To solve this problem, we will calculate the divergence of the given vector field $\mathbf{F}(x, y, z) = -xyz \, \mathbf{i} + xz^2 \, \mathbf{j} + 3yz^2 \, \mathbf{k}$ and evaluate it at the point $(2, 5, 1)$.

Step 1: Formula for Divergence

The divergence of a vector field $\mathbf{F}(x, y, z) = F_1(x, y, z)\mathbf{i} + F_2(x, y, z)\mathbf{j} + F_3(x, y, z)\mathbf{k}$ is: $\text{div} \, \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}.$

Step 2: Compute Partial Derivatives

For $\mathbf{F}(x, y, z) = -xyz \, \mathbf{i} + xz^2 \, \mathbf{j} + 3yz^2 \, \mathbf{k}$:

1. $F_1 = -xyz$, $F_2 = xz^2$, $F_3 = 3yz^2$.

• $\frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}(-xyz) = -yz$,
• $\frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y}(xz^2) = 0$ (since there is no $y$-dependence in $F_2$),
• $\frac{\partial F_3}{\partial z} = \frac{\partial}{\partial z}(3yz^2) = 6yz$.

Step 3: Add the Partial Derivatives

$\text{div} \, \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} = -yz + 0 + 6yz = 5yz.$

Step 4: Evaluate at $(x, y, z) = (2, 5, 1)$

Substitute $y = 5$ and $z = 1$: $\text{div} \, \mathbf{F} = 5 \cdot 5 \cdot 1 = 25.$

Final Answer:

The divergence of $\mathbf{F}$ at $(2, 5, 1)$ is: $\boxed{25}.$

Would you like further details on any part of this process? Here are some questions to consider:

1. What is the significance of divergence in vector calculus?
2. How would this problem change if the vector field had additional components?
3. What are the units of divergence if the vector field represents a physical quantity?
4. Could we visualize divergence using a graph, and how?
5. How is divergence related to the concept of flux through a surface?

Tip: Always verify your vector field's components to ensure that the partial derivatives are taken with respect to the correct variable.