# 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?

Mar 31, 2018

$S \approx 151.4$

#### 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

$\mathrm{dS} = 2 \pi r \left(x\right) \cdot \mathrm{dl}$

The small length $\mathrm{dl}$ is the arclength:
${\mathrm{dl}}^{2} = {\mathrm{dy}}^{2} + {\mathrm{dx}}^{2} \implies {\mathrm{dl}}^{2} = \left(1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}\right) {\mathrm{dx}}^{2}$
$\mathrm{dl} = \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

So, we integrate, since the radius of the solid is just $f \left(x\right)$:

$S = \int \mathrm{dS} = {\int}_{1}^{2} \setminus 2 \pi f \left(x\right) \mathrm{dl}$
$= 2 \pi {\int}_{1}^{2} \left({e}^{- x} + {e}^{x}\right) \sqrt{1 + {\left({e}^{x} - {e}^{-} x\right)}^{2}} \mathrm{dx}$
Using u-sub, with $u = {e}^{x} - {e}^{- x} = 2 \sinh \left(x\right)$,

$S = 2 \pi {\int}_{2 \sinh \left(1\right)}^{2 \sinh \left(2\right)} \sqrt{1 + {u}^{2}} \mathrm{du}$
$S = \pi \left[u \sqrt{1 + {u}^{2}} + {\sinh}^{-} 1 \left(u\right)\right] {|}_{2 \sinh \left(1\right)}^{2 \sinh \left(2\right)}$
$S \approx 151.4$