# Surface area generated by revolving about the y-axis

I have to find the surface area which is generated by revolving the curve about the y-axis found below:

$$x=\frac{1}{2}(e^{y} + e^{-y}) \ ; 0<=y<=ln2$$

I know how to solve the question, when it is referring to x-axis. However, i am not sure on how to approach the question when it is referring to the y-axis. Should i move the values around till its y equal to x, then solve it? Or is there any other way to approach these kinds of problem?

All help and suggestions are appreciated. Thank you very much!

• Do we have any constraints on the z axis or is this purely in $\mathbb{R}^2$? Apr 20 '14 at 10:39
• @ellya There is no info given on z-axis Apr 20 '14 at 10:40

$\sigma(\phi,t)=(\frac{1}{2}(e^t+e^{-t})\cos\phi,t,(\frac{1}{2}(e^t+e^{-t})\sin\phi)$
For $0\le t \le \ln 2, 0\le\phi\le 2\pi$