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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!

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

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  • $\begingroup$ Sorry, i don't really understand these terms. Aren't we suppose to differentiate and then plug in the values into the formula for integration? $\endgroup$
    – Phantom
    Apr 20 '14 at 10:53
  • $\begingroup$ Correct but I believe you need the parametrization first, which is what I've given you $\endgroup$
    – Ellya
    Apr 20 '14 at 11:07
  • $\begingroup$ Hmm..ok. Will try it out :) $\endgroup$
    – Phantom
    Apr 20 '14 at 11:28

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