Mat 241 Semester Final
Fall, 2008
Name KEY
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. Evaluate
 F  dr
exactly where F  x, y, z  
xyi  e y j  xzk and C is
C
the curve given by: r  t   t , t , t
4
2
3
; 0  t  1.
b
 F  dr   F  r  t    r '  t dt
C
a
r '  t   4t 3 , 2t ,3t 2 ; F  r  t   
 F  dr   
t 6 , et , t 7  t 3 , et , t 7  t 3 , et , t 7 ; t  0
2
1
C
2
dt    4t
1
t2
3
t ,e ,t
7
 4t , 2t ,3t
3
0
2
0
1


1
6
2

1
9
6
0
1
2
4t 7 3t10
4t 7 3t10
4 3 


  2tet dt 

 eu       e  1
0
7
10 0 0
7
10 0
 7 10 
 e
9
70
1
1


 2te  3t dt    4t  3t dt   2tet dt
t2
9
0
2
#2. Find divF and curlF exactly at the point
1, , 2
if
F  x, y, z   xy 2 z 2 , z 2 sin y, x 2e y
divF =
divF  x, y, z   y 2 z 2  z 2 cos( y )
divF 1,  , 2    2  2    2  cos( )  4 2  4
2
2
curlF =
i
j

x
xy 2 z 2

y
2
z sin y
k

  x 2e y  2 z sin y  i   2 xe y  2 xy 2 z  j   0  2 xyz 2  k
z
x 2e y
curlF 1,  , 2    e  i   2e  4 2  j  8  k
 e , 2e  4 2 , 8
#3. Use the Divergence Theorem to calculate the surface integral
exactly where F  x, y, z   x , y , e
3
3
 F  dS
S
z
and the surface is the solid in the
first octant bounded by the cylinder x  y  1 and the planes z = 0 & z = 1.
2
2
divF  3x 2  3 y 2  e z

1 1 x 2 1
 F  dS      3x
S
0
0

2 1 1
2
2 1 1
 3 y  e dzdydx      3r  e rdzdrd      3r 3  re z dzdrd
2
z
2
0
z
0 0 0


0 0 0


1
2
 3r 4 er 2 r 2 
3 e 1
3
3
   3r z  re drd     3r  er  r drd   

  d      d
0
4
2
2 0
4 2 2
0 0
0 0
0
0
2 1
2 1
z 1

5 e
5 e
    d     
4 2
4 2
0
2

2

2
0
 5
e 
     5  2e 
2 4 2 8
#4. Use Stokes’ Theorem to evaluate
 F  dr
where
C
F  x, y, z   sin( x )  xz , x  yz ,cos( z ) and C is the boundary of the
3
4
surfaced defined by z  x 2  y 2 for 0  x  2 and 0  y  1 . C is oriented
counterclockwise as viewed from the positive z – axis.
curlF 
i
j
k

x

y

z
sin  x 3   xz
  0    y   i   0  x  j  1  0  k
x  yz cos  z 4 
 y, x,1
z  g  x, y   x 2  y 2 ;
g
g
 2 x,
 2y
x
y
1 2
 F  dr   curlF  dS   curlF  ndS      y  2 x   x  2 y   1dxdy
C
S
S
1 2
1
0 0
0
0 0
1
    4 xy  1dxdy    2 x 2 y  x  dy    8 y  2  dy   4 y 2  2 y 
 2
2
0
0
1
0
#5. Let v  x, y  
2x
2y
, 2
be a velocity vector field.
2
x  y  1 x  y2  1
2
A. Show that v is a conservative vector field.
2
2
P  x  y  1  0  2 x  2 y 
4 xy


;
2
2
y
 x 2  y 2  1
 x 2  y 2  1
2
2
Q  x  y  1  0  2 y  2 x 
4 xy


;
2
2
2
2
2
2
x
x

y

1
x

y

1




 Since
P Q

, F is conservative.
y x
B. Find a potential function for v .
2x
2y
; fy  2
2
x  y 1
x  y2 1
2x
2y
f1  x, y    2
dx; f 2  x, y    2
dy
2
x  y 1
x  y2 1
fx 
2
u  x2  y 2  1
du  2 xdx
u  x2  y 2  1
du  2 ydy
1
1
f1  x, y    du; f 2  x, y    du
u
u
2
2
f1  x, y   ln x  y  1  ln  x 2  y 2  1 ; f 2  x, y   ln x 2  y 2  1  ln  x 2  y 2  1
 f ( x, y )  ln  x 2  y 2  1
C. Find the work performed by the vector field on a particle that moves
1 
 t  1   t  
along the curve defined as r  t  
 t  sin  3   ,
 sin    t 
2
  2   3  
from t = - 6 seconds to t = 6 seconds. The curve and field are shown
below.
r  6  
r  6 
1 
 6

6

sin


2
 3

  1   6 
sin

6
 ,



  3 2,3 2
 2   3 

1 
 6   1   6
 6  sin  3   ,
 sin 
2

 2   3
6


6 F  r  t    r ' t  dt  ln  3 2
 ln  37   ln  37 
0
 
2
 3 2

2


  6   3 2, 3 2



 1  ln  3 2


 
2
 3 2

2
 1

#6. Consider the interesting shape shown below. It is made by the
intersection of the curves x  1  sin( y ) and x  1  sin( y ) .
Compute the line integral over this closed region given that
F  x, y   x 2  xy  y 2 , x 2  2 xy
and assuming it is traversed
counterclockwise.
Using Green’s Theorem:
Q
P
Q P
 2 x  2 y;
 x  2 y;

 2x  2 y  x  2 y  x
x
y
x y
1 sin y


 Q P 
1 2 1sin y
F

dr


dA

xdxdy

x
dy

R  x y 
0 1sin y
2 0 1sin y
C



1
1
2
2
  1  sin y   1  sin y   dy   4sin ydy  2  sin ydy


20
20
0

 2 cos y 0
 2  1  2(1)
4
#7. Consider the solid egg formed by the intersection of the lower half of
the sphere given by x  y  z  4 and the paraboloid z  4  x
Determine the Flux through this fun solid if the vector field given is:
F  x, y, z   ( y 2  cos( z ))i  (3 y  sin( z )) j  ( z  x 2 ) k .
2
2
2
divF  0  3  1  4
2
4 x2  y 2
4 x2
 F  dS   divFdV   
S
2
4 x2
 
4 x2  y 2
2 2

2  4  x  4  x  y
2
2
  4r  r
4dzdydx  4 
3
 
4r 2
rdzdrd  4   rz 
2
0 0

 r 4  r 2 drd  u  4  r 2 , du  2rdr
0 0
2 2
 4
4dzdydx
2 2
4r 2
0 0  4 r
2
2 2
 4

2  4  x 2  4  x 2  y 2
E
2 4
3
  4r  r drd  2 
0 0
  u dud
0 0
2
2
4
2
2
2

r4 
2 2 u3
32
 4   2 r 2   d  2 
d  16  d 
d
4 0
3
3 0
0 
0
0
0
 32 

160
3
64
3
4 r 2
drd
2
 y2