Fall 2014
Math 265 – Exam 3A – solutions
√
Problem 1 Find the area of the region S inside the circle r = 2 3 and
outside of the circle r = 4 cos θ (in polar coordinates).
Solution.
Both circles are symmetric with respect to the x-axis, so we can find the
area of the upper half of this region and then multiply by 2. If 0 ≤ θ ≤ π/2,
w get bounds for r given θ as
√
4 cos θ ≤ r ≤ 2 3
But this only allows any value for r if
√
4 cos θ ≤ 2 3
√
so cos θ ≤ 3/2, therefore θ ≥ π/6. This value θ = π/6 describes the point
in the upper half-plane where the two circles meet.
√
For π/2 ≤ θ ≤ π, we just get the quarter circle 0 ≤ r ≤ 2 3 with area 3π.
So the desired area is
Z
√
π/2 Z 2 3
A(S) = 2
!
r dr dθ + 3π
π/6
Z
4 cos θ
π/2
=
12 − 16 cos2 θ dθ + 6π
π/6
π/2
Z
12 − (8 + 8 cos 2θ) dθ + 6π
=
π/6
π/2
= [4θ − 4 sin 2θ]π/6 + 6π
√
22π
+ 2 3.
=
3
We used the formula cos2 (θ) = (1 + cos(2θ))/2 to integrate the squared
cosine.
It is also possible to find this area using formulas for the area of a triangle
and sector of a circle, much like Problem 3. The lines for cutting up the
region S appropriately are indicated in the plot above.
As a third option, you could find the area of the intersection of the two
circles and subtract it from the area of the bigger circle (which is 12π).
Solid of problem 2. The red ball indicates the center of mass.
Problem 2 Consider the solid bounded by the coordinate planes, the
surface x = y 2 , and the plane x + z = 4, with density δ(x, y, z) = 2x.
a) Set up, but do NOT evaluate, an integral for the z-coordinate of the
center of mass of the solid in order dz dy dx.
b) Set up, but do NOT evaluate, an integral for the x-coordinate of the
center of mass of the solid in order dy dx dz
Solution.
The inequalities describing the given solid are 0 ≤ x ≤ 4,
√
0 ≤ y ≤ x, and 0 ≤ z ≤ 4 − x. For part b), we need to use the last
inequality for z together with x ≥ 0 to get 0 ≤ z ≤ 4 − x ≤ 4, so 0 ≤ z ≤ 4
for the integral with z on the outside.
Z Z √x Z 4−x
1 4
z̄ =
2xz dz dy dx
m 0 0
0
Z Z
Z √x
1 4 4−z
=
2x2 dy dx dz
m 0 0
0
Problem 3 A function f (x, y) is continuous on a triangle ABC. Call
D, E, F the midpoints of AB, BC, CA, respectively, and M the intersection
of line segments AE, BF , CD. It is given that
ZZZ
ZZZ
f (x, y) dV = 10,
f (x, y) dV = 10,
ABE
ABF
ZZZ
ZZZ
ZZZ
f (x, y) dV = 7,
f (x, y) dV = 3 =
f (x, y) dV.
ABM
ECM
CF M
Determine the integral of f (x, y) over the whole triangle ABC.
Solution. The integral over BEM equals 10 − 7 = 3, same for the triangle
AM F . Together with triangles ECM and CF M , the integral over ABC
equals
ZZZ
f (x, y) dV = 7 + 3 + 3 + 3 + 3 = 19.
ABM
Problem 4 Consider the solid inside the cylinder x2 + y 2 = 9, inside the
sphere x2 + y 2 + z 2 = 18, and above the xy-plane.
a) Set up the triple integral for its volume in cylindrical coordinates dr dθ dz.
b) Set up the triple integral for its volume in sphericalpcoordinates dρ dθ dφ.
c) Find the mass of the solid if the density is δ = 1/ x2 + y 2 .
2
2
2
Solution. a) The√limits on x, y, z translate
√ to r ≤ 9, r + z ≤ 18, and
z ≥ 0. So 0 ≤ z ≤ 18 and 0 ≤ r ≤ min{3, 18 − z 2 }. Then
√
Z
18 Z 2π
Z
V =
√
min{3, 18−z 2 }
r dr dθ dz.
0
0
0
b) In spherical coordinates, this translates to ρ sin φ ≤ 3, 0 ≤ ρ ≤
0 ≤ φ ≤ π/2. So
Z π/2 Z 2π Z min{3/ sin φ,√18}
V =
ρ2 sin φ dρ dθ dφ.
0
0
√
18, and
0
c) If you use spherical coordinates,
we need to break up the integral accord√
ing to whether 3/
sin
φ
or
18
provides
the minimum. The tipping point is
√
when sin φ = 1/ 2, or φ = π/4. This is similar to cylindrical coordinates
in the given order. However, if you switch to dz dr dθ, you get a simpler
integral. We will do the integration over θ on the fly, since there is no θ
in the
root of 18 − r2 is integrated using the substitution
√ integrand. The √
r = 18 sin u, so dr = 18 cos u du.
Z 2π Z 3 Z √18−r2
m =
1 dz dr dθ
0
0
0
π/4
Z
18 cos2 u du = 18π
= 2π
0
Z
π/4
1 + cos(2u) du
0
π/4
= 9π [2u + sin(2u)]0
=
9π(π + 2)
.
2
For comparison, we do the integral in spherical coordinates. The density is
1/r = 1/(ρ sin φ), so our integrand becomes simply ρ.
Z π/2 Z 2π Z min{3/ sin φ,√18}
m =
ρ dρ dθ dφ
0
0
0
√
π/4 Z 18
Z
= 2π
ρ dρ dφ
0
Z
0
π/2 Z 3/ sin φ
+2π
=
=
ρ dρ dφ
π/4
0
√
2
π
18
ρ2 0 +
4
9π 2
2
Z
π/2
π/4
π/2
9π
dφ
sin2 φ
− 9π [cot φ]π/4 =
9π 2
− 9π.
2
Problem 5 Let R be the region defined by (2x − y)2 ≤ 4 and |4x + 5y| ≤ 3
(this is indeed a finite region). Evaluate the double integral
Z Z
4x + 5y dA
R
using the transformation u = 2x − y and v = 4x + 5y.
Solution.
u, v:
Limits are −2 ≤ u ≤ 2, −3 ≤ v ≤ 3. Express x, y in terms of
5u + v
v − 2u
,
y=
.
14
7
This leads to J(u, v) = 1/14 and the integral equals
x=
1
14
Z
2
Z
3
v dv du = 0.
−2
−3