Let's analyze the problem step-by-step:

Function Definition:

The function is defined as: $u = f(x, y, z) = \sqrt{x^2 + y^2 + z}$

Domain:

The square root function $\sqrt{A}$ is defined only when $A \geq 0$. Here: $A = x^2 + y^2 + z$ So the domain of the function is given by: $x^2 + y^2 + z \geq 0$ In other words, the function is defined for all points $(x, y, z)$ in $\mathbb{R}^3$ such that $z \geq -x^2 - y^2$. This describes a region in 3D space that lies above or on the paraboloid $z = -x^2 - y^2$.

Evaluation:

To compute $f(-1, 2, 0)$, substitute $x = -1$, $y = 2$, and $z = 0$ into the function: $f(-1, 2, 0) = \sqrt{(-1)^2 + 2^2 + 0}$ $f(-1, 2, 0) = \sqrt{1 + 4 + 0} = \sqrt{5}$

Final Answer:

1. Domain: $z \geq -x^2 - y^2$
2. Value at $(-1, 2, 0)$: $f(-1, 2, 0) = \sqrt{5}$

Would you like additional clarification or further exploration of related concepts? Here are some questions to consider:

1. What does the domain of this function look like geometrically in 3D space?
2. How would you evaluate the gradient of $f(x, y, z)$?
3. What happens if $z < -x^2 - y^2$? Why is the function undefined there?
4. Can we extend this function to make it valid for all real numbers?
5. How does this function behave asymptotically as $x, y, z$ become very large?

Tip: When working with square roots, always consider the non