What is the area of the region inside the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$ and above the line $y = 1$?

Explanation

1. 01

   $y = 1$ meets the ellipse where $\frac{x^{2}}{16} = \frac{3}{4}$, at $x = \pm 2\sqrt{3}$; the upper arc is $y = \frac{1}{2}\sqrt{16 - x^{2}}$.

2. 02

   $\frac{1}{2}\int_{-2\sqrt{3}}^{2\sqrt{3}} \sqrt{16 - x^{2}}\,dx = \left[\frac{x}{4}\sqrt{16 - x^{2}} + 4\sin^{-1}\frac{x}{4}\right]_{-2\sqrt{3}}^{2\sqrt{3}} = 2\sqrt{3} + \frac{8\pi}{3}$.

3. 03

   Take away the rectangle under $y = 1$, of area $4\sqrt{3}$: $\frac{8\pi}{3} - 2\sqrt{3}$ square units.