What is the area of the region bounded by the parabola $x^{2} = -16y$ and the line $y = -4$?

Explanation

1. 01

   The line meets the parabola where $x^{2} = 64$, at $x = \pm 8$; between them the parabola $y = -\frac{x^{2}}{16}$ lies above the line $y = -4$.

2. 02

   $\int_{-8}^{8} \left(4 - \frac{x^{2}}{16}\right)dx = 64 - \frac{64}{3} = \frac{128}{3}$ square units, which is $\frac{8}{3}a^{2}$ with $a = 4$.