- Thread starter
- #1

#### aleksbooker

##### New member

- Sep 10, 2013

- 5

Find the area of the surface generated by revolving the curve

\(\displaystyle x=\frac{e^y + e^{-y} }{2}\)

from 0 \(\displaystyle \leq\) y \(\displaystyle \leq\) ln(2) about the y-axis.

I tried the normal route first...

g(y) = x = \(\displaystyle \frac{1}{2} (e^y + e^{-y})\)

g'(y) = dx/dy = \(\displaystyle \frac{1}{2} (e^y - e^{-y}) \)

S = \(\displaystyle \int 2\pi \frac{1}{2} (e^y + e^{-y}) \sqrt{1+(e^y - e^{-y})^2} dy\)

S = \(\displaystyle \pi \int (e^y + e^{-y}) \sqrt{\frac{1}{4}e^{2y}+\frac{1}{2}+\frac{1}{4}e^{-2y} } dy\)

S = \(\displaystyle \pi \int (e^y + e^{-y}) \sqrt{\frac{1}{4} (e^y+e^{-y})^2} dy\)

S = \(\displaystyle \pi \int (e^y + e^{-y}) \frac {1}{2} (e^y+e^{-y}) dy\)

But then I got stuck here...

S = \(\displaystyle \frac{1}{2} \pi \int (e^y + e^{-y})^2 dy \)

How should I proceed? Thanks in advance.