Math 6C – Chapter 14 Quiz – SOLUTIONS
1. Calculate the line integral,


2 y 
 , over the elliptical path
  x 
F dr , with F ( x, y)  

parametrized by
 cos t 
 from t = 0 to t =  .
 2sin t 
r (t )  
Since
  sin t 
 and
 2 cos t 
r '(t )  
 4sin t 
 , we get:
  cos t 
F (r (t ))  

 4sin t 


  cos t 
0
 F dr  


  sin t 
2
2

 dt  4sin t  2cos t
 2 cos t 
0

dt

 2  2sin t  cos t dt  2 sin 2 t  sin 2 t  cos 2 t dt
0
2
0
2



 2  sin t  1dt  2 sin t dt  2 dt
0
2

0
2
0

1  cos(2t )
 2
dt  2   1  cos(2t )  dt  2
2
0
0
   2  3
2. Calculate the line integral,


 2t 
from t = 0 to t = 1.
2
t 
r (t )  
2
Since r '(t )    and
 2t 

y
 , over the path  parametrized by
 xy 
F dr , with F ( x, y)  
 t2 
F (r (t ))   3  , we get:
 2t 

1
F

1
2
2
4
dr 
  dt  2t  4t
 2t 
0
1
 t3
t5 
2 4 22
  2  4    
5
3 5 15
 3
0
 t2 
 3
 2t 
0

3. Calculate

1
1
dt  2 t dt  4 t dt
2
4
0
0
 F dr with F (x, y)  g(x, y) , where g(x, y)  2x  xy , over the path 
2
t 2 
parametrized by r (t )    from t = 0 to t = 1.
t 3 
Since
F ( x, y)  g ( x, y) , we have:
 F dr  g (r (1))  g (r (0))  g (1,1)  g (0,0)
 1 0  1
4. Suppose that
 2 xy 
with
2 
 x 
g ( x, y)  
g (0,0)  0 .
Then
g ( x, y)  ?
 2 xy   g x 


2  
 x   g y 
g  

g x  2xy  g  2xy dx  x2 y  c( y)
g y  x2  x2  c '( y)  c( y)  C
Then
5. Calculate


g ( x, y)  x2 y  c , so g(0,0)  0  c  0 , and so
g ( x, y)  x2 y .
 2 xy 
, over the path
2 
 x 
F dr with F ( x, y)  g ( x, y)  
1,1 along the circle

from the point
x2  y 2  2 .
 F dr  g (r (1))  g (r (0))  g 
1 0  1

2, 0  g (1,1)


2, 0 to
6.
Suppose that
 y2  2x
 with
 2 xy 
g ( x, y)  
g (0,0)  0 .
Then
g ( x, y)  ?
 y2  2x  g x 
 
 2 xy   g y 
g  
g x  y 2  2x  g 
 y
2
 2x  dx  xy 2  x2  c( y)
g y  2xy  2xy  c '( y)  c( y)  C
Then
7. Calculate


g ( x, y)  xy 2  x2  c , so g(0,0)  0  c  0 , and so
g ( x, y)  xy 2  x2 .
 2  2x
 , over the path
 2 xy 
y
F dr with F ( x, y)  g ( x, y)  

from the point ( 1, 2 ) to

from the point ( 0, 1 ) to
( 3, 1 ) along the line segment connecting these two points.
 F dr  g 3,1  g (1,2)  6  3  9
8. Calculate
(


 y2  2x
 , over the path
 2 xy 
F dr with F ( x, y)  g ( x, y)  
 , –1 ) along the curve
y = cos x.
 F dr  g  , 1  g(0,1)   
 xy 
 M ( x, y)   
9. If F ( x, y)  
   x  and
 N ( x, y)   y 
 
(1,3), then

2
 0    2
is the boundary of the square with vertices (1,1), (3,1), (3,3),
 Mdx  Ndy  ?
By Green’s theorem we have that
 Mdx  Ndy 

 N

dx
R 


M 
 dxdy
dy 
3 3
1

1

  x  dxdy 
  x  dxdy
y
y 

1 1
R 
 Mdx  Ndy  

3 3


1 1
1
dxdy 
y
3
 2ln y  x
10. Suppose
N M

5
x y
3
1 1
1
  x dxdy  
3
2
1
3 3
3

3
3
1
1
 
1
dy dx  dy x dx
y
1
 2ln3  8
1
and region R has an area of 2 and has boundary
 that is a simple closed curve.
 Mdx  Ndy  ?
Then

By Green’s theorem,
 Mdx  Ndy 

11. Suppose
 N

dx
R 

F is a gradient field, and so

M 
 dxdy 
dy 
R 5dxdy  5R dxdy  5 2  10 .
F  g , where g  x, y, z   x2 cos y  sin( z y 2 ) . If
r t   5 cos t, 3 sin t, 2 cos t 
12. If

is the circle
with
0  t  2 , then

is the ellipse parametrized by
 F dr =
 F dr  0 over any closed path such as  .
x2  y 2  9 , then
 y dx  x dy = ?
2
2
 y2  M 
    about the given circle of radius 3.
This integral is the circulation of F ( x, y)  
  x 2   N 
By Green’s theorem,
 N M 
Mdx  Ndy  

2 x  2 y dxdy  2  x  y  dxdy .
 dxdy 
dx
dy


R 
R
R
Changing to poplar coordinates, we get


 


2 3

0
2 3
  r cos  r sin  r drd  
0
2



13. If F ( x, y)  


interior of

0
3
0
cos d r dr 

r cos drd 
2
0 0
2
0
2
2 3
0 0
r 2 sin drd
3
0
sin  d r 2 dr  0

 M 
2     and  is the unit circle centered at the origin. If R is the circular
x2  y 2   N 

 N M 
, then
= ?
[ Hint: this time it’s easier to compute the line integral! ]



 x y 
x2  y 2



dA
R
By Green’s theorem,
 N

dx
R 


M 
 dxdy   Mdx  Ndy .
dy 

We will use the unit-circle parametrization,
cos t 
 , from t = 0 to t =
sin
t


r (t )  
2
1
and F (r (t ))    , for all t, we get
1
 Mdx  Ndy     sin t  cos t  dt  0 .
0
 y2 
 
14. Compute the flux of the field F ( x, y, z )   x 2  through the portion of the surface
 
 z 
 
above (or below) the rectangle in the xy-plane with corners at (0,0), (3,0), (3,2), (0,2).
The flux is given by the integral
described by
  sin t 

 cos t 
2 . Since r '(t )  
S F nˆ d .
Let
z  y 2  x2
G( x, y, z)  x2  y 2  z , so the S is the surface
G( x, y, z)  0 . Then,
d 
G
G  kˆ
dxdy 
F nˆ  F ( x, y, y 2  x 2 )
 2 x, 2 y,1
1
dxdy  4 x2  4 y 2 1 dxdy , and
 2x 


1
2 y  

4 x2  4 y 2 1  1 


 y2 


1
 x2 
4 x2  4 y 2 1  2 2 
y  x 


 2x 


 2 y 
 1 


that is
2 xy 2  2 x2 y  y 2  x2
.
F nˆ 
4 x2  4 y 2 1
Thus,
2 3
S F nˆ d   
0 0
2 3
2 xy 2  2 x2 y  y 2  x2
4 x2  4 y 2 1 dxdy 
4 x2  4 y 2 1
0
2 xy 2  2 x 2 y  y 2  x 2  dxdy


2 3
2
2 3
0
2 3
2 3
0 0 xy dxdy  20 0 x ydxdy  0 0 y dxdy  0 0 x dxdy
2
2
0
2
3
 2 y dy
2
2
2
3
2
2
3
2
3
0 xdx  20 ydy0 x dx  0 y dy 0 dx  0 dy 0 x dx
2
y3
2
3
2
2
0
3
2
0
0
y2
x2
2
2
2
2
2
3
2
0
0
y3
x3

3
3
x3
3 2
3
3
0
23 32
22 33 23
33
2
 3  2  22
3 2
2 3 3
3
 yz 


15. Compute the divergence of the vector field F ( x, y )    xz  .
 xy 



 
 x   yz 

 F ( x, y)     xz   0  0  0 
y
   xy 

 
 z 
0
 yz 


16. Compute the outward flux of F ( x, y )    xz  through the surface x2  2 y 2  3z 2
 xy 


Since the divergence of this field is identically zero, the divergence theorem gives us:
0dV  0 .
 F nˆ d    FdV  
V
S
V
 yz 


17. Compute the curl of the vector field F ( x, y )    xz  .
 xy 


1 .

 

 x   yz  

 F ( x, y)      xz   
y

   xy  
 
  
 z 

y

z

x
,

xy
z
 xz

x
,

xy
y
yz

yz 

 xz 

  x  x,  y  y,  z  z    2 x, 0,  2 z 
 yz 


18. Compute the circulation of F ( x, y )    xz  around the square curve
 xy 


(2,2,1), (0,2,1).
 with corners (0,0,1), (2,0,1),
 F dr  ?
By Stoke’s theorem,
 F dr  S F nˆ d , S being the interior of the given square.
0 
 
parallel to the xy-plane, it follows that n̂  0  , and that d
1 
 
the results from problem #17, the above integral becomes
Since S is
 dxdy . Since z=1 on this square, and using
S F nˆ d  S (2x,0, 2 1) (0,0,1) d
2 2
 2  dxdy  8
0 0
19. Suppose that
xyz
F  g with g ( x, y, z) 
. Then
 F  ?
1 x6  y8  z10
The curl of any gradient field is zero:
 F  0
20. Compute the circulation of the field F ( x, y) featured in #19, around the square curve
(0,0,1), (2,0,1), (2,2,3), (0,2,3).
 F dr  ?
The circulation around any closed curve is zero, for any gradient field:
 F dr  0
 with corners