The area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

1. By symmetry $A = 4\int_0^a y\,dx$ where $y = \frac{b}{a}\sqrt{a^2 - x^2}$ in the first quadrant. 2. Thus $A = \frac{4b}{a}\int_0^a \sqrt{a^2 - x^2}\,dx = \frac{4b}{a}\left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a$. 3. The bracket evaluates to $\frac{a^2}{2}\cdot\frac{\pi}{2}$. 4. So $A = \frac{4b}{a}\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi a b$. (Setting $a = b$ recovers $\pi a^2$, ruling out $\pi a^2 b$.) _Source: syllabus Mathematics Ch 8 "Application of Integrals", p.3_