Mat 241 Semester Final - Key
Fall, 2007
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) Evaluate
 F  dr
exactly where F  x, y, z   e i  xzj   x  y  k
z
C
and C is the curve given by: r  t   t , t , t ; 0  t  1 .
2
3
F  x, y, z   e z i  xzj   x  y  k ; r  t   t 2 , t 3 , t ;0  t  1
F  r  t    e  t i  t 3 j   t 2  t 3  k  e  t , t 3 , t 2  t 3 ; r   t   2t ,3t 2 , 1
F  r  t    r   t   e  t , t 3 , t 2  t 3  2t ,3t 2 , 1  2te  t  3t 5  t 2  t 3
 t 1 1 t 
t6
t3
t4
F

dr

2
te

3
t

t

t
dt

2

te

e
dt

3


  0   6 3 4

0 
C
0


0
0
1
1 1 1
2 2
13
 1
 2   et         2 
0
e e
12
e
 2 3 4
11 4
 
12 e
1
1
t
5
2
1
1
3
0


#2. (2 pts) Find curlF at the point  2,
 5 
,  exactly if
2 6 
F  x, y, z   e x sin y cos z , e x cos y cos z , e x sin y sin z
i
curlF 
j
k



x
y
z
x
x
x
e sin y cos z e cos y cos z e sin y sin z
 e x cos y sin z  e x cos y sin z , e x sin y sin z  e x sin y sin z, e x cos y cos z  e x cos y cos z
curlF  0,0,0  0   x, y, z 
#3. (5 pts) Use the Divergence Theorem to calculate the surface integral
 F  dS
where F  x, y, z   x , y , z
3
3
3
and the surface, S, is the solid
S
bounded by the cylinder x  y  1 and the planes z = 0 and z = 2.
2
2
 F  dS   divFdV
S
E
divF  3x 2  3 y 2  3z 3  3  x 2  y 2  z 2 
2 1 2
2 1
2
2 1
z 3r
 3 8 
F

dS

3
r

z
rdzdrd


3
r
z

drd


3


 2r  r drd
S
0 




3
3 
0 0
0 0
0 0
0
2
2
1
2
2
3
2
r 4 4r 2
33
66
1 4
 3 
d  3    d 
d


2
3 0
2 3
6 0
6
0
0 
 11
#4. (5 pts) Use Stokes’ Theorem to evaluate
 F  dr
where
C
F  x, y, z   xy, yz , zx and C is the triangle with vertices (1,0,0), (0,1,0),
and (0,0,1) oriented counterclockwise as viewed from above.
F  x, y, z   xy, yz , zx 
curlF 
i
j

x
xy

y
yz
k

  0  y  i   z  0 j   0  x  k
z
zx
F *   y,  z ,  x ; P   y, Q   z , R   x
g  x, y   z  1  x  y;
g
g
 1,
 1
x
y


g
g
  P x  Q y  R     y  z  x  ,  F  dr   curlF  dS


C
S




g
g
g
g
    P
Q
 R  dA     P
Q
 R  dA
x
y
x
y


D 
D 
1 1 x
1 1 x
1 1 x
0 0
0 0
0 0

   y  z  x dydx      y  1  x  y   x dydx     1dydx
1

x2 
   1  x dx    x  
2 0

0
1

2
1

#5. Let F  x, y   ye
xy
 1 i  xe xy j .
A. (1 pt) Show that F is a conservative vector field.
F  x, y    ye xy  1 i  xe xy j
P
Q
 yxe xy  e xy ;
 yxe xy  e xy
y
x
P Q


 F conservative on the simply connected curve
y x
B. (1 pt) Find a potential function for F .
f x  ye xy  1; f y  xe xy
By inspection : f ( x, y )  e xy  x
C. (2 pts) Find the work performed by the vector field on a particle that
moves along the sawtooth curve represented by the parametric
equations:
x(t )  t  sin 1  sin t 
y (t ) 
2sin 1  sin t 

0  t  8
We use the fundamental theorem for line integrals.
x(0)  0, y (0)  0 : x(8 )  8 , y  0  ;
( x, y )  (0,0) &  8 ,0 
fy  8 ,0   f  0, 0   e0  8   e0  0 
 8
1
x
5
-1
10
15
20
25
  xy 4 
2 3
#6. (5 pts) Compute  
dx   x y  dy, where C is the closed curve
2 
C 
figure shown below (all tick marks represent 1 unit).
This is an example where Green's Theorem may be applied.
  xy 4 
 Q P 
2 3
dx

x
y
dy





C  2 
D  x  y dA
1

1
  2 xy
1  1 x 2
1
3
1
 2 xy dydx  4 

3
1  1 x 2
1

1
2
1
   2 x 3  x 5 dx  0
1
The integrand is odd! see page 420.
4
2
-5
5
-2
-4
-6

xy dydx   x  x 1  x 2  dx
3
#7. ( 5 pts) In HW set 9 you found the volume of an ice-cream cone shaped
solid bounded by the hemisphere z  8  x  y and the cone
2
2
z  x2  y2 .
Determine the Flux through the solid if F  x, y, z   xy i  yz j  x zk .
2
2
2
We ' ll use the divergence theorem.
F  x, y, z   xy 2i  yz 2 j  x 2 zk
divF  y 2  z 2  x 2
The intersection of the two surfaces is:
z  8  x2  y 2 ; z  x2  y 2  x2  y 2  8  x2  y 2
C : x2  y 2  4
now,
 F  dS  divFdV
S
E

 divFdV    x
E
2
 y 2  z 2  dV 
E



4
sin  d  d d 
0 0 0
2 4

0 0
128 2

5

2
 2 sin  d  d d

8

5
5
sin  d  d d 
0
128 2
5
2 4
  sin  d d
0 0
2
128 2  2 
256 2  2 
cos

d




1
d



 1



0
0
5
2
5
2




0
2
4
256 2 256

5
5
256

2 1
5



0 0 0

2 4 8
2 4 8
