Which integral gives the area of the part of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{25} = 1$ above the $x$-axis?

Explanation

1. 01

   Above the $x$-axis, $y = 5\sqrt{1 - \frac{x^{2}}{16}} = \frac{5}{4}\sqrt{16 - x^{2}}$, for $-4 \le x \le 4$.

2. 02

   So the area is $\frac{5}{4}\int_{-4}^{4} \sqrt{16 - x^{2}}\,dx = \frac{5}{4} \times 8\pi = 10\pi$, half of $\pi \times 4 \times 5$.