The problem gives an equation $z = \sqrt{16 - x^2 - y^2}$. We are asked to solve the problem. The specific question is not given. Let's find the domain.

1. Problem Description

The problem gives an equation $z = \sqrt{16 - x^2 - y^2}$. We are asked to solve the problem. The specific question is not given. Let's find the domain.

2. Solution Steps

To find the domain of the function, the expression under the square root must be non-negative.

Thus, we need to satisfy the inequality:

$16 - x^2 - y^2 \ge 0$

$16 \ge x^2 + y^2$

$x^2 + y^2 \le 16$

This inequality describes the set of all points $(x, y)$ within a circle centered at the origin with radius $r = \sqrt{16} = 4$. The circle itself is included.

Therefore, the domain of the function is the set of all points $(x, y)$ such that $x^2 + y^2 \le 16$.

The range of $z$ is the set of all possible values of $z$.

Since $x^2 + y^2 \le 16$, we have $16 - (x^2 + y^2) \ge 0$. The maximum value of $16 - x^2 - y^2$ is 16, which occurs when $x = 0$ and $y = 0$.

Then the maximum value of $z = \sqrt{16 - x^2 - y^2}$ is $\sqrt{16} = 4$.

The minimum value of $16 - x^2 - y^2$ is 0, which occurs when $x^2 + y^2 = 16$.

Then the minimum value of $z = \sqrt{16 - x^2 - y^2}$ is $\sqrt{0} = 0$.

So the range of $z$ is $[0, 4]$.

3. Final Answer

The domain is $x^2 + y^2 \le 16$.

The range is $0 \le z \le 4$.