Greetings I want to find the surface given by the curve $\Gamma:(x^2+y^2)^3=x^2y\,$ The graph looks pretty nice( see the picture bellow) and in my book it says that after the parametrization the Area is just $$I=\int_0^{\frac{\pi}{2}} r^2d\theta=\int_0^{\frac{\pi}{2}}\cos^{\frac{4}{3}}(\theta)\sin^{\frac{2}{3}}(\theta)d\theta$$ The last one I know how to evaluate using beta function $$I=\frac{1}{2}\beta\left(\frac{7}{6},\frac{5}{6}\right)=\frac{1}{2}\frac{1}{6}\frac{\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})}{\Gamma(2)}=\frac{1}{12}\frac{\pi}{\sin(\frac{\pi}{6})}=\frac{\pi}{6}$$ However I dont understand how did they parametrize that in order to get that integral. My thoughts are that they used polar coordinates and they got $0$ and $\frac{\pi}{2}$ as bounds because the graph is in the upper semiplane. However I know that the Area given by a curve is given by $A=\frac{1}{2}\int_{\Gamma} (xdy-ydx)$ How did thet got there?
Plug $x=r\cos\theta$, $y=r\sin\theta$ into the equation to obtain $$ (r^2)^3 = r^2\cos^2\theta \cdot r\sin\theta$$ or (after gettiung rid of the trival solution $r=0$) $$ r^3=\cos^2\theta\cdot\sin \theta$$ $$ r = \cos^{2/3}\theta\cdot\sin^{1/3} \theta$$ $$ r^2 = \cos^{4/3}\theta\cdot\sin^{2/3} \theta$$ Now recall that the formula to obtain the area from polar coordinates is $$ I=\frac12\int_\alpha^\beta r^2\,\mathrm d\theta.$$ They should have used this with $\alpha=0$, $\beta=\pi$. However, using the symmetr< of the curve, they used $\beta=\frac\pi2$ instead and multiplied by $2$ (thus getting rid of the $\frac 12$)