Math 254, Homework #4, Spring 2011 Name _______________________________________________
Instructions: Write your work up neatly and attach to this page. Record your final answers (only)
directly on this page. Use exact values unless specifically asked to round.
1. Evaluate the integral. Sketch or describe the region
1 2
a.
  ( x  y)dydx
0 0
4
b.
x
  2 ye
x
3
2 5

e e4
dydx
1 1
2
2 2 y y
c.
 
3 ydxdy
0 3 y2 6 y

d.
4 cos
 
0
3r 2 sin  drd
1/16
0
3
e.
16
x
2
  1 y
2
dydx
9π/2
0 0
x
4
f.
6 6 x
2
  dydx    dydx
0 0
4 0
6
2 4
g.

x sin xdxdy (It will be necessary to switch the order of integration here.) -
0 y2
8cos(4)+2sin(4)… in radians!
ln10 10
h.
1
  ln y dydx (It will be necessary to switch the order of integration here.) 9
0 ex
2. Set up an integral for both orders of integration and use the more convenient one to evaluate
the integral over the region R.
a.
 sin x sin ydA ; R: rectangle with vertices (-π,0), (π,0),(π,π/2),(-π,π/2)
R


2
  sin x sin ydydx
4, since limits are constants, just switch them or the other
 0
b.
 xe dA ; R: triangle bounded by y  4  x, y  0, x  0
y
R
4 4 x

xe y dydx
(easier)
0 0
4 4 y
or
c.

e 4  13
0 0
 ( x
R
xe y dxdy
2
2
 y )dA ; R: semicircle bounded by y  4  x , y  0
2
2
  r d dr
3
0 0
4π
3. Set up a double integral to find the volume of the solid bounded by the graphs of the equations.
1 y
  xydxdy
z  xy, z  0, y  x, x  1 , first octant
a.
0 0
r 2  y2
r
 
x2  y 2  z 2  R2
b.
2 R
2 r 2  x 2  y 2 dxdy
 r  r 2  y2
 2
1
z
, x  0, x  2, y  0
1 y2
c.
or
1
  1 y
2
  2r
R 2  r 2 drd
0 0
dxdy
0 0
4. Use polar coordinates to evaluate the double integral. For integrals in rectangular coordinates,
you will first need to convert to polar. Sketch the region.

2 3
  re
a.
r2
 1
1  
4  e9 
drd
0 0

2 1 cos
 
b.
0
2
c.
sin( )rdrd
1
6
x 2  y 2 dxdy
4
2
3
0
8 y 2
 
0
y
2 2 8 x2
2 x
d.

x  y dydx 
2
2
0 0
 
2
( x
e. f ( x, y)  e
2
 y 2 )/2
x 2  y 2 dydx (combine into a single integral in polar)
0

 1 
; R : x2  y 2  25, x  0

2
2
2
2
f. f ( x, y)  9  x  y ; R : x  y  9, x  0, y  0
8 2
3
1 

e 2
81
4
25
5. Use polar coordinates to find the volume of the solid bounded by the graphs of the equations.
a. z  xy, x 2  y 2  1, y  0, x  0, z  0
1/8
b.
z  ln  x 2  y 2  , z  0, x 2  y 2  1, x 2  y 2  4
c.
inside z  16  x 2  y 2 , outside x 2  y 2  1
3

 16 ln 4  
2

3
128
2

15 2  8
3
3


6. Find the mass and center of mass of the lamina bounded by the graphs of the equations for the
given density. Some of the integrals may be easier in polar coordinates.
a.
x 2  y 2  a 2 , x  0, y  0,   k ( x 2  y 2 )
b.
xy  4, x  1, x  4,   kx 2
c.
y  cos
x
L
, y  0, x  0, x 
L
,  k
2
 8a 8a 
 , 
 5 5 
 14 4 
 , 
 5 5
 L  
  , 
2  4
d.
 13 
 ,0
6 
r  1  cos  ,   k
7. Find the area of the surface given by f (or z) over the region R. (Some of the integrals are simpler
in polar coordinates).
a.
b.
c.
f ( x, y )  15  2 x  3 y ; R: square with vertices (0,0), (3,0), (0,3), (3,3)
9 14
2 3
4
f ( x, y )  2  y 2 ; R  ( x, y ) : 0  x  2, 0  y  2  x
9 3 1
3
15
3

17 2  1
f ( x, y )  9  x 2  y 2 ; R  ( x, y ) : x 2  y 2  4
6



2
2
d. The portion of the cone z  2 x  y inside the cylinder x 2  y 2  4 .
e.
f ( x, y)  2 x  y ; R: triangle with vertices (0,0), (2,0), (2,2)
f.
f ( x, y)  x  3xy  y ; R: square with vertices (1,1), (-1,1), (-1,-1),(1,-1)
2
3
≈5.6797
3
1 1

1 1
g.
h.
1  (3x 2  3 y)2  (3x  3 y 2 ) 2 dxdy
set up only


f ( x, y )  cos( x 2  y 2 ); R  ( x, y ) : x 2  y 2  
2

x
f ( x, y)  e sin y; R  ( x, y) : 0  x  4,0  y  x
≈15.8206
≈8.118

4 5