Can't integrate the surface area of revolving curve the normal way

1. Feb 18, 2014

aleksbooker

1. The problem statement, all variables and given/known data

Find the area of the surface generated by revolving the curve

$x=\frac{e^y + e^{-y} }{2}$

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

3. The attempt at a solution

I tried the normal route first...

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

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

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

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

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

But then I got stuck here...

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

How should I proceed? Thanks in advance.

2. Feb 18, 2014

pasmith

I think this should be
$$S = \int_0^{\ln 2} 2\pi \frac 12 (e^y + e^{-y}) \sqrt{1 + \frac14 (e^y - e^{-y})^2}\,dy$$

And now missing factor of 1/4 has appeared (but the limits are still missing).

$(e^y + e^{-y})^2 = e^{2y} + 2 + e^{-2y}$...

3. Feb 18, 2014

aleksbooker

I forgot how much I hate working with e. Thanks for pointing out how simple that really was.

Here's what I was doing:

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

Then, breaking it up into three separate integrals and working with just the first one...

S = $\frac{1}{2}\pi \int (e^{2y})$

S = $\frac{1}{2}\pi \frac{1}{2y+1} e^{2y+1}$ from 0 to ln(2)

4. Feb 19, 2014

pasmith

No.
$$\int e^{ay}\,dy = \frac{e^{ay}}a$$