The area of the region, inside the circle (x-2√3)^2+y^2=12 and outside the parabola y^2=2√…

Question

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 :

Accepted Answer

$\begin{aligned} & 	ext { Required area }=2 \int_0^{2 \sqrt{3}}\left(\sqrt{4 \sqrt{3} x-x^2}-\sqrt{2 \sqrt{3} x}ight) d x \\ & =2 \int_0^{2 \sqrt{3}}\left(\sqrt{12-(x-2 \sqrt{3})^2}-\sqrt{2 \sqrt{3} x}ight) d x\end{aligned}$

$\begin{gathered} =2\left[\frac{x-2 \sqrt{3}}{2} \sqrt{12-(x-2 \sqrt{3})^2}+\frac{12}{2} \sin ^{-1}\left(\frac{x-2 \sqrt{3}}{2 \sqrt{3}}ight)ight. \left.-\frac{\sqrt{2 \sqrt{3}} x^{\frac{3}{2}}}{\frac{3}{2}}ight]_0^{2 \sqrt{3}} \end{gathered}$

$\begin{aligned} & =2\{3 \pi-8\} \\ & =6 \pi-16 	ext { sq. units. }\end{aligned}$

Solutions · 1