Answers to Math 261 SP15 Exam 3
The following are just the final numerical (or otherwise) answers for Math 261 SP15 Exam 3.
Your solutions on the exam should include full details to allow for partial credit.
If you are having trouble coming up with these answers, please go to a review session or office
hours.
NOTE: If you believe this practice exam seems quite difficult, you are correct. Students
had difficulty finishing this exam in time. They also had difficulty with drawing sketches in
1, 2, and 3, integrating in 2, and the component test in 5a. Sketches will be provided on
this semester’s Exam 3 as necessary (and not graded if not provided), and heavy integration
(requiring anything beyond u-substitution) as in 2a will not be required. We will also provide
you with the following list of coordinate conversions on the cover page (exactly as listed here):
x = r cos(θ)
y = r sin(θ)
x = ρ sin(φ) cos(θ)
y = ρ sin(φ) sin(θ)
z = ρ cos(φ)
r = ρ sin(φ)
Z
1
2
Z
4−y 2
Z
1. (a)
xyz dzdydz.
0
Z
2x
1
0
4−4x2
Z
√
Z
4−z
xyz dydzdx.
(b)
0
Z
0
2
2x
4−y 2
Z
Z
(c)
1
y
2
xyz dxdzdy.
0
0
0
2. (a) 10π.
Z 2π Z
(b)
0
2
Z
0
2−r sin(θ)
r2 cos(θ)z dzdrdθ.
0
(c) There was a typo in this exam. The question should have been about Myz , not Mxz .
Myz
= 0, due to the symmetry in δ and in the region, so Myz = 0.
M
Z 2π Z π Z 2 cos(φ)
2
3.
ρ2 sin(φ) dρdφdθ.
0
0
cos(φ)
(
x = 21 u + 12 w,
4. (a)
y = 12 u − 12 w.
(b) Should be a square, with corners (u, w) = (−1, −1), (−1, 1), (1, −1), (1, 1).
(c) − 12 .
(d)
R1 R1
−1
−1
u2 euw 12 dudw.

2

My = z (cos(xy) − xy sin(xy)) = Nx .X
1
5. (a) Mz = 2yz cos(xy) + z+1
= Px .X


Nz = 2xz cos(xy) + 3 cos(z)e3y sin(z) + 9y sin(z) cos(z)e3y sin(z) = Py .X
(b) z 2 sin(xy) + x ln(z + 1) + e3y sin(z) + C.
(c) 0.