# What is the surface area of the solid created by revolving #f(x) =ln(2x) , x in [1,3]# around the x axis?

##### 1 Answer

Nov 3, 2017

The surface area of revolution is

#### Explanation:

The surface arte of revolution is given by:

# A = int_a^b \ 2piy \ dS # where#dS = sqrt(1+(dy/dx)^2) \ dx #

We have

# dy/dx = 1/(2x) * 2 = 1/x #

Thus we have:

# A = int_1^3 \ 2piy \ sqrt(1+(dy/dx)^2) \ dx #

# \ \ \ = 2pi \ int_1^3 \ (ln2x) \ sqrt(1+(1/x)^2) \ dx #

# \ \ \ = 2pi \ int_1^3 \ ln2x \ sqrt(1+1/x^2) \ dx #

We cannot find an elementary solution to this integral, but we can get a numerical solution:

# A = 19.04981414777664 ... #