[tex]x=2y[/tex] and [tex]x=y^3[/tex] intersect when
[tex]2y=y^3\implies y^3-2y=y(y^2-2)=y(y-\sqrt2)(y+\sqrt2)=0[/tex]
[tex]\implies y=0\text{ or }y=\sqrt2\text{ or }y=-\sqrt2[/tex]
We omit the negative root, since we only care for [tex]y\ge0[/tex].
In the interval (0, √2), we have [tex]y^3<2y[/tex]. Then the volume is given by the integral
[tex]\displaystyle\pi\int_0^{\sqrt2}(2y)^2-(y^3)^2\,\mathrm dy=\pi\int_0^{\sqrt2}4y^2-y^6\,\mathrm dy[/tex]
and so the volume is (32√2)/21 π.
