Answer
$$V = {\pi ^2} - 2\pi $$
Work Step by Step
$$\eqalign{
& y = \cos x{\text{ and }}y = 0{\text{ for the interval 0}} \leqslant x \leqslant \pi /2 \cr
& {\text{The volume is given by}} \cr
& V = \int_a^b {2\pi xf\left( x \right)dx} \cr
& V = 2\pi \int_0^{\pi /2} {x\cos xdx} \cr
& {\text{Integrating by parts}} \cr
& u = x,{\text{ }}du = dx \cr
& dv = \cos xdx,{\text{ }}v = \sin x \cr
& \int {x\cos xdx} = x\sin x - \int {\sin x} dx \cr
& \int {x\cos x} dx = x\sin x + \cos x + C \cr
& {\text{Therefore,}} \cr
& V = 2\pi \int_0^{\pi /2} {x\cos xdx} \cr
& V = 2\pi \left[ {x\sin x - \sin x} \right]_0^{\pi /2} \cr
& V = 2\pi \left[ {\frac{\pi }{2}\sin \left( {\frac{\pi }{2}} \right) + \cos \left( {\frac{\pi }{2}} \right)} \right] - 2\pi \left[ {0\sin \left( 0 \right) + \cos \left( 0 \right)} \right] \cr
& V = 2\pi \left[ {\frac{\pi }{2}} \right] - 2\pi \left[ 1 \right] \cr
& V = {\pi ^2} - 2\pi \cr} $$
