Answer
$$V = 2\sqrt 3 \pi - \frac{2}{3}{\pi ^2}$$
Work Step by Step
$$\eqalign{
& {\text{Let the functions }} \cr
& y = \sin x{\text{ and }}y = 1 - \sin x \cr
& {\text{From the graph of the region shown below}} \cr
& \sin x > 1 - \sin x{\text{ on the interval }}\left[ {\frac{\pi }{6},\frac{{5\pi }}{6}} \right]{\text{ }} \cr
& {\text{Using the washer method about the }}x{\text{ - axis}} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {\text{Let }}f\left( x \right) = \sin x{\text{ and }}g\left( x \right) = 1 - \sin x \cr
& V = \int_{\pi /6}^{5\pi /6} {\pi \left[ {{{\left( {\sin x} \right)}^2} - {{\left( {1 - \sin x} \right)}^2}} \right]} dx \cr
& V = \int_{\pi /6}^{5\pi /6} {\pi \left( {{{\sin }^2}x - 1 + 2\sin x - {{\sin }^2}x} \right)} dx \cr
& V = \pi \int_{\pi /6}^{5\pi /6} {\left( {2\sin x - 1} \right)} dx \cr
& {\text{Integrating}} \cr
& V = - \pi \left[ {2\cos x + x} \right]_{\pi /6}^{5\pi /6} \cr
& V = - \pi \left[ {2\cos \left( {\frac{{5\pi }}{6}} \right) + \left( {\frac{{5\pi }}{6}} \right)} \right] + \pi \left[ {2\cos \left( {\frac{\pi }{6}} \right) + \left( {\frac{\pi }{6}} \right)} \right] \cr
& {\text{Simplifying}} \cr
& V = - \pi \left[ { - \sqrt 3 + \frac{{5\pi }}{6}} \right] + \pi \left[ {\sqrt 3 + \frac{\pi }{6}} \right] \cr
& V = \sqrt 3 \pi - \frac{{5{\pi ^2}}}{6} + \sqrt 3 \pi + \frac{{{\pi ^2}}}{6} \cr
& V = 2\sqrt 3 \pi - \frac{2}{3}{\pi ^2} \cr} $$
