Answer
$V = \pi \int_0^\pi {\left[ {{{\left( {\sin x + 2} \right)}^2} - 4} \right]} dx$
Work Step by Step
$$\eqalign{
& y = \sin x,{\text{ }}y = 0,{\text{ }}0 \leqslant x \leqslant \pi ;{\text{ about }}y = - 2 \cr
& {\text{Graph of the region shown below }}\left( {{\text{Using Geogebra}}} \right) \cr
& \cr
& {\text{For this exercise the inner radius is given by:}} \cr
& {r_i} = f\left( x \right) - \left( { - 2} \right) \cr
& {r_i} = 0 - \left( { - 2} \right) \cr
& {r_i} = 2 \cr
& {\text{And the outer radius is given by:}} \cr
& {r_o} = f\left( x \right) - \left( { - 2} \right) \cr
& {r_o} = \sin x - \left( { - 2} \right) \cr
& {r_o} = \sin x + 2 \cr
& {\text{The cross - sectional area is}} \cr
& A\left( x \right) = \pi {\left( {{r_o}} \right)^2} - \pi {\left( {{r_i}} \right)^2} \cr
& A\left( x \right) = \pi {\left( {\sin x + 2} \right)^2} - \pi {\left( 2 \right)^2} \cr
& A\left( x \right) = \pi {\left( {\sin x + 2} \right)^2} - 4\pi \cr
& {\text{The volume is given by}} \cr
& V = \int_a^b {A\left( x \right)} dx \cr
& V = \int_0^\pi {\left[ {\pi {{\left( {\sin x + 2} \right)}^2} - 4\pi } \right]} dx \cr
& V = \pi \int_0^\pi {\left[ {{{\left( {\sin x + 2} \right)}^2} - 4} \right]} dx \cr} $$
