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 #C = 2pir#.

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 #1/(e^(x^2)#, we can say that the radius #r=e^(-x^2)#.

Therefore, our integral changes to:

#S = 2pi int_-1^1 e^(-x^2)dx#

Hmm... #e^(-x^2)# doesn't really have an elementary function as its antiderivative. However, its antiderivative IS given by the error function. Click the link to find out more about that and how it's used in statistics.

The error function #"erf"(x)# is defined so that:

#"erf"(z) = 1/sqrtpi int_-z^ze^(-x^2)dx#

Our function is in the right form to replace the integral with the #"erf"# function like this:

#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#

Final Answer