Let $\vec{F}(x,y,z) = y \hat{i} + z \hat{j} + x \hat{k}$. Using Stokes' theorem, evaluate the line integral $\oint_C \vec{F} \cdot d\vec{r}$ around the boundary curve $C$ of the triangle formed by intersecting the plane $x + y + z = 1$ with the coordinate planes in the first octant, oriented counter-clockwise when viewed from high above on the positive $z$-axis. (Enter the value with appropriate sign)

Answer Key:

-1.5

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Curl of Vector Field:** * $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ y & z & x \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(1 - 0) + \hat{k}(0 - 1) = -\hat{i} - \hat{j} - \hat{k}$. 2. **Stage 2: Surface Orientation & Normal Vector:** * Plane equation: $g(x,y,z) = x + y + z - 1 = 0$. * Unit normal pointing in the positive direction: $\hat{n} = \frac{\nabla g}{\|\nabla g\|} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$. 3. **Stage 3: Dot Product Integrand:** * $(\nabla \times \vec{F}) \cdot \hat{n} = (-\hat{i} - \hat{j} - \hat{k}) \cdot \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{-1 - 1 - 1}{\sqrt{3}} = -\frac{3}{\sqrt{3}} = -\sqrt{3}$. 4. **Stage 4: Surface Area of Triangular Plane in Octant 1:** * Vertices are at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. * Sides are $\sqrt{2}$. Area of equilateral triangle with side $s = \sqrt{2}$: * $\text{Area} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (\sqrt{2})^2 = \frac{\sqrt{3}}{4} (2) = \frac{\sqrt{3}}{2}$. 5. **Stage 5: Surface Integral Computation:** * $\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \, dS = (-\sqrt{3}) \times \left(\frac{\sqrt{3}}{2}\right) = -\frac{3}{2} = -1.5$. * Official Reference: Erwin Kreyszig, *Advanced Engineering Mathematics*, Chapter 10.

← Prev Question Next Question →