The region inside the ellipse $\frac{x^{2}}{64} + \frac{y^{2}}{16} = 1$ and above the line $y = 2$ is found with vertical strips. Between which limits of $x$ is the integral taken?

Explanation

1. 01

   On $y = 2$: $\frac{x^{2}}{64} = 1 - \frac{4}{16} = \frac{3}{4}$, so $x^{2} = 48$ and $x = \pm 4\sqrt{3}$.

2. 02

   Between these the arc $y = \frac{1}{2}\sqrt{64 - x^{2}}$ lies above the line: $\int_{-4\sqrt{3}}^{4\sqrt{3}} \left(\frac{1}{2}\sqrt{64 - x^{2}} - 2\right)dx$.