# What is the surface area of the solid created by revolving f(x) = lnx , x in [2,3] around the x axis?

Apr 25, 2017

Not integrable

#### Explanation:

Area of surface of revolution of y=f(x ) , between x=a to x=b, about x axis is

$S = {\int}_{a}^{b} 2 \pi y \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

$S = {\int}_{2}^{3} 2 \pi \ln x \sqrt{1 + \frac{1}{x} ^ 2} \mathrm{dx}$

This function is not integrable