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\usepackage{epsfig}
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{\bf Question}

Evaluate the following double integral (you should first sketch
the region of integration): $$\int_{x=0}^{x=1} \!
\int_{y=x}^{y=\sqrt x}x^2y \,dy dx.$$ Reverse the order of
integration and show that this gives the same answer.


{\bf Answer}

The region of integration is defined as follows:
$$\begin{array}{rl} {\rm{first\ keep\ x\ constant\ and\ vary\ y:}}
& {x\leq y \leq \sqrt x}\\ {\rm{then\ vary\ x:}} & {0\leq x\leq 1}
\end{array}$$

\begin{center}
$ \begin{array}{c}
\epsfig{file=158-9-14.eps, width=50mm}
\end{array}
\ \ \
\begin{array}{c}
(\rm{Note:\ } y=\sqrt{x} \textrm{is one branch}\\
\textrm{of the parabola } x=y^2)
\end{array} $
\end{center}

\begin{eqnarray*}  {\int_{x=0}^{x=1} \left\{\int_{y=x}^{y=\sqrt x}
x^2y \,dy \right\} dx} & =
& {\int_0^1 \left[\frac{x^2y^2}{2}\right]_{y=x}^{y=\sqrt x}dx}
\\ & = &
\int_0^1\frac{x^3}{2}-\frac{x^4}{2}\,dx\\ & = &
\left[\frac{x^4}{8}-\frac{x^5}{10}\right]_0^1 \\ & = &
\frac{1}{8}-\frac{1}{10} \\ & = & \frac{1}{40}\end{eqnarray*}

To reverse the order of integration, take horizontal lines ($y$
fixed):

\begin{center}
$ \begin{array}{c}
\epsfig{file=158-9-15.eps, width=50mm}
\end{array}
\ \ \
\begin{array}{lrll}
\mathrm{first\ vary}\ x &: y^2 & \leq & x \leq y\\
\mathrm{then\ vary}\ y &:  0 & \le & y \le 1
\end{array} $
\end{center}

Integral becomes:

\begin{eqnarray*}
 {\int_{y=0}^{y=1} \left\{\int_{x=y^2}^{x=y} x^2y \,dx \right\}
dy} & = & {\int_0^1 \left[\frac{x^3y}{3}\right]_{x=y}^{x=y^2}dy}\\
& = & \int_0^1\frac{y^4}{3}-\frac{y^7}{3} dy\\ & = &
\left[\frac{y^5}{15}-\frac{y^8}{24}\right]_0^1 \\ & = &
\frac{1}{15}-\frac{1}{24} \\ & = & \frac{3}{120}\\ & = &
\frac{1}{40}\end{eqnarray*}



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