What is the area of the region in the first quadrant that lies below the line $y = 2$ and inside the circle $x^{2} + y^{2} = 16$?

Explanation

1. 01

   The region lies against the $y$-axis: horizontal strips from $x = 0$ to $x = \sqrt{16 - y^{2}}$, for $0 \le y \le 2$.

2. 02

   $\int_0^{2} \sqrt{16 - y^{2}}\,dy = \left[\frac{y}{2}\sqrt{16 - y^{2}} + 8\sin^{-1}\frac{y}{4}\right]_0^{2} = \sqrt{12} + 8 \times \frac{\pi}{6}$.

3. 03

   $= 2\sqrt{3} + \frac{4\pi}{3}$ square units.