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 $A=\frac{3\pi}{32}$.
 A: You only graphed the first quadrant part of the curve. There are $3$ other parts, obtained by reflections in the axes. For note that $(x,y)$ is on the curve if and only if $(-x,y)$ is on the curve if and only if $\dots$.
The area of the region they enclose is $\frac{3\pi}{8}$. Your calculation is correct. If you only want the first quadrant area (but that is not what is asked for) divide by $4$, or integrate from $0$ to $\pi/2$ instead of $0$ to $2\pi$.
A: 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.
A: Notice that we have symmetry about the $y$ and $x$ axes. Hence we can use the formula
$$y=(1-x^{2/3})^{3/2}$$
Which gives the top half, integrate from $x=0$ to $1$, and then multiply by $4$, exploiting the symmetry. Hence we have
$$A=4\int_0^1(1-x^{2/3})^{3/2}dx$$
Letting $x=\cos(u)^{6/2}=\cos^3(u)$, $dx=-3\cos^2(u)\sin(u)du$ we have
$$\begin{align}
A &=4\cdot 3\int_0^{\pi/2}\sin^4(u)\cos^2(u)du \\
&= 4\cdot\frac{3\pi}{32}
\end{align}$$
Hence we have the area $3\pi/32$ like you got, but we must account for the other pieces.
