MVC
Chapter 5 Quiz
NAME:
Instructions: Though a calculator is allowed, you must show all steps leading to your answer.
SHOW ALL WORK !!
#1. Evaluate the following integral (show all steps in calculating integral by hand):
1 x

1
e x dy dx   ye x
2
0 0
0
2
x
dx
0
1
  xe x dx
2
0
1  x2 1 
e
2  0 
1
  e  1
2

#2. Rewrite the integral as an integral or sum of integrals with the order of integration reversed.
You need not (should not) evaluate the integral.
l ey

F ( x, y ) dx dy   F ( x, y ) dy dx   F ( x, y ) dy dx
0 y
A
B
l x
e 1
0 0
1 ln x
   F ( x, y ) dy dx  

F ( x, y ) dy dx
B
A
y  e x or x  ln(y)
MVC
#3. Rewrite

4  x 2  y 2 dA using cylindrical coordinates, where D is the region in the first
D
quadrant bounded by the graphs of y  0, y  x, and x 2  y 2  4 .


D
#4. Write
4 2
4  x 2  y 2 dA =  
4  r 2 r dr d
0 0
 f ( x, y, z) dV as an iterated integral in rectangular coordinates, where D is the set in
D
the first octant bounded by the graphs of y  2 x 2 and y  4 z  8.
2
8
1
2 8 2 4 y
 f ( x, y, z ) dV    
D
MVC
0 2 x2
0
f ( x, y, z ) dz dy dx
2 x
#5 Calculate
  ( x  y) dy dx , by making the change of variables
x  u  v and y  u  v.
0 0
(2,2)
(1,1)
Under the transformation:
 ( x, y ) 1 1

2
 (u, v) 1 1
2 x
1 2v
  ( x  y) dy dx  
0 0
 (u  v)  (u  v) 2 du dv
0 v
 4 (using calculator)
MVC
#6. 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
2
 y  dV 
2
D
2 2 r
  r
0 0 0
2 2
2
dz r dr d   
0
4 r 2
 
2
r 2 dz r dr d 
0
b. Write this integral as an iterated integral (or integrals) in spherical or rectangular
coordinates (your choice).
Since,
2
2
x 2  y 2    sin( ) cos( )     sin( )sin( ) 
  2 sin 2 ( )
  x
D
MVC
2
 y 2  dV 

2 2 2
   
0  0
4
2
sin 2 ()  2 sin()d  d  d 
Concepts
1

#7. 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 triangle and x  y  2 is the vertex ( x, y )  (1,1) (see figure of D
1
below), we conclude 0  u  1 . The line segment y  1,0  x  1 goes to v  . The line segment
u
1
x  1,0  y  1 goes to v  1  . So we get D* as shown below:
u
v
y
1
D*
1
1
Finally,
u
2
1
( x, y) 1  v u

 (1  v)u  uv  u .
v
u
(u, v)
Therefore:
1

1
1 x
0
MVC
2
y ln(1  x  y ) dy dx 
D

1
1
u
1 u1
2
uv ln(1  u )u dv du 

1
1
u
1 u1
u 2v ln(1  u ) dv du
x