# How do you find the volume of the solid generated by revolving the region bounded by the graphs of the equations y=sqrtx, y=0, and x=4 about the y-axis?

Mar 15, 2018

V=$8 \pi$ volume units

#### Explanation:

Essentially the problem you have is:

V=$\pi {\int}_{0}^{4} {\left(\left(\sqrt{x}\right)\right)}^{2} \mathrm{dx}$

Remember, the volume of a solid is given by:

V=$\pi \int {\left(f \left(x\right)\right)}^{2} \mathrm{dx}$

Thus, our original Intergral corresponds:

V=$\pi {\int}_{0}^{4} \left(x\right) \mathrm{dx}$

Which is in turn equal to:

V=$\pi \left[{x}^{2} / \left(2\right)\right]$ between x=0 as our lower limit and x=4 as our upper limit.

Using The fundamental theorem of Calculus we substitute our limits into our integrated expression as subtract the lower limit from the upper limit.

V=$\pi \left[\frac{16}{2} - 0\right]$

V=$8 \pi$ volume units