Ma 227 Homework Solutions Fall 2012
Due 9/7/2012
12.4-pg. 857 #3, 5, 7, 9, 11, 15, 21, 23, 25, 29
3.
R
The region R is more easily described by the rectangular coordinates:
x, y| − 1 ≤ x ≤ 1, 0 ≤ y ≤ 12 x  12
1
1
Thus   fx, ydA    2
x 12
fx, ydydx
−1 0
R
2 7
   4 rdrd represents the area of the region R 
5.
the lower half of a ring as shown below.
r4
-7
-6
-5
-4
-3
-2
-1
0
y
1
2
r, | 4 ≤ r ≤ 7,  ≤  ≤ 2 ,
3
4
5
6
7
x
-2
-4
-6
2
7
2
   4 rdrd   
r2
2
7
4
d   1 49 − 16  33
2
2
 D xydA, where D is the disk with center at the origin and radius 3.
7.
D can be described as D  r, | 0 ≤ r ≤ 3, 0 ≤  ≤ 2

2
3
2
r4
4
2
3
 D xydA   0  0 r cos r sin rdrd   0  0 r 3 cos  sin drd

0
3
cos  sin drd  81
4
0
1 sin 2 
2
2
0
0
9.   cosx 2  y 2 dA where R is the region that lies above the x-axis within the circle
R
x2  y2  9
1
D can be described as D 
11.  e −x
2 −y 2
dA.
r, | ≤ r ≤, 0 ≤  ≤
The region of integration is shown below.
D
4 − y2
y
2
1
0
0.5
1.0
1.5
2.0
x
-1
-2
Switching to polar coordinates we have

2
2

2
 e −x −y dA   −  0 e −r rdrd   −
2
2
2

2
D


2
−
− 1 e −r
2

2
2
2
0
d


2
− 1 e −4 − e 0  d  − 1 e −4 − e 0   −2 
2
2
2
  1 − e 4 
2
Find the volume under the cone z 
15.
V  
2 2
x 2  y 2 dA   
x 2 y 2 ≤4
0
0
x 2  y 2 and above the disk x 2  y 2 ≤ 4.
2
2
0
0
r 2 rdrd   d  r 2 dr   2
0
1
3
r3
2
0
 2
8
3

21.
The cone z  x 2  y 2 intersects the sphere x 2  y 2  z 2  1 when
x 2  y 2   x 2  y 2  2  1 or x 2  y 2  12 . So:
V  
1
2
2
1 − x 2 − y 2 − x 2  y 2 dA   
x 2 y 2 ≤ 12
1
2
2
  d 
0
  − 13
0
0
0
3
 1 − r 2 − rrdrd
r 1 − r 2 − r 2 dr   2
− 13 1 − r 2  2 −
0
1
2
−1


3
1
3
r3
1
2
0
2− 2
2
16
3
23.
Use polar coordinates to find the volume inside both the cylinder x 2  y 2  4 and the
ellipsoid 4x 2  4y 2  z 2  64.
SOLUTION
The ellipsoid intersects the x, y-plane in the circle x 2  y 2  16. Thus, our region is
bounded by the circle x 2  y 2  4. So , in polar coordinates we have the equation r  2. Next,
we can solve the equation of the ellipsoid 4x 2  4y 2  z 2  64 for z, i.e.,
z  2 −x 2 − y 2  16 which can be rewritten in polar coordinates as z  2 16 − r 2 . The
volume of the solid can now be written as:
2 2
2   2 16 − r 2 rdrd  −64 3   512 
3
0
0
25.
Use a double integral to find the area of one loop of the rose r  cos 3.
r  cos 3
y
0.8
0.6
0.4
0.2
-0.4
-0.2
-0.2
0.2
0.4
0.6
0.8
1.0
x
-0.4
-0.6
-0.8
3
r  0  cos 3  0 or 3  2 or   6 . Thus for the loop in the first and fourth quadrants
we have − 6 ≤  ≤ 6 . Using symmetry we get
Area  2 

6
0
cos 3
0
rdrd  
12
29. Evaluate
1
0 y
2−y 2
x  ydxdy
by converting to polar coordinates.
The region of integration is the part of the circle of radius 2 centered at the origin in the
first quadrant below the line y  x.
2 − y2
y 1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.2

4
2
0.4
0.6
0.8
1.0
1.2
1.4
x
Thus
1
0 y
2−x 2
x  ydydx 

0 0
r cos   r sin rdrd 

2 2
2 2
sin  − cos  04 
3
3

2
0
r3
3
2
0
cos   sin d
2
2
−
−01
2
2

2 2
3
12.7-pg. 880 #5, 9, 11, 17, 19
5.
4
3
1
0 0 0
1−z 2
ze y dxdzdy 

3
1
3
1 − z 2  2
−
3
0 0
0
 1
3
9.
1 − z 2 ze y dzdy
3
1
e y dy
0
3
 0 e y dy 
1 e 3 − 1
3
Evaluate    2xdV where E  x, y, z| 0 ≤ z ≤ y, 0 ≤ y ≤ 2, 0 ≤ x ≤
E
4 − y 2 .
  E 2xdV
2
 
0 0
4−y 2
y
2
 0 2xdzdxdy   0  0
4−y 2
2
2xydxdy   4 − y 2 ydy  2y 2 −
0
1
4
2
y4 0  4
11.
Evaluate    6xydV where E lies under the plane z  1  x  y and above the
E
region in the xy − plane bounded by the curves y  x , y  0, and x  1
y 2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
x
Here E  x, y, z|0 ≤ x ≤ 1, 0 ≤ y ≤
1
x
1xy
  E 6xydV   0  0  0
x , 0 ≤ z ≤ 1  x  y, so
1
x
1
x
6xydzdydx    6xyz z1xy
dydx    6xyx  y  1dydx
z0
0 0
0 0
5
1
1
x
  3xy 2  3x 2 y 2  2xy 3  y
dx   3x 2  3x 3  2x 2 dx  x 3 
y0
0
0
5
3
4
x4 
4
7
7
x2
0
1

65
28
  E xdV, where E is bounded by the parabaloid x  4y 2  4z 2 and the plane x  4
17.
The projection on the yz − plane is the disk y 2  z 2 ≤ 1. Using polar coordinates y  r cos 
and z  r sin  we get:
4
  E xdV   D  4y 2 4z 2 xdx dA 
2
1
 8  d  r − r 5 dr  82
0
0
1
2
1
2
2 1
 D 4 2 − 4y 2  4z 2  2 dA  8  0  0 1 − r 4 rdrd
r2 −
1
6
r6
1
0

16
3

19.
Find the volume of the tetrahedron bounded by the coordinate planes and the plane
2x  y  z  4.
The plane 2x  y  z  4 intersects the xy-plane in the line y  4 − 2x,
6
y4
3
2
1
0
0
1
2
x
So E  x, y, z|0 ≤ x ≤ 2, 0 ≤ y ≤ 4 − x, 0 ≤ z ≤ 4 − 2x − y
2 4−2x 4−2x−y

V 
dzdydx  16
3
0 0
0
7