Solução_Calculo_Stewart_6e

F.
300 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
S 1
F · dS = S 1
F · n dS = S 1
xdS = 2π
0
On S 2 , z =0, n = −k,andF = xy i so S 2
F · dS = S 2
0 dS =0.
1 (r cos θ) rdrdθ= sin θ 2π 1 r3 1
=0.
0 0 3 0
S 3 is given by r(θ, z) =cosθ i +sinθ j + z k, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 1. Thenr θ × r z =cosθ i +sinθ j and
S 3
F · dS = F · (r D θ × r z ) dA = 2π
1
0 0 (cos2 θ sin θ + z sin 2 θ) dz dθ
= 2π
0
Thus S F · dS =0+0+π 2 = π 2 .
cos 2 θ sin θ + 1 2 sin2 θ dθ =
− 1 3 cos3 θ + 1 4 θ −
1
sin 2θ 2π
= π 2 0 2
5. div F = ∂
∂x (ex sin y)+ ∂
∂y (ex cos y)+ ∂ ∂z (yz2 )=e x sin y − e x sin y +2yz =2yz, so by the Divergence Theorem,
S F · dS = E div F dV = 1
0
1
0
2
0 2yz dz dy dx =2 1
0 dx 1
0 ydy 1
0 zdz=2 x 1
0
1 y2 1 1 z2 2
=2.
2 0 2 0
7. div F =3y 2 +0+3z 2 , so using cylindrical coordinates with y = r cos θ, z = r sin θ, x = x we have
S F · dS = E (3y2 +3z 2 ) dV = 2π
1
2
0 0 −1 (3r2 cos 2 θ +3r 2 sin 2 θ) rdxdrdθ
=3 2π
dθ 1
0 0 r3 dr 2
dx =3(2π)
1
−1 4 (3) =
9π
2
9. div F = y sin z +0− y sin z =0, so by the Divergence Theorem, F · dS = 0 dV =0.
S E
11. div F = y 2 +0+x 2 = x 2 + y 2 so
S F · dS = E (x2 + y 2 ) dV = 2π
0
2
0
4
r 2 · rdzdrdθ= 2π
2
r 2 0 0
= 2π
0
dθ 2
0 (4r3 − r 5 ) dr =2π r 4 − 1 6 r6 2
0 = 32 3 π
r3 (4 − r 2 ) dr dθ
13. div F =12x 2 z +12y 2 z +12z 3 so
S F · dS = E 12z(x2 + y 2 + z 2 ) dV = 2π
π
R 12(ρ cos 0 0 0
φ)(ρ2 )ρ 2 sin φdρdφdθ
=12 2π
dθ π
sin φ cos φdφ R
0 0 0 ρ5 dρ =12(2π) 1
2 sin2 φ π 1 ρ6 R
=0
0 6 0
15. F · dS = √
S E 3 − x2 dV = 1
1
−1 −1
2 − x
4 − y
4
0
√
3 − x2 dz dy dx = 341
60
√ √3
2+
81
20 sin−1 3
17. For S 1 we have n = −k,soF · n = F · (−k) =−x 2 z − y 2 = −y 2 (since z =0on S 1 ). So if D is the unit disk, we get
S 1
F · dS = S 1
F · n dS = D (−y2 ) dA = − 2π 1
0 0 r2 (sin 2 θ) rdrdθ = − 1 π.NowsinceS 4 2 is closed, we can use
the Divergence Theorem. Since div F = ∂
∂x (z2 x)+ ∂ 1
∂y 3 y3 +tanz + ∂ ∂z (x2 z + y 2 )=z 2 + y 2 + x 2 , we use spherical
coordinates to get S 2
F · dS = E div F dV = 2π
0
S F · dS = S 2
F · dS − S 1
F · dS = 2 5 π − − 1 4 π = 13
20 π.
π/2
1
0 0 ρ2 · ρ 2 sin φdρdφdθ = 2 π.Finally
5
19. The vectors that end near P 1 are longer than the vectors that start near P 1 ,sothenetflow is inward near P 1 and div F(P 1 ) is
negative. The vectors that end near P 2 are shorter than the vectors that start near P 2, so the net flow is outward near P 2 and
div F(P 2 ) is positive.