TITLE: Calculating area of astroid $x^{2/3}+y^{2/3}=a^{2/3}$ for $a>0$ using Green's theorem
QUESTION [6 upvotes]: question as follows.
Show that for any planar region $\Omega$, $$\mathrm{area}\left(\Omega\right)=\frac{1}{2}\oint_{\partial\Omega}(xdy-ydx).$$
Use this result to find the area enclosed by the astroid $x^{2/3}+y^{2/3}=a^{2/3}$ for $a>0$.
The first part's easy:
$$\begin{array}{l c l} 
\mathrm{RHS}&=&\frac{1}{2}\oint_{\partial\Omega}(xdy-ydx)\\
&=&\frac{1}{2}\oint_{\partial\Omega}(-ydx+xdy)\\
&=&\frac{1}{2}\iint_{\Omega}(1+1)dA\\
&=&\iint_{\Omega}1dA\\
&=&\mathrm{area}(\Omega)\\
&=&\mathrm{LHS}.
\end{array}$$
But the next part really has me stumped. In this case $\Omega=\{ (x,y) : x^{2/3}+y^{2/3}\le a^{2/3} \}$ and $$\mathrm{area}(\Omega)=\iint_{\Omega}1dA=\frac{1}{2}\oint_{\partial\Omega}(xdy-ydx)$$ but I have a lot of problems coming up with $\partial\Omega$ to make it work. Any ideas would be welcome! Thanks.

REPLY [5 votes]: Hint: You need to parametrize $x^{2/3}+y^{2/3}=a^{2/3}$ to compute the integral on the boundary.
Consider that the circle
$$
x^2+y^2=a^2
$$
is parametrized by
$$
x=a\cos(t)\qquad y=a\sin(t)
$$
Modifying slightly, we get that
$$
x=a\cos^3(t)\qquad y=a\sin^3(t)
$$
parametrizes $x^{2/3}+y^{2/3}=a^{2/3}$.