Answer
$$V = \frac{{{\pi ^2}}}{2}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \sin x{\text{ and }}g\left( x \right) = 0 \cr
& {\text{The volume is given by}} \cr
& V = \pi \int_a^b {\left( {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right)dx} \cr
& V = \pi \int_0^{\pi /4} {\left( {{{\sin }^2}x - 0} \right)dx} \cr
& V = \pi \int_0^\pi {{{\sin }^2}x} dx \cr
& {\text{Use the identity }}{\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right) \cr
& V = \frac{\pi }{2}\int_0^\pi {\left( {1 - \cos 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \frac{\pi }{2}\left[ {x - \frac{1}{2}\sin 2x} \right]_0^\pi \cr
& V = \frac{\pi }{2}\left[ {\pi - \frac{1}{2}\sin 2\pi } \right] - \frac{\pi }{2}\left[ {0 - \frac{1}{2}\sin 0} \right] \cr
& V = \frac{{{\pi ^2}}}{2} \cr} $$
