# How do you find the domain and range of f(x,y) = sqrt(9-x^2-y^2)?

##### 1 Answer

D: $- 3 \le x \le 3$
R: $- 3 \le y \le 3$

#### Explanation:

For the function:

$f \left(x , y\right) = \sqrt{9 - {x}^{2} - {y}^{2}}$

What is the list of allowable $x$ values - the domain?

We can't have a value under the square root sign that is less than 0, and so we can see that with $- 3 \le x \le 3$ we have acceptable values.

What then is the resulting values of $y$ - the range?

With $x = - 3 , y = 0$
With $x = 0 , y = \pm 3$
With $x = 3 , y = 0$

And so the range is $- 3 \le y \le 3$