We can use Archimedes' method to determine the center of gravity of the paraboloid. By symmetry, we know that the center of gravity lies at some point \(X\) on its axis (not indicated on the applet). Archimedes' Proposition 5, illustrated below, shows a paraboloid and a cone inscribed inside a cylinder.

The sketch illustrates that the paraboloid, left where it is, balances the cone, moved to \(H\), where \(|AH|=|AD|\). So, by the law of the lever,

\[(\mbox{Volume of paraboloid})\times |AX|=(\mbox{Volume of cone})\times |AH|.\]

Since the volume of the paraboloid is half the volume of the cylinder (by the previous exercise), and the volume of the cone is one-third the volume of the cylinder, this implies that \[\frac12|AX|=\frac13|AH|,\,\,\,\,{\rm{hence}}\,\,\,{\rm{that}}\,\,\,\, |AX|=\frac23|AH|.\] So the center of gravity of the paraboloid lies on its axis, twice as far from the vertex as from the base.