It's so cute!

This problem can be easily solved if one uses polar coordinates.

Substitute $x=r\cos t$ and $y=r\sin t$ . I am using $t$ instead of $\theta$ . The equation changes to:

$\displaystyle r\cos^4t+\sin^3t=\cos^2t\sin t$

$\displaystyle \Rightarrow r=\tan t \sec t(1-\tan^2t)$

Before setting up the area integral, we need to find the limits for integration. Substitute $r=0$ to get $t=0$ and $t=\pi/4$ .

The area bounded by the curve is hence

$\displaystyle A=2\times \left(\int_0^{\pi/4} \frac{1}{2}r^2(t)\,dt\right)$

$\displaystyle \Rightarrow A=\int_0^{\pi/4} \tan^2t \sec^2t (1-\tan^2t)^2\,dt$

The above integral can solved by using the substitution $\tan t=x$ and comes out to be:

$\displaystyle \boxed{A=\dfrac{8}{105}}$

I tried polar first since it looked like a polar graph haha but I got the identities mixed up so the integral wasn't too clean.

Another way that I found is by parametrizing the curve using the usual $y = tx$ . Plugging in, we get $x^4 + x^3t^3 = x^2(xt)$ whence, $x=t-t^3$ and $y=t^2-t^4$ for $t\in\mathbb{R}$

Since we want the area of two copies of the loop in the first quadrant, we solve $x=y=0$ and get the first positive solution for $t$ , which is $1$ . We get the area by integrating about the first loop using Green's Theorem

$2A = 2\cdot\left(\dfrac{1}{2} \displaystyle\oint_C x\;dy-y\;dx\right)$

$=\displaystyle\int_0^1(t-t^3)(2t-4t^3) -(t^2-t^4)(1-3t^2)\;dt$

$=\dfrac{8}{105}$