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17Calculus - Calculus 3 - Practice Exam 4 - Final Exam (Semester A)

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This is the fourth and final exam for third semester (multi-variable) calculus. This exam contains 6 questions on vector fields covering line integrals, surface integrals, Green's Theorem, Stokes' Theorem, the Divergence Theorem and related topics. This exam is comprehensive in that all the material from calculus 3 is used including vector and vector operations, partial derivatives, partial integrals and vector functions.

Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
- Use your calculator as little as possible except for graphing and checking your calculations.
- Work the entire exam before checking any solutions.
- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

2 hours

Questions

6

Total Points

85

Tools

Calculator

no

Formula Sheet(s)

2 pages, 8.5x11 or A4

Other Tools

ruler for drawing graphs

Instructions:
- Show all your work.
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).
- Give exact, simplified answers.

(10 points) Evaluate \(\oint\limits_C {2 \arctan(y/x)~dx+\ln(x^2+y^2)~dy}\) on the curve \(C:~x=4+2\cos(\theta), y=4+\sin(\theta) \).
[Note: \(d[\arctan(t)]/dt=1/(1+t^2)\)]

Problem Statement

(10 points) Evaluate \(\oint\limits_C {2 \arctan(y/x)~dx+\ln(x^2+y^2)~dy}\) on the curve \(C:~x=4+2\cos(\theta), y=4+\sin(\theta) \).
[Note: \(d[\arctan(t)]/dt=1/(1+t^2)\)]

Final Answer

Since \(\displaystyle{ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} }\), the vector field is conservative and, since the curve is closed, the line integral is zero.

Problem Statement

(10 points) Evaluate \(\oint\limits_C {2 \arctan(y/x)~dx+\ln(x^2+y^2)~dy}\) on the curve \(C:~x=4+2\cos(\theta), y=4+\sin(\theta) \).
[Note: \(d[\arctan(t)]/dt=1/(1+t^2)\)]

Solution

Noticing that the curve is closed, we will first check to see if the vector field is conservative.
Let \(M=2\arctan(y/x)\) and \(N=\ln(x^2+y^2)\).
\(\displaystyle{ \frac{\partial M}{\partial y} = \frac{2(1/x)}{1+(y/x)^2}}\)
Multiplying the numerator and denominator by \(x^2\) we get \(\displaystyle{\frac{\partial M}{\partial y} = \frac{2x}{x^2+y^2}}\)

\(\displaystyle{\frac{\partial N}{\partial x} = \frac{1}{x^2+y^2}(2x)}\)

Notice that \(\displaystyle{ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} }\), which means that the vector field is conservative. We know that when we have a conservative vector field, the line integral depends only on the endpoints. For closed curves, this means that the line integral is zero.

Final Answer

Since \(\displaystyle{ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} }\), the vector field is conservative and, since the curve is closed, the line integral is zero.

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(20 points) Verify Green's Theorem (by evaluating both integrals) for \(\int\limits_C {y^2~dx + x^2~dy}\) where C is the boundary lying between the graphs of \(y=x\) and \(y=x^2\).

Problem Statement

(20 points) Verify Green's Theorem (by evaluating both integrals) for \(\int\limits_C {y^2~dx + x^2~dy}\) where C is the boundary lying between the graphs of \(y=x\) and \(y=x^2\).

Final Answer

Both integrals evaluate to \(1/30\).

Problem Statement

(20 points) Verify Green's Theorem (by evaluating both integrals) for \(\int\limits_C {y^2~dx + x^2~dy}\) where C is the boundary lying between the graphs of \(y=x\) and \(y=x^2\).

Solution

plot of curve and area for question 2

We are asked to verify Green's Theorem. So we need to evaluate both integrals \(\displaystyle{ \oint\limits_C {M~dx+N~dy} = \iint\limits_R { \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} ~dA } }\) and verify that they are equal.
Here is the graph of the area and the curves. The black curve is \(y=x^2\) and the blue curve is \(y=x\). The area is shaded.

Area Integral
For our problem, \(M=y^2\) and \(N=x^2\) and we get \(M_y=2y\) and \(N_x=2x\). The area integral is

\(\displaystyle{ \iint\limits_R { \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} ~dA } }\)

\(\displaystyle{ \int_{0}^{1}{ \int_{x^2}^{x}{ 2x-2y ~ dy } ~dx } }\)

\(\displaystyle{ \int_{0}^{1}{ \left[ 2xy-y^2 \right]_{y=x^2}^{y=x} ~dx } }\)

\(\displaystyle{ \int_{0}^{1}{ x^4-2x^3+x^2 ~ dx } }\)

\(\displaystyle{ \left[ \frac{x^5}{5} - \frac{2x^4}{4} + \frac{x^3}{3} \right]_{0}^{1} }\)

\(\displaystyle{ \frac{1}{30} }\)

Line Integrals
Now we need to evaluate the line integrals. The direction of the curve must be counter-clockwise, so we will start at the origin and follow the arrows along segment 1 (\(y=x^2\)) to the point \((1,1)\) and then back to the origin along segment 2 (\(y=x\)).
Segment 1 - To parameterize the curve, we let \(x=t\) to get \( \langle t,t^2\rangle, 0 \leq t \leq 1\).
\(x=t ~~ \to ~~ dx=dt\)
\(y=t^2 ~~ \to ~~ dy=2t~dt\)
Now we set up the integral and evaluate it.

\(\displaystyle{ \int\limits_C { y^2~dx + x^2~dy } }\)

\(\displaystyle{ \int_{0}^{1}{ t^4~dt + t^2(2t~dt) } }\)

\(\displaystyle{ \int_{0}^{1}{ t^4+2t^3 ~ dt } }\)

\(\displaystyle{ \left[ \frac{t^5}{5} + \frac{2t^4}{4} \right]_{0}^{1} = \frac{7}{10} }\)

Segment 2 - Initially, we set up the parametric curve with \(x=t\) giving us \(\langle t,t \rangle, 0 \leq t \leq 1\). However, this curve is in the wrong direction. To switch directions, we replace t with 1-t to get \(\langle 1-t, 1-t \rangle, 0 \leq t \leq 1\).
\(x=1-t \to dx = -dt\)
\(y=1-t \to dy = -dt\)
Now set up the integral and evaluate.

\(\displaystyle{ \int\limits_C { y^2~dx + x^2~dy } }\)

\(\displaystyle{ \int_{0}^{1}{ (1-t)^2~(-dt) + (1-t)^2~(-dt) } }\)

\(\displaystyle{ \int_{0}^{1}{ -2(1-t)^2~dt } }\)

\(\displaystyle{ -2\int_{0}^{1}{ t^2-2t+1~dt} }\)

\(\displaystyle{ -2 \left[ \frac{t^3}{3}-t^2+t \right]_{0}^{1} = \frac{-2}{3} }\)

Now we add the result of the two line integrals to get the total around the entire closed curve.
\(\displaystyle{ \frac{7}{10} - \frac{2}{3} = \frac{1}{30} }\)
which matches the result of the area integral.

Final Answer

Both integrals evaluate to \(1/30\).

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(20 points) Use a surface integral to find the area of the hemisphere \(x^2+y^2+z^2=9\) for \(z \geq 0\) (excluding the base).

Problem Statement

(20 points) Use a surface integral to find the area of the hemisphere \(x^2+y^2+z^2=9\) for \(z \geq 0\) (excluding the base).

Final Answer

\( 18 \pi \)

Problem Statement

(20 points) Use a surface integral to find the area of the hemisphere \(x^2+y^2+z^2=9\) for \(z \geq 0\) (excluding the base).

Solution

If we stop and think about this one for a minute, this is just the top half of a sphere. So we anticipate that our answer should be \((1/2)(4\pi r^2) = (1/2)(4\pi 3^2)=18\pi\).
The surface integral to calculate area is \(\displaystyle{ \iint\limits_S {dS} = \iint\limits_R { \| \vec{r}_u \times \vec{r}_v \| ~ dA } }\). Note that this is just the general surface integral equation with \(F(x,y,z)=1\).
First, let's get the parametric equations for the surface. To do this, we use the equations for spherical coordinates
\(x=\rho\sin\phi \cos\theta, y=\rho\sin\phi \sin\theta, z=\rho\cos\phi\)
For our problem, \(\rho=3, 0 \leq \phi \leq \pi/2, 0 \leq \theta \leq 2\pi\).
In order to avoid problems later, we set \(u=\phi\) and \(v=\theta\). So the parametric equations for the surface are

\(\vec{r}=3\sin u \cos v \hat{i} + 3\sin u \sin v \hat{j} + 3\cos u \hat{k}, ~~~ 0 \leq u \leq \pi/2, 0 \leq v \leq 2\pi\)

\(\vec{r}_u = 3\cos u \cos v \hat{i} + 3\cos u \sin v \hat{j} -3\sin u \hat{k}\)

\(\vec{r}_v = -3\sin u \sin v \hat{i} + 3\sin u \cos v \hat{j} + 0\hat{k} \)

\( \vec{r}_u \times \vec{r}_v \)

\(\displaystyle{ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3\cos u \cos v & 3\cos u \sin v & -3\sin u \\ -3\sin u \sin v & 3\sin u \cos v & 0 \end{vmatrix}}\)

\( 9\sin^2u\cos v\hat{i} + 9\sin^2u\sin v \hat{j} + 9\sin u\cos u\hat{k} \)

\( \| \vec{r}_u \times \vec{r}_v \| \)

\( \sqrt{ 81\sin^4u\cos^2v + 81\sin^4u\sin^2v+81\sin^2u\cos^2u} \)

\( 9\sqrt{\sin^4u+\sin^2u\cos^2u} \)

\( 9\sin u \sqrt{ \sin^2u+\cos^2u} = 9\sin u \)

\( \int_{0}^{2\pi}{ \int_{0}^{\pi/2}{ 9\sin u ~ du } ~dv } \)

\( \int_0^{2\pi}{ \left[ -9\cos u \right]_{0}^{\pi/2} ~dv } \)

\( \int_{0}^{2\pi}{ 9 ~ dv} = 18\pi \)

Final Answer

\( 18 \pi \)

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(10 points) Use the Divergence Theorem to compute the net outward flux of the vector field \(\vec{F}=(x^2+y)\hat{i}+z^2\hat{j}+(e^y-z)\hat{k}\) across the surface bounded by the coordinates planes and the planes \(x=3, y=1, z=2\) in the first octant.

Problem Statement

(10 points) Use the Divergence Theorem to compute the net outward flux of the vector field \(\vec{F}=(x^2+y)\hat{i}+z^2\hat{j}+(e^y-z)\hat{k}\) across the surface bounded by the coordinates planes and the planes \(x=3, y=1, z=2\) in the first octant.

Final Answer

\( 12 \)

Problem Statement

(10 points) Use the Divergence Theorem to compute the net outward flux of the vector field \(\vec{F}=(x^2+y)\hat{i}+z^2\hat{j}+(e^y-z)\hat{k}\) across the surface bounded by the coordinates planes and the planes \(x=3, y=1, z=2\) in the first octant.

Solution

The Divergence Theorem equation is \(\displaystyle{\iint\limits_S{ \vec{F} \cdot \vec{n}~dS} = \iiint\limits_D {\vec{\nabla} \cdot \vec{F} ~dV}}\). Considering the surface, we would need to do 6 surface integrals (one for each face of the rectangular box) or we could do one volume integral. So it seems easier to do the volume integral.

\( \vec{\nabla} \cdot \vec{F} \)

\(\displaystyle{ \frac{\partial}{\partial x}[x^2+y] + \frac{\partial}{\partial y}[z^2] + \frac{\partial}{\partial z}[e^y-z] }\)

\(2x+0-1 = 2x-1\)

\( \int_{0}^{2}{\int_{0}^{1}{ \int_{0}^{3}{ 2x-1~dx }~dy}~dz} \)

\( \int_{0}^{2}{\int_{0}^{1}{ \left[ x^2-x \right]_{0}^{3}~dy}~dz} \)

\( \int_{0}^{2}{\int_{0}^{1}{6~dy}~dz} \)

\( \int_{0}^{2}{6~dz} = 12 \)

Final Answer

\( 12 \)

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(15 points) Use Stokes' Theorem to calculate the circulation of the vector field \( \vec{F} = (x-y)\vec{i} + (y-z)\hat{j} + (z-x)\hat{k} \) around the surface bounded by \(x+y+z=1\) in the first octant.

Problem Statement

(15 points) Use Stokes' Theorem to calculate the circulation of the vector field \( \vec{F} = (x-y)\vec{i} + (y-z)\hat{j} + (z-x)\hat{k} \) around the surface bounded by \(x+y+z=1\) in the first octant.

Final Answer

\( 3/2 \)

Problem Statement

(15 points) Use Stokes' Theorem to calculate the circulation of the vector field \( \vec{F} = (x-y)\vec{i} + (y-z)\hat{j} + (z-x)\hat{k} \) around the surface bounded by \(x+y+z=1\) in the first octant.

Solution

The Stokes' Theorem equation is \(\displaystyle{ \oint\limits_C{\vec{F}\cdot d\vec{r}} = \iint\limits_S{ \left( \vec{\nabla} \times \vec{F} \right) \cdot \vec{n}~dS } }\). Looking closely at the equations, we would have to evaluate more than one line integral but only one surface integral. So, we will try the surface integral.
To get \(\vec{n}\), we can use \(\vec{r}_u \times \vec{r}_v\) where \(\vec{r}\) is the parameterized surface. So we set \(u=x, y=v\) giving us \(z=1-x-y = 1-u-v\). The parameterized surface is \(\vec{r}=u\hat{i}+v\hat{j}+(1-u-v)\hat{k}, 0 \leq u \leq 1, 0 \leq v \leq 1-x\).

\(\vec{\nabla} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x-y & y-z & z-x \end{vmatrix} = \) \(\displaystyle{\hat{i}+\hat{j}+\hat{k}}\)

\(\vec{r}_u = \hat{i}+0\hat{j} -\hat{k}\)
\(\vec{r}_v = 0\hat{i}+\hat{j}-\hat{k}\)
\(\vec{r}_u \times \vec{r}_v = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{vmatrix} = \hat{i}+\hat{j}+\hat{k}\)

\( \left( \vec{\nabla} \times \vec{F} \right) \cdot \left( \vec{r}_u \times \vec{r}_v \right) = 3\)

\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{1-x}{ 3~dy } ~dx } = \int_{0}^{1}{ 3(1-x) ~dx } = 3/2 }\)

Final Answer

\( 3/2 \)

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(10 points) Determine whether the vector field \(\vec{F} = y\hat{i} + (x+z^2)\hat{j} + 2yz\hat{k} \) is conservative. If it is, find a potential function.

Problem Statement

(10 points) Determine whether the vector field \(\vec{F} = y\hat{i} + (x+z^2)\hat{j} + 2yz\hat{k} \) is conservative. If it is, find a potential function.

Final Answer

\(\phi=xy+yz^2\)

Problem Statement

(10 points) Determine whether the vector field \(\vec{F} = y\hat{i} + (x+z^2)\hat{j} + 2yz\hat{k} \) is conservative. If it is, find a potential function.

Solution

First, we need to determine if the vector field is conservative.

\(\displaystyle{ \frac{\partial}{\partial y}[y] = 1 }\)

\(\displaystyle{ \frac{\partial}{\partial x}[x+z^2] =1 }\)

\(\displaystyle{ \frac{\partial}{\partial z}[y]=0 }\)

\(\displaystyle{ \frac{\partial}{\partial x}[2yz]=0 }\)

\(\displaystyle{ \frac{\partial}{\partial z}[x+z^2]=2z }\)

\(\displaystyle{ \frac{\partial}{\partial y}[2yz]=2z }\)

In the table above, each set of partial derivatives are equal, so the vector field is conservative. So we now calculate a potential function.

\(\phi = \int{ y~dx} = xy+f(y,z)\)

\(\phi = \int{ x+z^2 ~dy} = xy+yz^2+g(x,z)\)

\(\phi = \int{ 2yz~dz} = yz^2+h(x,y)\)

Combining the results in the above table, a potential function is \(\phi=xy+yz^2\).

Final Answer

\(\phi=xy+yz^2\)

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Topics You Need To Understand For This Page

vector fields and all related topics

Related Topics and Links

complete exam list

other exams from calculus 3

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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