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

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

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.

2. ## Re: Can't integrate the surface area of revolving curve the normal way

Multiply out:

$\displaystyle S = \dfrac{\pi}{2}\int_0^{\ln 2}(e^{2y} + 2 + e^{-2y})dy = \dfrac{\pi}{2}\left. \left(\dfrac{e^{2y}}{2} + 2y - \dfrac{e^{-2y}}{2} \right) \right]_0^{\ln 2}$

3. ## Re: Can't integrate the surface area of revolving curve the normal way

Another way to do this problem:

$\displaystyle x = \cosh y$ so $\displaystyle \dfrac{dx}{dy} = \sinh y$. This simplifies the formula to:

$\displaystyle S = 2\pi \int_0^{\ln 2}\cosh y \sqrt{1+\sinh^2 y}dy$

Since $\displaystyle \cosh^2 y = 1 + \sinh^2 y$, you can rewrite:

$\displaystyle S = 2\pi \int_0^{\ln 2} \cosh^2 y dy$

Using the double angle formula for hyperbolic cosine: $\displaystyle \cosh^2 y = \dfrac{1}{2} + \dfrac{1}{2}\cosh(2y)$

$\displaystyle S = \left. \pi y + \dfrac{\pi}{2}\sinh(2y) \right]_0^{\ln 2} = \left. \pi y + \dfrac{\pi}{4}\left(e^{2y} - e^{-2y}\right) \right]_0^{\ln 2}$

4. ## Re: Can't integrate the surface area of revolving curve the normal way

Originally Posted by SlipEternal
Another way to do this problem:

$$x = \cosh y$$ so $$\dfrac{dx}{dy} = \sinh y$$. This simplifies the formula to:

$$S = 2\pi \int_0^{\ln 2}\cosh y \sqrt{1+\sinh^2 y}dy$$

Since $$\cosh^2 y = 1 + \sinh^2 y$$, you can rewrite:

$$S = 2\pi \int_0^{\ln 2} \cosh^2 y dy$$

Using the double angle formula for hyperbolic cosine: $$\cosh^2 y = \dfrac{1}{2} + \dfrac{1}{2}\cosh(2y)$$

$$S = \left. \pi y + \dfrac{\pi}{2}\sinh(2y) \right]_0^{\ln 2} = \left. \pi y + \dfrac{\pi}{4}\left(e^{2y} - e^{-2y}\right) \right]_0^{\ln 2}$$
redone for new LaTex

5. ## Re: Can't integrate the surface area of revolving curve the normal way

Son of a monkey. I forgot how much I hate working with e. Thanks, @SlipEternal, for pointing out how simple that really was.

Here's what I was doing:

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

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

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

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