Answer
$$V = 64\pi $$
Work Step by Step
$$\eqalign{
& y = \sqrt x ,{\text{ }}y = - \frac{1}{2}x + 4,{\text{ }}x = 0,{\text{ }}x = 8 \cr
& {\text{From the graph }} - \frac{1}{2}x + 4 \geqslant \sqrt x {\text{ on }}\left[ {0,4} \right] \cr
& {\text{From the graph }}\sqrt x \geqslant - \frac{1}{2}x + 4{\text{ on }}\left[ {4,8} \right] \cr
& {\text{Apply the washer method}} \cr
& V = \pi \int_a^b {\left( {{{\left[ {R\left( x \right)} \right]}^2} - {{\left[ {r\left( x \right)} \right]}^2}} \right)} dx,{\text{ then}} \cr
& V = \pi \int_0^4 {\left( {{{\left[ { - \frac{1}{2}x + 4} \right]}^2} - {{\left[ {\sqrt x } \right]}^2}} \right)} dx \cr
& + \pi \int_4^8 {\left( {{{\left[ {\sqrt x } \right]}^2} - {{\left[ { - \frac{1}{2}x + 4} \right]}^2}} \right)} dx \cr
& {\text{Simplifying}} \cr
& V = \pi \int_0^4 {\left( {\frac{{{x^2}}}{4} - 4x + 16 - x} \right)} dx + \pi \int_4^8 {\left( {x - \frac{{{x^2}}}{4} + 4x - 16} \right)} dx \cr
& V = \pi \int_0^4 {\left( {\frac{{{x^2}}}{4} - 3x + 16} \right)} dx + \pi \int_4^8 {\left( {5x - \frac{{{x^2}}}{4} - 16} \right)} dx \cr
& V = \pi \left[ {\frac{{{x^3}}}{{12}} - \frac{3}{2}{x^2} + 16x} \right]_0^4 + \pi \left[ {\frac{{5{x^2}}}{2} - \frac{{{x^3}}}{{12}} - 16x} \right]_4^8 \cr
& V = \pi \left[ {\frac{{{{\left( 4 \right)}^3}}}{{12}} - \frac{3}{2}{{\left( 4 \right)}^2} + 16\left( 4 \right)} \right] + \pi \left[ {\frac{{5{{\left( 8 \right)}^2}}}{2} - \frac{{{{\left( 8 \right)}^3}}}{{12}} - 16\left( 8 \right)} \right] \cr
& - \pi \left[ {\frac{{5{{\left( 4 \right)}^2}}}{2} - \frac{{{{\left( 4 \right)}^3}}}{{12}} - 16\left( 4 \right)} \right] \cr
& V = \frac{{136}}{3}\pi - \frac{{32}}{3}\pi + \frac{{88}}{3}\pi \cr
& V = 64\pi \cr} $$
