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

Aug 3, 2017

Well, the surface area of a function $f \left(x\right)$ is given by

$S = \int 2 \pi f \left(x\right) \mathrm{ds}$

= 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.

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{2 x}{{x}^{2} + 1} ^ 2$

And square it to get:

${\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)}^{2} = {\left(2 x\right)}^{2} / {\left({x}^{2} + 1\right)}^{4}$

So, the surface area integral is:

$S = 2 \pi {\int}_{0}^{3} \frac{1}{{x}^{2} + 1} \sqrt{1 + \frac{{\left(2 x\right)}^{2}}{{x}^{2} + 1} ^ 4} \mathrm{dx}$

$= 2 \pi {\int}_{0}^{3} \sqrt{\frac{1}{{x}^{2} + 1} ^ 2 + \frac{{\left(2 x\right)}^{2}}{{x}^{2} + 1} ^ 6} \mathrm{dx}$

There is no elementary solution to this, so we can only get the numerical integration result, $\textcolor{b l u e}{2.72708 \pi}$ $\textcolor{b l u e}{{\text{u}}^{2}}$, or about $8.56737$.