# Stuck trying to integrate the surface area of a curve

#### aleksbooker

##### New member
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

Staff member
Hello and welcome to MHB! Next, I would expand the integrand, and then apply the given limits in the FTOC...

#### aleksbooker

##### New member
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 = $$\displaystyle \frac{1}{2}\pi \frac{1}{3y} e^{3y}$$ ?

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#### MarkFL

Staff member
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
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

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

#### aleksbooker

##### New member
Thanks @MarkFL and @HallsofIvy, I'll add that to my mental cheatsheet for antiderivatives. 