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!
I believe the parametrization would be:
$\sigma(\phi,t)=(\frac{1}{2}(e^t+e^{-t})\cos\phi,t,(\frac{1}{2}(e^t+e^{-t})\sin\phi)$
For $0\le t \le \ln 2, 0\le\phi\le 2\pi$
Can you take it from here?