Math 20C Homework 10 Solutions
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Problem 1 (15.4.1). Sketch the region D indicated and integrate f (x, y) over D using polar
coordinates.
p
f (x, y) = x2 + y 2 ;
x2 + y 2 ≤ 2
√
Solution. The region D indicated is the disk of radius 2 centered at the origin:
y
D
x
√
2
In polar coordinates D is given by
√
0 ≤ θ ≤ 2π.
0 ≤ r ≤ 2,
p
Moreover, the integrand f (x, y) = x2 + y 2 becomes
q
q
√
f (r, θ) = (r cos θ)2 + (r sin θ)2 = r2 cos2 θ + sin2 θ = r2 = r
after changing to polar coordinates. Therefore our integral is
ZZ
Z 2π Z √2
Z 2π Z √2
f (x, y) dA =
f (r, θ) r dr dθ =
r2 dr dθ
D
0
0
0
! 0
√ !
Z 2π
Z √2
3 2
r
=
1 dθ
r2 dr = 2π
3
0
0
0
√
4 2π
=
.
3
1
Problem 2 (15.4.3). Sketch the region D indicated and integrate f (x, y) over D using polar
coordinates.
f (x, y) = xy;
x ≥ 0, y ≥ 0, x2 + y 2 ≤ 4
Solution. The region D indicated is the portion of the disk of radius 2 centered at the origin
(x2 + y 2 ≤ 4) restricted to the first quadrant (x ≥ 0 and y ≥ 0):
y
D
2
x
In polar coordinates D is given by
0 ≤ r ≤ 2,
π
.
2
0≤θ≤
Moreover, the integrand f (x, y) = xy becomes
r2 sin(2θ)
2
f (r, θ) = (r cos θ)(r sin θ) = r2 cos θ sin θ =
after changing to polar coordinates and using the trig identity
sin(2θ) = 2 cos θ sin θ.
Therefore our integral is
ZZ
Z
π/2
Z
2
f (x, y) dA =
D
0
=
1
2
Z
0
π/2
Z
f (r, θ) r dr dθ =
! Z
sin(2θ) dθ
0
Z
0
2
r3 dr
0
π/2 !
2 !
r4 1
cos(2θ) =
−
2
2 0
4 0
1
cos π cos 0 24
=
−
+
2
2
2
4
= 2.
2
π/2
0
2
r3 sin(2θ)
dr dθ
2
Problem 3 (15.4.7). Sketch the region of integration and evaluate by changing to polar
coordinates.
Z 2 Z √4−x2
x2 + y 2 dy dx
−2
0
Solution. From the outer integral, we can tell that the x values of the region of integration
range from −2 to 2. From the inner integral, we can√tell that for a given value of x (with
−2 ≤√ x ≤ 2), the y-values range between 0 and 4 − x2 . As x ranges from −2 to 2,
y = 4 − x2 traces out the upper half of the circle of radius of 2 centered at the origin.
Therefore the region of integration looks like the following:
y
2
x
Thus we see that the region is described in polar coordinates as
0 ≤ r ≤ 2,
0 ≤ θ ≤ π,
and so our integral is
Z
2
−2
√
Z
4−x2
r2
z
Z
}| {
2
2
x + y dy dx =
0
π
2
Z
0
r2 r dr dθ
0
π
Z
=
Z
1 dθ
0
2
3
r dr
0
2
r4 = π = 4π.
4 0
Problem 4 (15.4.9). Sketch the region of integration and evaluate by changing to polar
coordinates.
Z 1/2 Z √1−x2
x dy dx
√
0
3x
Solution. From the outer integral, we can tell that the x values of the region of integration
range from 0 to 1/2. From the inner integral, √
we can tell that for a given
√ value of x (with
0 ≤ x ≤ 1/2),
√ the y-values range between y = 3x (the line with slope 3 and y-intercept
0) and y = 1 − x2 (the upper half of the circle of radius 1 centered at the origin). Therefore
the region of integration looks like the following:
3
y
y
√
√
3/2
=⇒
1
y
=⇒
1
√
y = 3x
3/2
1
π/3
1/2
x
x
1/2
1/2
x
Thus we see that the region is described in polar coordinates as
0 ≤ r ≤ 1,
π/3 ≤ θ ≤ π/2,
and so our integral is
Z
0
1/2
√
Z
√
1−x2
Z
π/2
1
Z
x dy dx =
3x
π/3
Z
0
π/2
r cos θ r dr dθ
! Z
1
cos θ dθ
=
0
π/3
=
r2 dr
π/2 = sin θ
π/3
√ !
3 1
.
1−
2
3
1 !
r3 3 0
Problem 5 (15.4.17). Calculate the integral over the given region by changing to polar
coordinates.
f (x, y) = |xy|;
x2 + y 2 ≤ 1.
Solution. The region D over which we’re integrating is the unit disk centered at the origin.
In polar coordinates, D is given by
0 ≤ r ≤ 1,
0 ≤ θ ≤ 2π.
Moreover, our integrand f (x, y) = |xy| is
r2
|sin(2θ)|
f (r, θ) = |(r cos θ) (r sin θ)| = r |cos θ sin θ| =
2
2
after changing to polar coordinates and using the trig identity
sin(2θ) = 2 cos θ sin θ.
We can determine the sign of sin 2θ = 2 cos θ sin θ, and hence the value of |sin(2θ)|, in each
of the four quadrants in the plane:
4
y
π/2 ≤ θ ≤ π
cos θ ≤ 0
sin θ ≥ 0
cos θ sin θ ≤ 0
0 ≤ θ ≤ π/2
cos θ ≥ 0
sin θ ≥ 0
cos θ sin θ ≥ 0
x
3π/2 ≤ θ ≤ 2π
cos θ ≥ 0
sin θ ≤ 0
cos θ sin θ ≤ 0
π ≤ θ ≤ 3π/2
cos θ ≤ 0
sin θ ≤ 0
cos θ sin θ ≥ 0

sin(2θ),



− sin(2θ),
=⇒ |sin(2θ)| =

sin(2θ),



− sin(2θ),
if
if
if
if
0 ≤ θ ≤ π/2
π/2 ≤ θ ≤ π
π ≤ θ ≤ 3π/2
3π/2 ≤ θ ≤ 2π
Finally, our integral is
ZZ
Z 2π Z 1
Z 2π Z 1 3
r
|sin(2θ)| dr dθ
f (x, y) dA =
f (r, θ) r dr dθ =
D
0
0
0
0 2
Z 2π
Z 1
1
3
=
|sin(2θ)| dθ
r dr
2
0
0
!
1 ! Z π/2
Z 2π
Z 3π/2
Z π
1 r4 sin(2θ) dθ
sin(2θ) dθ −
=
sin(2θ) dθ +
sin(2θ) dθ −
2 4 0
3π/2
π
π/2
0

π/2
π
3π/2
2π 
1  cos(2θ) cos(2θ) cos(2θ) cos(2θ) 
=
−
+
+
−
8
2 2 2 2 0
π/2
π
3π/2
1
(− cos π + cos 0 + cos(2π) − cos π − cos(3π) + cos(2π) + cos(4π) − cos(3π))
16
1
1
=
(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) = .
16
2
=
Problem 6 (15.4.21). Find the volume of the wedge-shaped region contained in the cylinder
x2 + y 2 = 9, bounded above by the plane z = x and below by the xy-plane.
Solution. See your text for the picture of the region D. From the picture (and the description
of the region) you can conclude that D is given in cylindrical coordinates as
0 ≤ r ≤ 3,
−π/2 ≤ θ ≤ π/2,
Therefore, the volume of D is
ZZZ
Z 3 Z π/2 Z
1 dV =
D
−π/2
0
Z
=
Z
Z
3
Z
r dz dθ dr =
0
!
−π/2
r cos θ dθ
Z
dr =
−π/2
0
3 ! π/2 r3 sin θ
= 18.
3 0
−π/2
5
π/2
r
0
π/2
r
0
=
3
r cos θ
0 ≤ z ≤ x = r cos θ.
3
r2 dr
!
r cos θ
z
dθ dr
0
Z
!
π/2
cos θ dθ
−π/2
Problem 7 (15.4.23). Evaluate
RR p
x2 + y 2 dA where D is the region
D
y
2
x
Solution. First, we consider the inner circle. The inner circle Dinn is centered at (1, 0) and
has radius 1. Therefore, in rectangular coordinates it is given by the equation
(x − 1)2 + y 2 = 1.
Changing to polar coordinates, we have
(r cos θ − 1)2 + r2 sin2 θ = 1
=⇒ r2 cos2 θ − 2r cos θ + 1 + r2 sin2 θ = 1
=⇒ r2 − 2r cos θ = 0
=⇒ r = 2 cos θ,
and π ranges between −π/2 and
p π/2 (since these are the values of θ that make r = 0). Thus
the integral of the integrand x2 + y 2 = r over the inner circle Dinn is given by
!
Z π/2
Z π/2 Z 2 cos θ
ZZ
3 2 cos θ
p
r
r r dr dθ =
x2 + y 2 dA =
dθ
3 0
Dinn
−π/2 0
−π/2
Z
Z
8 π/2
8 π/2
3
=
cos θ dθ =
(1 − sin2 θ) cos θ dθ.
3 −π/2
3 −π/2
Now let u = sin θ, so that du = cos θ dθ. Also, when θ = −π/2, we have u = sin(−π/2) = −1,
and when θ = π/2, we have u = sin(π/2) = 1. Therefore our integral over the inner circle
Dinn becomes
1
ZZ
Z
p
8 1
8
u3 8
2
32
2
2
2
x + y dA =
(1 − u ) du =
u−
=
2−
= .
3 −1
3
3 −1 3
3
9
Dinn
This finishes our consideration of the inner circle.
Next we consider the outer circle Dout . It is given in polar coordinates as
0 ≤ r ≤ 2,
0 ≤ θ ≤ 2π.
6
p
x2 + y 2 = r over Dout is
Z 2π
Z
Z 2π Z 2
p
2
2
x + y dA =
r r dr dθ =
1 dθ
Therefore the integral of the integrand
ZZ
Dout
0
0
0
2
2
r dr
=
0
16π
.
3
This finishes our consideration
RR pof the outer circle.
We can now compute D x2 + y 2 dA and finish the problem. Since D is obtained by
removing Dinn from Dout , it follows that
ZZ p
ZZ
ZZ
p
p
16π 32
− .
x2 + y 2 dA =
x2 + y 2 dA −
x2 + y 2 dA =
3
9
D
Dout
Dinn
RRR
Problem 8 (15.4.27). Use cylindrical coordinates to calculate
f (x, y, z) dV for the
W
given function and region.
f (x, y, z) = x2 + y 2 ;
x2 + y 2 ≤ 9,
0≤z≤5
Solution. In three dimensions, the inequality x2 + y 2 ≤ 9 is the filled-in cylinder of radius
3 centered along the z-axis, and in cylindrical coordinates it is given by the inequalities
0 ≤ r ≤ 3 and 0 ≤ θ ≤ 2π. Moreover, the inequality 0 ≤ z ≤ 5 gives us the portion of this
cylinder that we need to consider. The integrand f (x, y, z) = x2 + y 2 is given in cylindrical
coordinates as f (r, θ, z) = r2 , and so our integral is
ZZZ
Z 5 Z 2π Z 3
f (x, y, z) dV =
r2 r dr dθ dz
W
0
Z 3
Z 2π
0Z 50 34
405π
3
r dr = 5 · 2π ·
1 dθ
1 dz
=
=
.
4
2
0
0
0
RRR
Problem 9 (15.4.29). Use cylindrical coordinates to calculate
f (x, y, z) dV for the
W
given function and region.
x2 + y 2 ≤ 1,
f (x, y, z) = y;
x ≥ 0,
y ≥ 0,
0≤z≤2
Solution. The region W indicated here is the cylinder of radius 1 centered along the z-axis,
restricted to the first octant, and with maximum height 2. In polar coordinates, it is given
by
0 ≤ r ≤ 1,
0 ≤ θ ≤ π/2,
0 ≤ z ≤ 2.
The integrand f (x, y, z) = y is given in cylindrical coordinates as f (r, θ, z) = r sin θ, and so
our integral is
ZZZ
Z
2
Z
π/2
1
Z
f (x, y, z) dV =
W
r sin θ r dr dθ dz
! Z
Z 2
Z π/2
=
1 dz
sin θ dθ
0
0
0
0
0
7
0
1
2
r dr
2
= .
3
Problem 10 (15.4.31). Use cylindrical coordinates to calculate
given function and region.
RRR
W
f (x, y, z) dV for the
x2 + y 2 ≤ z ≤ 9.
f (x, y, z) = z;
Solution. From the description of the region, we can deduce two inequalities:
x2 + y 2 ≤ 9,
x2 + y 2 ≤ z ≤ 9.
The first inequality is the cylinder of radius 3 centered along the z-axis; in polar coordinates
it is 0 ≤ r ≤ 3 and 0 ≤ θ ≤ 2π. The second inequality means that the values of z that we
want to integrate over are above the paraboloid z = x2 + y 2 and below the plane z = 9; in
polar coordinates this means that r2 ≤ z ≤ 9. Therefore we can already integrate:
9 ! !
Z 2π
Z 3
ZZZ
Z 2π Z 3 Z 9
z 2 dr
f (x, y, z) dV =
z r dz dr dθ =
1 dθ
r
2 r 2
W
0
0
r2
0
0
Z 3 Z 3
81 r4
= 2π
r
−
dr = π
81r − r5 dr = 243π.
2
2
0
0
8