Question

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

Answer 1

Because `f(x,y)=sqrt(9-x^2-y^2)` we must have that

`9-x^2-y^2>=0=>9>=x^2+y^2=>3^2>=x^2+y^2`

The domain of `f(x,y)` 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(x,y)>=0` and `f(x,y)<=3` we find that the range of the function is the interval `[0,3]`