<s>[INST] 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?
 [/INST] $$
2\pi \int_0^3 {\frac{{x^3 }}{3}\sqrt {1 + x^4 } dx}  = \frac{{2\pi }}{3}\int_0^3 {x^3 (1 + x^4 )^{1/2} dx}  = \frac{{2\pi }}{{3 \cdot 4}}\int_0^3 {4x^3 (1 + x^4 )^{1/2} dx}$$
$$
= \frac{\pi }{6}\int_0^3 {(1 + x^4 )^{1/2} 4x^3 dx} .
$$
Hope this helps.
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