The volume is
[tex]\displaystyle2\pi\int_0^2x^3\,\mathrm dx=\frac\pi2 x^4\bigg|_0^2=8\pi[/tex]
Each shell has a height of [tex]x^2[/tex] (distance between [tex]y=x^2[/tex] and [tex]y=0[/tex]) and width equal to the distance to the axis of revolution ([tex]x-0=x[/tex]), hence contributing a surface area of [tex]2\pi x^3[/tex]. Sum (integrate) over all [tex]x[/tex] in the region of interest to get the volume.
