Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 15

In cylindrical coordinates, $x=r\cos \theta$ and $y=r\sin \theta$, so $x^2+y^2=r^2$. Thus the circular elliptic paraboloid has equationAt height $h\ge 0$, the rim radius is $\sqrt{ah}$.

For $h>0$, parametrize the curved surface by $(r\cos \theta ,r\sin \theta ,r^2/a)$, with $0\le r\le \sqrt{ah}$ and $0\le \theta \le 2\pi$. The magnitude of the cross product of its coordinate tangents gives the area element of a surface:Integrating givesThis is the curved surface area, not the area of a lid across the horizontal opening. The expression tends to zero at $h=0$ and has dimensions of area. The volume and centroid of a parabolic cup use the same geometry.

The liquid occupies $0\le z\le h$ and $0\le r\le \sqrt{az}$. A horizontal slice is a disk of area $\pi az$. Integrating these areas givesEquivalently, the triple integral is ${\int }_0^{2\pi }{\int }_0^h{\int }_0^{\sqrt{az}}rdrdzd\theta$, including the cylindrical Jacobian determinant factor $r$.

For positive constant mass density ${\rho }_0$, reflection symmetry makes the horizontal coordinates of the centre of mass zero. The total mass and the vertical first moment areTheir ratio givesThis is the centre of the liquid volume, rather than of the curved wall. The density cancels because it is uniform; the larger disks near the top place the centre above half-height.