Range of a Multivariable Function

How do I find the range of f(x,y)=sqrt(9-x^2-y^2)

$\sqrt{9-x^2-y^2}\\ =\sqrt{9-(x^2+y^2)}\\ \text{So for real solutions}\\ x^2+y^2\le9\\ -3\le x \le3\qquad and \qquad -3\le y\le3\\ \text{The range is } -3\le y\le 3$

f (x,y)  = √ [ 9 - x ^2 - y^2 ]  = √ [ 9 - ( x ^2 + y^2) ]

We have a 3D surface here

Let  f(x,y)  = z

In real terms, the minimum for z, 0, will be achieved when x^2 + y^2   = 9

And the maximum for z, 3, will be achieved when   x , y   = 0 

And the range of z will be      0 ≤ z ≤ 3