\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}

Sketch the domain of integration, and calculate the iterated integral
for
$$\int_0^1 \,dx \int_x^{x^{1/3}} \sqrt{1-y^4} \,dy$$


\textbf{Answer}

$$ \ $$
\begin{center}
\epsfig{file=MI-2A-4.eps, width=50mm}
\end{center}

\begin{eqnarray*}
I & = & \int_0^1 \,dx \int_x^{x^{1/3}} \sqrt{1-y^4} \,dy \\
& = & \int\!\!\!\int_R \sqrt{1-y^4} \,dA\\
& = & \int_0^1 y\sqrt{1-y^4} \,dy - \int_0^1 y^3\sqrt{1-y^4} \,dy
\end{eqnarray*}
$$\textrm{Let }u=y^2 \ \ \ \ \ \textrm{Let }v=1-y^4$$
$$\Rightarrow du=2ydy \ \ \ \ \ \Rightarrow dv=-4y^3dy$$
\begin{eqnarray*}
\Rightarrow I & = & \frac{1}{2} \int_0^1 \sqrt{1-u^2} \,du +
\frac{1}{4} \int_1^0 v^{1/2} \,dv\\
& = & \frac{1}{2} \left ( \frac{\pi}{4} \times 1^2 \right ) +
\left. \frac{1}{6} v^{3/2} \right |_1^0\\
& = & \frac{\pi}{8}-\frac{1}{6}
\end{eqnarray*}

\end{document}