Answer
$$V = 10\pi $$
Work Step by Step
$$\eqalign{
& {\text{We have }}x = 1 - {y^2},\,\,\,\,x = 2 + {y^2},\,\,\,y = - 1,\,\,\,y = 1 \cr
& {\text{Let }}w\left( y \right) = 1 - {y^2}{\text{ and }}v\left( y \right) = 2 + {y^2} \cr
& w\left( 0 \right) = 1{\text{ and }}v\left( 0 \right) = 2 \cr
& v\left( y \right) \geqslant w\left( y \right){\text{ on the interval }} - 1 \leqslant y \leqslant 1 \cr
& \cr
& {\text{The volume of the solid can be calculated using}} \cr
& V = \int_c^d {\pi \left( {{{\left[ {v\left( y \right)} \right]}^2} - {{\left[ {w\left( y \right)} \right]}^2}} \right)dy} \cr
& {\text{Thus}}{\text{,}} \cr
& V = \pi \int_{ - 1}^1 {\left( {{{\left[ {2 + {y^2}} \right]}^2} - {{\left[ {1 - {y^2}} \right]}^2}} \right)dy} \cr
& V = \pi \int_{ - 1}^1 {\left( {4 + 4{y^2} + {y^4} - 1 + 2{y^2} - {y^4}} \right)dy} \cr
& V = \pi \int_{ - 1}^1 {\left( {3 + 6{y^2}} \right)dy} \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {3y + 2{y^3}} \right]_{ - 1}^1 \cr
& V = \pi \left[ {3\left( 1 \right) + 2{{\left( 1 \right)}^3}} \right] - \pi \left[ {3\left( { - 1} \right) + 2{{\left( { - 1} \right)}^3}} \right] \cr
& V = \pi \left( 5 \right) - \pi \left( { - 5} \right) \cr
& V = 10\pi \cr} $$
