Answer
$ \oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} = -\frac{1}{2}$
Work Step by Step
Given Vector Field and Rotation Given vector field: \[ \vec{F}(\mathbf{r}) = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \] The rotation (curl) of this vector field is given by: \[ \begin{aligned} \nabla \times \vec{F} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & z \end{vmatrix} \\ &= \left(\frac{\partial}{\partial y}(z) - \frac{\partial}{\partial z}(y)\right) \mathbf{i} - \left(\frac{\partial}{\partial x}(z) - \frac{\partial}{\partial z}(x)\right) \mathbf{j} + \left(\frac{\partial}{\partial x}(y) - \frac{\partial}{\partial y}(x)\right) \mathbf{k} \\ &= (-y) \mathbf{i} - (-x) \mathbf{j} + (0) \mathbf{k} \\ &= -y \mathbf{i} + x \mathbf{j} \end{aligned} \] \section*{Use of Stokes' Theorem} Let $\mathcal{C}$ be the triangle with vertices $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$ in the $xy$-plane with counterclockwise orientation looking from the first octant to the origin. $\Sigma$ (the surface bounded by $\mathcal{C}$) is the portion of the plane $x + y + z = 1$ in the first octant. Using Stokes' Theorem, we have: \[ \begin{aligned} \oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} &= \iint_{\Sigma} (\nabla \times \vec{F}) \cdot \mathbf{n} \, dS \\ &= \iint_{R} (\nabla \times \vec{F}) \cdot (\nabla G) \, dA \end{aligned} \] 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: \[ \oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} = -\frac{1}{2} \] Therefore: \[ \oint_{\mathcal{C}} \vec{F} \cdot d\vec{r} = -\frac{1}{2} \]
