area of sector $O A C B=\frac{r^{2}}{2} \theta=\frac{1}{2} \cdot \frac{\pi}{3}=\frac{\pi}{6}$
area of shaded region $=\frac{\pi}{6}-$ area of $\triangle O A B$
$$
=\frac{\pi}{6}-\frac{\sqrt{3}}{4}
$$
Hence area of line $=$ Area of semi-circle $-$ area of shaded region
$$
\begin{array}{l}
=\frac{1}{2} \pi\left(\frac{1}{2}\right)^{2}-\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \\
=\frac{\sqrt{3}}{4}+\frac{\pi}{8}-\frac{\pi}{6} \\
=\frac{\sqrt{3}}{4}-\frac{\pi}{24}
\end{array}
$$