The area bounded by the curve ${\mathrm{x}}^{2/3}+{\mathrm{y}}^{2/3}={\mathrm{a}}^{2/3}$ and $\overrightarrow{\mathrm{OX}}, \overrightarrow{\mathrm{OY}}$ is

$$\begin{gathered} \mathrm{R}.\mathrm{A}={\int }_0^{\mathrm{a}} \mathrm{xdy} ={\int }_0^{\mathrm{a}} {\left(a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-y^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{dy} ; \\ put x= a {\cos }^3\theta ,y=a {\sin }^3\theta \Rightarrow dy=3a{\sin }^2\theta \cos \theta d\theta \end{gathered}$$

$R.A = {\int }_0^{\mathrm{\pi }/2}a{\left(1-{\sin }^2\theta \right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} 3a{\sin }^2\theta \cos \theta d\theta$

$= {\int }_0^{\mathrm{\pi }/2} {\mathrm{acos}}^3\mathrm{\theta }\cdot 3{\mathrm{asin}}^2\mathrm{\theta }\cdot \cos \mathrm{\theta d\theta }$

$=3a^2{\int }_0^{\mathrm{\pi }/2} {\cos }^4\mathrm{\theta }\cdot {\sin }^2\mathrm{\theta }\cdot \mathrm{d\theta }$

$=3a^2\frac{3.1.1}{6.4.2}\frac{\pi }{2}=\frac{3\pi a^2}{32}$