Find the surface area generated when the curve is revolved around the x-axis Find the surface area generated when the curve is revolved around the x-axis
$y=\frac{x^3}{10}$ on $[0,\sqrt{10}]$
This is what I have so far:
$$f'(x)=\frac{3x^2}{10}$$ $$f'(x)^2=\frac{9x^4}{100}$$
formula:
$$S=\int_{0}^{\sqrt{10}} 2π \left(\frac{x^3}{10}\right)\sqrt{1+\frac{9x^3}{25}} $$
Now using substitution:
$$u=1+\frac{9x^4}{100}$$ $$du=\frac{9x^3}{25}$$
Then you need to find the new bounds:
$$u=1+\frac{9(0)^4}{100}=1$$ $$u=1+\frac{9\sqrt{10}^4}{100}=\frac{9*100}{100}=10$$ 
New formula with new bounds:
$$\int_{1}^{10} 2π\sqrt{u} $$
$$\int_{1}^{10} 2π \left(\frac{2}{3}u^\frac{3}{2}\right)$$
This is where I get confused. Also, is $2π$ supposed change?
 A: You are on the right track, but are a little off. This can be remedied by paying more attention to the infinitesimal quantities and their corresponding coefficients. Given a function $f(x)$ to be rotated about the $x$-axis from $a$ to $b$ we have $$2\pi\int_a^bf(x)\sqrt{1+f'(x)^2}\space \text{d}x$$ as the surface area. For you, this means $$S =  2 \pi \int_{0}^{\sqrt{10}} \left(\frac{x^3}{10}\right)\sqrt{1+\frac{9x^4}{100}}\space \text{d}x $$ You did make the right $u$-sub of $u = 1+\frac{9x^4}{100}$ (even though this choice of $u$ didn't appear in your equation for $S$ up above). This means $\text{d}u = \frac{9x^3}{25}\text{d}x$. Or more usefully, $\frac{x^3}{10} \text{d}x = \frac{5}{18}\text{d}u$. You did correctly find the lower bound of the integral in terms of $u$, but the upper bound should be $10$, not $9$. Hence,  $$\begin{align}2 \pi \int_{0}^{\sqrt{10}} \left(\frac{x^3}{10}\right)\sqrt{1+\frac{9x^4}{100}}\space \text{d}x =  2 \pi \int_{1}^{10} \left(\frac{5}{18}\right)\sqrt{u}\space \text{d}u  \\ =   \frac{5\pi}{9} \int_{1}^{10}\sqrt{u}\space \text{d}u \\ =  \frac{10\pi}{27}\left[u^{3/2} \right]_1^{10} \end{align}$$
To answer your question at the end, no, $2\pi$ is not supposed to "change", other than how it may interact with the coefficients that appear from the $u$-sub. I pulled it out of the integral from the get-go, which is allowed by the integral property that $\int cf(x)dx = c\int f(x)dx$ for all constants $c \in \Bbb{R}$. I personally prefer to pull constants in front of the integrand whenever possible because it allows me to focus more on the quantity that is to be integrated.
