Question

What is the surface area of the solid created by revolving `f(x) = e^-x+e^(x) , x in [1,2]` around the x axis?

Answer 1

`S approx 151.4`

Explanation:

If we imagine this solid being broken into small cylindrical slices (as we do for finding volumes of solids of revolution), we realize that the surface area of each of these solids is

`dS = 2pi r(x) * dl`

The small length `dl` is the arclength:
`dl^2 = dy^2 + dx^2 implies dl^2 = (1 + (dy/dx)^2)dx^2`
`dl = sqrt(1 + (dy/dx)^2)dx`

So, we integrate, since the radius of the solid is just `f(x)`:

`S = int dS = int_1^2\ 2pi f(x) dl`
`= 2pi int_1^2 (e^(-x) + e^x)sqrt(1 + (e^x - e^-x)^2)dx`
Using u-sub, with `u = e^x - e^(-x) = 2sinh(x)`,

`S = 2pi int_(2sinh(1))^(2sinh(2)) sqrt(1+u^2)du`
`S = pi [usqrt(1+u^2) + sinh^-1(u)]|_{2sinh(1)}^(2sinh(2))`
`S approx 151.4`