Assignment 7 - Solutions
Math 209 – Fall 2008
1. (Sec. 15.4, exercise 8.) Use polar coordinates to evaluate the double integral
ZZ
(x + y) dA,
R
where R is the region that lies to the left of the y-axis between the circles x2 + y 2 = 1 and
x2 + y 2 = 4.
Solution: This region R can be described in polar coordinates as the set of all points
(r, θ) with 1 ≤ r ≤ 2 and π/2 ≤ θ ≤ 3π/2. Then
ZZ
Z
3π/2
2
Z
Z
3π/2
Z
(x + y) dA =
R
(r cos θ + r sin θ) r dr dθ =
π/2
1
π/2
! Z
Z 3π/2
2
r2 dr
(cos θ + sin θ) dθ
=
2
r2 (cos θ + sin θ) dr dθ
1
1
π/2
3π/2
= sin θ − cos θ|π/2
2 !
r3 14
= ··· = −
3 1
3
2. (Sec. 15.4, exercise 16.) Use a double integral to find the area of the region enclosed
by the curve r = 4 + 3 cos θ.
Solution: This region, call it D, can be described in polar coordinates as consisting of
all points (r, θ) with 0 ≤ r ≤ 4 + 3 cos θ and 0 ≤ θ ≤ 2π. Then
Z
ZZ
2π
Z
dA =
Area(D) =
D
4+3 cos θ
Z
r dr dθ =
0
0
2π
0
2π
r2
2
r=4+3 cos θ
dθ
r=0
Z
Z
1
1 2π
2
=
(4 + 3 cos θ) dθ =
(16 + 24 cos θ + 9 cos2 θ) dθ
2 0
2 0
Z 1 2π
1 + cos(2θ)
16 + 24 cos θ + 9 ·
=
dθ
2 0
2
2π
1
9
9
41π
=
16θ + 24 sin θ + θ + sin(2θ)
=
.
2
2
4
2
0
3. (Sec. 15.4, exercise 22.) Use polar coordinates to find he volume of the solid inside
the sphere x2 + y 2 + z 2 = 16 and outside the cylinder x2 + y 2 = 4.
Solution: The sphere x2 + y 2 + z 2 = 16 intersects the xy-plane along the circle with
equation x2 + y 2 = 16. Since the solid is symmetric about the xy-plane, we may compute
its total volume as twice the volume of the part that
p lies above the xy-plane, and this
latter is the solid that lies below the graph of z = 16 − x2 − y 2 and above the annular
region D = {(x, y) | 4 ≤ x2 + y 2 ≤ 16}. Hence, changing polar coordinates,
Z 2π Z 4 √
ZZ p
2
2
16 − x − y dA = 2
16 − r2 r dr dθ
Vol = 2
2
0
D
Z 4√
Z 2π
dθ
16 − r2 r dr
=2
2
0
4 !
√
1
2π
− (16 − r2 )3/2 = · · · = 32 3π.
= 2 θ|0
3
2
4. (Sec. 15.4, part of exercise 36.) Use polar coordinates to evaluate the double integral
ZZ
2
2
e−(x +y ) dA,
D(a)
where D(a) = {(x, y) | x2 + y 2 ≤ a2 } is the disk of radius a centered at the origin.
Solution:
ZZ
e
D(a)
−(x2 +y 2 )
Z
2π
Z
dA, =
a
−r2
e
0
0
r dr dθ = 2π
a 1 −r2 2
− e = π(1 − e−a ).
2
0
5. (Sec. 15.5, exercise 2.) Electric charge is distributed over the disk x2 + y 2 ≤ 4 so that
the charge density at (x, y) is σ(x, y) = x + y + x2 + y 2 (in coulombs per square meter).
Find the total charge on the disk.
Solution: Call Q the total charge on the disk. We have
ZZ
ZZ
Q=
σ(x, y) dA =
(x + y + x2 + y 2 ) dA
D
Z 2πDZ 2
Z 2π Z 2
2
=
(r cos θ + r sin θ + r ) r dr dθ =
(r2 (cos θ + sin θ) + r3 ) dr dθ
0
Z0 2π Z0 2
Z 2π Z 2 0
=
r2 (cos θ + sin θ) dr dθ +
r3 dr dθ
0
0Z 2π 0
Z 2
0 Z 2
=
(cos θ + sin θ) dθ
r2 dr + 2π
r3 dr
0
0
! 0
!
2
2
4
r3 + 2π r = · · · = 8π Coulombs.
= sin θ − cos θ|2π
0
3 0
4 0
6. (Sec. 15.5, exercise 10.) Find the mass and center of mass of the lamina that occupies
the region√D bounded by the parabolas y = x2 and x = y 2 and has density function
ρ(x, y) = x.
Solution:√ We may describe D as a region of type I: it is the set of all points (x, y) with
x2 ≤ y ≤ x and 0 ≤ x ≤ 1 (draw a picture!). Then, calling m the mass of the lamina,
we have
Z 1 Z √x
ZZ
Z 1
Z 1
√
√ √
3
2
ρ(x, y) dA =
(x − x5/2 ) dx = .
m=
x dy dx =
x( x − x ) dx =
14
D
0
x2
0
0
Next let’s compute the (first) moments of the lamina:
Z 1
Z
Z 1 Z √x
√
√ 1
1 1 3/2
6
4
y x dy dx =
(x − x9/2 ) dx =
x · (x − x ) dx =
Mx =
2
2 0
55
x2
0
0
and
Z
1
√
Z
Mx =
x
√
Z
1
√ √
x x( x − x2 ) dx =
x x dy dx =
0
x2
0
Z
0
1
1
(x2 − x7/2 ) dx = .
9
14 28
Finally, the center of mass of the lamina is (x, y) = ( Mmy , Mmx ) = ( 27
, 55 ).
7. (Sec. 15.5, exercise 12.) A lamina occupies the part of the disk x2 + y 2 ≤ 1 in the first
quadrant. Find its center of mass if the density at any point is proportional to the square
of its distance from the origin.
Solution: The region of integration (part of the disk x2 + y 2 ≤ 1 in the first quadrant) is
described easily in polar coordinates as the set of all (r, θ) with 0 ≤ r ≤ 1 and 0 ≤ θ ≤ π2 .
p
Also, the density is ρ(x, y) = k( x2 + y 2 )2 = k(x2 + y 2 ) = kr2 , where k is a constant of
proportionality. Then the mass m of the lamina is
Z π/2 Z 1
Z π/2 Z 1
kπ
2
kr r dr dθ =
kr3 dr dθ =
;
m=
8
0
0
0
0
and its first moments are
ZZ
Z
Mx =
y · ρ(x, y) dx dy =
region
π/2
0
ZZ
Z
1
Z
0
π/2
Z
1
x · ρ(x, y) dx dy =
My =
region
0
0
π/2
k
kr sin θ dr dθ =
5
Z
k
kr cos θ dr dθ =
5
Z
4
4
sin θ dθ =
k
,
5
cos θ dθ =
k
.
5
0
0
π/2
8
8
, 5π
).
Finally, the center of mass is (x, y) = ( 5π
8. (Sec. 15.5, exercise 20.) Consider a square fan blade with sides of length 2 and the
lower left corner placed at the origin. If the density of the blade is ρ(x, y) = 1 + 0.1x, is
it more difficult to rotate the blade about the x-axis or about the y-axis?
Solution: We can determine in which direction rotation will be more difficult by comparing the moments of inertia about the x-axis and about the y-axis. Calling D the region
of the xy-plane occupied by the fan blade, we have
Z 2
Z 2
ZZ
Z 2Z 2
2
2
2
Ix =
y ρ(x, y) dA =
y (1 + 0.1x) dy dx =
(1 + 0.1x) dx
y dy
D
0
0
0
0
x=2 !
y=2 !
x2 y 3 8
∼ 5.87,
= x + 0.1 · = (2 + 0.2) ·
2 x=0
3 y=0
3
and similarly
Z 2
Z 2 Z 2Z 2
ZZ
2
2
3
2
x (1 + 0.1x) dy dx =
(x + 0.1x ) dx
dy
x ρ(x, y) dA =
Iy =
0
0
0
0
D
x=2 ! 8
x3
x4 y=2
=
+ 0.1 · y|y=0 =
+ 0.4 · (2) ∼ 6.13.
3
4 x=0
3
Since Iy > Ix , more force is required to rotate the blade about the y-axis; in other words,
it is easier to rotate it about the x-axis.
9. (Sec. 15.6, exercises 4 and 8.) Evaluate the iterated integrals
1
Z
Z
2x
√
y
Z
Z
2xyz dz dy dx
(a)
0
x
0
x
Z
x
Z
(b)
0
0
Solution: For (a) we have
Z
Z 1 Z 2x Z y
2xyz dz dy dx =
π
1
0
xz
x2 sin y dy dz dx
0
2x
Z 1 Z 2x
2 z=y
xyz z=0 dy dx =
xy 3 dy dx
0
x
0
x
Z 1 4 y=2x
Z 1
xy
15 5
5
=
dx =
x dx =
4 y=x
8
0
0 4
0
Z
while for (b) we have
√
Z
π
Z
x
Z
√
xz
Z
2
π
Z
π
Z
x
x sin y dy dz dx =
0
0
0
0
√
Z
0
=
0
√
Z
0
=
0
√
x
x2 − x2 cos(xz) dz dx
0
π
=
Z
2
y=xz
−x cos y y=0 dz dx
2
z=x
x z − x sin(xz) z=0 dx
π
(x3 − x sin(x2 )) dx =
π2
− 1.
4
RRR
10. (Sec. 15.6, exercise 10.) Evaluate the tripple integral
E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, x ≤ z ≤ 2x}.
E
yz cos(x5 ) dV, where
Solution: Set this triple integral as an iterated integral:
z=2x
yz 2
5
cos(x )
yz cos(x ) dz dy dx =
dy dx
yz cos(x ) dV =
2
0
0
0
0
x
E
z=x
y=x
Z Z
Z 1 1 3 2 2
1 1 x 2
5
5
3x y cos(x ) dy dx =
x y cos(x )
=
dx
2 0 0
2 0 2
y=0
1
Z
3 1
3
3 1 4
5
5
x cos(x ) dx =
sin(x ) =
sin(1).
=
4 0
4 5
20
0
Z Z Z
5
Z
1
Z
x
Z
2x
5
Z
1
Z
x