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QUESTION


\begin{description}

\item[(a)]
Sketch the region $R$ defined by the inequalities $x^2+y^2\leq1,
x\geq0$ and $y\geq0$. Express $R$ in terms of plane polar
coordinates. Hence evaluate the following double integral using
plane polar coordinates:

$$\int\!\!\!\int_Rxy\,d(x,y).$$

(HINT: You may assume the identity
$\sin(2\theta)=2\sin(\theta)\cos(\theta).$)

\item[(b)]
Sketch the region $A$ defined by the inequalities $y\leq4-x^2$ and
$y\geq(2-x)^2$. Write the double integral

$$\int\!\!\!\int_A2x\sqrt{y}\,d(x,y)$$

so that the integration with respect to $x$ is performed first.
Hence evaluate the integral, and show that it is equal to

$$\frac{2^5}{5}.$$

\end{description}



ANSWER


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\item[(a)]

DIAGRAM

For plane polar coordinates:
$\left.\begin{array}{c}x=r\cos\theta\\y=r\sin\theta\end{array}\right\}$

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\begin{picture}(9,7)

\put(1,1){\vector(1,0){7}} \put(1,1){\vector(0,1){5}}
\put(1,1){\line(3,2){7}}

\put(1,6.2){$y$} \put(8,.2){$x$} \put(4,3.5){$r$}
\put(3,1.5){$\theta$}

\end{picture}

$R$ is defined by $\left.\begin{array}{c}0\leq
r\leq1\\0\leq\theta\leq\frac{\pi}{2}\end{array}\right\}$

An area element in plane polar coordinates $rdrd\theta$. So the
integral becomes

\begin{eqnarray*}
&amp;&amp;\int_{\theta=0}^{\theta=\frac{\pi}{2}}\!\int_{r=0}^{r=1}(r\cos\theta)(r\sin\theta)rdrd\theta\\
&amp;=&amp;\int_0^\frac{\pi}{2}\int\theta\cos\theta\,d\theta\int_0^1r^3\,dr\\
&amp;=&amp;\int_0^\frac{\pi}{2}\frac{1}{2}\sin2\theta\,d\theta\int_0^1r^3\,dr
\textrm{ (using }\sin2\theta=2\sin\theta\cos\theta)\\
&amp;=&amp;\left[-\frac{1}{4}\cos2\theta\right]_0^\frac{\pi}{2}\left[\frac{1}{4}r^4\right]_0^1\\
&amp;=&amp;\left(-\frac{1}{4}\cos\pi+\frac{1}{4}\cos0\right)\left(\frac{1}{4}\right)=\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)=\frac{1}{8}
\end{eqnarray*}

\item[(b)]

DIAGRAM

To perform integration with respect to $x$ first the limits become

\begin{eqnarray*}
2-\sqrt{y}\leq&amp;x&amp;\leq\sqrt{4-y}\\ 0\leq&amp;y&amp;\leq4
\end{eqnarray*}

So the integral is

\begin{eqnarray*}
&amp;&amp;\int_{y=0}^{y=4}\!\int_{x=2-\sqrt{y}}^{x=\sqrt{4-y}}2xy^\frac{1}{2}\,dxdy\\
&amp;=&amp;\int_{y=0}^{y=4}\left[x^2y^\frac{1}{2}\right]_{x=2-\sqrt{y}}^{x=\sqrt{4-y}}\,dy\\
&amp;=&amp;\int_0^4(4-y)y^\frac{1}{2}-(2-y^\frac{1}{2})^2y^\frac{1}{2}\,dy\\
&amp;=&amp;\int_0^44y^\frac{1}{2}-y^\frac{3}{2}-y^\frac{1}{2}(4+y-4y^\frac{1}{2})\,dy\\
&amp;=&amp;\int_0^44y-2y^\frac{3}{2}\,dy\\
&amp;=&amp;\left[2y^2-\frac{4}{5}y^\frac{5}{2}\right]_0^4\\
&amp;=&amp;2.2^4-\frac{4}{5}(2^2)^\frac{5}{3}\\
&amp;=&amp;2^5-\frac{4}{5}.2^5=2^5\left(1-\frac{4}{5}\right)=\frac{2^5}{5}
\end{eqnarray*}

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