Area bound by algebraic curve $x^4 + y^3 = x^2y$ in first quadrant The area of the loop formed by the algebraic curve $x^4 + y^3 = x^2y$ in the first quadrant can be expressed in the form $A/B$, where $A$ and $B$ are relatively prime. Find $A+B$.
My attempt: I tried using a polar substitution $x = (rcosa)^{1/2}$ and $y = (rsina)^{2/3}$, but the resulting equation ended up being too difficult to integrate. Any help would be appreciated.
 A: HINT:
If we draw the curve with some software ( for instance with WA ) then we see three branches through the point $(0,0)$. Therefore, a line $y = t x$ intersects the curve at $(0,0)$ with multiplicity $3$. There should be one more point of intersection, it's 
$$(x,y) = (t-t^3, t^2 - t^4)=\gamma(t)$$
The desired loop is described by $0\le t \le 1$. The area equals
$$\int_{\gamma} x d y $$
Obs: In general, every curve of form $P_n(x,y) + P_{n-1}(x,y)=0$, where $P_n$, $P_{n-1}$ are homogenous of degree $n$, $n-1$, is parametrizable in a similar way. 
A: From the expressions I would try with polar coordinates. You get the equation
$$
r^4\cos^4\theta+r^3\sin^3\theta=r^3\cos^2\theta\sin\theta
$$
from where you can conclude that you are interested in
$$
r(\theta)=\frac{\cos^2\theta\sin\theta-\sin^3\theta}{\cos^4\theta}
$$
You will find easily that the correct limits for $\theta$ are $0$ and $\pi/4$. I leave it to you to find the area, i.e. to integrate (it is elementary, but perhaps not the easiest way)
$$
\frac{1}{2}\int_0^{\pi/4}r(\theta)^2\,d\theta.
$$
