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:
Step-by-Step Solution
Key Concept: Find intersection points of the circle and parabola, then compute the area as $2\int_0^{2\sqrt{3}}\!\left(\sqrt{12-(x-2\sqrt{3})^2} - \sqrt{2\sqrt{3}\,x}\right)dx$ using the standard formula for a circular arc area and a parabolic area.
Intersection: substitute $y^2 = 2\sqrt{3}\,x$ into $(x-2\sqrt{3})^2+y^2=12$. Solving gives $x=0$ and $x=2\sqrt{3}$, with $y=\pm 2\sqrt{3}$.
Required area $= 2\int_0^{2\sqrt{3}}\!\left(\sqrt{12-(x-2\sqrt{3})^2}-\sqrt{2\sqrt{3}\,x}\right)dx.$
Using the standard circular-arc formula and power rule:
$= 2\left[\frac{x-2\sqrt{3}}{2}\sqrt{12-(x-2\sqrt{3})^2}+6\sin^{-1}\frac{x-2\sqrt{3}}{2\sqrt{3}}-\frac{\sqrt{2\sqrt{3}}\,x^{3/2}}{\tfrac{3}{2}}\right]_0^{2\sqrt{3}}$
$= 2(3\pi-8) = 6\pi-16.$
Correct Answer: 2