\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\un}{\underline}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

\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 over the $xy$-plane and below the surface $z=1-x^2-2y^2$.


\textbf{Answer}

\begin{eqnarray*}
\textrm{Vol} & = & \int\!\!\!\int_E (1-x^2-2y^2) \,dA\\
& = & 4\int_0^1 \,dx \int_0^{\sqrt{(1-x^2)/2}} (1-x^2-2y^2) \,dy\\
& = & 4 \int_0^1 \left ( \frac{1}{\sqrt{2}}^{3/2} - \frac{2}{3}
\frac{(1-x^2)^{3/2}}{2\sqrt{2}} \right ) \,dx\\
& = & \frac{4\sqrt{2}}{3}\int_0^1 (1-x^2)^{3/2} \,dx\\
\textrm{Let }x & = & \sin \theta\\
du & = & \cos\theta d\theta\\
& & \\
\Rightarrow V & = & \frac{4\sqrt{2}}{3} \int_0^{\pi/2} \cos^4\theta\\
& = & \frac{4\sqrt{2}}{3} \int_0^{\pi/2} \left ( \frac{1+\cos
2\theta}{2} \right )^2 \,d\theta\\
& = & \frac{\sqrt{2}}{3} \int_0^{\pi/2} \left ( 1 + 2\cos\theta +
\frac{1 + \cos\theta}{2} \right ) \,d\theta\\
& = & \frac{\sqrt{2}}{3} \left. \left [ \frac{3\theta}{2} + \sin\theta +
\frac{1}{8} \sin 4\theta \right ) \right |_0^{\pi/2}\\
& = & \frac{\pi}{2\sqrt{2}} \textrm{cu. units}
\end{eqnarray*} 

\end{document}











