JEE Main 2025 (Online) 22nd January Morning ShiftQuestion 1
The area of the region, inside the circle $\left(x-2\sqrt{3}\right)^{2}+y^{2}=12$ and outside the parabola $y^{2}=2\sqrt{3}x$ is :
Application of Integrals
Solution
The curves meet at $x=0$ and $x=2\sqrt3$ (the circle's centre line). Common area $=2\int_0^{2\sqrt3}\sqrt{2\sqrt3x}\,dx+\frac12\cdot12\pi=16+6\pi$, so the required area is $12\pi-(16+6\pi)=6\pi-16$.