What is the surface area produced by rotating f(x)=e^(x^2), x in [-1,1]f(x)=ex2,x∈[−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)S=2π∫1−1r(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
