We have an ellipse $x=a\cos(\theta)$ and $y=b\sin(\theta)$. We will find the area by adding sectors $\frac12r^2\;d\theta$. Now,
$$ \begin{align} \int_0^{2\pi}\frac12r^2\;d\theta&=\frac12\int_0^{2\pi}\left(a^2\cos^2\theta+b^2\sin^2\theta\right)\;d\theta\\ &=\frac12\int_0^{2\pi}\left(\frac{a^2+b^2}2+\frac{a^2-b^2}2\cos2\theta\right)\;d\theta\\ &=\left.\frac12\left(\frac{a^2+b^2}2\theta+\frac{a^2-b^2}4\sin2\theta\right)\right|_0^{2\pi}\\ &=\frac\pi2\left(a^2+b^2\right) \end{align} $$
Why is it not $\pi a b$?
