x = \sqrt [4]{4y} \\ \\ \indent x^{2} = (4y)^{\frac {1}{8}} \\ \\ \indent V = \pi \int_{0}^{2} (4y)^{\frac {1}{8}} $ dy $ \\ \\ \indent V = \pi \left[\frac {8(4y)^{\frac {9}{8}}}{9}}\right]_0^2 \\ \\ \indent V = \frac {8\pi \sqrt [8]{8^{9}}}{9} $ units$^{3}
Pretty sure that might be wrong, but...