What is the area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$?

Explanation

1. 01

   By symmetry the area is 4 times the part in the first quadrant: $4\int_0^3 \frac{2}{3}\sqrt{9 - x^2}\,dx$.

2. 02

   $\int_0^3 \sqrt{9 - x^2}\,dx$ is a quarter circle of radius 3, $\frac{9\pi}{4}$, so the area is $4 \cdot \frac{2}{3} \cdot \frac{9\pi}{4} = 6\pi$, that is $\pi ab$ with $a = 3$, $b = 2$.