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

$$\begin{gathered} \mathrm{R}.\mathrm{A}={\int }_0^{\mathrm{a}} \mathrm{xdy} ={\int }_0^{\mathrm{a}} {\left(a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-y^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{dy} ; put x= a {\cos }^3\theta ,y=a {\sin }^3\theta \Rightarrow dy=3a{\sin }^2\theta \cos \theta d\theta \\ R.A = {\int }_0^{\mathrm{\pi }/2}a{\left(1-{\sin }^2\theta \right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} 3a{\sin }^2\theta \cos \theta d\theta \\ = {\int }_0^{\mathrm{\pi }/2} {\mathrm{acos}}^3\mathrm{\theta }\cdot 3{\mathrm{asin}}^2\mathrm{\theta }\cdot \cos \mathrm{\theta d\theta } \\ =3a^2{\int }_0^{\mathrm{\pi }/2} {\cos }^4\mathrm{\theta }\cdot {\sin }^2\mathrm{\theta }\cdot \mathrm{d\theta } \\ =3a^2\frac{3.1.1}{6.4.2}\frac{\pi }{2}=\frac{3\pi a^2}{32} \end{gathered}$$