domain of multivariable function

Mar 2014
909
2
malaysia
from the graph of x and y shown , we know that the x and y value cant be exceed 3 , since it's a circle , why the domain for x and y is x and y can be any number (R) ?

 

romsek

MHF Helper
Nov 2013
6,838
3,079
California
from the graph of x and y shown , we know that the x and y value cant be exceed 3 , since it's a circle , why the domain for x and y is x and y can be any number (R) ?
It's not.

It's $x \in \mathbb{R}, ~~y \in \mathbb{R}$ SUCH THAT $x^2 + y^2 \leq 9$
 
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Mar 2014
909
2
malaysia
It's not.

It's $x \in \mathbb{R}, ~~y \in \mathbb{R}$ SUCH THAT $x^2 + y^2 \leq 9$
why not ? x^2 + y^2 is a graph of circlre with radius 3 , am i right ? beyond r= 3 , the function is undefined , right ?
 

romsek

MHF Helper
Nov 2013
6,838
3,079
California
you asked why the domain was $x \in \mathbb{R}, ~~y \in \mathbb{R}$

the domain is not this

it's this plus the constraint that $(x,y)$ be a point on the interior or boundary of a circle centered at the origin of radius 3.
 
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Mar 2014
909
2
malaysia
you asked why the domain was $x \in \mathbb{R}, ~~y \in \mathbb{R}$

the domain is not this

it's this plus the constraint that $(x,y)$ be a point on the interior or boundary of a circle centered at the origin of radius 3.
so the domain is x<3 and y<3 ?
 
Sep 2012
838
90
Canada
Just as Romsek said, less than and equal to 3. You can have x=3 and y=0 or y=3 and x=0
 
Mar 2014
909
2
malaysia
But the problem is when x or y more than 3, then it would be outside the shaded region.... How can domain of x be any number true??

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Dec 2013
2,002
757
Colombia
It's not. \(\displaystyle x=4\) violates the second part of the condition \(\displaystyle x^2+y^2 \le 9\) as does \(\displaystyle x=y=3\).
 
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Mar 2014
909
2
malaysia
It's not. \(\displaystyle x=4\) violates the second part of the condition \(\displaystyle x^2+y^2 \le 9\) as does \(\displaystyle x=y=3\).
Since you said x=4 viiolates the second condition, so , x should be less than 3 , am I right???