Elliptic paraboloid ($z=x^2/A^2+y^2/B^2$)

An elliptic paraboloid is a smooth bowl-shaped quadratic surface with elliptical horizontal sections. In the circular case $A=B$, it is a surface of revolution. For $z=x^2+y^2$, polar coordinates give the parametrization $(r\cos \theta ,r\sin \theta ,r^2)$ and upward vector area element $(-2r^2\cos \theta ,-2r^2\sin \theta ,r)drd\theta$. A band between two positive heights has two boundary circles and no cap.

Suppose a solid is above the elliptic paraboloid $z=x^2+y^2$ and below a plane whose vertical separation from it is $A-|\mathbf{x}-\mathbf{p}{|}^2$, where $A>0$. A second lower plane exceeds the elliptic paraboloid by $B-|\mathbf{x}-\mathbf{q}{|}^2$, with $B>0$. If $|\mathbf{p}-\mathbf{q}|+\sqrt{B}\le \sqrt{A}$, the disk where the lower plane matters lies within the upper projection. Raising the bottom to this lower plane therefore removes exactly its excess volume, without changing the outer projection.

Translate polar coordinates separately to the two centres. The double integral over a disk of squared radius $A$ isSubtracting the analogous integral with $B$ proves the displayed volume formula. The containment condition is essential: if the two disks overlap only partly, subtract only over their intersection.

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.

For $h,R>0$ and $0<\alpha \le 2\pi$, the curved sector $z=h(1-r^2/R^2)$, $0\le r\le R$, $0\le \theta \le \alpha$ has volume $\alpha h{R}^2/4$ beneath it. Its surface area isTo derive this, parametrize the surface by $(r\cos \theta ,r\sin \theta ,h(1-r^2/R^2))$. The upward vector area element is $(2h{r}^2\cos \theta /R^2,2h{r}^2\sin \theta /R^2,r)drd\theta$; its magnitude integrates to the stated surface area. Its integral gives the vector areaIn the shallow limit $h/R\to 0$, the surface area is $\alpha R^2/2+\alpha h^2/2+O(h^4/R^2)$, approaching the area of its planar sector projection.

For the rotational elliptic paraboloid $z=H(1-(x^2+y^2)/R^2)$ above $z=0$, allow inscribed shapes to touch the boundary. A coaxial cylinder of radius $r$ and height $h$ obeys $h+H{r}^2/R^2=H$. Applying a Lagrange multiplier to $\pi r^{2}h$ gives $h=H{r}^2/R^2$, hence $h=H/2$ and $r=R/\sqrt{2}$.

A centred rectangular parallelepiped with horizontal side lengths $a,b$ and height $c$ obeys $c+H(a^2+b^2)/(4R^2)=H$. The Lagrange multiplier equations for $abc$ give $a=b$ and $c=H{a}^2/(2R^2)$, hence $a=b=R,c=H/2$. These yield the displayed maximal volumes; both positive stationary shapes are interior optima among the dimension variables, while their degenerate boundary shapes have zero volume. The cylinder's volume is $\pi /2$ times the box's. If strict containment in the open region is required, these are suprema approached by slightly smaller shapes rather than attained maxima.