If ${\int }_0^{2a}{x}^2\sqrt{2ax-x^2}dx=k{a}^4$, then $k:\pi =$

Given that, ${\int }_0^{2a}{x}^2\sqrt{2ax-x^2}dx=k{a}^4$

Let, $I={\int }_0^{2a}{x}^2\sqrt{2ax-x^2}dx$

Put, $x=2a{\sin }^2\theta \Rightarrow dx=4a\sin \theta \cos \theta d\theta$

And, $\sqrt{2ax-x^2}=\sqrt{4a^2{\sin }^2\theta {\cos }^2\theta }=2a\sin \theta \cos \theta$.

Therefore, $I={\int }_0^{\frac{\mathrm{\pi }}{2}}4a^2{\sin }^4\theta ·2a\sin \theta \cos \theta ·4a\sin \theta \cos \theta d\theta$

$=32a^4{\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^6\theta {\cos }^2\theta d\theta$

$=32a^4\frac{5·3·1·1}{8·6·4·2}·\frac{\mathrm{\pi }}{2}=\frac{5{\mathrm{\pi a}}^4}{8}$.

Hence, we get, $k=\frac{5\mathrm{\pi }}{8}\Rightarrow k:\mathrm{\pi }=5:8$.