Answer
$$V = 24\pi $$
Work Step by Step
$$\eqalign{
& {\text{From the graph of the region shown below}}{\text{, we have the points}} \cr
& \left( {0,8} \right){\text{ and }}\left( {3,0} \right) \cr
& {\text{The equation of the line is }} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 8 = - \frac{8}{3}\left( {x - 0} \right) \cr
& y = - \frac{8}{3}x + 8 \cr
& {\text{Use the Shell method about the }}y{\text{ - axis}} \cr
& V = \int_c^d {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{From the graph we can see that}} \cr
& - \frac{8}{3}x + 8 > 0{\text{ on the interval }}\left[ {0,8} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^3 {2\pi x\left( { - \frac{8}{3}x + 8} \right)} dx \cr
& V = 2\pi \int_0^3 {\left( { - \frac{8}{3}{x^2} + 8x} \right)} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ { - \frac{8}{9}{x^3} + 4{x^2}} \right]_0^3 \cr
& V = 2\pi \left[ { - \frac{8}{9}{{\left( 3 \right)}^3} + 4{{\left( 3 \right)}^2}} \right] - 2\pi \left[ { - \frac{8}{9}{{\left( 0 \right)}^3} + 4{{\left( 0 \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& V = 2\pi \left[ {12} \right] - 2\pi \left[ 0 \right] \cr
& V = 24\pi \cr} $$
