Solução_Calculo_Stewart_6e

F.
294 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
13. Using x and z as parameters, we have r(x, z) =x i +(x 2 + z 2 ) j + z k, x 2 + z 2 ≤ 4. Then
r x × r z =(i +2x j) × (2z j + k) =2x i − j +2z k and |r x × r z | = √ 4x 2 +1+4z 2 = 1+4(x 2 + z 2 ).Thus
ydS =
S
x 2 +z 2 ≤4
(x 2 + z 2 ) 1+4(x 2 + z 2 ) dA = 2π
=2π 2
0 r2 √ 1+4r 2 rdr
1
0
let u =1+4r
2
=2π 17 1
(u − 1)√ u · 1 du = 1 π 17
1 4 8 16 1 (u3/2 − u 1/2 ) du
17
= 1 π 2
16 5 u5/2 − 2 3 u3/2 = 1 π 2
16 5 (17)5/2 − 2 3 (17)3/2 − 2 + 2 5 3
2
0 r2 √ 1+4r 2 rdrdθ= 2π
0
dθ 2
0 r2 √ 1+4r 2 rdr
⇒
r 2 = 1 (u − 1) and 1 du = rdr
4 8
= π √
391 17 + 1
60
15. Using spherical coordinates and Example 17.6.10 [ ET 16.6.10] we have r(φ, θ) =2sinφ cos θ i +2sinφ sin θ j +2cosφ k
and |r φ × r θ | =4sinφ. Then S (x2 z + y 2 z) dS = 2π π/2
(4 sin 2 φ)(2 cos φ)(4 sin φ) dφ dθ =16π sin 4 φ π/2
=16π.
0 0 0
17. S is given by r(u, v) =u i +cosv j +sinv k, 0 ≤ u ≤ 3, 0 ≤ v ≤ π/2. Then
r u × r v = i × (− sin v j +cosv k) =− cos v j − sin v k and |r u × r v | = cos 2 v +sin 2 v =1,so
S (z + x2 y) dS = π/2
3 (sin v + 0 0 u2 cos v)(1) du dv = π/2
(3 sin v +9cosv) dv
0
=[−3cosv +9sinv] π/2
0
=0+9+3− 0=12
19. F(x, y, z) =xy i + yz j + zxk, z = g(x, y) =4− x 2 − y 2 ,andD is the square [0, 1] × [0, 1], so by Equation 10
S F · dS = [−xy(−2x) − yz(−2y)+zx] dA = 1
1
D 0 0 [2x2 y +2y 2 (4 − x 2 − y 2 )+x(4 − x 2 − y 2 )] dy dx
= 1
1
0 3 x2 + 11 x −
3 x3 + 34
15 dx =
713
180
21. F(x, y, z) =xze y i − xze y j + z k, z = g(x, y) =1− x − y,andD = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x}. SinceS has
downward orientation, we have
S F · dS = − D [−xzey (−1) − (−xze y )(−1) + z] dA = − 1
1−x
(1 − x − y) dy dx
0 0
= − 1
1
0 2 x2 − x + 1 2 dx = −
1
6
23. F(x, y, z) =x i − z j + y k, z = g(x, y) = 4 − x 2 − y 2 and D is the quarter disk
(x, y)
0 ≤ x ≤ 2, 0 ≤ y ≤
√
4 − x
2 . S has downward orientation, so by Formula 10,
S F · dS = − D
= −
D
−x · 1 (4 − 2 x2 − y 2 ) −1/2 (−2x) − (−z) · 1 (4 − 2 x2 − y 2 ) −1/2 (−2y)+y
x 2
4 − x2 − y 2 − 4 − x 2 − y 2 ·
y
4 − x2 − y 2 + y
dA
= − D x2 (4 − (x 2 + y 2 )) −1/2 dA = − π/2
2 (r cos 0 0 θ)2 (4 − r 2 ) −1/2 rdrdθ
dA
= − π/2
cos 2 θdθ 2
0 0 r3 (4 − r 2 ) −1/2 dr
= − π/2
0
let u =4− r
2
1
2 + 1 2 cos 2θ dθ 0
4 − 1 2 (4 − u)(u)−1/2 du
⇒
r 2 =4− u and − 1 du = rdr
2
= − 1
θ + 1 sin 2θ π/2
2 4 0 −
1
8 √ 0
u − 2 2 3 u3/2 = − π 4
4
−
1
2
−16 +
16
3 = −
4
π 3