Answer
$$V = \frac{{{\pi ^2}}}{8}$$
Work Step by Step
$$\eqalign{
& y = \cos 2x,{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = \frac{\pi }{4} \cr
& V = \pi \int_a^b {{{\left[ {R\left( x \right)} \right]}^2}} dx \cr
& {\text{Let }}R\left( x \right) = \cos 2x \cr
& {\text{So}},{\text{ the volume of the solid of revolution is}} \cr
& V = \pi \int_0^{\pi /4} {{{\left( {\cos 2x} \right)}^2}} dx \cr
& V = \pi \int_0^{\pi /4} {{{\cos }^2}2x} dx \cr
& {\text{Using trigonometric identities}} \cr
& V = \pi \int_0^{\pi /4} {\frac{{1 + \cos 4x}}{2}} dx \cr
& {\text{Integrate}} \cr
& V = \frac{\pi }{2}\left[ {x + \frac{1}{4}\sin 4x} \right]_0^{\pi /4} \cr
& V = \frac{\pi }{2}\left[ {\frac{\pi }{4} + \frac{1}{2}\sin \pi } \right] - \frac{\pi }{2}\left[ {0 - \frac{1}{2}\sin 0} \right] \cr
& V = \frac{{{\pi ^2}}}{8} \cr} $$
