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.
Best Answer
$$x^{2/3}+y^{2/3}=a^{2/3}\longleftrightarrow\;x=a\cos^3t\;,\;\;y=a\sin^3t\;,\;\;0\le t\le 2\pi$$
So
$$\frac12\int\limits_0^{2\pi}\left[(a\cos^3t\cdot3a\sin^2t\cos t)-(a\sin^3t\cdot(-3a\cos^2t\sin t)\right]dt =$$
$$=\frac{3a^2}2\int\limits_0^{2\pi}(\cos^4t\sin^2t+\cos^2t\sin^4t)dt=\frac{3a^2}2\int\limits_0^{2\pi}\cos^2t\sin^2t\,dt=$$
$$=\left.\frac{3a^2}{64}\left(4x-\sin 4x\right)\right|_0^{2\pi}=\frac{3a^2}{64}8\pi=\frac38a^2\pi$$