Answer
$\displaystyle{V=\frac{16\pi}{15}}$
Work Step by Step
$\displaystyle{-x^2+6x-8=0}\\
\displaystyle{x^2-6x+8=0}\\
\displaystyle{x=2\qquad x=4}$
$\displaystyle{A(x)=\pi\left(-x^2+6x-8\right)^2}\\
\displaystyle{A(x)=\pi\left(x^4-12x^3+52x^2-96x+64\right)}$
$\displaystyle{V=\int_{2}^{4}A(x)\ dx}\\
\displaystyle{V=\int_{2}^{4}\pi\left(x^4-12x^3+52x^2-96x+64\right)\ dx}\\
\displaystyle{V=\pi\int_{2}^{4}x^4-12x^3+52x^2-96x+64\ dx}\\
\displaystyle{V=\pi\left[\frac{1}{5}x^5-3x^4+\frac{52}{3}x^3-48x^2+64x\right]_{2}^{4}}\\
\displaystyle{V=2\pi\left(\left(2(4)^3-\frac{1}{4}(4)^4-4(4)^2\right)-\left(2(2)^3-\frac{1}{4}(2)^4-4(2)^2\right)\right)}\\
\displaystyle{V=\frac{16\pi}{15}}$
