Solving Multivariable Domain: x^2+y^2<=|z^2|

[lukatwo]

Homework Statement

I'm having a problem with solving the domain of |x^2+y^2|<=|z^2|.

Homework Equations

The Attempt at a Solution

From what I got this should be separated into two expressions:
x^2+y^2<=z^2, and x^2+y^2>=-z^2. The later doesn't have a real solution because of the negative root. The first one should then be z>=+-sqrt(x^2+y^2). I can understand that z>=+sqrt(x^2+y^2) is above the positive cone(inside), but where is z>=-sqrt(x^2+y^2) supposed to be? I'm thinking above the negative cone(outside), but I'm not sure.

 

[Mark44]

Homework Statement

I'm having a problem with solving the domain of |x^2+y^2|<=|z^2|.

x, y, and z are all real numbers, right?

If so, you can get rid of the absolute value symbols, since x² + y² ≥ 0 for any real numbers x and y. Similarly, z² ≥ 0 for any real number z.

Homework Equations

The Attempt at a Solution

From what I got this should be separated into two expressions:
x^2+y^2<=z^2, and x^2+y^2>=-z^2. The later doesn't have a real solution because of the negative root. The first one should then be z>=+-sqrt(x^2+y^2). I can understand that z>=+sqrt(x^2+y^2) is above the positive cone(inside), but where is z>=-sqrt(x^2+y^2) supposed to be? I'm thinking above the negative cone(outside), but I'm not sure.

 

[lukatwo]

Maybe should have put the part that was troubling me in the question.

I can understand that z>=+sqrt(x^2+y^2) is above the positive cone(inside), but where is z>=-sqrt(x^2+y^2) supposed to be? I'm thinking above the negative cone(outside), but I'm not sure.

So basically, where is z>=-sqrt(x^2+y^2) supposed to be?

 

[Mark44]

If you strip out the unnecessary stuff, your inequality is x² + y² ≤ z².

First off, look at the equation x² + y² = z². This is part of your solution set. What does it look like?

Once you have that, then tackle the rest, which is x² + y² < z². The absolute smallest that x² + y² can be is 0.