Center for Academic Resources in Engineering (CARE)
Peer Exam Review Session
Math 241 − Calculus III
Midterm 4 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: Nov. 28, 5:00-6:30 pm in 2039 CIF, Cami, Jaycob, Rose
Session 2: Nov. 29, 4:30-6:00 pm in 3025 CIF, Gabe, Matthew
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 4 Exam Review
1. Consider the following vector fields F⃗ (x, y, z). Are they conservative? If so, find a function
f (x, y, z) so that ∇f = F⃗ . If not, justify your response.
(a) F⃗ (x, y, z) = ⟨yz, xz, xy + 2z⟩
(b) F⃗ (x, y, z) = ⟨y + ex , x − cos y, 4 + z⟩
(c) F⃗ (x, y, z) = ⟨y, z 2 , x⟩
Conservative vector field test: a vector field F⃗ is conservative if the curl is the zero vector.
x̂ ŷ ẑ D ∂F
⃗ × F⃗ = ∂ ∂ ∂ z ∂Fy ∂Fx ∂Fz ∂Fx ∂Fy E ⃗
∇ ∂x ∂y ∂z = − , − , − =0
∂y ∂z ∂z ∂x ∂y ∂x
Fx Fy Fz
(a)
x̂ ŷ ẑ
∂ ∂ ∂
∂x ∂y ∂z = ⟨x − x, y − y, z − z⟩ = ⃗0
yz xz xy + 2z
The vector field is conservative, therefore, a potential function exists. To find it, we must find the
necessary terms from each component (We neglect the constant for now, we’ll add it back later).
Z Z
Fx dx = yz dx = xyz
Z Z
Fy dy = xz dy = xyz
Z Z
Fz dz = xy + 2z dz = xyz + z 2
We see that the necessary terms are xyz and z 2 , therefore
The field is conservative and has potential function f (x, y, z) = xyz + z 2 + C
(b)
x̂ ŷ ẑ
∂ ∂ ∂
∂x ∂y ∂z = ⟨0 − 0, 0 − 0, 1 − 1⟩ = ⃗0
x
y+e x − cos y 4 + z
The vector field is conservative, therefore, a potential function exists. To find it, we must integrate
each component (We neglect the constant for now, we’ll add it back later).
Z Z
Fx dx = y + ex dx = xy + ex
Z Z
Fy dy = x − cos y dy = xy − sin y
2 of 10
Peer Exam Review Session
Math 241 − Calculus III
Midterm 4 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: Nov. 28, 5:00-6:30 pm in 2039 CIF, Cami, Jaycob, Rose
Session 2: Nov. 29, 4:30-6:00 pm in 3025 CIF, Gabe, Matthew
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 4 Exam Review
1. Consider the following vector fields F⃗ (x, y, z). Are they conservative? If so, find a function
f (x, y, z) so that ∇f = F⃗ . If not, justify your response.
(a) F⃗ (x, y, z) = ⟨yz, xz, xy + 2z⟩
(b) F⃗ (x, y, z) = ⟨y + ex , x − cos y, 4 + z⟩
(c) F⃗ (x, y, z) = ⟨y, z 2 , x⟩
Conservative vector field test: a vector field F⃗ is conservative if the curl is the zero vector.
x̂ ŷ ẑ D ∂F
⃗ × F⃗ = ∂ ∂ ∂ z ∂Fy ∂Fx ∂Fz ∂Fx ∂Fy E ⃗
∇ ∂x ∂y ∂z = − , − , − =0
∂y ∂z ∂z ∂x ∂y ∂x
Fx Fy Fz
(a)
x̂ ŷ ẑ
∂ ∂ ∂
∂x ∂y ∂z = ⟨x − x, y − y, z − z⟩ = ⃗0
yz xz xy + 2z
The vector field is conservative, therefore, a potential function exists. To find it, we must find the
necessary terms from each component (We neglect the constant for now, we’ll add it back later).
Z Z
Fx dx = yz dx = xyz
Z Z
Fy dy = xz dy = xyz
Z Z
Fz dz = xy + 2z dz = xyz + z 2
We see that the necessary terms are xyz and z 2 , therefore
The field is conservative and has potential function f (x, y, z) = xyz + z 2 + C
(b)
x̂ ŷ ẑ
∂ ∂ ∂
∂x ∂y ∂z = ⟨0 − 0, 0 − 0, 1 − 1⟩ = ⃗0
x
y+e x − cos y 4 + z
The vector field is conservative, therefore, a potential function exists. To find it, we must integrate
each component (We neglect the constant for now, we’ll add it back later).
Z Z
Fx dx = y + ex dx = xy + ex
Z Z
Fy dy = x − cos y dy = xy − sin y
2 of 10
