Multivariable domain

  • Thread starter lukatwo
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  • #1
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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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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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?
 
  • #4
35,140
6,892
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.
 

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