Surface area of revolution Given the function $y=\frac{1}{3}x^3$ on the interval [0, 3], find the surface area of the revolution on the x-axis.
The derivative is $y'=x^2$, so plugging everything in the integral gives
$$2\pi\int_0^3\frac{x^3}{3}\sqrt{1+(x^2)^2}dx$$
$$2\pi\int_0^3\frac{x^3}{3}\sqrt{1+x^4}dx$$
I got a little stuck here, so looking at the book work, the next step is shown as
$$\frac{\pi}{6}\int_0^3(1+x^4)^{\frac{1}{2}}\cdot(4x^3)dx$$
After this step, I can figure out the integral, but this setup has me confused on two fronts.  
First, how did $2\pi$ transform into $\frac{\pi}{6}$?
Second, where did $4x^3$ come from?
 A: It's not just $2 \pi$ transforming into $\frac{\pi}{6}$; it's 
$\frac{2 \pi}{3}$ transforming into $\frac{4 \pi}{6}$.

As I mentioned in the comment, they are setting up a u-substitution. 
Let $u = 1 + x^4$. Then $du = 4x^3dx$, which they have placed all together in the integrand. 
A: 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
$$\int_{x=0}^{x=3} 2\pi\frac{x^3}{3}\sqrt{1+x^4}\,dx=\int_{u=0}^{u=81}\frac{2\pi}{(3)(4)}u^{1/2}\,du.$$
A: So what's going on is a u-substitution. You start with $$\frac{2\pi}{3}$$ out front (sorry, I don't know how to format the answer) and an integrand of $$x^3*\sqrt{1+x^4}$$. Now, because $$x^3$$ is almost the derivative of what's under the square root, you can add a factor of 4 in front of $$x^3$$, but you have to make up for it by putting a factor of 1/4 out front. Now $$\frac{2\pi}{3}$$ becomes $$\frac{\pi}{6}$$ and you end up with $$4x^3*\sqrt{1+x^4}$$ in the integrand.
Now if you set $$u = 1+x^4$$, then $$du = 4x^3dx$$. So the integrand can be written as $$\sqrt{u}*du$$. Antidifferentiating, you get $$2/3*x^\frac{3}{2}$$. Plugging back in for u, you get $$2/3*(1+x^4)^\frac{3}{2}$$. Now evaluating on [0,3], you get $$(2/3*(82^\frac{3}{2}-1))$$.
Hope that helps! Sorry again for the lack of formatting!
A: $$
2\pi \int_0^3 {\frac{{x^3 }}{3}\sqrt {1 + x^4 } dx}  = \frac{{2\pi }}{3}\int_0^3 {x^3 (1 + x^4 )^{1/2} dx}  = \frac{{2\pi }}{{3 \cdot 4}}\int_0^3 {4x^3 (1 + x^4 )^{1/2} dx}$$
$$
= \frac{\pi }{6}\int_0^3 {(1 + x^4 )^{1/2} 4x^3 dx} .
$$
Hope this helps.
