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
 A: \begin{aligned}
ds^2 &= dr^2+r^2 d\theta^2\\
&=\left(\frac{4 \sin^2 2\theta}{\cos 2 \theta}+\cos 2\theta\right)d\theta^2\\
ds &= \sqrt{\frac{1+3 \sin^2 2\theta}{{\cos 2\theta}} }d\theta \\
A &=2\int_0^{\pi/4}2\pi r \cos \theta ds\\
  &=4\pi \int_0^{\pi/4}\sqrt{1+3 \sin^2 2\theta} cos \theta d\theta\\
  &=\int_0^{\frac{1}{\sqrt 2}} \sqrt{1+12 t^2 (1-t^2)} dt
\end{aligned}
This is as simplified I was able to make it.
A: Note some useful relationships and identities:
$r^2 = x^2 + y^2$
${cos 2\theta} = cos^2\theta - sin^2\theta$
${sin \theta} = {y\over r} = {y\over{\sqrt{x^2 + y^2}}}$
${sin^2 \theta} = {y^2\over {x^2 + y^2}}$
These hint at the possibility of doing this in Cartesian coordinates.
A: 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.
