Q. The area of the region, inside the circle $(x-2\sqrt3)^2+y^2=12$ and outside the parabola $y^2=2\sqrt3x$ is :
A) $3\pi+8$
B) $6\pi-16$
C) $3\pi-8$
D) $6\pi-8$
Explanation
[Solution by Dr. Abubacker] The curves meet at $x=0$ and $x=2\sqrt3$. For $0<x<2\sqrt3$ the circle lies outside the parabola, so area $=2\int_0^{2\sqrt3}\left(\sqrt{12-(x-2\sqrt3)^2}-\sqrt{2\sqrt3x}\right)dx=2(3\pi-8)=6\pi-16$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.