Squashed ellipse

Relevant wiki: Green’s Theorem

Green's theorem gives the area as $\begin{aligned} \oint_C y \, dx &= \int_0^{2\pi} (a \sin^2(\theta) \cos(\theta)) (a \cos(\theta)) \, d\theta \\ &= a^2 \int_0^{2\pi} \sin^2(\theta) \cos^2(\theta) \, d\theta \\ &= a^2 \int_0^{2\pi} \frac14 \sin^2(2\theta) \, d\theta \\ &= \frac{a^2}4 \int_0^{2\pi} \left( \frac12 - \frac12 \cos(4\theta) \right) \, d\theta \\ &= \frac{\pi a^2}4 - \frac{a^2}8 \int_0^{2\pi} \cos(4\theta) \, d\theta \\ &= \frac{\pi a^2}4 - \frac{a^2}{32} \sin(4\theta) \bigg|_0^{2\pi} \\ &= \frac{\pi a^2}4. \end{aligned}$

Since the curve is symmetrical on the axes, the area it encloses is given by:

$\begin{aligned} A & = 4 \int_0^\frac \pi 2 y dx & \small \color{#3D99F6} x = a\sin \theta \implies dx = a \cos \theta \ d\theta \\ & = 4 \int_0^\frac \pi 2 a^2\sin^2 \theta \cos^2 \theta \ d\theta \\ & = 2a^2 B \left(\frac 32, \frac 32 \right) & \small \color{#3D99F6} B(m,n) \text{ is beta function.} \\ & = \frac {2a^2 \left(\Gamma \left(\frac 32 \right)\right)^2}{\Gamma \left(3 \right)} & \small \color{#3D99F6} \Gamma(x) \text{ is gamma function.} \\ & = \frac {2a^2 \left(\frac 12 \Gamma \left(\frac 12 \right)\right)^2}{2!} \\ & = a^2 \left(\frac 12 \sqrt \pi \right)^2 \\ & = \boxed{\dfrac 14 \pi a^2} \end{aligned}$