Question

What is the surface area produced by rotating `f(x)=1/(x^2+1), x in [0,3]` around the x-axis?

Answer 1

Well, the surface area of a function `f(x)` is given by

`S = int 2pi f(x)ds`

`= int overbrace(2pi f(x))^"Circumference"overbrace(sqrt(1 + ((df)/(dx))^2))^("Arc Length")dx`

This surface would look like:

First we get the derivative.

`(df)/(dx) = -(2x)/(x^2 + 1)^2`

And square it to get:

`((df)/(dx))^2 = (2x)^2/(x^2 + 1)^4`

So, the surface area integral is:

`S = 2pi int_(0)^(3) 1/(x^2 + 1)sqrt(1 + ((2x)^2)/(x^2 + 1)^4)dx`

`= 2pi int_(0)^(3) sqrt(1/(x^2 + 1)^2 + ((2x)^2)/(x^2 + 1)^6)dx`

There is no elementary solution to this, so we can only get the numerical integration result, `color(blue)(2.72708pi)` `color(blue)("u"^2)`, or about `8.56737`.