Find the area of the region in the first quadrant enclosed by

The area of the region bounded by the circle, $x^{2}+y^{2}=4, x=\sqrt{3} y$, and the $x$-axis is the area $\mathrm{OAB}$.

The point of intersection of the line and the circle in the first quadrant is $(\sqrt{3}, 1)$.

Area $\mathrm{OAB}=$ Area $\triangle \mathrm{OCA}+$ Area $\mathrm{ACB}$

Area of $O A C=\frac{1}{2} \times O C \times A C=\frac{1}{2} \times \sqrt{3} \times 1=\frac{\sqrt{3}}{2}$         ...(1)

Area of $\mathrm{ABC}=\int_{\sqrt{3}}^{2} y d x$

$=\int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} d x$

$=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}$

$=\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2} \sqrt{4-3}-2 \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right]$

$=\left[\pi-\frac{\sqrt{3} \pi}{2}-2\left(\frac{-}{3}\right)\right]$

$=\left[\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\right]$

$=\left[\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]$                            ...(2)

Therefore, required area enclosed $=\frac{\sqrt{3}}{2}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}$ square units