For $k > 0$, the region bounded by the parabola $y = kx^{2}$ and the line $y = 4$ has area $\frac{16}{3}$ square units. What is $k$?

Explanation

1. 01

   The line meets the parabola where $kx^{2} = 4$, at $x = \pm\frac{2}{\sqrt{k}}$.

2. 02

   Area $= 2\int_0^{2/\sqrt{k}} (4 - kx^{2})\,dx = 2\left(\frac{8}{\sqrt{k}} - \frac{8}{3\sqrt{k}}\right) = \frac{32}{3\sqrt{k}}$.

3. 03

   $\frac{32}{3\sqrt{k}} = \frac{16}{3}$ gives $\sqrt{k} = 2$, so $k = 4$.