Solutions: MATH 251, Exam 2
Jens Forsgård
March 23, 2016
1. Let D denote the type I domain {0 ≤ x ≤ 1, x3 ≤ y ≤ x}.
(a) Sketch D.
3 pts.
–
(b) Express D in polar coordinates.
4 pts.
We have that 0 ≤ θ ≤ π/4 and 0 ≤ r ≤
q
sin(θ)
cos3 (θ) .
To find the upper bound for r, notice that this is given
3
by y = x ; now substitute to polar coordinates and solve for r.
(c) Express D as a type II domain.
3 pts.
√
3
We have that 0 ≤ y ≤ 1 and y ≤ x ≤
y.
2. Let D = {0 ≤ x, 0 ≤ y, x + y ≤ 1}, and let E be a lamina covering D with the density described by
the function ρ(x, y) = x + y.
(a) Sketch the domain D.
2 pts.
–
(b) Find the total mass of the lamina.
4 pts.
The total mass m is given by
ZZ
Z
m=
1
1−x
Z
x + y dx dy =
Z
D
1
x2
xy +
2
=
0
x + y dy dx
0
0
1−x
1
Z
dx =
0
0
1 x2
1
−
dx = .
2
2
3
(c) Find the x-coordinate of the center of mass of the lamina.
4 pts.
The x-coordinate x̄ is given by x̄ = Mx /m, where Mx is the first moment
ZZ
Z
Mx =
x(x + y) dx dy =
Z
=
0
D
1
1
Z
x
0
1−x
x + y dy dx
0
1
x x3
−
dy = ,
2
2
8
where we re-used some of the computations from (b). All in all we find that x̄ = 3/8.
1
3. Compute integral
2
RR
D
ex
+y 2
dx dy where D = {x2 + y 2 ≤ 4}.
6 pts.
By switching to polar coordinates, we find that
ZZ
ex
2
+y
2
Z
2π
2
Z
dx dy =
0
0
D
"
2
er
rer dr dθ = 2π
2
#r=2
2
= π(e4 − 1).
r=0
4. Compute the volume of the ball B = {x2 + y 2 + z 2 ≤ 2} using a triple integral and spherical
coordinates.
6 pts.
By switching to spherical coordinates we obtain that
√
ZZZ
Z
V =
2
Z
2π
Z
dx dy dz =
B
r3
= 2π
3
r=√2
0
0
φ=π
[− cos(φ)]φ=0
r=0
π
r2 sin(φ) dr dθ dφ
0
√
√
2 2
8 2π
= 2π
(1 + 1) =
.
3
3
5. Let E consisting of all points in the closed ball B = {x2 + y 2 + z 2 ≤ 2} which lies above the plane
z = 1.
(a) Express E in cylindrical coordinates.
4 pts.
That z ≥ 1 gives that x2 + y 2 ≤ 1. Setting x = r cos(θ), y = r sin(θ) and z = z yields
p
{0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, 1 ≤ z ≤ 2 − r2 }.
(b) Express E in spherical coordinates.
4 pts.
2
2
Set x = r√cos(θ) sin(φ), y = r sin(θ) sin(φ) and √
z = r cos(φ). The intersection of z = 1 and x +y
√ +z = 2
has r = 2 and, therefor, cos(φ) = z/r = 1/ 2. Thus, the upper bound for φ is arccos(1/ 2) = π/4.
The lower bound for ris given by the plane z = 1, which gives r cos(φ) = 1 or r = 1/ cos(φ). We find
that
√
{0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4, 1/ cos(φ) ≤ r ≤ 2}
2
2