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

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Answer 1 · 2016-02-28T21:01:10

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

$9 - x^{2} - y^{2} \geq 0 \Rightarrow 9 \geq x^{2} + y^{2} \Rightarrow 3^{2} \geq x^{2} + y^{2}$ $9-x^2-y^2>=0=>9>=x^2+y^2=>3^2>=x^2+y^2$

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

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

Now hence $f (x, y) \geq 0$ $f(x,y)>=0$ and $f (x, y) \leq 3$ $f(x,y)<=3$ we find that the range of the function is the interval $[0, 3]$ $[0,3]$

What is the domain and range of f(x,y) = sqrt(9-x^2-y^2)f(x,y)=√9−x2−y2f(x,y) = sqrt(9-x^2-y^2)?