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

Question

Samantha Adkison · Accepted Answer

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 = 2piint_(-1)^(1) r(x) dS(x)#

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

#dS(x) = sqrt(1 + (r'(x))^2)dx#,

and:

#S = 2piint_(-1)^(1) r(x) sqrt(1 + (r'(x))^2)dx#

First, let's evaluate the squared derivative:

#d/(dx)[e^(x^2)] = 2xe^(x^2)#

#(r(x))^2 = 4x^2e^(2x^2)#

This gives:

#color(blue)(S) = 2piint_(-1)^(1) e^(x^2) sqrt(1 + 4x^2e^(2x^2))dx#

#= color(blue)(2piint_(-1)^(1) e^(2x^2) sqrt(e^(-2x^2) + 4x^2)dx)#

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

Cameron Alexander · Answer

undefined To find the surface area produced by rotating $ f(x) = e^{x^2} $ around the x-axis over the interval $[-1,1]$, we use the formula for the surface area of a solid of revolution:

$$ S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + \left( f'(x) \right)^2} \, dx $$

where $ f'(x) $ is the derivative of $ f(x) $.

First, we find $ f'(x) $:
$$ f'(x) = 2xe^{x^2} $$

Next, we plug $ f(x) $ and $ f'(x) $ into the formula and integrate over the interval $[-1,1]$:
$$ S = \int_{-1}^{1} 2\pi e^{x^2} \sqrt{1 + (2xe^{x^2})^2} \, dx $$

$$ = \int_{-1}^{1} 2\pi e^{x^2} \sqrt{1 + 4x^2e^{2x^2}} \, dx $$

This integral may not have a closed-form solution and may need to be evaluated numerically.