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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

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    $\begingroup$ $I=a^2\pi \int_0^{\sqrt2 a} (2u/a^2) (u^2/a^2+1)^{1/2}du$ $\endgroup$ – Tito Eliatron Jan 25 at 10:14
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    $\begingroup$ substitute $t = \frac{u^2}{a^2} + 1$ $\endgroup$ – Math Lover Jan 25 at 10:16
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    $\begingroup$ An observation to begin with: $$ I = \left( {\int_0^{2\pi } {dv} } \right)\left( {\int_0^{\sqrt 2 a} {\sqrt {\frac{{u^4 }}{{a^2 }} + u^2 } du} } \right). $$ $\endgroup$ – Gary Jan 25 at 10:24
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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 $a^2$.

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For

$$\int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du $$

use the substitution $t= \frac{u^2}{a^2}+1$. Then you get the integral

$$\frac{a^2}{2}\int_1^3 \sqrt{t} dt,$$

which is independent of $v$, hence

$$\int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du = 2 \pi \cdot \frac{a^2}{2}\int_1^3 \sqrt{t} dt.$$

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