= Solution
Use cylindrical radius $\rho=\sqrt{x^2+y^2}$. The upper <cone> is $z=1-\rho$ and the lower <elliptic paraboloid> is $z=\rho^2-1$, both with $0\leq\rho\leq1$. They meet along the unit <circle> in $z=0$. The enclosed solid is
$$
0\leq\rho\leq1,\qquad \rho^2-1\leq z\leq1-\rho.
$$
Thus the coordinate ranges on the complete surface are
$$
\boxed{-1\leq x\leq1,\qquad-1\leq y\leq1,\qquad-1\leq z\leq1}.
$$
These separate ranges do not mean every point of their Cartesian product lies on the surface. A vertical cross-section has a straight upper cone profile and a curved lower paraboloid profile.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2-q16-shape.png]
{title=A cone and a paraboloid joined along their unit-circle rim}
{description=The upper <cone> has apex $(0,0,1)$; the lower <elliptic paraboloid> has vertex $(0,0,-1)$. Both meet at cylindrical radius one and height zero. The meridian cross-section shows the enclosed vertical range at each radius.}
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