This problem asks us to calculate the work done by the force field $\mathbf{F}(x, y, z) = \langle -y, z, x \rangle$ in moving a particle along the curve $\mathbf{C}$, which is the line segment from $(2, -1, 3)$ to $(5, 1, 5)$.

Step-by-step solution:

1. Parameterize the curve $\mathbf{C}$: The curve $\mathbf{C}$ is a line segment, so we can parameterize it as: $\mathbf{r}(t) = (1-t)\mathbf{r}_0 + t\mathbf{r}_1,$ where $\mathbf{r}_0 = (2, -1, 3)$ and $\mathbf{r}_1 = (5, 1, 5)$, and $t \in [0, 1]$.

   Simplifying: $\mathbf{r}(t) = \langle 2 + 3t, -1 + 2t, 3 + 2t \rangle.$

2. Find $\mathbf{r}'(t)$: Differentiate $\mathbf{r}(t)$ with respect to $t$: $\mathbf{r}'(t) = \langle 3, 2, 2 \rangle.$

3. Substitute $\mathbf{r}(t)$ into $\mathbf{F}$: The force field at $\mathbf{r}(t)$ is: $\mathbf{F}(\mathbf{r}(t)) = \langle -y, z, x \rangle = \langle -(-1 + 2t), 3 + 2t, 2 + 3t \rangle.$ Simplify: $\mathbf{F}(\mathbf{r}(t)) = \langle 1 - 2t, 3 + 2t, 2 + 3t \rangle.$

4. Calculate the work: The work done is given by: $W = \int_C \mathbf{F} \cdot d\mathbf{r},$ which becomes: $W = \int_0^1 \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt.$

   Substitute $\mathbf{F}(\mathbf{r}(t)) = \langle 1 - 2t, 3 + 2t, 2 + 3t \rangle$ and $\mathbf{r}'(t) = \langle 3, 2, 2 \rangle$: $W = \int_0^1 \langle 1 - 2t, 3 + 2t, 2 + 3t \rangle \cdot \langle 3, 2, 2 \rangle \, dt.$

   Compute the dot product: $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = (1 - 2t)(3) + (3 + 2t)(2) + (2 + 3t)(2).$ Simplify: $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = 3 - 6t + 6 + 4t + 4 + 6t.$ Combine terms: $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = 13 + 4t.$

5. Integrate: Now, integrate $13 + 4t$ from $t = 0$ to $t = 1$: $W = \int_0^1 (13 + 4t) \, dt.$ Compute: $W = \left[ 13t + 2t^2 \right]_0^1.$ Evaluate: $W = 13(1) + 2(1)^2 - (13(0) + 2(0)^2).$ Simplify: $W = 13 + 2 = 15.$

Final Answer:

The work done by the force field $\mathbf{F}$ along the curve $\mathbf{C}$ is: $\boxed{15}$

Let me know if you'd like additional details or clarifications!

Related Questions:

1. How is a force field like $\mathbf{F}$ used in real-world applications?
2. What are other parameterization techniques for more complex curves?
3. What is the physical significance of the work done by a force field?
4. How would the calculation change if the curve $\mathbf{C}$ were not a straight line?
5. What are conservative vector fields, and how do they simplify work calculations?

Tip:

Always check if the force field is conservative. If it is, calculating work becomes easier using potential functions!