- #1

aleksbooker

- 22

- 0

## Homework Statement

Find the area of the surface generated by revolving the curve

[itex]x=\frac{e^y + e^{-y} }{2}[/itex]

from 0 [itex]\leq[/itex] y [itex]\leq[/itex] ln(2) about the y-axis.

## The Attempt at a Solution

I tried the normal route first...

g(y) = x = [itex]\frac{1}{2} (e^y + e^{-y})[/itex]

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

S = [itex]\int 2\pi \frac{1}{2} (e^y + e^{-y}) \sqrt{1+(e^y - e^{-y})^2} dy[/itex]

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

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

S = [itex]\pi \int (e^y + e^{-y}) \frac {1}{2} (e^y+e^{-y}) dy[/itex]

But then I got stuck here...

S = [itex]\frac{1}{2} \pi \int (e^y + e^{-y})^2 dy [/itex]

How should I proceed? Thanks in advance.