The area of the region, inside the circle $(x-2 \sqrt{3})^2+y^2=12$ and outside the parabola $y^2=2 \sqrt{3} x$ is :

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Required area }=2 \int_0^{2 \sqrt{3}}\left(\sqrt{4 \sqrt{3} x-x^2}-\sqrt{2 \sqrt{3} x}ight) d x \ & =2 \int_0^{2 \sqrt{3}}\left(\sqrt{12-(x-2 \sqrt{3})^2}-\sqrt{2 \sqrt{3} x}ight) d x \ & =2\left[\frac{x-2 \sqrt{3}}{2} \sqrt{12-(x-2 \sqrt{3})^2}+\frac{12}{2} \sin ^{-1}\left(\frac{x-2 \sqrt{3}}{2 \sqrt{3}}ight)ight. \ & \left.-\frac{\sqrt{2 \sqrt{3}} x^{\frac{3}{2}}}{\frac{3}{2}}ight]_0^{2 \sqrt{3}} \ & =2\{3 \pi-8\} \ & =6 \pi-16 	ext { sq. units. }\end{aligned}$

The area of the region, inside the circle $(x-2 \sqrt{3})^2+y^2=12$ and outside the parabola $y^2=2 \sqrt{3}…

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