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\begin{document}

\begin{center}
{\Large Key to TEST 3}
\end{center}

\begin{enumerate}


\item  Compute the triple integral below
\[
\iiint_E \sqrt{x^2+y^2} \; dV
\]
Where $E$ is the solid bounded by the paraboloid $z=9-x^2-y^2$ and the $xy$-plane.
\\
\sol: 
This is best put into cylindrical coordinate (or, you can think of just putting the $x$ and $y$ plane into polar coordinate).
Doing this transforms the integral into:
\begin{eqnarray*}
\iiint_E \sqrt{x^2+y^2} \; dV 
& = & \int_0^{2\pi} \int_0^{3} \int_0^{9-r^2} r \cdot r \; dz \; dr \; d\theta \\
& = & \int_0^{2\pi} \int_0^{3} r^2(9-r^2) \; dr \; d\theta \\
& = & \int_0^{2\pi} \frac{162}{5} \; d\theta \\
& = & \frac{324 \pi}{5}
\end{eqnarray*}

\item  Let $S$ be the region in space defined by the inequalities:
\[
z \geq 0 \qquad x \geq z \qquad x \leq -2z+4 \qquad y \leq 5 \qquad  y \geq z
\]
\begin{enumerate}
\item  Write a triple integral that represents the volume of $S$.
\\
\sol:  It is probably best to graph the region.
	I had a hard time getting a Maple graph that looked good enough to put into this file, so we don't get a pretty picture here.
	Instead, I will try to describe the picture to you.

	First, graph $x=z$, $x=-2z+4$ and $z=0$.
	These should give a infinite triangular cylinder in the direction of the $y$-axis.
	The plane $y=5$ cuts this triangular cylinder at a right angle and the plane $y=z$
	cuts it a funny angle.  You now need to write down an integral for this, here are a couple:
\[
\int_0^{4/3} \int_z^{-2z+4} \int_{z}^{5} \; dy \;dx \;dz
\]
\[
\int_0^{4/3} \int_{z}^{5} \int_z^{-2z+4} \; dx \;dy \;dz
\]

\item  Evaluate your triple integral to compute the volume of $S$.
\\
\sol: These work out to be $328/27$.
\end{enumerate}


\item  Compute the following line integral:
\[
\int_C (x^2+y)\;dx + (y^2+x)\; dy
\]
where $C$ is the curve $y=x^2$ from $(3,9)$ to $(-2,4)$.
\\
\sol:
We parametrize the curve as $\mathbf{r}(t) = (t, t^2)$ and we're interested in $t$ from $3$ to $-2$.
We make the substitutions:
\[
\begin{array}{ccc}
x = t & \qquad & y = t^2 \\
dx = dt & & dy = 2t\;dt
\end{array}
\]
which gives the integral
\[
\int_{3}^{-2} (t^2+t^2)\; dt + (t^4+t) (2t\; dt) 
= \int_{3}^{-2} 4t^2+2t^5\;dt
\]
which we compute to be $-805/3$.


\item  Consider the line integral below:
\[
\int_C y \; dx + (x+z^2)\; dy + 2yz\; dz
\]
where $C$ is a curve from $(1,2,3)$ to $(-1,-1,2)$.
\begin{enumerate}
\item  Find a vector field $\mathbf{F}$ so that the line integral can be written $\int_C F \dotprod d\mathbf{r}$.
\\
\sol:
$\mathbf{F} = (y, x+z^2, 2yz)$.

\item  Show that the above integral is independant of path.
\\
\sol: we just have to show that the curl of $\mathbf{F}$ is the zero vector:
\[
\mathbf{Curl }F = 
\left|
\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y & x+z^2 & 2yz
\end{array}
\right|
= (0,0,0)
\]
\item  Find a potential function for $\mathbf{F}$.
  \\
  Hint: Notice that $\mathbf{F}$ should be a vector in $\R^3$ and the potential function should be a function of three variables.
\\
\sol: This is now a matter of integrating.
We want to find $f$ so that 
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}  \right)
 = (y, x+z^2, 2yz)
\]
First, we integrate $\partial f/\partial x =y$ to get $f=xy + C(y,z)$.
We can take the partial of this with respect to $z$ (which must be equal to $2yz$).
$f_z = \partial C/\partial z (y,z) = 2yz$.
We can now integrate this with respect to $z$ (and remember that $C$ can not have any $x$'s in it) to get
$C=yz^2+C_2(y)$ (so $C_2$ is a function that can only have $y$'s).
So, now we have:
\[
f = xy+yz^2+C_2(y)
\]
We now take the partial of this with respect to $y$ (which must be equal to $x+z^2$) to get
$f_y = x+z^2+C_2'(y) = x+z^2$ and therefore $C_2'=0$ (and therefore $C_2$ is actually a constant.
So, here is $f$:
\[
f = xy+yz^2
\]

\item  Using the potential function, compute the line integral.
\\
\sol:
We can now just plug in the end points and subtract:
\begin{eqnarray*}
\int_C y \; dx + (x+z^2)\; dy + 2yz\; dz 
	& = & f(-1,-1,2) - f(1,2,3) \\
	& = & -3 - 20 = -23
\end{eqnarray*}


\end{enumerate}





\end{enumerate}

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