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

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.
