= Volume and centroid of a parabolic cup
{title2=$V=\pi ah^2/2,\quad \bar z=2h/3$}
The solid $0\leq z\leq h$, $x^2+y^2\leq az$, with $a,h>0$, has horizontal slice area $\pi az$. Thus its <volume> is $\int_0^h\pi az\,dz=\pi ah^2/2$. For a positive constant <mass density>, rotational symmetry puts its <centre of mass> on the vertical axis, and
$$
\bar z=\frac{\int_0^h z(\pi az)dz}{\int_0^h\pi az\,dz}=\frac{2h}{3}.
$$
The 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> $r\sqrt{1+4r^2/a^2}\,dr\,d\theta$, with $0\leq r\leq\sqrt{ah}$, proving this last formula. It is also a translated and reflected specialization of <paraboloid sector surface measures>.
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