Answer
$$V = \frac{{{\pi ^2}}}{2}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{\sqrt {1 + {x^2}} }},{\text{ }}y = 0,{\text{ }}x = - 1,{\text{ }}x = 1 \cr
& {\text{From the graph shown below, we can apply the disk method}} \cr
& V = 2\left[ {\pi \int_0^1 {{{\left[ {f\left( x \right)} \right]}^2}} dx} \right] \cr
& V = 2\pi \int_0^1 {{{\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right)}^2}} dx \cr
& {\text{Integrate}} \cr
& V = 2\pi \int_0^1 {\frac{1}{{1 + {x^2}}}} dx \cr
& V = 2\pi \left[ {\arctan x} \right]_0^1 \cr
& V = 2\pi \left[ {\arctan \left( 1 \right) - \arctan \left( 0 \right)} \right] \cr
& V = 2\pi \left( {\frac{\pi }{4}} \right) \cr
& V = \frac{{{\pi ^2}}}{2} \cr} $$
