Fall 2014
Math 265 - Practice Exam 3, Integrals - solutions
Problem 1 Find the area of the region S in the first quadrant enclosed by
the curve r = 3 cos 2θ (in polar coordinates).
Solution. The words ’first quadrant’ tell us 0 ≤ θ ≤ π/2. Since 0 ≤ cos 2θ,
we have to narrow down the limits to 0 ≤ θ ≤ π/4. Limits for r are
0 ≤ r ≤ 3 cos 2θ. So
Z π/4 Z 3 cos 2θ
A(S) =
r dr dθ
0
=
=
9
2
Z
9
4
Z
0
π/4
cos2 (2θ) dθ
(sub 2θ = α)
0
π/2
cos2 (α) dα =
0
9π
.
16
The last step used the formula cos2 (α) = (1 + cos(2α))/2.
Problem 2 a) Find the volume of the solid in the first octant bounded by
x + 2y + 3z = 6.
b) Find the mass of the solid from part a) if the density is given by δ(x, y, z) =
y.
c) Find the x-coordinate of the center of mass of the solid above.
Solution.
a) Limits for dz dy dx are 0 ≤ x ≤ 6, 0 ≤ y ≤ (6 − x)/2,
0 ≤ z ≤ (6 − x − 2y)/3. The same limits are used for a), b), and c).
Z
V
6 Z (6−x)/2 Z (6−x−2y)/3
=
dz dy dx
0
=
=
=
=
0
0
6 Z (6−x)/2
Z
1
6 − x − 2y dy dx
3 0 0
Z 6
1
(6−x)/2
[(6 − x)y − y 2 ]0
dx
3 0
Z 6
1
(6 − x)2 dx
12 0
1
[−(6 − x)3 ]60 = 6.
36
Note in particular that I never expanded (6 − x)2 . For the mass, the same
trick will produce even bigger savings.
Z
6 Z (6−x)/2 Z (6−x−2y)/3
m =
y dz dy dx
0
=
=
=
=
0
0
6 Z (6−x)/2
Z
1
y(6 − x − 2y) dy dx
3 0 0
Z
1 6
(6−x)/2
[(6 − x)y 2 /2 − 2y 3 /3]0
dx
3 0
Z 6
1
(6 − x)3 dx
72 0
9
1
[−(6 − x)4 ]60 = .
288
2
... and even for the center of mass, we can do without expanding if we use
integration by parts.
Z
6 Z (6−x)/2 Z (6−x−2y)/3
Myz =
xy dz dy dx
0
=
=
=
=
0
0
6 Z (6−x)/2
Z
1
xy(6 − x − 2y) dy dx
3 0 0
Z
1 6
(6−x)/2
x[(6 − x)y 2 /2 − 2y 3 /3]0
dx
3 0
Z 6
1
x(6 − x)3 dx
72 0
Z
1
1 6
63
27
4
6
4
[−x(6 − x) /4]0 +
(6 − x) dx =
= .
72
4 0
40
5
So, finally, x̄ = Myz /m = 56 .
Problem 3 A function f (x, y) is integrated over the square S with vertices
A = (0, 0), B = (2, 0), C = (2, 2), and D = (0, 2). Denote a triangle by its
three vertices. With M being the point (1, 1), it is given that
ZZZ
ZZZ
f (x, y) dV = 10,
f (x, y) dV = 4,
S
ABC
ZZZ
ZZZ
f (x, y) dV = 7
f (x, y) dV = 3.
BCD
BCM
Determine the integral of f (x, y) over the triangle AM D.
Solution. The integral over ABM equals 4 − 3 = 1 (make a sketch! – take
away triangle BCM from ABC). The triangle AM D is the whole square
with triangles BCD and ABM taken away, so the integral over AM D equals
10 − 7 − 1 = 2.
Problem 4 Rewrite the integral below with the order of integration dz dy dx.
2/3 Z 1−3y/2 Z 2−3y−2z
Z
f (x, y, z) dx dz dy
I=
0
0
0
Solution.
Z
2 Z (2−x)/3 Z (2−3y−x)/2
I=
f (x, y, z) dz dy dx.
0
0
0
Problem 5 Consider the solid S bounded by the coordinate planes and by
the planes z = 3 − y, z = 2, and 4 − x = z.
a) Find inequalities for x, y, z describing S (for each of the given equations,
you have to pick ≤ or ≥).
b) Set up an integral for the volume of S in order dz dy dx.
c) Set up an integral for the volume of S in order dx dy dz.
d) Find the volume of S using the integral of your choice (one is easier than
the other one!).
Solution. a) 0 ≤ x, 0 ≤ y, 0 ≤ z ≤ 2, z ≤ 3 − y, 4 − x ≥ z. The last two
inequalities have to be this way because otherwise, y and x could grow to
infinity (the solid would not be bounded, and the problem would not make
sense).
b) Biggest possible value for x is 4. Given x, 0 ≤ y ≤ 3. Given x, y,
0 ≤ z ≤ min{2, 3 − y, 4 − x}.
So the volume is
Z
4 Z 3 Z min{2,3−y,4−x}
Vol(S) =
1 dz dy dx.
0
0
0
Horrible! has to be split twice if we want to evaluate it.
c) First, 0 ≤ z ≤ 2. Given z, 0 ≤ x ≤ 4 − z and 0 ≤ y ≤ 3 − z. So
Z 2 Z 3−z Z 4−z
Vol(S) =
1 dx dy dz.
0
0
0
d) The setup in c) is way easier! so we use that,
Z 2 Z 3−z
4 − z dx dy dz
Vol(S) =
0
0
Z
Z 2
(3 − z)(4 − z) dy dz =
=
2
12 − 7z + z 2 dz
0
0
2
7z 2 z 3
38
= 12z −
+
= .
2
3 0
3
Problem 6 A solid is defined by x2 + y 2 + z 2 ≤ 4 and 0 ≤ x ≤ y.
a) Set up the triple integral for its volume in cylindrical coordinates (r, θ, z).
b) Set up the triple integral for its volume in spherical coordinates (ρ, φ, θ).
c) Find the mass of the solid if the density is δ = 1/ρ.
Solution. a) The limits on x, y, z translate to r2 + z 2 ≤ 4, 0 ≤ cos θ ≤ sin θ
so π/4 ≤ θ ≤ π/2. The volume is then
Z
2 Z π/2 Z
V =
0
√
4−r2
√
− 4−r2
π/4
r dz dθ dr.
b) In spherical coordinates, we get ρ ≤ 2, and again π/4 ≤ θ ≤ π/2, no
restriction on φ. So
Z 2 Z π/2 Z π
V =
ρ2 sin φ dφ dθ dρ.
0
π/4
0
c) Using spherical coordinates since the density is using them, the mass is
Z 2 Z π/2 Z π
m =
ρ sin φ dφ dθ dρ = π.
0
π/4
0
Problem 7 Let R be the region defined by 0 ≤ 2(x+y) ≤ π and (6x−2y)2 ≤
1 (this is indeed a finite region). Evaluate the double integral
Z Z
cos(x + y)e3x−y dA
R
using the transformation u = x + y and v = 3x − y.
Solution. Limits are 0 ≤ u ≤ π/2, −1/2 ≤ v ≤ 1/2. Express x, y in terms
of u, v:
x = (u + v)/4, y = (3u − v)/4.
This leads to J(u, v) = −1/4 and the integral
1
4
Z
0
π/2 Z 1/2
1
cos(u)ev dv du = (e1/2 − e−1/2 ).
4
−1/2