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{\bf Question}

If $R$ is the region defined by the inequalities  $x^2+y^2\leq 1$,
$x\geq 0$ and $y\geq 0$, evaluate the double integral $$\int
\!\!\! \int_R (xy+1)\,d(x,y)$$ by first transforming it into plane
polar co-ordinates.



{\bf Answer}

\begin{center}
$ \begin{array}{c}
\epsfig{file=158-9-16.eps, width=50mm}
\end{array}
\ \ \
\begin{array}{c}
\textrm{R is defined by the } r,\theta \rm{inequalities}\\
 \begin{array} {ccccc} 0 & \leq & r & \leq & 1\\ 0 & \leq & \theta
 & \leq & \frac{\pi}{2}
 \end{array}
\end{array}$
\end{center}

Since $x=r\cos \theta$ and $y=r\sin \theta$ we have:

$$xy=(r\cos \theta)(r\sin \theta)=r^2\sin \theta \cos
\theta=\frac{1}{2}r^2\sin 2 \theta$$

The integral becomes:
\begin{eqnarray*} \int \!\!\! \int_R (xy+1)\,d(x,y)& = &
\int_{\theta=0}^{\theta=\frac{\pi}{2}} \!
\int_{r=0}^{r=1}\left(\frac{1}{2}r^2\sin 2\theta+1\right)r \,dr
d\theta\\ & = & \int_{\theta =0}^{\theta = \frac{\pi}{2}}
\int_{r=0}^{r=1}\frac{1}{2}r^3\sin 2\theta \,dr
d\theta+\int_{\theta = 0}^{\theta = \frac{\pi}{2}}
\int_{r=0}^{r=1}r \,dr d\theta\\ & = &
\left\{\int_0^{\frac{\pi}{2}}\frac{1}{2}\sin 2\theta \,d \theta
\right\} \left\{\int_0^1 r^3 \,dr \right\} +
\left\{\int_0^{\frac{\pi}{2}} d \theta \right\} \left\{\int_0^1 r
\,dr \right\}\\ & {} & ({\rm since\ all\ limits\ are\ independent\
of\ } r\ {\rm and}\ \theta)\\ & = & \left[ -\frac{\cos 2\theta}{4}
\right]_0^{\frac{\pi}{2}} \left[ \frac{r^4}{4} \right]_0^1 +
\left[\theta
\right]_0^{\frac{\pi}{2}}\left[\frac{r^2}{2}\right]_0^1
\\& = & \left(\frac{1}{4}+\frac{1}{4}\right)\left(\frac{1}{4}\right)+\left(
\frac{\pi}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{8}+\frac{\pi}{8}
\end{eqnarray*}


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