As to how the $2\pi$ become $\pi/6$, the explanation is simple. The person wanted a $4$ on the "inside". But if you multiply by $4$, you also have to divide by $4$, in order to leave the integral unchanged. This division was done on the outside. Earlier, the $1/3$ had been brought outside.
I think the layout of the calculation was not as good as it might be, it tends to create confusion. First of all, I would write the area formula as $\int_0^3 2\pi\frac{x^3}{3}\sqrt{1+x^4}\,dx.$
The reason for this is that the $2\pi$ "belongs with" $x^3/3$, the expression $2\pi x^3/3$ represents the circumference of a certain circle.
Then for the indefinite integral, I would probably do the substitution more mechanically and less magically, by letting $u=x^4$.
Then $\dfrac{du}{dx}=4x^3$, and therefore $du=4x^3\,dx$, so $x^3\,dx$ (which appears in the expression, if we rearrange a bit) is $(1/4)\,du$. Thus $\int 2\pi\frac{x^3}{3}\sqrt{1+x^4}\,dx=\int\frac{2\pi}{(3)(4)}u^{1/2}\,du.$ Now if you like bring the constant to the outside, though it looks pretty happy inside to me.
I would expect that by now you have learned how to substitute for the limits of integration. So actually I would write $\int_{x=0}^{x=3} 2\pi\frac{x^3}{3}\sqrt{1+x^4}\,dx=\int_{u=0}^{u=81}\frac{2\pi}{(3)(4)}u^{1/2}\,du.$