Solução_Calculo_Stewart_6e

F.
298 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
5. C is the square in the plane z = −1. By(3), S 1
curl F · dS = C F · dr = S 2
curl F · dS where S 1 is the original cube
without the bottom and S 2 is the bottom face of the cube. curl F = x 2 z i +(xy − 2xyz) j +(y − xz) k. ForS 2 , we choose
n = k so that C has the same orientation for both surfaces. Then curl F · n = y − xz = x + y on S 2 ,wherez = −1. Thus
S 2
curl F · dS = 1
−1
1
−1 (x + y) dx dy =0so S 1
curl F · dS =0.
7. curl F = −2z i − 2x j − 2y k and we take the surface S to be the planar region enclosed by C,soS is the portion of the plane
x + y + z =1over D = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x}. SinceC is oriented counterclockwise, we orient S upward.
Using Equation 17.7.10 [ ET 16.7.10], we have z = g(x, y) =1− x − y, P = −2z, Q = −2x, R = −2y,and
F · dr = curl F · dS = [−(−2z)(−1) − (−2x)(−1) + (−2y)] dA
C S D
= 1
0
1−x
0
(−2) dy dx = −2 1
0
(1 − x) dx = −1
9. curl F =(xe xy − 2x) i − (ye xy − y) j +(2z − z) k and we take S to be the disk x 2 + y 2 ≤ 16, z =5.SinceC is oriented
counterclockwise (from above), we orient S upward. Then n = k and curl F · n =2z − z on S,wherez =5. Thus
F · dr =
S curl F · n dS= S (2z − z) dS = S (10 − 5) dS =5(areaofS) =5(π · 42 )=80π
11. (a) The curve of intersection is an ellipse in the plane x + y + z =1with unit normal n = 1 √
3
(i + j + k),
curl F = x 2 j + y 2 k,andcurl F · n = 1 √
3
(x 2 + y 2 ).Then
C F · dr = S
1
√
3
x 2 + y 2 dS = x 2 + y 2 ≤ 9
x 2 + y 2 dx dy = 2π
0
3
0 r3 dr dθ =2π
81
4 =
81π
2
(b)
(c) One possible parametrization is x =3cost, y =3sint,
z =1− 3cost − 3sint, 0 ≤ t ≤ 2π.
13. The boundary curve C is the circle x 2 + y 2 =1, z =1oriented in the counterclockwise direction as viewed from above.
We can parametrize C by r(t) =cost i +sint j + k, 0 ≤ t ≤ 2π, andthenr 0 (t) =− sin t i +cost j. Thus
F(r(t)) = sin 2 t i +cost j + k, F(r(t)) · r 0 (t) =cos 2 t − sin 3 t,and
C F · dr = 2π
0 (cos2 t − sin 3 t) dt = 2π 1
(1 + cos 2t) dt − 2π
(1 − 0 2 0 cos2 t)sintdt
= 1 2 t +
1
sin 2t 2π
−
2
− cos t + 1 0 3 cos3 t 2π
= π
0
Now curl F =(1− 2y) k, and the projection D of S on the xy-plane is the disk x 2 + y 2 ≤ 1, so by Equation 17.7.10
[ ET 16.7.10] with z = g(x, y) =x 2 + y 2 we have
S curl F · dS = D (1 − 2y) dA = 2π
0
1
0 (1 − 2r sin θ) rdrdθ= 2π
0
1
2 − 2 3 sin θ dθ = π.
15. The boundary curve C is the circle x 2 + z 2 =1, y =0oriented in the counterclockwise direction as viewed from the positive
y-axis. Then C can be described by r(t) =cost i − sin t k, 0 ≤ t ≤ 2π,andr 0 (t) =− sin t i − cos t k. Thus