Explanation

(c)
$Required\ area = \int_{0}^{1}{\left( x_{2} - x_{1} \right)dy}$
$= \int_{0}^{1}\left( \sqrt{4 - y^{2}} - \sqrt{3}\ y \right)dy$
$= \left\lbrack \frac{1}{2}y\sqrt{4 - y^{2}} + \frac{1}{2}(4)\sin^{- 1}\frac{y}{2} - \frac{\sqrt{3}\ y^{2}}{2} \right\rbrack_{0}^{1}$
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$= \frac{\sqrt{3}}{2} + \sin^{- 1}\ \left( \frac{1}{2} \right) - \frac{\sqrt{3}}{2} - 2\ \sin^{- 1}0$
$= \frac{\pi}{3}sq\ units$
Alternate
$= Area = \frac{\theta}{360{^\circ}} \times {\pi r}^{2}$
$= \frac{30}{360} \times \pi(2)^{2}$
$= \frac{\pi}{3}sq\ units$