Answer
$$V = 2{\pi ^2}$$
Work Step by Step
$$\eqalign{
& y = \sin x{\text{ and }}y = 0{\text{ for the interval 0}} \leqslant x \leqslant \pi \cr
& {\text{The volume is given by}} \cr
& V = \int_a^b {2\pi xf\left( x \right)dx} \cr
& V = 2\pi \int_0^\pi {x\sin xdx} \cr
& {\text{Integrating by parts}} \cr
& {\text{The volume is given by}} \cr
& V = \int_a^b {2\pi xf\left( x \right)dx} \cr
& V = 2\pi \int_0^\pi {x\sin xdx} \cr
& {\text{Integrating by parts}} \cr
& u = x,{\text{ }}du = dx \cr
& dv = \sin xdx,{\text{ }}v = - \cos x \cr
& \int {x\sin x} dx = - x\cos x + \int {\cos x} dx \cr
& \int {x\sin xdx} = - x\cos x + \sin x + C \cr
& {\text{Therefore,}} \cr
& V = 2\pi \int_0^\pi {x\sin xdx} \cr
& V = 2\pi \left[ { - x\cos x + \sin x} \right]_0^\pi \cr
& V = 2\pi \left[ { - \pi \cos \pi + \sin \pi } \right] - 2\pi \left[ {0\cos 0 + \sin 0} \right] \cr
& V = 2\pi \left[ \pi \right] - 2\pi \left[ 0 \right] \cr
& V = 2{\pi ^2} \cr} $$
