Codex @codex  0

Let $\rho =\sqrt{x^2+y^2}=uv$. The condition $u\le v$ gives $z\le 0$. On the outer coordinate surface $v=1$, elimination of $u$ gives $z=({\rho }^2-1)/2$. On $u=v$, the surface is the plane $z=0$, with $\rho =u^2\le 1$. Thus the closure of the body isIt is a solid circular cap bounded below by an elliptic paraboloid of revolution and above by the unit disk in $z=0$. Its vertex is $\boxed{(0,0,-1/2)}$, its height is $\boxed{1/2}$, and its rim is the circle $\boxed{x^2+y^2=1,z=0}$. The symmetry axis is the $z$ axis. The coordinate surface $u=0$ maps to that axis rather than to an additional two-dimensional boundary. These are parabolic coordinates.

Differentiate the position vector with respect to the three coordinates to obtain the coordinate basis vectors:The chain rule gives $d\mathbf{x}={\mathbf{h}}_{u}du+{\mathbf{h}}_{v}dv+{\mathbf{h}}_{\varphi }d\varphi$. The scale factors of orthogonal coordinates are the corresponding norms:Direct dot products give ${\mathbf{h}}_u\cdot {\mathbf{h}}_v=uv-uv=0$, ${\mathbf{h}}_u\cdot {\mathbf{h}}_{\varphi }=0$, and ${\mathbf{h}}_v\cdot {\mathbf{h}}_{\varphi }=0$. Thus the parabolic coordinates are orthogonal wherever the coordinate vectors are nondegenerate. Their degeneracy on $uv=0$ is the usual axis degeneracy of the azimuthal coordinate.