Answer
$V = 4\pi $
Work Step by Step
$$\eqalign{
& y = \frac{1}{2}{x^2} + 1 \cr
& {\text{Applying the shell method about the }}y{\text{ - axis}} \cr
& V = 2\pi \int_a^b {x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{From the graph}} \cr
& {\text{Let }}f\left( x \right) = 3{\text{ and }}g\left( x \right) = \frac{1}{2}{x^2} + 1{\text{ on the interval }}\left[ {0,2} \right],{\text{ then}} \cr
& V = 2\pi \int_0^2 {x\left[ {3 - \left( {\frac{1}{2}{x^2} + 1} \right)} \right]} dx \cr
& V = 2\pi \int_0^2 {x\left( {3 - \frac{1}{2}{x^2} - 1} \right)} dx \cr
& V = 2\pi \int_0^2 {x\left( {2 - \frac{1}{2}{x^2}} \right)} dx \cr
& V = 2\pi \int_0^2 {\left( {2x - \frac{1}{2}{x^3}} \right)} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {{x^2} - \frac{1}{8}{x^4}} \right]_0^2 \cr
& V = 2\pi \left[ {{{\left( 2 \right)}^2} - \frac{1}{8}{{\left( 2 \right)}^4}} \right] - 2\pi \left[ {{{\left( 0 \right)}^2} - \frac{1}{8}{{\left( 0 \right)}^4}} \right] \cr
& V = 2\pi \left[ 2 \right] - 2\pi \left[ 0 \right] \cr
& V = 4\pi \cr} $$
