# How do you find the volume of the solid obtained by rotating the curve about the y-axis over [0,1]?

## y=x$\sqrt{1 - {x}^{2}}$

May 13, 2018

We must imagine that the graph has radius $r$. Then the volume will be $\pi {r}^{2} h$. In this case we have:

$V = \pi {\int}_{0}^{1} {\left(x \sqrt{1 - {x}^{2}}\right)}^{2} \mathrm{dx}$

Therefore

$V = \pi {\int}_{0}^{1} {x}^{2} \left(1 - {x}^{2}\right) \mathrm{dx}$

$V = \pi {\int}_{0}^{1} {x}^{2} - {x}^{4} \mathrm{dx}$

$V = \pi {\left[\frac{1}{3} {x}^{3} - \frac{1}{5} {x}^{5}\right]}_{0}^{1}$

$V = \pi \left(\frac{1}{3} - \frac{1}{5}\right)$

$V = \frac{2}{15} \pi$ cubic units

Hopefully this helps!