Solutions to Assignment #09 –MATH 2421
Puhalskii/Kawai
Section 11.3
(I) Complete Problem #6 on p. 954.
When = 0; we start on the x-axis. For what positive value of
the origin)? Solve:
cos (2 ) = 0 ) 2 = ) = :
2
4
is r = 0 (and we return to
[There are other values which give a zero cosine, but this is the …rst positive value.]
The area in one-half of the petal is
1
2
The trig. identity is cos2 (2 ) =
1
2
Z
=4
0
Z
=4
(cos (2 ))2 d =
0
16
:
1
+ cos (4 ) :
2
=4
1
1
sin (4 )
+ cos (4 ) d =
+
2
2 2
4
=
0
1
2
8
=
16
We double this to obtain the area in one full leaf, ( =8) sq. units.
(II) Complete Problem #10 on p. 954.
The disk is the circular region
x2 + y 2
16
2
16
r
4:
r
[We always assume that r is nonnegative. We never allow r to be negative.]
Thus, our polar order of integration must be:
r dr d
inner ! outer curve
r: 0 ! 4
: 0!2 :
We convert the integrand to polar also. x2 + y 2 = r2 :
p
p
x2 + y 2 + 1 ) r2 + 1:
The polar double integral represents a volume:
Z 2 Z 4p
2
V =
r2 + 1 r dr d =
3
0
0
1
p
17 17
1 :
:
1
Inner: u = r2 + 1; du = 2r dr ) r dr = du:
2
Z 4p
Z
1
1 2 3=2
1
r2 + 1 r dr )
u1=2 du =
u
= u3=2 :
2
2 3
3
0
h
i
p
1
1
1
3=2 4
r2 + 1
=
173=2 13=2 =
17 17
=
3
3
3
0
p
We think of 173=2 as 171 171=2 = 17 17:
1 :
The Outer integral multiplies the previous result by 2 :X
(III) Complete Problem #11 on p. 954.
For a disk of radius 2, we have:
r dr d
r: 0 ! 2
: 0!2 :
The integrand becomes
e
x2 y 2
)e
r2
:
The polar double integral represents a volume:
Z 2 Z 2
2
V =
e r r dr d =
0
Inner: u =
r2 ; du =
Z
0
2
e
r2
=
e
4
:
1
du:
2
2r dr ) r dr =
r dr )
1
0
1
2
Z
1h
e
2
eu du =
1 u
e
2
i2
1
e
2
r2
0
=
4
e0 =
1
1
2
e
4
:
The Outer integral multiplies the previous result by 2 :X
(IV) Complete Problem #18 on p. 954. Requires integration by parts!
This time, the disk has radius 3.
r dr d
r: 0 ! 3
: 0!2 :
We have r =
p
x2 + y 2 :
Z
V =
0
2
Z
3
cos (r) r dr d = 2 (3 sin (3) + cos (3)
1)
0
Inner: Integration by parts.
Z 3
r cos (r) dr = [r sin (r) + cos (r)]30 = 3 sin (3) + cos (3)
0
= 3 sin (3) + cos (3)
1:
The Outer integral multiplies the previous result by 2 :X
2
(0 + cos (0))
(V) Complete Problem #21 on p. 954. Annulus.
We want the region outside r = 1 and inside r = 2: This gives us 1
r dr d
inner ! outer polar curve
r: 1 ! 2
: 0!2
The integrand is z = f (x; y) =
p
r
2:
x2 + y 2 = r:
The trapped volume is
V =
Z
0
Inner:
2
Z
2
r r dr d =
1
Z
2
r2 dr =
1
1 3
r
3
14
3
2
1
cubic units.
7
= :
3
The Outer integral multiplies the previous result by 2 :X
Here is the solid.
(VI) Complete Problem #22 on p. 954. Circle centered at (0; 1) with radius 1.
If R is inside this circle then we have (in polar)
1)2 = 1
x2 + (y
x2 + y 2
2y + 1 = 1
x2 + y 2 = 2y
r2 = 2r sin ( )
Since we already know that the origin is on this curve, we can divide both sides by r and get
all the other points where r 6= 0:
r = 2 sin ( ) ; 0
3
:
[We showed that this domain traces out the circle exactly once in the Polar Coordinates
Handout!]
r dr d
r: 0 ! 2 sin ( )
: 0! :
p
So we have the right circular cone, z = x2 + y 2 = r and we will trap the volume beneath
this surface above the circle.
Z Z 2 sin( )
32
V =
r r dr d =
cubic units.
9
0
0
Inner:
Z
2 sin( )
r2 dr =
0
Outer:
8
3
Z
1 3
r
3
2 sin( )
0
=
8
sin3 ( ) :
3
sin3 ( ) d =???
0
We break up the sine cubed into
sin2 ( ) sin ( ) = 1
Thus, we have
Z
3
sin ( ) d =
Z
cos2 ( ) sin ( ) :
Z
sin ( ) d
cos2 ( ) sin ( ) d :
The …rst integral is easy. The second requires u = cos ( ) ; du =
du:
Z
Z
u3
2
cos ( ) sin ( ) d )
u2 ( du) =
3
0
=
Thus, we have
8
3
Z
1
cos3 ( )
3
sin3 ( ) d =
0
8
3
2
0
=
2
3
1
( 1)3
3
=
32
:X
9
(VII) Complete Problem #34 on p. 954. Sketch the region R carefully!
This one is horizontal …rst.
dx dy
lef t ! right curve
p
x: y ! 2 y 2
y: 0 ! 1:
4
sin ( ) d ) sin ( ) d =
2
13 = :
3
The left curve is the line y = xpand the right curve is the
circle, x2 + y 2 = 2 radius = 2 :
This gives us:
r dr d p
r: 0 ! 2
: 0 ! =4:
Why = =4? Recall that the slope of the line is equal
to the tangent of the angle created with the x-axis.
Since m = 1; we have tan 1 (1) = =4:
r dr d
r: 0 !
p
2
: 0 ! =4:
The integrand is certainly polar friendly...
1
p
x2
So the volume is
Z
V =
0
=4 Z
p
2
0
1
r
r dr d =
+
Z
0
1
! :
r
y2
=4 Z
p
2
1 dr d =
p
2
0
We note that the surface
4
cubic units.
1
z = f (x; y) = p
x2
+ y2
approaches +1 as (x; y) ! (0; 0) : So this technically should fall in the category of improper
integrals. We see, however, that the transformed integral gives us a nice, …nite volume!
(VIII) Complete Problem #42 on p. 954. This is a cardioid.
This lamina is symmetric with respect to the x-axis and the density function
(x; y) = x2 + y 2
is also symmetric with respect to the x-axis (substitute ( y) for y). So we know that y =
0 (because Mx will be equal to zero).
For this polar region R; we have:
r dr d
r: 0 ! 2
2 cos ( )
: 0!2 :
The density function is
The mass integral is
(x; y) = x2 + y 2 ! r2 :
m=
Z
0
2
Z
2 2 cos( )
0
5
r2 r dr d = 35 :
Inner:
Z
2 2 cos( )
r3 dr =
0
1 4
r
4
2 2 cos( )
0
=
1
(2
4
2 cos ( ))4 :
The Outer integral is too terrible to evaluate by hand.
Z
1 2
(2 2 cos ( ))4 d = 35 :
4 0
Now we need the …rst moment with respect to the y-axis.
My =
ZZ
R
x
(x; y) dA =
Z
0
2
Z
2 2 cos( )
r cos ( ) r2 r dr d =
84 :
0
From the diagram, it makes sense that the x-coordinate of the centroid is located left of the
origin.
My
( 84 )
12
x=
=
=
:
m
35
5
6