Answer
\[V=\pi \left( 6\sqrt{3}-2\pi \right)\]
Work Step by Step
\[\begin{align}
& \text{Using the disk and washer method about the }x\text{-axis} \\
& V=\int_{a}^{b}{\pi \left( f{{\left( x \right)}^{2}}-g{{\left( x \right)}^{2}} \right)}dx \\
& \text{When the graph of }f\text{ is revolved about }y=-1,\text{ it sweeps out a } \\
& \text{solid of revolution whose radius at a point }x\text{ is}\text{.} \\
& f\left( x \right)+1=\sin x+1. \\
& \text{Similarly},\text{ when the graph ofgisrevolved about }y=-1,\text{it sweeps } \\
& \text{out a solid of revolution whose radius at a point }x\text{ is } \\
& g\left( x \right)+1=1-\sin x+1=2-\sin x \\
& \text{Therefore,} \\
& V=\int_{\pi /6}^{5\pi /6}{\pi \left( {{\left( \sin x+1 \right)}^{2}}-{{\left( 2-\sin x \right)}^{2}} \right)}dx \\
& V=\int_{\pi /6}^{5\pi /6}{\pi \left( {{\sin }^{2}}x+2\sin x+1-4+4\sin x-{{\sin }^{2}}x \right)}dx \\
& V=\pi \int_{\pi /6}^{5\pi /6}{\left( 6\sin x-3 \right)}dx \\
& \text{Integrating} \\
& V=\pi \left[ -6\cos x-3x \right]_{\pi /6}^{5\pi /6} \\
& V=\pi \left[ -6\cos \left( \frac{5\pi }{6} \right)-3\left( \frac{5\pi }{6} \right) \right]-\pi \left[ -6\cos \left( \frac{\pi }{6} \right)-3\left( \frac{\pi }{6} \right) \right] \\
& \text{Simplifying} \\
& V=\pi \left[ 3\sqrt{3}-\frac{15\pi }{6} \right]-\pi \left[ -3\sqrt{3}-\frac{\pi }{2} \right] \\
& V=\pi \left[ 3\sqrt{3}-\frac{15\pi }{6}+3\sqrt{3}+\frac{\pi }{2} \right] \\
& V=\pi \left( 6\sqrt{3}-2\pi \right) \\
\end{align}\]
