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

##### 17Calculus

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.

Okay, so here are a few videos we recommend that expand on the some of the above tips and also provide some insight on taking exams. This guy has lots of other videos about how to succeed in college, so we recommend his YouTube channel, Thomas Frank.

### Thomas Frank - 5 Rules (and One Secret Weapon) for Acing Multiple Choice Tests [9mins-42secs]

video by Thomas Frank

### Thomas Frank - 10 Study Tips for Earning an A on Your Next Exam [7mins-49secs]

video by Thomas Frank

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:
- 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).

(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: $$d[\arctan(t)]/dt=1/(1+t^2)$$]

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.

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.

Log in to rate this practice problem and to see it's current rating.

(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 $$y=x^2$$.

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

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.

Both integrals evaluate to $$1/30$$.

Log in to rate this practice problem and to see it's current rating.

(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).

$$18 \pi$$

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

 $$\vec{r}=3\sin u \cos v \hat{i} + 3\sin u \sin v \hat{j} + 3\cos u \hat{k}, ~~~ 0 \leq u \leq \pi/2, 0 \leq v \leq 2\pi$$ $$\vec{r}_u = 3\cos u \cos v \hat{i} + 3\cos u \sin v \hat{j} -3\sin u \hat{k}$$ $$\vec{r}_v = -3\sin u \sin v \hat{i} + 3\sin u \cos v \hat{j} + 0\hat{k}$$ $$\vec{r}_u \times \vec{r}_v$$ $$\displaystyle{ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3\cos u \cos v & 3\cos u \sin v & -3\sin u \\ -3\sin u \sin v & 3\sin u \cos v & 0 \end{vmatrix}}$$ $$9\sin^2u\cos v\hat{i} + 9\sin^2u\sin v \hat{j} + 9\sin u\cos u\hat{k}$$ $$\| \vec{r}_u \times \vec{r}_v \|$$ $$\sqrt{ 81\sin^4u\cos^2v + 81\sin^4u\sin^2v+81\sin^2u\cos^2u}$$ $$9\sqrt{\sin^4u+\sin^2u\cos^2u}$$ $$9\sin u \sqrt{ \sin^2u+\cos^2u} = 9\sin u$$ $$\int_{0}^{2\pi}{ \int_{0}^{\pi/2}{ 9\sin u ~ du } ~dv }$$ $$\int_0^{2\pi}{ \left[ -9\cos u \right]_{0}^{\pi/2} ~dv }$$ $$\int_{0}^{2\pi}{ 9 ~ dv} = 18\pi$$

$$18 \pi$$

Log in to rate this practice problem and to see it's current rating.

(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.

$$12$$

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.

 $$\vec{\nabla} \cdot \vec{F}$$ $$\displaystyle{ \frac{\partial}{\partial x}[x^2+y] + \frac{\partial}{\partial y}[z^2] + \frac{\partial}{\partial z}[e^y-z] }$$ $$2x+0-1 = 2x-1$$ $$\int_{0}^{2}{\int_{0}^{1}{ \int_{0}^{3}{ 2x-1~dx }~dy}~dz}$$ $$\int_{0}^{2}{\int_{0}^{1}{ \left[ x^2-x \right]_{0}^{3}~dy}~dz}$$ $$\int_{0}^{2}{\int_{0}^{1}{6~dy}~dz}$$ $$\int_{0}^{2}{6~dz} = 12$$

$$12$$

Log in to rate this practice problem and to see it's current rating.

(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.

$$3/2$$

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 $$\vec{r}=u\hat{i}+v\hat{j}+(1-u-v)\hat{k}, 0 \leq u \leq 1, 0 \leq v \leq 1-x$$.

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

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

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

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

$$3/2$$

Log in to rate this practice problem and to see it's current rating.

(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.

$$\phi=xy+yz^2$$

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 $$\phi=xy+yz^2$$.

$$\phi=xy+yz^2$$

Log in to rate this practice problem and to see it's current rating.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.