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QUESTION

\begin{description}

\item[(a)]
By considering $\ds\int_0^Kxe^{-x}\,dx$ determine whether or not
the integral\\ $\ds\int_0^\infty xe^{-x}\,dx$ is defined, and if
it is find its value.

\item[(b)]
A double integral is defined by

$$I=\int_{y=0}^{y=1}\!\int_{x=0}^{x=\sqrt{1-y^2}}y\,dx\,dy.$$

\begin{description}

\item[(i)]
Sketch the region of integration.

\item[(ii)]
State the expression for an element of area in terms of plane
polar coordinates $(r,\theta)$.

\item[(iii)]
Rewrite the double integral $I$ in terms of plane polar
coordinates and hence evaluate the integral.

\end{description}

\end{description}

\bigskip

ANSWER

\begin{description}

\item[(a)]

\begin{eqnarray*}
\int_0^Kxe^{-x}\,dx
&=&\left[\frac{xe^{-x}}{-1}\right]_0^K-\int_0^K\left(\frac{e^{-x}}{-1}\right).1\,dx\\
&=&-\frac{K}{e^K}-0+\left[\frac{e^{-x}}{-1}\right]_0^K\\
&=&-\frac{K}{e^K}-\frac{1}{e^K}+(e^{-0})\\
&=&1-\frac{K}{e^K}-\frac{1}{e^K}
\end{eqnarray*}

$\ds\int_0^\infty
xe^{-x}\,dx=\lim_{K\to\infty}\left(\int_0^Kxe^{-x}\,dx\right)=1$
since both $\ds\frac{1}{e^K}$ and $\ds\frac{K}{e^K}$ tend to 0 as
$K$ tends to infinity.

\item[(b)]
$\ds I=\int_{y=0}^{y=1}\!\int_{x=0}^{x=\sqrt{1-y^2}}y\,dx\,dy$

\begin{description}

\item[(i)]
$x=\sqrt{1-y^2},\ x^2=1-y^2,$ i.e. $x^2+y^2=1$ (a circle)

\epsfig{file=160-2001-6.eps, width=40mm}

\item[(ii)]
$dA=rdrd\theta$

\item[(iii)]

\begin{eqnarray*}
I&=&\int_{\theta=0}^\frac{\pi}{2}\!\int_{r=0}^1(r\sin\theta)(r\,dr\,d\theta)\\
&=&\int_0^\frac{\pi}{2}\sin\theta\,d\theta\int_0^1r^2\,dr\\
&=&\left[-\cos\theta\right]_0^\frac{\pi}{2}\left[\frac{r^3}{3}\right]_0^1\\
&=&\left(-\cos\frac{\pi}{2}-(-\cos0)\right)\left(\frac{1}{3}-0\right)\\
&=&\frac{1}{3}
\end{eqnarray*}

\end{description}

\end{description}



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