What is the surface area produced by rotating #f(x)=1/e^(x^2), x in [-1,1]# around the x-axis?
Just like the volume of a shape is found by summing up each infinitesimal cross section of area, the surface area is found by summing up each infinitesimal cross section of perimeter.
In this case, the perimeter of each cross section is circular, so we can find it using the circumference formula
Therefore, our integral becomes:
#S = int_-1^1 2pirdx#
Now we need to find our radius. Since the circle's center is the x-axis, and a point on the outside of the circle lies on the graph of
Therefore, our integral changes to:
#S = 2pi int_-1^1 e^(-x^2)dx#
The error function
#"erf"(z) = 1/sqrtpi int_-z^ze^(-x^2)dx#
Our function is in the right form to replace the integral with the
#S = 2pi(sqrtpicolor(white)"-""erf"(1))#
#S = 2pi^(3/2)color(white)"-""erf"(1)#
This is a good stopping point, unless the problem wants the decimal form of the answer, in which case plugging this into a calculator gives:
#S ~~ 9.385#