MHB Stuck trying to integrate the surface area of a curve

[aleksbooker1]

Here's the problem I was given:

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.

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.

 

[MarkFL]

Hello and welcome to MHB! :D

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

 

[aleksbooker1]

Hi @MarkFL,

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

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

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 = $$ \frac{1}{2}\pi \int e^{2y} + \frac{1}{2}\pi \int 2 + \frac{1}{2}\pi \int e^{-2y} $$

But then I get stuck trying to integrate even $$ e^{2y} $$ by itself.

S = $$ \frac{1}{2}\pi \frac{1}{3y} e^{3y} $$ ?

 

[MarkFL]

What you actually have is:

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

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

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

where $$F'(x)=f(x)$$.

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.

 

[aleksbooker1]

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: $$ e^{2y} $$.

The best I can come up with is:

$$ \frac{1}{2 \ln{2} +1} e^{2 \ln{2} +1 } $$

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 $$ e^{2y} [/MATH} is $$ \frac{1}{2} e^{2y} $$, not some twisted $$ \frac{1}{2y+1}e^{2y+1} $$ like I was thinking...

 

[MarkFL]

There's really no need to break it up into three integrals (although you can).

If you have the indefinite integral:

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

$$u=ax$$

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

 

[HallsofIvy]

Its not that difficult to remember the slightly more general formula
\int e^{ax}dx= \frac{1}{a}e^{ax}+ C

 

[aleksbooker1]

Thanks @MarkFL and @HallsofIvy, I'll add that to my mental cheatsheet for antiderivatives. :)