 Volume and centroid of a parabolic cup

The solid $0\le z\le h$, $x^2+y^2\le az$, with $a,h>0$, has horizontal slice area $\pi az$. Thus its volume is ${\int }_0^h\pi azdz=\pi a{h}^2/2$. For a positive constant mass density, rotational symmetry puts its centre of mass on the vertical axis, andThe curved surface area, excluding the horizontal opening, is $\pi a^2[(1+4h/a{)}^{3/2}-1]/6$. Parametrizing the elliptic paraboloid by $(r\cos \theta ,r\sin \theta ,r^2/a)$ gives the area element of a surface $r\sqrt{1+4r^2/a^2}drd\theta$, with $0\le r\le \sqrt{ah}$, proving this last formula. It is also a translated and reflected specialization of paraboloid sector surface measures.