Answer
$ \oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} = \pi a^2$
Work Step by Step
Given Vector Field and Rotation Given vector field: \[ \vec{F}(\mathbf{r}) = \begin{bmatrix} x - y \\ y - z \\ z - x \end{bmatrix} \] The rotation (curl) of $\vec{F}$ is given by: \[ \begin{aligned} \nabla \times \vec{F} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x - y & y - z & z - x \end{vmatrix} \\ &= \left(\frac{\partial}{\partial y}(z - x) - \frac{\partial}{\partial z}(y - z)\right) \mathbf{i} - \left(\frac{\partial}{\partial x}(z - x) - \frac{\partial}{\partial z}(x - y)\right) \mathbf{j} + \left(\frac{\partial}{\partial x}(y - z) - \frac{\partial}{\partial y}(x - y)\right) \mathbf{k} \\ &= (-1) \mathbf{i} + (1) \mathbf{j} - (1) \mathbf{k} \\ &= -\mathbf{i} + \mathbf{j} - \mathbf{k} \end{aligned} \] Use of Stokes' Theorem Given that $\mathcal{C}$ is the circle $x^2 + y^2 = a^2$ in the $xy$-plane with counterclockwise orientation looking down the positive $z$-axis. Using Stokes' Theorem: \[ \begin{aligned} \oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} &= \iint_{\Sigma} (\nabla \times \vec{F}) \cdot \mathbf{n} \, dS \\ &= \iint_{R} (\nabla \times \vec{F}) \cdot (\nabla G) \, dA \end{aligned} \] where: \[ \begin{aligned} \mathbf{r} &= \begin{bmatrix} x \\ y \end{bmatrix}, \quad R: \{(x, y) \mid 0 \leq x \leq 1, 0 \leq y \leq 1 - x\} \\ G &= z + x + y - 1 \end{aligned} \] Then, the line integral simplifies to: \[ \oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} = \pi a^2 \] Therefore: \[ \oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} = \pi a^2 \]
