Answer
Elliptic paraboloid with vertex at $(0,0,4)$ intercepted at $z=4$.
Work Step by Step
Given: $z=4-r^2$
We know that $r^2=x^2+y^2 \implies r=\sqrt{x^2+y^2}$
Thus, we have
$z=4-(x^2+y^2)$
This can be written as:
$x^2+y^2+z=4$
Thus, the surface is an elliptic paraboloid with vertex at $(0,0,4)$ intercepted at $z=4$.