Answer
$\displaystyle{V=\frac{40\pi}{3}}$
Work Step by Step
$\displaystyle{y=4-2x}\\
\displaystyle{0=4-2x}\\
\displaystyle{2x=4}\\
\displaystyle{x=2}$
$\displaystyle{V=\int_{0}^{2}(2\pi (x+1))\left(4-2x\right)\ dx}\\
\displaystyle{V=2\pi\int_{0}^{2}2x-2x^2+4\ dx}\\
\displaystyle{V=2\pi\left[x^2-\frac{2}{3}x^3+4x\right]_{0}^{2}}\\
\displaystyle{V=2\pi\left(\left((2)^2-\frac{2}{3}(2)^3+4(2)\right)-(0)\right)}\\
\displaystyle{V=\frac{40\pi}{3}}$
