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

Note that $((\sqrt{\frac{y-1}{3}})')^2=(\frac{1}{2\sqrt{\frac{y-1}{3}}} (1/3))^2=\frac{1}{4(y-1)}$
$$=\int_1^{13} 2\pi\sqrt\frac{y-1}3 \cdot \sqrt{1+\frac{1}{4(y-1)}}dy
\\= \int_1^{13} 2\pi\sqrt{\frac{y-1}3+\frac{1}{12}}dy
\\= \int_1^{13} 2\pi\sqrt{\frac{y}3-\frac{1}{4}}dy
\\=  6\pi\{(\frac{y}3-\frac{1}{4})^\frac{3}{2}\}_1^{13}$$
