294
CHAPTER FIFTEEN /SOLUTIONS
Solutions for Section 15.5
1.
Z 3=4 Z
=4
2.
=2
4.
2
f rdr d
1
p2
Z 2 Z
0
f rdr d
0
Z 3=2 Z
3.
2
0
f rdr d
Z =2 Z 1=2
0
0
y
5.
f rdr d
r =2
r=1
1
2
x
Figure 15.36
y
6.
r=1
R
1
x
,1
Figure 15.37
15.5 SOLUTIONS
y
7.
= =3
r=1
= =6
x
Figure 15.38
y
8.
= 3=4
x
r=4
= 3=2
r=3
Figure 15.39
9.
y
= =4
x
r = 1= cos or r cos = 1
or x = 1
Figure 15.40
295
296
CHAPTER FIFTEEN /SOLUTIONS
y
10.
= =4
r = 2= sin or r sin = 2
or y = 2
x
Figure 15.41
y
11.
= =2
r=4
x
= ,=2
Figure 15.42
12. By using polar coordinates, we get
Z
R
sin(x2 + y2 )dA =
Z 2 Z
0
Z 2
=
0
=
, 12
2
0
sin(r2 )r dr d
2
, 12 cos(r2 ) d
0
Z 2
(cos 4
0
, cos 0) d
, 12 (cos 4 , 1) 2
= (1 , cos 4)
=
15.5 SOLUTIONS
297
13. The region is pictured in Figure 15.43.
y
2
1
1
2
x
Figure 15.43
By using polar coordinates, we get
Z
R
x2 , y2 )dA =
Z =2 Z
2
(
0
1
r2 (cos2 , sin2 )rdr d =
=
=
15
4
15
4
Z =2
2
(cos
0
Z =2
0
Z =2
(cos
0
2
2
, sin2 ) 14 r4 d
1
, sin2 ) d
cos 2 d
=2
15 1
=
sin 2
4 2
0
=0
14.
y
(a)
y = x=3
1
3
x
Figure 15.44
R 1 R 3y
(b) 0 0 f (x; y) dx dy:
(c) For polar coordinates, on the line y = x=3, tan = y=x = 1=3, so = tan,1 (1=3). On the y-axis,
= =2. The quantity r goes from 0 to the line y = 1, or r sin = 1, giving r = 1= sin and
f (x; y) = f (r cos ; r sin ). Thus the integral is
Z =2
tan,1 (1=3)
Z 1= sin 0
f (r cos ; r sin )r dr d:
298
15.
CHAPTER FIFTEEN /SOLUTIONS
By the given limits 0
p
p
x ,1, and , 1 , x2 y 1 , x2, the region of integration is in Figure 15.45.
y
x
,1
Figure 15.45
In polar coordinates, we have
Z
3=2
=2
Z
1
r cos r dr d =
0
Z
3=2
=2
cos Z 3=2
1
cos 3 =2
3=2
1
=
sin 3
=2
=
=
16.
1
1 3 r 3
0
d
, , 1) = , 23
1
( 1
3
From the given limits, the region of integration is in Figure 15.46.
y
2
p
x=y
2
=4
Figure 15.46
p
2
2
x
d
15.5 SOLUTIONS
299
So, in polar coordinates, we have,
Z
=4 Z 2
r
(
0
cos sin )r dr d =
2
Z
0
=4
cos sin 0
Z
=
4
sin(2)
2
0
=4
cos(2)
=4
,
= 0 , (,1)
=
=
17.
2
1 4 r d
4
0
d
0
1:
From the given limits, the region of integration is in Figure 15.47.
y
p
6
y=x
px
6
y = ,x
p
,
6
Figure 15.47
In polar coordinates,
becomes
p
,=4 =4. Also,
Z
0
p6 Z
x
,x
dy dx =
Z
,=4
Z
=
=4
=4
,=4
6 =
p6=cos Z
r cos . Hence, 0 r 2 0
=
p
6= cos . The integral
r dr d
r2 ,=4
p
=
p6=cos !
0
=4
= 3 tan p
x
d =
Z
=4
6
2
2
cos
,=4
d
, (,1)) = 6:
3 (1
Notice that we can check this answer because the integral gives the area of the shaded triangular region which is
1
6 (2 6) = 6.
2
300
CHAPTER FIFTEEN /SOLUTIONS
18. The graph of f (x; y) = 25 , x2 , y2 is an upside down bowl, and the region whose volume we want is
contained between the bowl (above) and the xy-plane (below). We must first find the region in the xy-plane
where f (x; y) is positive. To do that, we set f (x; y) 0 and get x2 + y2 25. The disk x2 + y2 25 is the
region R over which we integrate.
Z
Volume =
R
(25
, x2 , y2 ) dA =
Z 2 Z
(25
0
Z 2 25
=
0
Z
r2 , 14 r4
2
=
625 2
4 0
625
2
=
5
0
, r2 ) rdr d
5
d
0
d
z
19.
2
y
2
R
x
Figure 15.48
First, let’s find where the two surfaces intersect.
p
8 , x2 , y2
8 , x2 , y2
p
x2 + y2
2
= x +y
=
2
x2 + y2 = 4
So z = 2. The volume of the ice cream cone has two parts. The first part (which
is the volume of the cone)
p
2 + y 2 . Hence, this volume is
is the volume
of
the
solid
bounded
by
the
plane
z
=
2
and
the
cone
z
=
x
Z
given by
R
(2
p
, x2 + y2 ) dA, where R is the disk of radius 2 centered at the origin, in the xy-plane. Using
polar coordinates, one has:
Z R
2,
p
x2 + y2 dA =
Z 2 Z
2
(2
0
0
, r) r dr d
15.5 SOLUTIONS
Z 2
"
r2 ,
=
0
Z
4 2
3 0
= 8=3
The second part isZthe volume of the region above the plane z
which is given by
Z
R
R
(
p
(
8,x
2
301
d
0
d
=
p
#
r3 2
3 =
2 but inside the sphere x2 + y2 + z 2
, y , 2) dA where R is the same disk as before. Now
=
8,
2
8 , x2 , y2 , 2) dA =
Z 2 Z
p
2
8 , r2 , 2)rdr d
(
0
Z 2 Z
0
2
r
=
0
Z 2
0
=
0
Z
p
8 , r2 dr d ,
2 !
, 13 (8 , r2)3=2 0
Z 2 Z
0
d ,
1 2 3=2
(4
, 83=2) d ,
=,
3 0
p
1
= , 2 (8 , 16 2) , 8
3
p
2
=
(16 2 , 8) , 8
3 p
8 (4 2 , 5 )
=
3
0
2
Z
Z 2
0
2
2r dr d
0
2
2
r d
0
4 d
p
Thus, the total volume is the sum of the two volumes, which is 32( 2 , 1)=3.
20. The density function is given by
(r) = 10 , 2r
where r is the distance from the center of the disk. So the mass of the disk in grams is
Z
R
(r) dA =
Z 2 Z
(10
0
Z 2 =
0
=
21.
0
5r 2 ,
Z 2
125
0
=
5
3
, 2r)rdr d
2 3
r
3
5
d
0
d
250
(grams)
3
(a)
Z 3=2 Z
Total Population =
=2
4
1
(r; ) rdr d:
(b) We know that (r; ) decreases as r increases, so that eliminates (iii). We also know that (r; ) decreases
as the x-coordinate decreases, but x = r cos . With a fixed r, x is proportional to cos . So as the
x-coordinate decreases, cos decreases and (i) (r; ) = (4 , r)(2 + cos ) best describes this situation.
302
CHAPTER FIFTEEN /SOLUTIONS
(c)
Z 3=2 Z
=2
4
(4
1
, r)(2 + cos ) rdr d =
Z 3=2
=2
(2 +
cos )(2r
Z 3=2
=
9
=
9 2 + sin =
18( , 1)
=2
2
,
4
1 3 r ) d
3 1
d
(2 + cos )
3=2
=2
39
Thus, the population is around 39; 000.
22. A rough graph of the base of the spring is in Figure 15.49, where the coil is roughly of width 0:01 inches.
The volume is equal to the product of the base area and the height. To calculate the area we use polar
coordinates, taking the following integral:
Z 4 Z 0:26+0:04
Area =
0:25+0:04
0
=
=
1
2
1
2
Z 4
(0
0
Z 4
0
rdrd
:26 , 0:04)2 , (0:25 , 0:04)2d
0:01 (0:51 + 0:08)d
=
0:0051 2 +
=
0:0636
4
1
2
(0:0008 )
4
0
Therefore Vol = 0:0636 0:2 = 0:0127in3 .
Figure 15.49