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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.

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2 Answers

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$$x^{2/3}+y^{2/3}=a^{2/3}\longleftrightarrow\;x=a\cos^3t\;,\;\;y=a\sin^3t\;,\;\;0\le t\le 2\pi$$

So

$$\frac12\int\limits_0^{2\pi}\left[(a\cos^3t\cdot3a\sin^2t\cos t)-(a\sin^3t\cdot(-3a\cos^2t\sin t)\right]dt =$$

$$=\frac{3a^2}2\int\limits_0^{2\pi}(\cos^4t\sin^2t+\cos^2t\sin^4t)dt=\frac{3a^2}2\int\limits_0^{2\pi}\cos^2t\sin^2t\,dt=$$

$$=\left.\frac{3a^2}{64}\left(4x-\sin 4x\right)\right|_0^{2\pi}=\frac{3a^2}{64}8\pi=\frac38a^2\pi$$

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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 $x^{2/3}+y^{2/3}=a^{2/3}$.

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