Find the surface area obtained by rotating $y=1+3x^2$ from $x=0$ to $x=2$ about the y-axis. Find the surface area obtained by rotating $y= 1+3 x^2$ from $x=0$ to $x = 2$ about the $y$-axis.
Having trouble evaluating the integral: 
Solved for $x$:


*

*$x=0, y=1$

*$x=2, y=13$


$$\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+\sqrt\frac{y-1}3'}^2\,dy$$
I got stuck at 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+(1/12)+(1/(y-1))}
any help would be great thanks

This can be done using Pappus's Theorem and integrating in $x$:
$$
\begin{align}
\int_0^22\pi x\,\overbrace{\sqrt{y'^2+1}\,\mathrm{d}x}^{\mathrm{d}s}
&=\int_0^22\pi x\sqrt{36x^2+1}\,\mathrm{d}x\\
&=\frac\pi{36}\int_0^2\sqrt{36x^2+1}\,\mathrm{d}(36x^2+1)\\
&=\frac\pi{36}\int_1^{145}\sqrt{u}\,\mathrm{d}u\\
&=\frac\pi{36}\left[\frac23u^{3/2}\right]_1^{145}\\[4pt]
&=\frac\pi{54}\left(145^{3/2}-1\right)
\end{align}
$$
where we used the substitution $u=36x^2+1$.
