# 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.

• @B.Mehta In addition to my previous problem, it would be nice if you could provide a complete solution to these problems. I don't mind approving your answers and upvoting them, but I don't understand how parametric would even help in this and would rather see a solution to fully understand your thought process and how you solved this problem. Commented Apr 14, 2018 at 19:14

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.

• Nice solution! From the diagram, is it obvious that the substitution $y=tx$ gives a convenient parametrisation - how does the multiplicity help? Commented Apr 14, 2018 at 19:22
• @B. Mehta: It guarantees that the one more point of intersection has rational coordinates in $t$. Commented Apr 14, 2018 at 19:27
• Your solution is easier than mine. Very nice! (I could swear that I actually tried to put $y=t x$ first, but I must have done some error in my calculation.) Commented Apr 14, 2018 at 19:51
• @mickep: Thanks! I use software for calculations whenever possible... so by hand I might have done some error too :) Commented Apr 14, 2018 at 20:01
• I keep getting 4/105 Commented Apr 14, 2018 at 20:30

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) $$\frac{1}{2}\int_0^{\pi/4}r(\theta)^2\,d\theta.$$