Points for a cone surface region

Question

The surface $z^2=x^2+y^2$ is a cone with center at $(0,0,0)$, the surface $z=-(x^2+y^2)^{1/2}$ is the bottom part of that cone, now, the region $z\leq-(x^2+y^2)^{1/2}$ are the points outside of that cone, but if I pick the point $(1,1,0)$ which is outside of the cone and plug it into the inequality I get nonsense,$0≤-(2)^{-1/2}$, what is happening here? Where am I wrong? If my inequality is wrong, can you tell me why? Thanks

Answer 1

You have a few things to clean up:

The equation of a cone: $z^2 = x^2 + y^2$

$z = \pm \sqrt {x^2 + y^2}$

The points outside the cone:

$|z| < \sqrt {x^2 + y^2}$

Show that the point $(1,1,0)$ it outside the cone...

$|0| < \sqrt {2}$

and your remaing problem

Consider: $z > - (x^2 + y^2)^{\frac 12}$

at the point $(x,y) = (1,1)$

$z > -(2)^\frac 12 = -\sqrt 2$

Exponentiation happens before the multiplication of the $-1.$ The negative is "on the outside."