What is the area enclosed by the ellipse?

Explanation

1. 01

   $a = \sqrt{36} = 6$ and $b = \sqrt{16} = 4$; in the first quadrant $y = \frac{2}{3}\sqrt{36 - x^{2}}$.

2. 02

   Area $= 4\int_0^{6} \frac{2}{3}\sqrt{36 - x^{2}}\,dx = 4 \times \frac{2}{3} \times 9\pi = 24\pi$ square units, which is $\pi ab$.