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

Nov 3, 2017

The surface area of revolution is $A = 19.04981414777664 \ldots$

#### Explanation:

The surface arte of revolution is given by:

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

We have $y = \ln \left(2 x\right)$ so differentiating wrt $x$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 x} \cdot 2 = \frac{1}{x}$

Thus we have:

$A = {\int}_{1}^{3} \setminus 2 \pi y \setminus \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \pi \setminus {\int}_{1}^{3} \setminus \left(\ln 2 x\right) \setminus \sqrt{1 + {\left(\frac{1}{x}\right)}^{2}} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \pi \setminus {\int}_{1}^{3} \setminus \ln 2 x \setminus \sqrt{1 + \frac{1}{x} ^ 2} \setminus \mathrm{dx}$

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

$A = 19.04981414777664 \ldots$