Question

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

Answer 1

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 2pi y sqrt (1+(dy/dx)^2 ) dx`

`S= int_2^3 2pi ln x sqrt (1+1/x^2) dx`

This function is not integrable