Answer
$$V = \frac{{16}}{3}\pi $$
Work Step by Step
$$\eqalign{
& y = {x^2}{\text{ and }}y = 2 - {x^2},{\text{ revolved about the }}x{\text{ - axis}} \cr
& {\text{Using the washer method about the }}x{\text{ - axis}} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {\text{From the graph of the region shown below}} \cr
& 2 - {x^2} \geqslant {x^2}{\text{ on the interval }}\left[ { - 1,1} \right]{\text{ }} \cr
& {\text{Let }}f\left( x \right) = 2 - {x^2}{\text{ and }}g\left( x \right) = {x^2} \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_{ - 1}^1 {\pi \left[ {{{\left( {2 - {x^2}} \right)}^2} - {{\left( {{x^2}} \right)}^2}} \right]} dx \cr
& V = \pi \int_{ - 1}^1 {\left( {4 - 4{x^2} + {x^4} - {x^4}} \right)} dx \cr
& V = \pi \int_{ - 1}^1 {\left( {4 - 4{x^2}} \right)} dx \cr
& {\text{By symmetry}} \cr
& V = 2\pi \int_0^1 {\left( {4 - 4{x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {4x - \frac{4}{3}{x^3}} \right]_0^1 \cr
& V = 2\pi \left[ {4\left( 1 \right) - \frac{4}{3}{{\left( 1 \right)}^3}} \right] - 2\pi \left[ {4\left( 0 \right) - \frac{4}{3}{{\left( 0 \right)}^3}} \right] \cr
& V = 2\pi \left( {\frac{8}{3}} \right) - 0 \cr
& V = \frac{{16}}{3}\pi \cr} $$
