## 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.

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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:
- 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.

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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 $$y=x^2$$.

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 $$1/30$$.

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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).

$$18 \pi$$

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

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

$$18 \pi$$

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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.

$$12$$

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.

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

$$12$$

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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.

$$3/2$$

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

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

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

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

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

$$3/2$$

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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.

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

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

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

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### 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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