Answer
The inequality x^2 + y^2 + z^2 > 2z
this implies x^2+y^2+(z-1)^2 >1 is equivalent to \sqrt x^2+y^2+(z-1)^2 is strictly greater than 1, so the region consists of those points whose distance from the point (0, 0, 1) is greater than 1. This is the set of all points outside the sphere with radius 1 and center (0, 0, 1).
Work Step by Step
The inequality x^2 + y^2 + z^2 > 2z
this implies x^2+y^2+(z-1)^2 >1 is equivalent to \sqrt x^2+y^2+(z-1)^2 is strictly greater than 1, so the region consists of those points whose distance from the point (0, 0, 1) is greater than 1. This is the set of all points outside the sphere with radius 1 and center (0, 0, 1).
