# Calc5.3

Find the volume of the solid generated by revolving the region bounded by $y=x^{2}$ and $y=x^{3}$ around the x-axis.
First, we need to find the intersection points of the two graphs. These are easy to find with these particular points. The intersections come at $x=0,1$. Since $x^{2}$ is greater than $x^{3}$ between 0 and 1, and because we are revolving this region about the x-axis, our outer function is $y=x^{2}$. So, we will find the volume of the solid of revolution using $y=x^{2}$ and subtract from this the volume of the solid of revolution using $y=x^{3}$. That is,
$V=\pi \int _{{0}}^{{1}}\left[(x^{2})^{2}-(x^{3})^{2}\right]\,dx=\pi \int _{{0}}^{{1}}(x^{4}-x^{6})\,dx=\pi \left[{\frac {1}{5}}x^{5}-{\frac {1}{7}}x^{7}\right]_{{0}}^{{1}}={\frac {2\pi }{35}}$