Answer
$$V = \frac{{{\pi ^2}}}{2} - \pi $$
Work Step by Step
$$\eqalign{
& {\text{Let the functions}}{\text{, }}y = \frac{1}{{\sqrt {{x^2} + 1} }},{\text{ }}y = \frac{1}{{\sqrt 2 }} \cr
& {\text{The graph of the region }}R{\text{ is shown below}}{\text{.}} \cr
& {\text{From the graph we can note that the interval of integration}} \cr
& {\text{is from }}\underbrace {\left[ { - 1,1} \right]}_{\left[ {a,b} \right]} \cr
& \frac{1}{{\sqrt {{x^2} + 1} }} \geqslant \frac{1}{{\sqrt 2 }}{\text{ on the interval }}\left[ { - 1,1} \right] \cr
& {\text{Revolving the region about the }}x{\text{ - axis}}{\text{, using the}} \cr
& {\text{Washer Method about the }}x{\text{ - axis }} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {\text{Let }}f\left( x \right) = \frac{1}{{\sqrt {{x^2} + 1} }}{\text{ and }}g\left( x \right) = \frac{1}{{\sqrt 2 }}{\text{ on the interval }}\left[ { - 1,1} \right] \cr
& V = \int_{ - 1}^1 {\pi \left[ {{{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2} - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right]} dx \cr
& {\text{By symmetry}} \cr
& V = 2\int_0^1 {\pi \left[ {{{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2} - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} \right]} dx \cr
& V = 2\int_0^1 {\pi \left( {\frac{1}{{{x^2} + 1}} - \frac{1}{2}} \right)} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {{{\tan }^{ - 1}}x - \frac{1}{2}x} \right]_0^1 \cr
& V = 2\pi \left[ {{{\tan }^{ - 1}}\left( 1 \right) - \frac{1}{2}\left( 1 \right)} \right] - 2\pi \left[ {{{\tan }^{ - 1}}\left( 0 \right) - \frac{1}{2}\left( 0 \right)} \right] \cr
& V = 2\pi \left( {\frac{\pi }{4} - \frac{1}{2}} \right) - 2\pi \left( 0 \right) \cr
& V = \frac{{{\pi ^2}}}{2} - \pi \cr} $$
