\documentclass[12pt]{article}
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\begin{document}
\begin{large}
\noindent
\makebox[\textwidth]{
	\textbf{Quiz 9 | MA261 | July 28, 2017}
}
\makebox[\textwidth]{Christina Jamroz, Alden Bradford}
\begin{enumerate}
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%%% PROBLEMS

\item \pts{10} Find a parametric representation of the part of the surface $x^2+y^2=4z^2$ that lies above the $xy$-plane.

\key{Switch to cylindrical coordinates. Then the equation becomes $r^2=4z^2$. Since we are looking above the $xy$-plane, this means $r=2z$, or $z=r/2$. $r$ and $\theta$ are the free variables, and the parametric representation is $\langle r\cos\theta,r\sin\theta,r/2\rangle$ where $0\leq \theta<2\pi$, $r>0$.}

%16.7.9
\item \pts{10} Evaluate $\iint_S x^2yz\;dS$, where $S$ is the part of the plane $z=1+2x+3y$ that lies above the rectangle $[0,3]\times[0,2]$.

\key{
Here $\pdv{z}{x}=2$ and $\pdv{z}{y}=3$ so $\sqrt{(\pdv{z}{x})^2+(\pdv{z}{y})^2+1}=\sqrt{14}$.\begin{align*}
\iint_S x^2yz\;dS&=\sqrt{14}\int_0^2\int_0^3 x^2y(1+2x+3y)\;dx\;dy\\
&=\sqrt{14}\int_0^2\int_0^3 x^2y + 2x^3y + 3x^2y^2\;dx\;dy\\
&=\sqrt{14}\int_0^2\frac{1}{3} x^3 y +\frac{1}{2}x^4y + x^3y^2\Big|_0^3\;dy\\
&=\sqrt{14}\int_0^2 9y+\frac{81}{2} y+27y^2\;dy\\
&=\sqrt{14}(\frac{9}{2}y^2+\frac{81}{4}y^2+9y^3)\Big|_0^2\\
&=\sqrt{14}(18+81+72)\\
&=171\sqrt{14}
\end{align*}
}



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\end{document}

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