# What is the domain and range of f(x,y) = sqrt(9-x^2-y^2)?

Because $f \left(x , y\right) = \sqrt{9 - {x}^{2} - {y}^{2}}$ we must have that

$9 - {x}^{2} - {y}^{2} \ge 0 \implies 9 \ge {x}^{2} + {y}^{2} \implies {3}^{2} \ge {x}^{2} + {y}^{2}$

The domain of $f \left(x , y\right)$ is the border and the interior of the circle
${x}^{2} + {y}^{2} = {3}^{2}$
or

The domain is represented by the disc whose center is the origin of the coordinates system and the radius is 3.

Now hence $f \left(x , y\right) \ge 0$ and $f \left(x , y\right) \le 3$ we find that the range of the function is the interval $\left[0 , 3\right]$