Solução_Calculo_Stewart_6e

F.
250 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
25. The paraboloid z =4x 2 +4y 2 intersects the plane z = a when a =4x 2 +4y 2 or x 2 + y 2 = 1 4
a. So, in cylindrical
coordinates, E = (r, θ, z) | 0 ≤ r ≤ 1 2
√ a, 0 ≤ θ ≤ 2π, 4r 2 ≤ z ≤ a .Thus
m =
2π
√ a/2 a
= K
0 0
2π
0
4r 2 Kr dz dr dθ = K
1
2 ar2 − r 4 r= √ a/2
r=0
Since the region is homogeneous and symmetric, M yz = M xz =0and
M xy =
Hence (x, y, z) = 0, 0, 2 3 a .
2π
√ a/2 a
= K
0 0
2π
0
4r 2 Krzdzdrdθ= K
1
4 a2 r 2 − 4 r6 r= √ a/2
dθ = K
3
r=0
2π
√ a/2
0
0
(ar − 4r 3 ) dr dθ
2π
1
dθ = K
16 a2 dθ = 1 8 a2 πK
0
2π
√ a/2
0 0
2π
0
1
2 a2 r − 8r 5 dr dθ
1
24 a3 dθ = 1 12 a3 πK
27. The region of integration is the region above the cone z = x 2 + y 2 ,orz = r, and below the plane z =2.Also,wehave
−2 ≤ y ≤ 2 with − 4 − y 2 ≤ x ≤ 4 − y 2 which describes a circle of radius 2 in the xy-plane centered at (0, 0). Thus,
2
−2
√ 4−y 2 2
√
xz dz dx dy =
− 4−y
√x 2 2 +y 2
2π
2
2
0
0
r
(r cos θ) z r dz dr dθ =
= 2π 2
0 0 r2 (cos θ) 1
z2 z=2
dr dθ = 1
2 z=r 2
4r 2 − r 4 dr = 1 2
= 1 2
2π
cos θdθ 2
0 0
2π
2
2
0
2π
0
0
[sin θ]2π
0
r
r 2 (cos θ) zdzdrdθ
2
0 r2 (cos θ) 4 − r 2 dr dθ
4
3 r3 − 1 5 r5 2
0 =0
29. (a) The mountain comprises a solid conical region C. The work done in lifting a small volume of material ∆V with density
g(P ) to a height h(P ) above sea level is h(P )g(P ) ∆V . Summing over the whole mountain we get
W = h(P )g(P ) dV .
C
(b) Here C is a solid right circular cone with radius R =62,000 ft, height H =12,400 ft,
and density g(P )=200lb/ft 3 at all points P in C. We use cylindrical coordinates:
W = 2π H
0 0
=400π
R(1−z/H)
z · 200rdrdzdθ=2π H
200z 1
r2 r=R(1−z/H)
dz
0 0 2 r=0
H
0
z R2
2
1 − z H
2
dz = 200πR
2
H
z
=200πR 2 2
H
2 − 2z3
3H +
z4
H
= 200πR 2 2
4H 2 0
2 − 2H2
3
= 50 3 πR2 H 2 = 50
3 π(62,000)2 (12,400) 2 ≈ 3.1 × 10 19 ft-lb
0
z − 2z2
H + z3
H 2
dz
+ H2
4
r
R = H − z
H
=1− z H