Solution:

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 OAB.

The point of intersection of the line and the circle in the first quadrant is $(\sqrt{3}, 1)$. Area $O A B=$ Area $\triangle O C A+$ Area $A C B$
Area of OAC $=\frac{1}{2} \times \mathrm{OC} \times \mathrm{AC}=\frac{1}{2} \times \sqrt{3} \times 1=\frac{\sqrt{3}}{2}$
$\ldots(1)$
Area of $A B C=\int_{\sqrt{3}}^{2} y d x$
$\begin{array}{l}
=\int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} d x \\
=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2} \\
=\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2} \sqrt{4-3}-2 \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right] \\
=\left[\pi-\frac{\sqrt{3} \pi}{2}-2\left(\frac{1}{3}\right)\right] \\
=\left[\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}\right] \\
=\left[\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]
\end{array}$
As a result, area enclosed by $x$-axis, the line $x=\sqrt{3} y$, and the circle $x^{2}+y^{2}=4$ in the first quadrant $=\frac{\sqrt{3}}{2}+\left( \frac{\pi }{3}-\frac{\sqrt{3}}{2} \right)=\frac{\pi }{3}$sq. units