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

[aleksbooker]

Homework Statement

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.

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.

 

[pasmith]

Homework Statement

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.

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

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

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

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

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}) dyBut then I got stuck here...

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

How should I proceed? Thanks in advance.

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

 

[aleksbooker]

@pasmith, thanks for your response.

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)

 

[pasmith]

@pasmith, thanks for your response.

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)

No.
<br /> \int e^{ay}\,dy = \frac{e^{ay}}a<br />