Explanation
\begin{align*}
& \text{Circle: } (x - 2\sqrt{3})^2 + y^2 = 12 \implies x = 2\sqrt{3} \pm \sqrt{12 - y^2} \\
& \text{Parabola: } y^2 = 2\sqrt{3}x \\
& \text{Point of intersection } \implies y^2 = 12 \implies y = \pm 2\sqrt{3} \\
& \therefore x = +2\sqrt{3} \\
& \text{Area of required region } = 2 \int_{0}^{2\sqrt{3}} (x_1 - x_2) \, dy \\
& = 2 \int_{0}^{2\sqrt{3}} \left( \frac{y^2}{2\sqrt{3}} - 2\sqrt{3} + \sqrt{12 - y^2} \right) \, dy \\
& = 2 \left[ \frac{y^3}{6\sqrt{3}} - 2\sqrt{3}y + \frac{y}{2}\sqrt{12 - y^2} + \frac{12}{2}\sin^{-1} \left( \frac{y}{2\sqrt{3}} \right) \right]_{0}^{2\sqrt{3}} \\
& = 2 \left[ 4 - 12 + 0 + 6 \times \frac{\pi}{2} \right] \\
& = 6\pi - 16
\end{align*}
