This is the fourth and final exam for third semester (multivariable) 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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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 

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
\(\displaystyle{ \iint\limits_R { \frac{\partial N}{\partial x}  \frac{\partial M}{\partial y} ~dA } }\) 
\(\displaystyle{ \int_{0}^{1}{ \int_{x^2}^{x}{ 2x2y ~ dy } ~dx } }\) 
\(\displaystyle{ \int_{0}^{1}{ \left[ 2xyy^2 \right]_{y=x^2}^{y=x} ~dx } }\) 
\(\displaystyle{ \int_{0}^{1}{ x^42x^3+x^2 ~ dx } }\) 
\(\displaystyle{ \left[ \frac{x^5}{5}  \frac{2x^4}{4} + \frac{x^3}{3} \right]_{0}^{1} }\) 
\(\displaystyle{ \frac{1}{30} }\) 
Line Integrals
Now we need to evaluate the line integrals. The direction of the curve must be counterclockwise, 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 1t to get \(\langle 1t, 1t \rangle, 0 \leq t \leq 1\).
\(x=1t \to dx = dt\)
\(y=1t \to dy = dt\)
Now set up the integral and evaluate.
\(\displaystyle{ \int\limits_C { y^2~dx + x^2~dy } }\) 
\(\displaystyle{ \int_{0}^{1}{ (1t)^2~(dt) + (1t)^2~(dt) } }\) 
\(\displaystyle{ \int_{0}^{1}{ 2(1t)^2~dt } }\) 
\(\displaystyle{ 2\int_{0}^{1}{ t^22t+1~dt} }\) 
\(\displaystyle{ 2 \left[ \frac{t^3}{3}t^2+t \right]_{0}^{1} = \frac{2}{3} }\) 
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 \(\vec{F}=(x^2+y)\hat{i}+z^2\hat{j}+(e^yz)\hat{k}\) 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 \(\vec{F}=(x^2+y)\hat{i}+z^2\hat{j}+(e^yz)\hat{k}\) 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 \(\vec{F}=(x^2+y)\hat{i}+z^2\hat{j}+(e^yz)\hat{k}\) 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.
\( \vec{\nabla} \cdot \vec{F} \) 
\(\displaystyle{ \frac{\partial}{\partial x}[x^2+y] + \frac{\partial}{\partial y}[z^2] + \frac{\partial}{\partial z}[e^yz] }\) 
\(2x+01 = 2x1\) 

\( \int_{0}^{2}{\int_{0}^{1}{ \int_{0}^{3}{ 2x1~dx }~dy}~dz} \) 
\( \int_{0}^{2}{\int_{0}^{1}{ \left[ x^2x \right]_{0}^{3}~dy}~dz} \) 
\( \int_{0}^{2}{\int_{0}^{1}{6~dy}~dz} \) 
\( \int_{0}^{2}{6~dz} = 12 \) 
Final Answer 

\( 12 \) 
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(15 points) Use Stokes' Theorem to calculate the circulation of the vector field \( \vec{F} = (xy)\vec{i} + (yz)\hat{j} + (zx)\hat{k} \) 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 \( \vec{F} = (xy)\vec{i} + (yz)\hat{j} + (zx)\hat{k} \) 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 \( \vec{F} = (xy)\vec{i} + (yz)\hat{j} + (zx)\hat{k} \) 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 \(z=1xy = 1uv\). The parameterized surface is \(\vec{r}=u\hat{i}+v\hat{j}+(1uv)\hat{k}, 0 \leq u \leq 1, 0 \leq v \leq 1x\).
\(\vec{\nabla} \times \vec{F} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
xy & yz & zx
\end{vmatrix} = \)
\(\displaystyle{\hat{i}+\hat{j}+\hat{k}}\)
\(\vec{r}_u = \hat{i}+0\hat{j} \hat{k}\)
\(\vec{r}_v = 0\hat{i}+\hat{j}\hat{k}\)
\(\vec{r}_u \times \vec{r}_v =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 1 \\
0 & 1 & 1
\end{vmatrix} = \hat{i}+\hat{j}+\hat{k}\)
\( \left( \vec{\nabla} \times \vec{F} \right) \cdot \left( \vec{r}_u \times \vec{r}_v \right) = 3\)
\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{1x}{ 3~dy } ~dx } = \int_{0}^{1}{ 3(1x) ~dx } = 3/2 }\)
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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You CAN Ace Calculus
vector fields and all related topics 
other exams from calculus 3 

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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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