To evaluate the line integral of the vector field $\mathbf{F} = (y - z)\mathbf{i} + (z - x)\mathbf{j} + (x - y)\mathbf{k}$ around the boundary of the surface $S$, where $S$ is the part of the plane $2x + y + z = 2$ that lies inside the cylinder $x^2 + y^2 = 1$, we can use Stokes' Theorem.

Step 1: Applying Stokes' Theorem

Stokes' Theorem relates the line integral of a vector field around a closed curve $C$ to the surface integral of the curl of the vector field over the surface $S$ bounded by $C$. Mathematically, it is given by: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$ where:

• $C$ is the boundary of the surface $S$,
• $d\mathbf{r}$ is the differential vector along $C$,
• $\nabla \times \mathbf{F}$ is the curl of the vector field $\mathbf{F}$,
• $d\mathbf{S}$ is the vector area element of the surface $S$.

Step 2: Calculate the Curl of $\mathbf{F}$

First, we compute the curl of the vector field $\mathbf{F}$:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y - z & z - x & x - y \end{vmatrix}$$ Expanding this determinant: $$\nabla \times \mathbf{F} = \mathbf{i} \left(\frac{\partial (x - y)}{\partial y} - \frac{\partial (z - x)}{\partial z}\right) + \mathbf{j} \left(\frac{\partial (y - z)}{\partial z} - \frac{\partial (x - y)}{\partial x}\right) + \mathbf{k} \left(\frac{\partial (z - x)}{\partial x} - \frac{\partial (y - z)}{\partial y}\right)$$ Evaluating these partial derivatives: $$\nabla \times \mathbf{F} = \mathbf{i} \left(-1 - (-1)\right) + \mathbf{j} \left(-1 - (-1)\right) + \mathbf{k} \left(-1 - (-1)\right)$$ $$\nabla \times \mathbf{F} = \mathbf{i}(0) + \mathbf{j}(0) + \mathbf{k}(0) = \mathbf{0}$$ Thus, the curl of $$\mathbf{F}$$ is zero: $$\nabla \times \mathbf{F} = \mathbf{0}$$ ### Step 3: Evaluate the Surface Integral Since $$\nabla \times \mathbf{F} = \mathbf{0}$$, the surface integral also evaluates to zero: $$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S \mathbf{0} \cdot d\mathbf{S} = 0$$ ### Conclusion By Stokes' Theorem, the line integral of $$\mathbf{F}$$ around the boundary $$C$$ of the surface $$S$$ is zero: $$\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$$ Would you like to explore the problem further or do you have any questions? Here are some related questions you might find interesting: 1. What if the vector field $$\mathbf{F}$$ was different, say $$\mathbf{F} = (x, y, z)$$? How would that affect the line integral? 2. How do we apply Stokes' Theorem to a surface that's not planar? 3. What is the physical interpretation of the curl of a vector field? 4. How would the result change if the surface $$S$$ were oriented downward? 5. Can you derive Stokes' Theorem from first principles? **Tip:** Remember that when using Stokes' Theorem, verifying that the curl of the vector field is zero can greatly simplify the computation.