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

Jan 20, 2017

$S = \sqrt{2} \pi$

#### Explanation:

If you consider the interval $\left(x , x + \mathrm{dx}\right)$ the length of the arc element of the curve $y = f \left(x\right)$ is:

$\mathrm{dl} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}} = \sqrt{{\mathrm{dx}}^{2} + f ' {\left(x\right)}^{2} {\mathrm{dx}}^{2}} = \mathrm{dx} \sqrt{1 + f ' {\left(x\right)}^{2}}$

so the area of the surface element generated by the rotation around the $x$-axis is:

$\mathrm{dS} = 2 \pi y \mathrm{dl} = 2 \pi f \left(x\right) \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$

Integrating over the interval, the surface is then calculated as:

$S = 2 \pi {\int}_{1}^{2} f \left(x\right) \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$

As $f ' \left(x\right) = 1$:

$S = 2 \sqrt{2} \pi {\int}_{1}^{2} \left(x - 1\right) \mathrm{dx} = \sqrt{2} \pi {\left[{\left(x - 1\right)}^{2}\right]}_{1}^{2} = \sqrt{2} \pi$