Mat 241 Chapter 16 exam BKEY
Fall, 2013
Name ___________________________
Directions: Show all work for each question and make sure your answers
are clearly identified. You may use the back side of pages if needed.
#1. (5 pts) Use the Divergence Theorem to calculate the surface integral
 F  dS
with the given vector field
S
F  x, y, z   e  y sin z  xy 2 , x 2 y  e  x cos z , tan 1  xy 
on the extremely uncool surface of the bounded solid created by the
functions:
z  x 2  y 2 and z  9
Assume outward pointing orientation.
 F  dS   divFdV    x
S
E
2 3
   r dzdrd    r z
3
3
0 0 r2
2
0 0
3
 9 4 r6 
   r   d
4
6 0
0 
 9  81 9  81 
0  4  6 d
243

2
2

 y 2 dV
E
2 3 9

2
9
r
2
drd 
2 3
   9r
0 0
3
 r 5 drd
#2. Suppose we wish to compute the work done on a particle as it
traverses the helical path shown in the figure defined by:
r  t   2 cos  t  ,2 sin  t  , t
from
2 , 0 , 0  to


1 , 3 , 3  .


This requires a line integral.
b
 
W   F  dr   F r  t   r'  t dt
C
a
If we are lucky, the field is a gradient field and is therefore
independent of path (conservative). Use the vector field, F , for
questions A, B, and C.
F  x, y, z   8 xy 3 z ,12 x 2 y 2 z ,4 x 2 y 3
A. (2pts) Compute curlF to show that F is a conservative vector field.
F  x, y, z   8 xy 3 z ,12 x 2 y 2 z ,4 x 2 y 3
i
curl F 
j


x
y
3
8 xy z 12 x 2 y 2 z
k

z
4x2 y3
 12 x 2 y 2  12 x 2 y 2  i   8 xy 3  8 xy 3  j   24 xy 2 z  24 xy 2 z  k  0
B. (2pts) Find a potential function, f , such that f  F .
F  x, y, z   8 xy 3 z ,12 x 2 y 2 z , 4 x 2 y 3
f x  8 xy 3 z   f x    8 xy 3 z dx  4 x 2 y 3 z  g  y , z 
f y  12 x 2 y 2 z   f y   12 x 2 y 2 z dy  4 x 2 y 3 z  h  x, z 
f z  4 x 2 y 3   f z    4 x 2 y 3 dz  4 x 2 y 3 z  j  x, y 
 f  x, y , z   4 x 2 y 3 z
C.(2pts) Compute the work using your result from part B and the
fundamental theorem of line integrals:
 f  dr  f r b    f r a 
where f is a potential function.
C



f

dr

f
1,
3,

  f  2,0,0  where f is a potential function.
C
3

3 

2
2
3

f  1, 3,   4 1 
3    4 3; f 2,0,0   4 2   0   0   0
3

3


f  1, 3,   f  2,0,0   4 3  0  4 3
3

 
#3. (6pts) I went over to the tutor center with the following very nasty vector
field
F  x, y, z   e z
2
2 z
x,sin  xyz   y  1, e z sin  z 2 
2
I wanted to compute the work along the very simple curve, C,
parameterized by:
r  t   cos  t  ,sin t  ,0 where 0  t  2
where C is oriented counterclockwise as viewed from the positive z – axis.
A tutor named Peter said, “Hey, you can use Stokes’ Theorem” with the
following surface.
Looking at the surface a light-bulb went off in my head! I can use any
surface which the curve bounds with Stokes’ Theorem.
Part1: (3 points) Instead of Peter’s ridiculous surface, choose a VERY
simple surface and compute the work around the curve.
F  x, y , z   e z
curl F 
ez
2
2 z
x,sin  xyz   y  1, e z sin  z 2 
2
i
j
k

x

y

z
2
2 z
x sin  xyz   y  1 e z sin  z 2 
2

  0  xy cos  xyz   i  0   2 z  2  e z
  xy cos  xyz  ,  2 z  2  e z
2
2 z
2
2 z

x j   yz cos  xyz   0  k
x, yz cos  xyz 
Since we really just have a circle in the xy-plane, let’s choose the surface to
be z = 0. Our space curve certainly contains it.
Then, the normal to the surface z = 0 pointing upward is given by:
n 
z z
,  ,1  0 , 0 ,1
x y
So by Stokes’ Theorem;
 F  dr   curlF  ndA
C
D
  xy cos  xyz  ,  2z  2  ez
2
2z
x , yz cos  xyz   0 , 0 ,1 dA
D
  yz cos  xyz  dA
D
  y  0  cos  xy  0   dA /** When I replace z = 0.
D
=  0dA
D
0
Part2: (3 points) Compute the line integral directly using.
b
 
W   F  dr   F r  t   r'  t dt
C
F  x, y , z   e z
F  r t   e
2
2 z
02  2 0 
a
x,sin  xyz   y  1, e z sin  z 2 
2
cos t ,sin  cos t sin t  0    sin t  1, e 0 sin  0 2  
2
cos t ,sin t  1,0
r '  t    sin t ,cos t ,0
2
b
 F  dr   F  r  t    r ' t  dt  
C
a
cos t ,sin t  1,0   sin t ,cos t ,0 dt
0
2

   cos t sin t  sin t cos t  0  dt
0
2

  0  dt
0
0
#4. (5pts)
I have decided to divulge the top-secret equation that draws the Hershey
Kiss candy. Here it is:
r  t 
sin2  t 
t
, t2  1
-1  t  1
Use Green’s Theorem in reverse to determine the cross-sectional area of
the Kiss!
Sorry, guys….wrong Kiss
We need to select a vector field, F , such that
Q P
  1 . There are three
x y
choices which our book presents:
y x
F  0 , x , F  y , 0 , and F   ,
2 2
We’ll use the first.
Then the line integral value will be equivalent to the area of the Kiss.
 Q P 
F
r
t

r'
t
dt





a
D  x  y dA  D 1dA  the area value
b
 
r'  t   stuff,2t
b
1
a
1
 F r  t    r'  t dt  

1
sin2  t 
0,
 stuff,2t dt   2sin2  t  dt
t
1

2
2u 1

  sin2 u  du    sin 2u  
 
2 4
 

2   1
   1

 sin  2      
  sin  2    

  2 4
  2 4

2       
 

  2   2  
2
