Final answer:
To show that the surface S lies in the upper half space (z≥0), we need to prove that for any point on S, the z-coordinate is greater than or equal to zero. By substituting z = 0 into the equation of the surface and applying properties of squares, we can derive the condition x² + y² + z² ≥ 0, which confirms the surface lies in the upper half space.
Step-by-step explanation:
To show that the surface S lies in the upper half space (z≥0), we need to prove that for any point on S, the z-coordinate is greater than or equal to zero. The equation of the surface S is z - (x² + y² + z²)² = 0. Substituting z = 0 into this equation, we get 0 - (x² + y² + 0)² = 0. Since a square is always non-negative, the left-hand side of the equation is non-positive. Therefore, the only way for the equation to be true is if the right-hand side is non-negative or zero. This means that (x² + y² + z²)² ≥ 0, which implies that x² + y² + z² ≥ 0. Since the sum of squares of any real numbers is always non-negative, we have x² + y² + z² ≥ 0. Therefore, S lies in the upper half space (z≥0).