ 5
1+r
2
e z rdzdrdθ= 2π
√ 5
r e z z=1+r 2
dr dθ = 2π
√ 5
0 0 0 0 z=0 0
= 2π
dθ √ 5
0 0
re 1+r2 − r
dr =2π
√
1
2 e1+r2 − 1 5
2 r2 0
= π(e 6 − e − 5)
0
r(e 1+r2 − 1) dr dθ
21. In cylindrical coordinates, E is bounded by the cylinder r =1, the plane z =0, and the cone z =2r. So
E = {(r, θ, z) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, 0 ≤ z ≤ 2r} and
E x2 dV = 2π 1
2r
r 2 cos 2 θ r dz dr dθ = 2π 1
0 0 0 0 0 r 3 cos 2 θz z=2r
dr dθ = 2π 1
z=0 0 0 2r4 cos 2 θdrdθ
= 2π
2
0 5 r5 cos 2 θ r=1
dθ = 2 2π
r=0 5
cos 2 θdθ= 2 2π
1+cos2θ
dθ = 1 θ + 1 2π
0
5 2 5 2 sin 2θ = 2π 5
23. (a) The paraboloids intersect when x 2 + y 2 =36− 3x 2 − 3y 2 ⇒ x 2 + y 2 =9, so the region of integration
is D = (x, y) | x 2 + y 2 ≤ 9 . Then, in cylindrical coordinates,
E = (r, θ, z) | r 2 ≤ z ≤ 36 − 3r 2 , 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π and
V = 2π 3
36 − 3r
2
0 0 r 2
rdzdrdθ= 2π 3
0 0
0
36r − 4r
3
dr dθ = 2π
0 18r 2 − r 4 r=3
dθ = 2π
81 dθ =162π.
r=0 0
(b) For constant density K, m = KV =162πK from part (a). Since the region is homogeneous and symmetric,
M yz = M xz =0and
M xy = 2π
3
0 0
= K 2
2π
0
36−3r
2
r 2
(zK) rdzdrdθ= K 2π
3 r 1
z2 z=36−3r 2
0 0 2
dr dθ
z=r 2
3
0 r((36 − 3r2 ) 2 − r 4 ) dr dθ = K 2
= K (2π) 8
2 6 r6 − 216
4 r4 + 1296 r2 3
= πK(2430) = 2430πK
2 0
Myz
Thus (x, y, z) =
m , M xz
m , M
xy
m
2π
0
dθ 3
0 (8r5 − 216r 3 +1296r) dr
=
0, 0, 2430πK
162πK =(0, 0, 15).
0
0