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\begin{document}
\begin{large}
\noindent
\makebox[\textwidth]{
	\textbf{Quiz 5 | MA261 | July 11, 2017}
}
\makebox[\textwidth]{Christina Jamroz, Alden Bradford}
\begin{enumerate}
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%%% PROBLEMS

%problem 15.4.9
\item Evaluate $\displaystyle\iint_R\sin(x^2+y^2)\;dA$, where $R$ is the region in the first quadrant between the circles with center the origin and radii 1 and 3. 

\key{ Substitute using $x=r\cos\theta$, $y=r\sin\theta$, and $dA=rdrd\theta$. Then the integral becomes \begin{align*}
\int_0^{\pi/2}\int_1^3\sin(r^2)r\;drd\theta&=\int_0^{\pi/2}d\theta\int_1^3r\sin(r^2)dr\\
&=\frac{\pi}{2}\qty[-\frac{1}{2}\cos(r^2)]_1^3\\
&=\frac{\pi}{4}\qty(\cos 1-\cos 9)
\end{align*} }

\item Evaluate $\displaystyle \int_0^2\int_0^{3}\int_0^{z} (2x-y)\;dxdydz$.

\key{
\begin{align*}
\int_0^2\int_0^{3}\int_0^{z} (2x-y)\;dxdydz&=\int_0^2\int_0^{3} (x^2-yx)\Big|_0^z\;dydz\\
&=\int_0^2\int_0^{3} (z^2-yz)\;dydz\\
&=\int_0^2(yz^2-\frac{1}{2}y^2z)\Big|_0^3\;dz\\
&=\int_0^2(3z^2-\frac{9}{2}z)\;dz\\
&=z^3-\frac{9}{4}z^2\Big|_0^2\\
&=-1
\end{align*}
}

\end{enumerate}
\end{large}
\end{document}

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