Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 17 REVIEW ET CHAPTER 16 ¤ 305
27. z = f(x, y) =x 2 + y 2 with 0 ≤ x 2 + y 2 ≤ 4 so r x × r y = −2x i − 2y j + k (using upward orientation). Then
zdS = (x
S 2 + y 2 ) 4x 2 +4y 2 +1dA = 2π 2 r3√ 1+4r
0 0 2 dr dθ = 1 π 391 √ 17 + 1
60
x 2 + y 2 ≤ 4
(Substitute u =1+4r 2 and use tables.)
29. Since the sphere bounds a simple solid region, the Divergence Theorem applies and
S F · dS = E (z − 2) dV = E zdV − 2 E dV = mz − 2 4
3 π23 = − 64 3 π.
Alternate solution: F(r(φ, θ)) = 4 sin φ cos θ cos φ i − 4sinφ sin θ j +6sinφ cos θ k,
r φ × r θ =4sin 2 φ cos θ i +4sin 2 φ sin θ j +4sinφ cos φ k, and
F · (r φ × r θ )=16sin 3 φ cos 2 θ cos φ − 16 sin 3 φ sin 2 θ +24sin 2 φ cos φ cos θ. Then
S F · dS = 2π π (16 0 0 sin3 φ cos φ cos 2 θ − 16 sin 3 φ sin 2 θ +24sin 2 φ cos φ cos θ) dφ dθ
= 2π
0
4
(−16 3 sin2 θ) dθ = − 64 π 3
31. Since curl F = 0, (curl F) · dS =0. We parametrize C: r(t) =cost i +sint j, 0 ≤ t ≤ 2π and
S
F · dr = 2π
(− C 0 cos2 t sin t +sin 2 t cos t) dt = 1 3 cos3 t + 1 3 sin3 t 2π
=0.
0
33. The surface is given by x + y + z =1or z =1− x − y, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x and r x × r y = i + j + k. Then
C F · dr = S curl F · dS = D (−y i − z j − x k) · (i + j + k) dA = D (−1) dA = −(area of D) =− 1 2 .
35. div F dV =
E
x 2 + y 2 + z 2 ≤ 1
3 dV =3(volume of sphere) =4π. Then
F(r(φ, θ)) · (r φ × r θ )=sin 3 φ cos 2 θ +sin 3 φ sin 2 θ +sinφ cos 2 φ =sinφ and
S F · dS = 2π
0
π
0
sin φdφdθ =(2π)(2) = 4π.
37. Because curl F = 0, F is conservative, and if f(x, y, z) =x 3 yz − 3xy + z 2 ,then∇f = F. Hence
F · dr = ∇f · dr = f(0, 3, 0) − f(0, 0, 2) = 0 − 4=−4.
C C
39. By the Divergence Theorem, F · n dS = div F dV =3(volumeofE) =3(8− 1) = 21.
S E
41. Let F = a × r = ha 1 ,a 2 ,a 3 i×hx, y, zi = ha 2 z − a 3 y, a 3 x − a 1 z, a 1 y − a 2 xi. Then curl F = h2a 1 , 2a 2 , 2a 3 i =2a,
and 2a · dS = curl F · dS = F · dr = (a × r) · dr by Stokes’ Theorem.
S S C C