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Stuck trying to integrate the surface area of a curve

aleksbooker

New member
Sep 10, 2013
5
Here's the problem I was given:

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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello and welcome to MHB! :D

Next, I would expand the integrand, and then apply the given limits in the FTOC...
 

aleksbooker

New member
Sep 10, 2013
5
Hi @MarkFL,

Thanks for the response. When you say expand the integrand, do you mean like this?

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

If so, how do I use the Fundamental Theorem of Calculus. I know what FTOC stands for but I never learned it as a thing by itself. I was taught how to look at a problem and break it down, but the definitions weren't really a part of my curriculum. :(

I've also tried splitting it into its components after expanding it, like so:

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

But then I get stuck trying to integrate even \(\displaystyle e^{2y} \) by itself.

S = \(\displaystyle \frac{1}{2}\pi \frac{1}{3y} e^{3y} \) ?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What you actually have is:

\(\displaystyle S=\frac{\pi}{2}\int_{0}^{\ln(2)} e^{2y}+2+e^{-2y}\,dy\)

The anti-derivative form of the FTOC states that if $f(x)$ is continuous on $[a.b]$:

\(\displaystyle \int_a^b f(x)\,dx=F(b)-F(a)\)

where \(\displaystyle F'(x)=f(x)\).

So, you need to find the anti-derivative and then evaluate it at the upper limit, and then subtract from this the evaluation at the lower limit.
 

aleksbooker

New member
Sep 10, 2013
5
So, you need to find the anti-derivative and then evaluate it at the upper limit, and then subtract from this the evaluation at the lower limit.
Oh, okay. I know how to do that. The trick is anti-deriving with a variable in the exponent. For example, if I expand it and then split it into three integrands and try tackle each one by itself, I trip up on the first one: \(\displaystyle e^{2y} \).

The best I can come up with is:

\(\displaystyle \frac{1}{2 \ln{2} +1} e^{2 \ln{2} +1 } \)

Which I don't think is a "legal move" in calculus. Is it?

EDIT: Wait... now, I get it. I forgot how much I hate working with e, sometimes. The antiderivative of \(\displaystyle e^{2y} [/MATH} is \(\displaystyle \frac{1}{2} e^{2y} \), not some twisted \(\displaystyle \frac{1}{2y+1}e^{2y+1} \) like I was thinking...\)
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
There's really no need to break it up into three integrals (although you can).

If you have the indefinite integral:

\(\displaystyle \int e^{ax}\,dx\) where $a$ is a non-zero constant, use the substitution:

\(\displaystyle u=ax\)

What do you need for your new differential to be and how can you get it?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Its not that difficult to remember the slightly more general formula
[tex]\int e^{ax}dx= \frac{1}{a}e^{ax}+ C[/tex]
 

aleksbooker

New member
Sep 10, 2013
5
Thanks @MarkFL and @HallsofIvy, I'll add that to my mental cheatsheet for antiderivatives. :)