# Calc5.2

Find the volume of the solid generated by revolving the region bounded by $y=x^{2}$ and $y=4x-x^{2}$ around the x-axis.
$x^{2}=4x-x^{2}\Rightarrow 2x^{2}-4x=0\Rightarrow 2x(x-2)=0\Rightarrow x=0,2$
The function $y=4x-x^{2}$ is the outermost function since we are revolving about the x-axis. Thus, we can find the volume by finding the volume of the solid generated by revolving $4x-x^{2}$ about the x-axis and subtracting the volume of the solid generated by revolving $y=x^{2}$ about the x-axis.
$V=\pi \int _{{0}}^{{2}}(4x-x^{2})^{2}\,dx-\pi \int _{{0}}^{{2}}(x^{2})^{2}\,dx=\pi \int _{{0}}^{{2}}(-8x^{3}+16x^{2})\,dx$
$=\pi \left[-2x^{4}+{\frac {16}{3}}x^{3}\right]_{{0}}^{{2}}=\pi \left[-32+{\frac {128}{3}}\right]={\frac {32\pi }{3}}$