Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 17.7 SURFACE INTEGRALS ET SECTION 16.7 ¤ 295
25. Let S 1 be the paraboloid y = x 2 + z 2 , 0 ≤ y ≤ 1 and S 2 the disk x 2 + z 2 ≤ 1, y =1.SinceS is a closed
surface, we use the outward orientation.
On S 1 : F(r(x, z)) = (x 2 + z 2 ) j − z k and r x × r z =2x i − j +2z k (since the j-component must be negative on S 1 ). Then
S 1
F · dS = [−(x 2 + z 2 ) − 2z 2 ] dA = − 2π 1
0 (r2 +2r 2 cos 2 θ) rdrdθ
x 2 + z 2 ≤ 1
= − 2π 1
(1 + 2 0 4 cos2 θ) dθ = − π
+ π
2 2 = −π
On S 2 : F(r(x, z)) = j − z k and r z × r x = j. Then S 2
F · dS =
(1) dA = π.
Hence F · dS = −π + π =0.
S
0
x 2 + z 2 ≤ 1
27. Here S consists of the six faces of the cube as labeled in the figure. On S 1:
F = i +2y j +3z k, r y × r z = i and S 1
F · dS = 1
1
dy dz =4;
−1 −1
S 2 : F = x i +2j +3z k, r z × r x = j and S 2
F · dS = 1
1
2 dx dz =8;
−1 −1
S 3: F = x i +2y j +3k, r x × r y = k and S 3
F · dS = 1
1
3 dx dy =12;
−1 −1
S 4: F = −i +2y j +3z k, r z × r y = −i and S 4
F · dS =4;
S 5 : F = x i − 2 j +3z k, r x × r z = −j and S 5
F · dS =8;
S 6: F = x i +2y j − 3 k, r y × r x = −k and S 6
F · dS = 1
1
3 dx dy =12.
−1 −1
Hence F · dS = 6
S i=1 S i
F · dS =48.
29. Here S consists of four surfaces: S 1 , the top surface (a portion of the circular cylinder y 2 + z 2 =1); S 2 , the bottom surface
(a portion of the xy-plane); S 3, the front half-disk in the plane x =2,andS 4, the back half-disk in the plane x =0.
On S 1: The surface is z = 1 − y 2 for 0 ≤ x ≤ 2, −1 ≤ y ≤ 1 with upward orientation, so
2
1
F · dS =
−x 2 (0) − y 2 y
2
1
− + z 2 y 3
dy dx = +1−
S 1 0 −1
1 − y
2
0 −1 1 − y
2 y2 dy dx
= 2
− y=1
1 − y
0
2 + 1 (1 − 3 y2 ) 3/2 + y − 1 3 y3 dx = 2 4
dx = 8
0 3 3
On S 2 : The surface is z =0with downward orientation, so
S 2
F · dS = 2
0
y=−1
1
−1
−z
2 dy dx = 2
0
1
(0) dy dx =0
−1
On S 3 :Thesurfaceisx =2for −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y 2 , oriented in the positive x-direction. Regarding y and z as
parameters, we have r y × r z = i and
S 3
F · dS = 1
√ 1−y 2
x 2 dz dy = 1
√ 1−y 2
4 dz dy =4A (S
−1 0 −1 0 3)=2π
On S 4 :Thesurfaceisx =0for −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y 2 , oriented in the negative x-direction. Regarding y and z as
parameters, we use − (r y × r z )=−i and
Thus S F · dS = 8 3 +0+2π +0=2π + 8 3 .
S 4
F · dS = 1
√ 1−y 2
x 2 dz dy = 1
√ 1−y 2
(0) dz dy =0
−1 0 −1 0