The area bounded by the circle $x^2 + y^2 = 4$ and the line $x = y\sqrt{3}$ in the first quadrant (in sq units) is

Circle: $x^2 + y^2 = 4 \Rightarrow r = 2$ Line: $x = \sqrt{3}y \Rightarrow \frac{x}{y} = \sqrt{3} \Rightarrow \tan\theta = \frac{y}{x} = \frac{1}{\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}$ Required region = sector of angle $\frac{\pi}{6}$ Area $= \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 4 \cdot \frac{\pi}{6} = \frac{\pi}{3}$