Answer
$$\frac{{386}}{{27}}\pi $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{3}x + 2,\,\,\,\,y = 0,\,\,\,\,\,x = 1,\,\,\,\,\,x = 3 \cr
& {\text{using the formula of a solid of revolution }} \cr
& {\text{the volume of the solid formed by rotating }}R{\text{ about thex - axis is given by}} \cr
& V = \int_a^b {\pi {{\left[ {f\left( x \right)} \right]}^2}} dx \cr
& {\text{for this exercise }}x = 1,\,\,\,\,\,x = 3{\text{ then }}a = 1{\text{ and }}b = 3{\text{ and }}f\left( x \right) = \frac{1}{3}x + 2 \cr
& V = \int_1^3 {\pi {{\left( {\frac{1}{3}x + 2} \right)}^2}} dx \cr
& {\text{expanding}} \cr
& = \pi \int_1^3 {\left( {\frac{1}{9}{x^2} + \frac{4}{3}x + 4} \right)} dx \cr
& {\text{integrating}} \cr
& = \pi \left( {\frac{1}{{27}}{x^3} + \frac{2}{3}{x^2} + 4x} \right)_1^3 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \pi \left[ {\frac{1}{{27}}{{\left( 3 \right)}^3} + \frac{2}{3}{{\left( 3 \right)}^2} + 4\left( 3 \right)} \right] - \pi \left[ {\frac{1}{{27}}{{\left( 1 \right)}^3} + \frac{2}{3}{{\left( 1 \right)}^2} + 4\left( 1 \right)} \right] \cr
& = \pi \left( {19} \right) - \pi \left( {\frac{{127}}{{27}}} \right) \cr
& = \frac{{386}}{{27}}\pi \cr} $$
