Question

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

Answer 1

The surface area of revolution is `A = 19.04981414777664 ...`

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 `y=ln(2x)` so differentiating wrt `x` 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 ...`