Answer
$\displaystyle{V=8\pi}\\$
Work Step by Step
$\displaystyle{x^2=6x-2x^2}\\
\displaystyle{3x^2-6x=0}\\
\displaystyle{3x(x-2)=0}\\
\displaystyle{x=0\qquad x=2}$
$\displaystyle{V=\int_{0}^{2}(2\pi x)\left(6x-2x^2-x^2\right)\ dx}\\
\displaystyle{V=2\pi\int_{0}^{2}( x)\left(6x-3x^2\right)\ dx}\\
\displaystyle{V=2\pi\int_{0}^{2}6x^2-3x^3\ dx}\\
\displaystyle{V=2\pi\left[2x^3-\frac{3}{4}x^4\right]_{0}^{2}}\\
\displaystyle{V=2\pi\left(\left(2(2)^3-\frac{3}{4}(2)^4\right)-(0)\right)}\\
\displaystyle{V=8\pi}\\$