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

Mar 17, 2016

24.93

#### Explanation:

Surface area of solid of revolution about x axis is given by the formula

$S = {\int}_{a}^{b} 2 \pi y \mathrm{ds} = {\int}_{a}^{b} 2 \pi y \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$. In the present case ,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2} {x}^{\frac{1}{2}}$, hence $\sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} = \sqrt{1 + 9 \frac{x}{4}}$

$S = 2 \pi {\int}_{1}^{2} {x}^{\frac{3}{2}} \sqrt{1 + 9 \frac{x}{4}} \mathrm{dx}$

=$\pi {\int}_{1}^{2} {x}^{\frac{3}{2}} \sqrt{9 x + 4} \mathrm{dx}$

This integration is too long to write it here. hence
use integral calculator to solve the integral

= 24.93