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The area (in square units) of the region enclosed by the ellipse $x^2+3 y^2=18$ in the first quadrant below the line $y=x$ is
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Verified Answer
The correct answer is:
$\sqrt{3} \pi$
$\frac{x^2}{18}+\frac{y^2}{6}=1$
$\begin{aligned}
& \frac{x^2}{18}+\frac{3 x^2}{18}=1 \Rightarrow 4 x^2=18 \Rightarrow x^2=\frac{9}{2} \\
& \int_{\frac{3}{\sqrt{2}}}^{\sqrt[3]{2}} \frac{\sqrt{18-x^2}}{\sqrt{3}} d x \\
& =\frac{1}{\sqrt{3}}\left(\frac{x \sqrt{18-x^2}}{2}+\frac{18}{2} \sin ^{-1} \frac{x}{3 \sqrt{2}}\right)_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\
& =\frac{1}{\sqrt{3}}\left(9 \times \frac{\pi}{2}-\frac{3}{2 \sqrt{2}} \times \frac{3 \sqrt{3}}{\sqrt{2}}-9 \times \frac{\pi}{6}\right)
\end{aligned}$
Required Area
$\begin{aligned}
& =\frac{1}{2} \times \frac{9}{2}+\left(\frac{18 \pi}{6}-\frac{9 \sqrt{3}}{4}\right) \frac{1}{\sqrt{3}} \\
& =\sqrt{3} \pi
\end{aligned}$
$\begin{aligned}
& \frac{x^2}{18}+\frac{3 x^2}{18}=1 \Rightarrow 4 x^2=18 \Rightarrow x^2=\frac{9}{2} \\
& \int_{\frac{3}{\sqrt{2}}}^{\sqrt[3]{2}} \frac{\sqrt{18-x^2}}{\sqrt{3}} d x \\
& =\frac{1}{\sqrt{3}}\left(\frac{x \sqrt{18-x^2}}{2}+\frac{18}{2} \sin ^{-1} \frac{x}{3 \sqrt{2}}\right)_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\
& =\frac{1}{\sqrt{3}}\left(9 \times \frac{\pi}{2}-\frac{3}{2 \sqrt{2}} \times \frac{3 \sqrt{3}}{\sqrt{2}}-9 \times \frac{\pi}{6}\right)
\end{aligned}$
Required Area
$\begin{aligned}
& =\frac{1}{2} \times \frac{9}{2}+\left(\frac{18 \pi}{6}-\frac{9 \sqrt{3}}{4}\right) \frac{1}{\sqrt{3}} \\
& =\sqrt{3} \pi
\end{aligned}$
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