Answer
$\displaystyle{V=\frac{8\pi}{3}}$
Work Step by Step
$\displaystyle{3=4x-x^2}\\ \displaystyle{x^2-4x+3=0}\\ \displaystyle{x=1\qquad x=3}$
$\displaystyle{V=\int_{1}^{3}(2\pi (1-x))\left(3-\left(4x-x^2\right)\right)\ dx}\\ \displaystyle{V=2\pi\int_{1}^{3}3-7x+5x^2-x^3\ dx}\\ \displaystyle{V=2\pi\left[3x-\frac{7}{2}x^2+\frac{5}{3}x^3-\frac{1}{4}x^4\right]_{1}^{3}}\\ \displaystyle{V=2\pi\left(\left(3(3)-\frac{7}{2}(3)^2+\frac{5}{3}(3)^3-\frac{1}{4}(3)^4\right)-\left(3(1)-\frac{7}{2}(1)^2+\frac{5}{3}(1)^3-\frac{1}{4}(1)^4\right)\right)}\\ \displaystyle{V=\frac{8\pi}{3}}$
