What is the area formed by the loop of the curve $x^3 - y^3= xy$?

Question

I know how to find area of elementary functions by integration. The only loop questions that I know how to solve are the ones that are of the petal type, which can be converted into polar form and then integrated using polar coordinates. In this question conversion into polar form was not much of a help.

I plotted the function on Desmos (Online Graphing Calculator) to see how it precisely looks. I have no idea how to proceed. I am trying to isolate y to form a piece wise function, but I am not able to it yet. Please help me with this, I would be grateful.

Answer 1

Suggestion / partial answer:

If you swap to polar coordinates, you get $r = \frac{\sin\theta\cos\theta}{\cos^3\theta - \sin^3\theta}$. This means that the area is $$ \int_{\pi/2}^\pi \left(\int_0^{\frac{\sin\theta\cos\theta}{\cos^3\theta - \sin^3\theta}}r\,dr\right)d\theta = \frac12\int_{\pi/2}^\pi\left(\frac{\sin\theta\cos\theta}{\cos^3\theta - \sin^3\theta}\right)^2\,d\theta $$ That's not an integral I like. I am sure there are trigonometric identities that can be used and simplifications that can be done, but I know of none that jump out to me as immediately convenient to use. Or it can be put into a computer to get...$\frac16$.

Answer 2

Here's how to do the integral $:= I$ from the answer of Arthur: Dividing numerator and denominator by $\cos^6(x)$, we get \begin{align} I = \int \frac{\frac{\sin(x)}{\cos^5(x)}}{(\cos^3(x) - \sin^3(x))^2\sec^6(x)} dx = \int \frac{\tan^2(x) \sec^2(x)}{(\tan^3(x) - 1)^2} dx \end{align} Now, using the substitution $u = \tan(x)$ and $du = \sec^2(x) dx$ we get \begin{align} I = \int \frac{u^2}{(u^3 - 1)^2} du \end{align} This you can evaluate easily with a second substitution: $v = u^3 - 1$.