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MATH 241 Week 3 Worksheet Solutions Assignment | Calculus III | University of Illinois Urbana-Champaign | Spring 2023 | 2026/27 Update

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MATH 241 Week 3 Worksheet Solutions Assignment | Calculus III | University of Illinois Urbana-Champaign | Spring 2023 | 2026/27 Update

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Center for Academic Resources in Engineering (CARE)
Peer Exam Review Session

Math 241 − Calculus III


Midterm 3 Worksheet Solutions


The problems in this review are designed to help prepare you for your upcoming exam. Questions pertain
to material covered in the course and are intended to reflect the topics likely to appear in the exam. Keep
in mind that this worksheet was created by CARE tutors, and while it is thorough, it is not comprehensive.
In addition to exam review sessions, CARE also hosts regularly scheduled tutoring hours.



Tutors are available to answer questions, review problems, and help you feel prepared for your
exam during these times:

Session 1: Wed, 11/1, 5-7 pm - Jakob, Camila, Gabe @ CIF 4025

Session 2: Thurs, 11/2, 5-7 pm - Matthew, Rose @ CIF 2039

Can’t make it to a session? Here’s our schedule by course:

https://care.grainger.illinois.edu/tutoring/schedule-by-subject

Solutions will be available on our website after the last review session that we host.

Step-by-step login for exam review session:

1. Log into Queue @ Illinois: https://queue.illinois.edu/q/queue/845

2. Click “New Question”

3. Add your NetID and Name

4. Press “Add to Queue”

Please be sure to follow the above steps to add yourself to the Queue.


Good luck with your exam!

, Math 241 − Calculus III Midterm 3 Exam Review


1. Compute the double integral over the indicated rectangle. Confirm your answer by switching the
order of integration and recomputing.
ZZ
2x − 4y 3 dA R = [−5, 4] × [0, 3]
R



Z 3 Z 4 Z 4 Z 3
3
2x − 4y dxdy 2x − 4y 3 dydx
0 −5 −5 0
Z 3 Z 4
−9 − 36y 3 dy 6x − 81 dx
0 −5
3 4
− 9y − 9y 4 0
= −756 3x2 − 81x −5
= −756




2. Making an appropriate change of variables, compute the following double integral over the region
bound by a circle of radius 2 and a circle of radius 5.
ZZ
2 +y 2
ex dA
D


Using polar coordinates gives the following integral

ZZ Z 2π Z 5
x2 +y 2 2
e dA = rer drdθ
D 0 2


Then using a u−substitution u = r2 for the r dependence


2π 25
1
Z Z
eu dudθ = π(e25 − e4 ) ≈ 2.26 × 1011
2 0 4




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