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

##### 1 Answer

What you'd do to determine the surface area for a solid of revolution around the

#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