# What is the image of $f(x,y,z)=\sqrt{16-x^2-y^2-z^2}$?

Considering the function $$f(x,y,z)=\sqrt{16 - x^2 - y^2 - z^2}$$ I found that its domain is

$$D(f)= \{ (x,y,z) \in \mathbb R \} \setminus \{ x^2 - y^2 - z^2 \leq 16 \}$$

I'm not sure if the image is $[0,16]$, considering that square root has a restriction where its content must be $\geq 0$. Is it right?

After a few years I'm studying this subject and can't remember much about it.

• The image is $[0,4]$ because, in $D(f)$, the expression $16-x^2-y^2-z^2$ is in $[0,16]$. Commented Mar 10, 2018 at 22:46
• But how did you get these values of [0, 4]. Commented Mar 10, 2018 at 22:48
• Consider $f(x,y,z)=\sqrt{16-x²-y²-z²}=\sqrt{16-r^2}$ and think to the function $y=\sqrt{16-x^2}$, what is the image?
– user
Commented Mar 10, 2018 at 22:52

We require $$16-x^2-y^2-z^2 \ge 0$$

Hence, we need $$x^2+y^2+z^2 \le 16$$

$$0\le \sqrt{16-x^2-y^2-z^2} \le \sqrt{16}=4$$

Note that in spherical coordinates

$$f(x,y,z)=\sqrt{16-x²-y²-z²}=\sqrt{16-r^2}$$

with $0\le r\le4$ thus of course the image is $[0,4]$ since

$$0\le \sqrt{16-r^2} \le 4$$

• Sorry for the question, but why you replaced $x^2-y^2-z^2$ with $r^2$? Commented Mar 10, 2018 at 22:52
• it is to see better what is the image, remember that we can express (x,y,z) in spherical coordinates and $x^2+y^2+z^2=r^2$. As an alternative consider that the domain for f is the inner space of the sphere/ball with radius 4.
– user
Commented Mar 10, 2018 at 22:54

Obviously $$f(x,y,z)=\sqrt{16 - x^2 - y^2 - z^2}\ge 0$$

On the other hand $$16 - x^2 - y^2 - z^2$$ is maximized when $$x=y=z=0$$

Thus the maximum range is $4$.

That is $$0\le f(x,y,z)\le 4$$