multivariable function- domain and range

• Jul 12th 2008, 03:12 PM
dataspot
multivariable function- domain and range
$f(x, y)=\sqrt{(4-x^2-y2)}$

domain 4x-x^2-y^2 has to be > or equal to 0, so x^2+y^2 < or equal to 4.

range: I have no idea how to find the range here.
• Jul 12th 2008, 03:18 PM
TKHunny
Quote:

Originally Posted by dataspot
x^2+y^2 < or equal to 4

Domain:

Well, yes, but what is that? Do you see the border of this region as a circle? The region contained within the circle is a ...?

Range:

What happens for x = y = 0? Is that significant?
What do you know about the "Principle" square root? Positive? Negative? Zero?
• Jul 12th 2008, 05:10 PM
Mathstud28
Quote:

Originally Posted by dataspot
$f(x, y)=\sqrt{(4-x^2-y2)}$

domain 4x-x^2-y^2 has to be > or equal to 0, so x^2+y^2 < or equal to 4.

range: I have no idea how to find the range here.

Yes you are right, the range is the interior and edge of the circle of radius of two centered at $(0,0)$

For the range think the max and min of $f(x,y)=\sqrt{4-(x^2+y^2)}$ bounded by $x^2+y^2\leq{4}$