What is the surface area produced by rotating f(x)=e^(x^2), x in [-1,1] around the x-axis?

Feb 13, 2017

What you'd do to determine the surface area for a solid of revolution around the $x$ axis is take the equation for the circumference and integrate over the surface.

$S = 2 \pi {\int}_{- 1}^{1} r \left(x\right) \mathrm{dS} \left(x\right)$

As it turns out, the differential surface can be treated as the arc length for an infinitesimal distance along the chosen axis. So:

$\mathrm{dS} \left(x\right) = \sqrt{1 + {\left(r ' \left(x\right)\right)}^{2}} \mathrm{dx}$,

and:

$S = 2 \pi {\int}_{- 1}^{1} r \left(x\right) \sqrt{1 + {\left(r ' \left(x\right)\right)}^{2}} \mathrm{dx}$

First, let's evaluate the squared derivative:

$\frac{d}{\mathrm{dx}} \left[{e}^{{x}^{2}}\right] = 2 x {e}^{{x}^{2}}$

${\left(r \left(x\right)\right)}^{2} = 4 {x}^{2} {e}^{2 {x}^{2}}$

This gives:

$\textcolor{b l u e}{S} = 2 \pi {\int}_{- 1}^{1} {e}^{{x}^{2}} \sqrt{1 + 4 {x}^{2} {e}^{2 {x}^{2}}} \mathrm{dx}$

$= \textcolor{b l u e}{2 \pi {\int}_{- 1}^{1} {e}^{2 {x}^{2}} \sqrt{{e}^{- 2 {x}^{2}} + 4 {x}^{2}} \mathrm{dx}}$

There is however, no result in terms of elementary functions. Numerically, this integral is $46.3958$.