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QUESTION

\begin{description}

\item[(a)]
Sketch the plane region $R$ defined by the inequalities
$3x^2+2y^2\leq1$ and $x\geq0$ and $y\geq0$ and evaluate the
following double integral:

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

\item[(b)]
Sketch the region $S$ defined by the inequalities $a^2\leq
x^2+y^2\leq b^2$ and $y\leq x$ and $y\geq0$, where $a$ and $b$ are
positive real numbers. Evaluate the integral
$\int\!\!\!\int_S\frac{y^2}{x^2}\,d(x,y)$. (HINT: You may use the
fact that $\tan^2\theta=\sec^2\theta-1$.)

\end{description}




ANSWER


\begin{description}

\item[(a)]

DIAGRAM

\begin{eqnarray*}
\int\!\!\!\int_Rxy\,d(x,y)&=&
\int_{x=0}^\frac{1}{\sqrt{3}}\!\int_{y=0}^{\sqrt{\frac{1-3x^2}{2}}}xy\,dydx\\
&=&\int_0^\frac{1}{\sqrt{3}}\left[\frac{zy^2}{2}\right]_{y=0}^{y=\sqrt{\frac{1-3x^2}{2}}}\,dx\\
&=&\int_0^\frac{1}{\sqrt{3}}\frac{4-3x^3}{4}\,dx\\
&=&\left[\frac{x^2}{8}-\frac{3x^4}{16}\right]_0^\frac{1}{\sqrt{3}}\\
&=&\left(\frac{1}{24}-\frac{1}{48}\right)=\frac{1}{48}
\end{eqnarray*}

\item[(b)]

DIAGRAM

\begin{eqnarray*}
\int\!\!\!\int_S\frac{y^2}{x^2}\,d(x,y)&=&
\int_{\theta=0}^\frac{\pi}{4}\!\int_{\rho=a
}^b\frac{\rho^2\sin^2\theta}{\rho^2\cos^2\theta}\rho\,d\rho
d\theta\\ &=&\int_{\theta=0}^\frac{\pi}{4}\left[\frac{
\rho^2}{2}\tan^2\theta\right]_{\rho=a}^b\,d\theta\\
&=&\frac{b^2-a^2}{2}\int_{\theta=0}^\frac{\pi}{4}\sec^2-1\,d\theta\\
&=&\frac{b^2-a^2}{2}\left[\tan\theta-\theta\right]_0^\frac{\pi}{4}\\
&=&\frac{b^2-a^2}{2}\left(1-\frac{\pi}{4}\right)\\
&=&\frac{(4-\pi)(b^2-a^2)}{8}
\end{eqnarray*}

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