# How do you find the volume of the region bounded by y=sqrt x, y=0, x=0, and x=4 is revolved about the x-axis?

Jun 30, 2016

$= 8 \pi$

#### Explanation:

The volume of the small strip of width dx is given by

$\mathrm{dV} = \pi {y}^{2} \mathrm{dx}$ where $y = \sqrt{x}$ represents the radius of the revolution at that given value for x

so $\mathrm{dV} = \pi \setminus x \setminus \mathrm{dx}$

and

$V = \pi {\int}_{0}^{4} \setminus x \setminus \mathrm{dx}$

$= \pi {\left[{x}^{2} / 2\right]}_{0}^{4} = 8 \pi$