Calculating area of astroid $x^{2/3}+y^{2/3}=a^{2/3}$ for $a>0$ using Green's theorem

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

So

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

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

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

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