91Ó°ÊÓ

Stokes' Theorem relates a surface integral of the curl of a vector field over a surface \(S\) to a line integral of the vector field over the boundary curve \(C\) of \(S\). It is given by the formula \(\iint_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}\). Our task is to evaluate this line integral over the boundary of the given surface.

The curl of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \[ \operatorname{curl} \mathbf{F} = abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}. \]Substitute \( P = -y^2 \), \( Q = x \), \( R = z^2 \) to find the curl.\[\operatorname{curl} \mathbf{F} = (0 - 0) \mathbf{i} + (0 - 0) \mathbf{j} + (1 - (-2y)) \mathbf{k} = 2y \mathbf{k}.\]

The surface \(S\) is part of the plane \(x+y+z=1\) in the positive octant. Thus, the boundary is the triangular path \(\triangle ABC\) where \(A = (1,0,0)\), \(B = (0,1,0)\), and \(C = (0,0,1)\). This curve is traversed counterclockwise.

The path consists of three line segments. Parameterize each segment:1. From \(A\to B\): \(\mathbf{r}_1(t) = (1-t)\mathbf{i} + t\mathbf{j}\) for \(t\in [0,1]\).2. From \(B\to C\): \(\mathbf{r}_2(t) = (1-t)\mathbf{j} + t\mathbf{k}\) for \(t\in [0,1]\).3. From \(C\to A\): \(\mathbf{r}_3(t) = (1-t)\mathbf{k} + t\mathbf{i}\) for \(t\in [0,1]\).

Calculate \(\oint_C \mathbf{F} \cdot d\mathbf{r}\) using the parameterized paths:1. For \( \mathbf{r}_1(t) \), \( \mathbf{F} = -y^2 \mathbf{i} + x \mathbf{j} + z^2 \mathbf{k} = -t^2 \mathbf{i} + (1-t) \mathbf{j} \). \(d\mathbf{r}_1 = (-1)\mathbf{i} + \mathbf{j}\). Integral over \( \mathbf{r}_1(t) \) = \( \int_0^1 [ -t^2(-1) + (1-t) ] dt = \int_0^1 [t^2 + 1-t] dt \).2. For \( \mathbf{r}_2(t) \), \( \mathbf{F} = (1-t) \mathbf{i} + t^2 \mathbf{k} \), \(d\mathbf{r}_2 = (-1)\mathbf{j} + \mathbf{k}\). Integral is zero as only \(\mathbf{j}\) component of \(\mathbf{F}\) contributes and is zero.3. For \( \mathbf{r}_3(t) \), \( \mathbf{F} = t^2 \mathbf{k} \), \(d\mathbf{r}_3 = (-1)\mathbf{k} + \mathbf{i}\). Integral over \( \mathbf{r}_3(t) \) is zero as \(\mathbf{i}\) component of \(\mathbf{F}\) contributes and is zero.

Perform integration:1. \( \int_0^1 [t^2 + 1 - t] dt = \left[ \frac{t^3}{3} + t - \frac{t^2}{2} \right]_0^1 = [\frac{1}{3} + 1 - \frac{1}{2}] - [0] = \frac{5}{6} \).Combine results from all segments, final answer is \(\frac{5}{6}\).