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\textbf{Multiple Integration}

\textit{\textbf{Iteration of Double Integrals}}
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\textbf{Question}

Find the volume of the given solid

Below $z=1-y^2$ and over $z=x^2$.


\textbf{Answer}

$z=1-y^2$ and $z=x^2$ interesct on the cylinder $x^2+y^2=1$, and so
the volume can be given by
\begin{eqnarray*}
V &amp; = &amp; \int\!\!\!\int_{x^2+y^2 \le 1} (1-y^2-x^2) \,dA\\
&amp; = &amp; 4 \int_0^1 \,dx \int_0^{\sqrt{1-x^2}} (1-x^2-y^2) \,dy\\
&amp; = &amp; 4 \int_0^1 \,dx \left. \left ((1-x^2)y - \frac{y^3}{3} \right )
\right |_{y=0}^{y=\sqrt{1-x^2}}\\
&amp; = &amp; \frac{8}{3} \int_0^1 (1-x^2)^{3/2} \,dx\\
\textrm{Let }x &amp; = &amp; \sin u\\
du &amp; = &amp; \cos u du\\
&amp; &amp; \\
\Rightarrow V &amp; = &amp; \frac{8}{3}\int_0^{\pi/2} \cos^4 u \,du\\
&amp; = &amp; \frac{2}{3} \int_0^{\pi/2} (1+\cos 2u)^2 \,du\\
&amp; = &amp; \frac{2}{3} \int_0^{\pi/2} \left ( 1 + 2\cos 2u + \frac{1+\cos
4u}{2} \right ) \,du\\
&amp; = &amp; \frac{2}{3} \frac{3}{2} \frac{\pi}{2} = \frac{\pi}{2}
\textrm{cu. units}
\end{eqnarray*}

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