Answer
$$V = \pi \left( {\pi - 2} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \cos x,\;{\text{interval }}\left[ {0,\frac{\pi }{2}} \right] \cr
& {\text{Use the shell method}}{\text{, revolved about }}y{\text{ - axis}} \cr
& V = 2\pi \int_a^b {xf\left( x \right)} dx \cr
& V = 2\pi \int_0^{\pi /2} {x\left( {\cos x} \right)} dx \cr
& {\text{Integrate by parts}} \cr
& u = x,{\text{ }}du = dx \cr
& dv = \cos x,{\text{ }}v = \sin x \cr
& {\text{Use integration by parts formula}} \cr
& V = 2\pi \left[ {x\sin x} \right]_0^{\pi /2} - 2\pi \int_0^{\pi /2} {\sin x} dx \cr
& V = 2\pi \left[ {\frac{\pi }{2}\sin \left( {\frac{\pi }{2}} \right) - 0\sin \left( 0 \right)} \right] - 2\pi \left[ { - \cos x} \right]_0^{\pi /2} \cr
& V = 2\pi \left[ {\frac{\pi }{2}} \right] + 2\pi \left[ {\cos \left( {\frac{\pi }{2}} \right) - \cos \left( 0 \right)} \right] \cr
& V = {\pi ^2} + 2\pi \left( { - 1} \right) \cr
& V = {\pi ^2} - 2\pi \cr
& V = \pi \left( {\pi - 2} \right) \cr} $$
