Download Ma 227 Homework 12 Solutions Fall 2011 Due 12/9/2011 (Not to be

Ma 227 Homework 12 Solutions Fall 2011
Due 12/9/2011 (Not to be turned in)
Page 965 # 3, 5, 9, Stokes’ Theorem
Section 13.7
 x, y, z  x 2 z 2i  y 2 z 2j  xyzk, S is the par of the paraboloid z  x 2  y 2 the lies inside
3. F
the cylinder x 2  y 2  4 oriented upward.
The parabloid intersects the cylinder in the circle x 2  y 2  4, z  4. The boundary curve C
should be oriented in the counterclockwise direction when viewed from above, so a vector
equation of C is
rt  2 cos ti  2 sin tj  4k, 0 ≤ t ≤ 2
Hence r ′ t  −2 sin ti  2 cos tj
Frt  4 cos 2 t16i  4 sin 2 t16j  2 cos t2 sin t4k
By Stokes’ Theorem
2
 curl F  dS   F  dr   0
S
 rt  r ′ tdt
F
C

2
 0 −18 cos 2 t sin t  128 sin 2 t cos t  0dt
 128 1 cos 3 t  1 sin 3 t
3
3
5. C is the square in the plane z  −1. By 3
 curl FdS   F  dr 
S1
C
2
0
0
 curl FdS
S2
where S 1 is the original cube without the bottom and S 2 is the bottom face of the cube.
  x 2 zi  xy − 2xyzj  y − xzk
curl F
n  k so that C has the same orientation for both surfaces. Then
For S 2 we choose 

curl F
n  y − xz  x  y
since z  −1. Thus
1
1
 curl F  dS   −1  −1 x  ydxdy  0
S2
so
 curl F  dS  0
S1
1
9.
  xe xy − yi − ye xy − yj − 2z − zk
curl F
Take the surface S to be the disk x 2  y 2 ≤ 16, z  5. Since C is oriented clockwise (from
above), we orient S upward. Then n  k and curlF  n  2z − z on S were z  5. Thus
 F  dr    curl F  ndS  2z − zdS
s
C
D
10 − 5dS
D
 5Area S  5  4 2   80
Page 971 #3, 5, 7, 9, 13
Section 13.8
3.Verify the divergence theorm is true for the vector field F on the region E.
  〈z, y, x E is the ball x 2  y 2  z 2 ≤ 161.
F
  0  1  0  1 so
divF
  E div FdV   1dV  VE  43 4 3
E
Let S is a sphere of radius 4 centered at the origin which can be parametrized by
r,   〈4 sin  cos . 4 cos  sin , 4 cos 
(similar to Example 12.6.1)
r   r   〈4 cos  cos , 4 cos  sin , −4 sin   〈−4 sin  sin , 4 sin  cos , 0
 〈16 sin 2  cos , 16 sin 2  sin , 16 cos  sin 
and
 r,   〈4 cos , 4 sin  sin , 4 sin  cos 
F
  r  r    64 cos  sin 2  cos   64 sin 3  sin 2   64 cos  sin 2  cos 
F
 128 cos  sin 2  cos   64 sin 3  sin 2 
Then
2
2

 F  dS   F  r  r  dA   0  0 128 cos  sin 2  cos   64 sin 3  sin 2 dd
S
D
2

0

0
2

128 sin 3  cos   64 − 1 2  sin 2  cos  sin 2 
3
3
256 sin 2 d  256
3
3
1  − 1 sin 2
2
4
2
0
0
d
 256 
3
  ∂/∂xxye z   ∂/∂yxy 2 z 3   ∂/∂z−ye z   ye z  2xyz 3 − ye z  2xyz 3
5. div F
by the Divergence theorem,
3
2
1
 S F  dS    E div FdV   0  0  0 2xyz 3 dzdydx  2
1 x2
2
3
0
1 y2
2
2
0
1
1 z4
4
0
 9
2
 x, y, z  3xy 2i  xe zj  z 3k. S is the surface of the soild bounded by the cylinder
7. F
y 2  z 2  1 and the planes x  −1 and x  2
div F  3y 2  0  3z 2  3y 2  3z 2
so using cylindrical coordinates where y  r cos , z  r sin  and x  x
2 1 2
  F  dS     3y 2  3z 2 dV     3r 2 cos 2   3r 2 sin 2 rdxdrd  92 
S
E
0
0 −1
 x, y, z  x 2 sin yi  x cos yj − xz sin yk. S is the "fat sphere" x 8  y 8  z 8  8
9. F
div F  2x sin y − x sin y − x sin y  0 so by Divergence The0rem
  F  dS     0dV  0
S
E
 x, y, z  4x 3 zi  4y 3 ze zj  3z 4k. S is the spherewith radius R and center the origin.
13. F
div F  12x 2 z  12y 2 z  12z 3 so using spherical coordinates
2

R
 S F  dS    E 12zx 2  y 2  z 2 dV   0  0  −1 12 cos  2  2 sin ddd
 12 
2
0

R
d  sin  cos d   5 d  122 1 sin 2 
2
0
0

0
1 6
6
R
0
0
3