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{\bf Question}

Evaluate the double integral $$\int \!\!\! \int_R y\,d(x,y)$$
where $R$ is the first quadrant of the ellipse $2x^2+3y^2=1$.



{\bf Answer}

\begin{center}
$ \begin{array}{c}
\epsfig{file=158-9-12.eps, width=70mm}
\end{array}
\ \ \
\begin{array}{c}
2x^2+3y^2=1
\end{array} $
\end{center}

We will write the integral so that the integration with respect to
$y$ is performed first:

\begin{center}
$\begin{array}{c}
\textrm{Take vertical lines} ($x$ \textrm{fixed})\\
\textrm{so the region is defined by}\\
{0 \leq y } \leq \sqrt{\frac{1-2x^2}{3}}\\ 
{0 \leq x }  \leq \frac{1}{\sqrt2}
\end{array}
\begin{array}{c}
\epsfig{file=158-9-13.eps, width=50mm}
\end{array} $
\end{center}


The integral becomes

\begin{eqnarray*}  \int\!\!\!\int_R y
\,d(x,y) & = &
\int_{x=0}^{x=\frac{1}{\sqrt2}}\left\{\int_{y=0}^{y=\sqrt{\frac{1-2x^2}{3}}}
y \,dy \right\}\,dx \\ & = &
\int_0^{\frac{1}{\sqrt2}}\left[\frac{y^2}{2}\right]_{y=0}^{y=\sqrt{\frac{1-2x^2}{3}}}\,dx
\\ & = & \int_0^{\frac{1}{\sqrt2}} \frac{1-2x^2}{6} \,dx\\ & = &
\left[\frac{x}{6}-\frac{x^3}{9}\right]_0^{\frac{1}{\sqrt2}} \\ & =
& \frac{1}{6\sqrt2}-\frac{1}{18\sqrt2} \\ & = & \frac{1}{9\sqrt2}
\end{eqnarray*}


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