Question

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

Answer 1

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`.