91Ó°ÊÓ

Stokes' theorem states that the line integral of a vector field \( \mathbf{v} \) over a closed curve \( C \) is equal to the surface integral of the curl of \( \mathbf{v} \) over the surface \( S \) bounded by \( C \). It is expressed as \( \oint_{C} \mathbf{v} \cdot d\mathbf{r} = \int\int_{S} (abla \times \mathbf{v}) \cdot d\mathbf{S} \).

The given vector field is \( \mathbf{v} = y \mathbf{i} + (x - 2xz) \mathbf{j} - xy \mathbf{k} \). The curl \( abla \times \mathbf{v} \) is given by the determinate:\[ abla \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ y & (x - 2xz) & -xy \end{vmatrix} \]Calculating, we have:\[ abla \times \mathbf{v} = \left( \frac{\partial(-xy)}{\partial y} - \frac{\partial(x - 2xz)}{\partial z} \right) \mathbf{i} - \left( \frac{\partial(-xy)}{\partial x} - \frac{\partial y}{\partial z} \right) \mathbf{j} + \left( \frac{\partial(x - 2xz)}{\partial x} - \frac{\partial y}{\partial y} \right) \mathbf{k} \]Simplifying gives:\[ abla \times \mathbf{v} = (-x - 2x) \mathbf{i} + y \mathbf{j} + 0 \mathbf{k} = -3x \mathbf{i} + y \mathbf{j} \]

The surface \( S \) is the portion \( z = 1 \) within the square in the \( xy \)-plane. For this, \( d\mathbf{S} = \mathbf{k}\,dx\,dy \) since the normal is upward, coinciding with the positive \( z \)-axis. The surface integral becomes:\[ \int\int_{S} (abla \times \mathbf{v}) \cdot d\mathbf{S} = \int_{0}^{1} \int_{0}^{1} (-3x) \cdot 0 + y \cdot 1 \, dx \, dy \]This simplifies to:\[ \int_{0}^{1} \int_{0}^{1} y \, dx \, dy = \int_{0}^{1} \left[ yx \right]_{0}^{1} \, dy = \int_{0}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1} = \frac{1}{2} \]

The curve \( C \) is parameterized in segments as follows:1. From \( (0,0,1) \) to \( (1,0,1) \) with \( \mathbf{r}_1(t) = t\mathbf{i} \), \( 0 \leq t \leq 1 \) 2. From \( (1,0,1) \) to \( (1,1,1) \) with \( \mathbf{r}_2(t) = \mathbf{i} + t\mathbf{j} \), \( 0 \leq t \leq 1 \) 3. From \( (1,1,1) \) to \( (0,1,1) \) with \( \mathbf{r}_3(t) = (1-t)\mathbf{i} + \mathbf{j} \), \( 0 \leq t \leq 1 \) 4. From \( (0,1,1) \) to \( (0,0,1) \) with \( \mathbf{r}_4(t) = (1-t)\mathbf{j} \), \( 0 \leq t \leq 1 \).Now compute the line integral across each segment, \( \oint_{C} \mathbf{v} \cdot d\mathbf{r} = \sum_{i=1}^{4} \int \mathbf{v} \cdot \frac{d\mathbf{r}_i}{dt} \, dt \).

1. For \( \mathbf{r}_1(t) \), \( d\mathbf{r}_1/dt = \mathbf{i} \). Then: \( \int_{0}^{1} y(0) \mathbf{i} \cdot \mathbf{i} \, dt = 0 \) since \( y = 0 \).2. For \( \mathbf{r}_2(t) \), \( d\mathbf{r}_2/dt = \mathbf{j} \). \( \int_{0}^{1} (x - 2x \cdot 1) \cdot 1 \, dt = \int_{0}^{1} (-1)dt = -1.\)3. For \( \mathbf{r}_3(t) \), \( d\mathbf{r}_3/dt = -\mathbf{i} \). \( \int_{0}^{1} (-y) \mathbf{j} \cdot (-\mathbf{i})dt = 0 \).4. For \( \mathbf{r}_4(t) \), \( d\mathbf{r}_4/dt = -\mathbf{j} \). \( \int_{0}^{1} (-1)\cdot0 dt = 0. \) Summing up all contributions: \( 0 - 1 + 0 + 0 = -1 \).

The line integral around \( C \) is \(-1\) and does not match the surface integral result of \( \frac{1}{2} \). This implies that Stokes' theorem is not satisfied as expected, due to the cube surface not being closed.