Math 273 Spring 2016
Exam 3 Practice Solutions
Where I give a numerical approximation for an integral, if you want to test your answer,
use Wolfram Alpha or some similar tool to evaluate the integral you were able to set up.
1. (a) T, (b) F, (c) F, (d) T, (e) T
Z π Z πZ 3
2.
ρ4 sin3 φ dρ dθ dφ ≈ 88.4
π/2
0
2
3. (a) I accidentally made this rely on a quadratic that isn’t the nicest ... but really we
should all be able to handle any quadratic equation life throws at us.
Lets √
do this in cylindrical coordinates. The lower and upper surfaces become r2 ≤
z ≤ 4 − r2 . The intersection, which forms the boundary for the projection into
the xy-plane, occurs where z + z 2 = 4.
The z-coordinate√ where the intersection occurs is positive, so by the quadratic
equation z = 1+2 17 . Substituting this back into one of the equations gives r2 =
q √
√
1+ 17
1+ 17
2
2
. The a is for
x + y = 2 . Lets call the square root of that a =
2
apologies that it’s ugly.
2π
Z
a
Z
√
Z
4−r2
1 · r dz dr dθ ≈ 2.83
vol =
0
r2
0
(b) By symmetry of the region (which is enough for the centroid/uniform density)
x̄ = ȳ = 0.
The great thing about centroid is that all the integrals have the same bounds. So
1
z̄ =
vol
Z
2π
Z
a
√
Z
4−r2
z · r dz dr dθ ≈ 0.75
0
0
4. A is (i), B is a typo and would be
R π/2
−π/2
r2
f (r, θ)r dr dθ, C is (iii).
Page 2 of ??
5. (a) The four sides u = 0, u = 2, v = 0, v = 2 become the lines y = −2x, y = 10 − 2x,
y = (1/2)x, y = (1/2)x + 5.
(b)
R2R2
0
0
−3ve5u du dv ≈ −26430
6. A little sketch reveals that this domain of integration is the tetrahedron with x, y and z
all ≥ 0, sitting under the plane 6 = 3x + 2y + z. To find the projection into any plane,
set the relevant variable equal to zero. For example in the yz-plane this is the triangle
under the line 6 = 2y + z.
(a)
Z3

2− 32 y 6−2y−3x
Z
Z


0


f (x, y, z) dz  dx dy

0

0
(b)
Z3


6−2y
Z


0
2− 23 y− 13 z


Z
 
f (x, y, z) dx dz  dy


0
0
(c)

Z6
2− 31 z

Z


0
Z
π/2
Z
3− 32 y− 12 z


Z
 
f (x, y, z) dy  dx dz


0
0
cos θ sin θ
7. Mass =
cos θ dr dθ
0
0
8. (a) Rectangular
Z
2
−2
√
Z
4−x2
√
− 4−x2
Z √100−x2 −y2
0
(x2 + y 2 + z 2 ) dz dy dx
Page 3 of ??
(b) Cylindrical
Z
0
2π
Z
0
2
√
Z
100−r2
(r2 + z 2 )r dz dr dθ
0
(c) Spherical: Must be done in two pieces!
!
Z Z −1 Z
sin
2π
1/5
10
ρ4 sin φ dρ dφ dθ
0
0
0
Z
2π
Z
π/2
Z
+
0
sin−1
1/5
0
!
2 csc φ
ρ4 sin φ dρ dφ dθ