= Solution
Let $\rho=\sqrt{x^2+y^2}=uv$. The condition $u\leq v$ gives $z\leq0$. 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 $\rho=u^2\leq1$. Thus the closure of the body is
$$
\boxed{0\leq\rho\leq1,\qquad\frac{\rho^2-1}{2}\leq z\leq0.}
$$
It 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>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-paraboloidal-cap.png]
{title=The solid cap bounded by an <elliptic paraboloid> and the disk $z=0$, with a filled meridian cross-section}
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