Find the area of a surface of revolution I'm a calculus II student and I'm completely stuck on one question:

Find the area of the surface generated by revolving the right-hand
  loop of the lemniscate $r^2 = \cos2 θ$ about the vertical line through
  the origin (y-axis).

Can anyone help me out?  
Thanks in advance

I stole the formula from a website for surfaces of revolution that was linked in the comment above:
http://tutorial.math.lamar.edu/Classes/CalcII/PolarSurfaceArea.aspx
They prove it more generally for parametric surfaces. I am not sure what you are allowed to assume in your calculus two course; I was unsuccessful in getting a correct formula from a direct polar slicing argument.
In what follows, I am going to be sloppy about whether I write $r$ as a variable or as a function $r(\theta)$ of $\theta$.
Since the right half of the lemniscate is traced out between $-\pi/4$ and $\pi/4$, the integral you want is
$$
2\pi\int_{-\pi/4}^{\pi/4} r(\theta)\cos\theta\sqrt{r(\theta)^2 + r'(\theta)^2}d\theta
$$
We have
$$
r^2 = \cos(2\theta)
$$
and so 
$$
2rr' = -2\sin(2\theta),
$$
so 
\begin{align*}
(r')^2 &= \frac{\sin^2(2\theta)}{r^2}\\
&=\tan^2(2\theta).
\end{align*}
The integrand isn't very pretty. I used wolfram alpha and it numerically approximated the integral (without the 2 $\pi$) to be 2.1028, which seems geometrically reasonable to me. I'm sorry for the unsatisfying conclusion.
