Evaluate the I? This  is question is inspired from this
find the value   of $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}\sqrt{ \frac{u^4}{a^2} +u^2} du dv$$
My attempt : $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du dv$$
after that im not able to solved this

Before I worked out a complete solution, @TitoEliatron has left a key step in his comment.
\begin{align}
I&=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} \,du \, dv\\
&=\int_0^{2\pi}dv \; \frac12 \int_0^{\sqrt2 a} \sqrt{ \frac{u^2}{a^2} +1} \, d(u^2) \\
&= \pi \int_0^{2a^2} \sqrt{x/a^2 + 1} \, dx \tag{$x = u^2$} \\
&= \frac{\pi}{a} \int_0^{2a^2} \sqrt{x + a^2} \, dx \\
&= \frac{2\pi}{3a} \left[ (x+a^2)^{3/2} \right]_0^{2a^2} \\
&= \frac{2\pi (3\sqrt3 - 1) a^2}{3}
\end{align}
Edit : The $a^3$ in the numerator cancelled with $a$ in the denominator to give $a^2$.
