# Thread: Volume of region under a catenary?

1. ## Volume of region under a catenary?

I'm having a difficult time with this problem and can't seem to make much progress:
Let $y = a cosh(\frac{x}{a})$ be a catenary. Find the volume of the solid obtained by revolving the region under the graph of the catenary on the interval [-b,b], (b>0) about the x-axis.

So, I have decided that $y = a cosh(\frac{x}{a}) = \frac{a}{2}(e^\frac{x}{a} + e^\frac{-x}{a})$ based on definitions of hyperbolic functions. The problem I am having is setting up a formula to figure out the area. By using shell or disk method I can't get an exact area, only a formula since there isn't a specific interval.

2. Originally Posted by Funkychemist
I'm having a difficult time with this problem and can't seem to make much progress:
Let $y = a cosh(\frac{x}{a})$ be a catenary. Find the volume of the solid obtained by revolving the region under the graph of the catenary on the interval [-b,b], (b>0) about the x-axis.

So, I have decided that $y = a cosh(\frac{x}{a}) = \frac{a}{2}(e^\frac{x}{a} + e^\frac{-x}{a})$ based on definitions of hyperbolic functions. The problem I am having is setting up a formula to figure out the area. By using shell or disk method I can't get an exact area, only a formula since there isn't a specific interval.

$S=\pi\int\limits_{-b}^b\left[\frac{a}{2}\left(e^{x/a}+e^{-x/a}\right)\right]^2dx$ $=\frac{a^2\pi}{4}\int\limits_{-b}^b\left(e^{2x/a}+e^{-2x/a}+2\right)dx=...$ etc.

Tonio