Math 20E A00 Fall 2021: Practice Midterm 2
Instructor: Brian Tran
Instructions: The points shown here are the same points that are given on the respective problems
for the actual midterm (20 points for each normal problem; 10 points for the extra credit problem).
These are the instructions for the actual midterm:
You can apply any theorem or rule discussed in class, but make sure to show that the assumptions of
said theorem or rule apply. Show all of your work; you will not receive credit for just stating an answer.
You are allowed one sheet of notes (front and back); do not use any outside sources such as the
internet, a calculator, the textbook, or communicating with others.
On your writeup, please write your name and PID at the top. There are four problems worth 20 points
each and one extra credit problem worth 10 points; clearly indicate which problem you are working on.
You have 90 minutes, from the time that you view the exam, to complete the exam, scan the exam, and
submit it to Gradescope; your submission should be submitted as a single PDF file. Of the 90 minutes
given for the exam, make sure to leave about 10 minutes to make sure that you can scan and upload the
exam in time. Please correctly assign your solution pages on Gradescope to the corresponding question
before submitting your midterm, just as you would do on the homework.
Problem 1 (20 points)
Let the surface S be the graph of the function z = g(x, y) = y3/3 + 1 over (x, y) ∈ [−1, 1] × [0, 1]. Let
f : R3 → R be given by f(x, y, z) = xy2. Evaluate∫∫
S
f dS.
Problem 2 (20 points)
Let S be the closed surface which is the union of three surfaces, S = S1 ∪S2 ∪S3, where S1 is the curved
part of the cylinder,
S1 = {(x, y, z) : x2 + y2 = 1 and 0 ≤ z ≤ 1},
S2 is the bottom “lid” of the cylinder,
S2 = {(x, y, z) : x2 + y2 ≤ 1 and z = 0},
and S3 is the top “lid” of the cylinder,
S3 = {(x, y, z) : x2 + y2 ≤ 1 and z = 1}.
Let S be oriented with the outward normal. Let ~F : R3 → R3 be given by ~F (x, y, z) = (x, y, z). Evaluate∫∫
S
~F · d~S.
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Math 20E A00 Fall 2021: Practice Midterm 2 Instructor: Brian Tran
Problem 3 (20 points)
Let S2 be the surface of the unit sphere, S = {(x, y, z) : x2 + y2 + z2 = 1}, oriented with the outward
normal. Let ~F : R3 → R3 be given by ~F (x, y, z) = (y2, z − xy,−y). Evaluate∫∫
S2
~F · d~S.
Hint: While you can in principle evaluate this directly by parametrizing S (e.g., with spherical coordi-
nates), it is easier to use the geometric formula for the surface integral: what is the unit normal vector
field to S2 and what is its dot product with ~F?
Problem 4 (20 points)
Let C be the following closed curve in the xy plane: C goes from (0, 0) to (0, 2) along a straight line,
from (0, 2) to (2, 2) along a straight line, from (2, 2) to (2, 0) along a straight line, and from (2, 0) back
to (0, 0) along a straight line.
Let P (x, y) = 2xy + sin(x4), Q(x, y) = sin(y3) + x. Evaluate the line integral∫
C
Pdx + Qdy.
Hint: Use Green’s theorem.
Problem 5 (Extra Credit: 10 points)
Prove the following statement:
Let the surface S be the graph of a differentiable function z = g(x, y) over the domain (x, y) ∈
[−a, a] × [b, c] (where a, b, c are fixed constants satisfying a > 0 and c > b). Furthermore, assume that
g only depends on y; that is, it can be expressed g(x, y) = h(y) for some differentiable function of one
variable h. Let f : R3 → R be a (continuous) function that is odd with respect to the x variable; i.e.,
f(−x, y, z) = −f(x, y, z) for all (x, y, z). Then,∫∫
S
fdS = 0.
Hint: Parametrize the surface using the usual parametrization of a graph,
Φ(x, y) = (x, y, g(x, y)) = (x, y, h(y)).
The domain of Φ is D = [−a, a] × [b, c]. Use the definition of the surface integral to write ∫∫
S
fdS as a
double integral over D. Split the domain into two pieces, D+ = [0, a]× [b, c] and D− = [−a, 0]× [b, c] so
that D = D+ ∪D−. Then, split the double integral over D into two double integral over the two pieces,
D+ and D−. For the D− double integral, make a change of variable x → −x, and you will see that the
D+ double integral and the D− double integral exactly cancel each other, using the fact that f is odd
with respect to x.
Remark. We’ve discussed this symmetry principle before for double and triple integrals: integrating an
odd function over a symmetric domain gives 0. This is a generalization of the double integral result to
surfaces (this result is only stated for a surface which is a graph, but it holds more generally).
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