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\begin{center}
\textbf{Multiple Integration}

\textit{\textbf{Iteration of Double Integrals}}
\end{center}

\textbf{Question}

Find the volume for the solid defined by 

The space inside the cylinder $x^2+2y^2=8$ that is over the plane
$z=y-4$ and under the plane $z=8-x$.


\textbf{Answer}

The part of $z=8-x$ that lies inside $x^2=2y^2=8$ lies over $z=0$. The
part of $z=y-4$ that lies inside the cylinder lies under $z=0$.
\begin{eqnarray*}
\Rightarrow V &amp; = &amp; \int\!\!\!\int_{x^2+2y^2 \le 8} (8-x-(y-4))\,dA\\
&amp; = &amp; \int\!\!\!\int_{x^2+2y^2 \le 8} 12 \,dA\\
&amp; = &amp; 12 \times (\textrm{area of } \frac{x^2}{8} + \frac{y^2}{4}=1)\\
&amp; = &amp; 12 \times \pi(2\sqrt{2})(2)\\
&amp; = &amp; 48\sqrt{2}\pi \textrm{cu. units}
\end{eqnarray*}

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