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