MVC
Sample Quiz Key
Instructions: SHOW ALL WORK !!
We don’t really need a calculator, do we?
#1. Rewrite

NAME:
4  x 2  y 2 dA using polar coordinates, where D is the region in the first quadrant
D
bounded by the graphs of y  0, y  x, and x 2  y 2  4 .


D
4 2
4  x  y dA    4  r 2 r dr d
2
2
0 0
2 x
#2 Calculate
  ( x  y) dy dx , by making the change of variables
x  u  v and y  u  v.
0 0
 ( x, y ) 1 1

 2 .
 (u, v) 1 1
D*
D
y
v
(1,1)
(1,1)
(1,0)
(2,0) u
So,
2 x
1 2v
0 0
0 v
  ( x  y) dy dx   
MVC
1
 2 2 v
u 2 du dv    u
 dv   (4  4v)dv  2
v 
0
0
1
x
#3 Look at the integral
  x
2
 y 2  dV where D is the region above the plane z  0 , inside the
D
sphere x  y  z  4 , and outside the cone z  x 2  y 2 .
2
2
2
a. Write this integral as an iterated integral (or integrals) in cylindrical coordinates.
  x
D
2
 y  dV 
2
2 2 r
  r
0 0 0
2 2
2
 r dz dr d   
0
4 r 2
 
2
r 2  r dz dr d 
0
b. Write this integral as an iterated integral (or integrals) in spherical coordinates.
  x
D
MVC

4 2 2
2
 y 2  dV     ( cos()sin()) 2  ( sin()sin()) 2  2 sin() d  d  d 
0  o
4
1

#4. Convert the integral
1
1 x
y ln(1  x  y ) dy dx to an integral in u and v using the
0
transformation x  u  uv, y  uv .
Solution:
Note that x  u  uv and y  uv  x  u  y or u  x  y.
y
y
.
Then, y  uv  v  =
u x y
Since, x  y  1 , is bottom of rectangle and x  y  2 is the vertex ( x, y )  (1,1) (see figure of D
3 1
below), we conclude 0  u  1 . The line segment y  1,0  x  1 goes to v   u . The line
2 2
1
3
segment x  1,0  y  1 goes to v  u  . So we get D* as shown below:
2
2
v
y
D*
1
1
Finally,
1
2 u
1
( x, y) 1  v u

 (1  v)u  uv  u .
v
u
(u, v)
Therefore:
1

1
x
0
MVC
2
y ln(1  x  y ) dy dx 

1
D
 u  32
u  12
2
uv ln(1  u )u dv du 

1
 u  32
u  12
u 2v ln(1  u ) dv du
x
If the curve C is given by x  t   t , y (t )  t 2 with 0  t  2 and f(x, y) = xy , set up the
#5.
integral

f ds as an integral in terms of t.
C
ds 

 x(t ) 2   y(t ) 2 dt 
f ds 
C

C
#6.
1  (2t ) 2 dt 
2
2
   1  (2t ) dt   t
xyds  (t ) t
0
2
2
3
1  (2t ) 2 dt
0
Find a parameterization for each of the following curves:
a. The line segment from (1, 2, 4) to (  2, 2, 0)
 x  3t  1

 y  2 0  t 1
 z  4t  4

Many other possible correct answers
b. The curve that starts at the point (2, 0, 0) and goes to the point ( 2, 0,3) along the cylinder
x2  y 2  4 .

 x  2 cos t

 y  2sin t 0  t  

3t
 z


MVC
Let C be a curve in the plane starting at (0,0), moving to (1,1) along the curve y = x3, and
then returning to the origin along the straight line y = x.
#7.
a. Parameterize the path (in two pieces, most likely) to express the line integral
2
 2 x y dx  xy dy as an integral or integrals in the single variable t. DON’T integrate.
C
 2x
2
y dx  xy dy 
C
 2x
2
C1
y dx  xy dy 
 2x
2
y dx  xy dy
 C2
1
1
0
0
  2t 2t 3 dt  tt 3 3t 2 dt   2t 2t dt  tt dt
1
  (2t 5  3t 6  2t 3  t 2 ) dt
0
Note, I’m moving from (0,0) to (1,1) along y = x, which is how I picked up the negative
sign on the second integral.
b. Apply Green’s Theorem to the integral in part a to obtain a double integral, making
sure to provide appropriate limits of integration. DON’T integrate.
1 x
 N M 
2
2
x
y
dx

xy
dy

C
0 3  x  y 
x
 N M 
 


x y 
0 x3 
1 x
1 x
    y  2 x 2 dydx
0 x3
MVC
#7. (continued) Let C be a curve in the plane starting at (0,0), moving to (1,1) along the curve
y = x3, and then returning to the origin along the straight line y = x.
c. Given the vector field F( x, y)  2 x 2 y i  xy j , write the integral(s) in the single variable
t you would need to evaluate to find the outward flux of F across the curve C. DON’T
integrate.
  N dx  M dy    xy dx  2 x
C
2
y dy
C

  xy dx  2 x
2
y dy 
C1
  xy dx  2 x
2
y dy
 C2
1
1
0
0
  tt 3 dt  2t 2t 3 3t 2 dt   tt dt  2t 2t dt
1
  (t 4  6t 7  t 2  2t 3 )dt
0
d. Apply the divergence form of Green’s theorem to obtain a double integral that would
calculate this flux. DON’T integrate.
1 x
  N dx  Mdy    div( F )dA
C
0 x3
 M N 


 dA
x y 
0 x3 
1 x
1 x
    4 xy  x dA
0 x3
MVC
#8. Determine whether Green’s theorem can be used to evaluate

y  x dx  y 2 dy where C is the
C
curve given
a. C : x  2cos t, y  2sin t;0  t  2 
NO. M ( x, y)  y  x is not C1 throughout the circular disk. We can see that M is
undefined when x  y , and M is not differentiable whenever x  y .
b. C: x  cos t  1, y  sin t  1 ; 0  t  2
YES. This circular path is oriented so as to keep the interior on the left. The path is
smooth, simple, and closed. Both M ( x, y)  y  x and N ( x, y)  y 2 are C1 throughout the
circular disk and its boundary, since the circle lies above the line y  x.
c. C: the square with vertices (1, 2), (4, 2), (4,5), and (1,5) traversed counter-clockwise.
NO. The line y  x runs through the region bounded by the square, so
M ( x, y)  y  x is not C1 throughout the square region.
MVC