Answer
$\displaystyle{V=\frac{27\pi}{2}}$
Work Step by Step
$\displaystyle{4x-x^2=x}\\ \displaystyle{x^2-3x=0}\\ \displaystyle{x(x-3)=0}\\ \displaystyle{x=0 \qquad x=3}$
$\displaystyle{V=\int_{0}^{3}(2\pi x)\left(4x-x^2-x\right)\ dx}\\ \displaystyle{V=2\pi\int_{0}^{3}3x^2-x^3\ dx}\\ \displaystyle{V=2\pi\left[x^3-\frac{1}{4}x^4\right]_{0}^{3}}\\ \displaystyle{V=2\pi\left(\left((3)^3-\frac{1}{4}(3)^4\right)-\left(0\right)\right)}\\ \displaystyle{V=\frac{27\pi}{2}}$
