Find Area Enclosed by Curve I want to find the area enclosed by the plane curve $x^{2/3}+y^{2/3}=1$. My attempt was to set $x=\cos^3t, \ y=\sin^3t$ so:$$x^{2/3}+y^{2/3}=\cos^2t+\sin^2t=1$$
Then the area is $$2A=\oint_Cxdy-ydx=3\oint_C\cos^3ty'dy+\sin^3tx'dx=3\int_0^{2\pi}\cos^2t\cdot \sin^2tdt=\frac{3\pi}{4}\implies A=\frac{3\pi}{8}$$
However, when I did a level curve plot I got the following figure:

so does "area enclosed by figure" even make sense? For the graph above, my calculator gives me $A=\frac{3\pi}{32}$.

In simple cartesian coordinates, through the substitution $x=z^{3/2}$ and Euler's Beta function,
$$ A = \int_{0}^{1}(1-x^{2/3})^{3/2}\,dx = \frac{3}{2}\int_{0}^{1}z^{1/2}(1-z)^{3/2}\,dz=\frac{3}{2}\,B\left(\frac{3}{2},\frac{5}{2}\right)\tag{1}$$
hence:
$$ A = \frac{3\,\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{5}{2}\right)}{2\,\Gamma(4)}=\color{red}{\frac{3\pi}{32}}\tag{2}$$
as wanted.
