What is the area of the region in the first quadrant that lies below the line $y = 4$ and inside the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{64} = 1$?

Explanation

1. 01

   The region lies against the $y$-axis, so solve for $x$: $x = 4\sqrt{1 - \frac{y^{2}}{64}} = \frac{1}{2}\sqrt{64 - y^{2}}$, for $0 \le y \le 4$.

2. 02

   $\frac{1}{2}\int_0^{4} \sqrt{64 - y^{2}}\,dy = \frac{1}{2}\left[\frac{y}{2}\sqrt{64 - y^{2}} + 32\sin^{-1}\frac{y}{8}\right]_0^{4} = \frac{1}{2}\left(8\sqrt{3} + \frac{16\pi}{3}\right)$.

3. 03

   $= 4\sqrt{3} + \frac{8\pi}{3}$ square units.