91Ó°ÊÓ

The curve \( C \) is a triangle with vertices at \( (0,0,2), (4,0,0), \) and \( (0,3,0) \). We will divide the curve into three segments and parameterize each segment.1. **Segment 1** (from \( (0,0,2) \) to \( (4,0,0) \)): \[ \vec{r}_1(t) = (4t, 0, 2 - 2t) \text{ for } 0 \leq t \leq 1 \]2. **Segment 2** (from \( (4,0,0) \) to \( (0,3,0) \)): \[ \vec{r}_2(t) = (4 - 4t, 3t, 0) \text{ for } 0 \leq t \leq 1 \]3. **Segment 3** (from \( (0,3,0) \) to \( (0,0,2) \)): \[ \vec{r}_3(t) = (0, 3 - 3t, 2t) \text{ for } 0 \leq t \leq 1 \]

We compute the line integral for each segment using \( \vec{F} = \langle y, -z, y \rangle \) and the parameterizations from Step 1.**Segment 1:** - \( \frac{d\vec{r}_1}{dt} = \langle 4, 0, -2 \rangle \) - \( \vec{F}(\vec{r}_1(t)) = \langle 0, - (2 - 2t), 0 \rangle = \langle 0, -2 + 2t, 0 \rangle \) - Integral: \( \int_0^1 \langle 0, -2 + 2t, 0 \rangle \cdot \langle 4, 0, -2 \rangle \, dt = \int_0^1 0 \, dt = 0 \)**Segment 2:** - \( \frac{d\vec{r}_2}{dt} = \langle -4, 3, 0 \rangle \) - \( \vec{F}(\vec{r}_2(t)) = \langle 3t, 0, 3t \rangle \) - Integral: \( \int_0^1 \langle 3t, 0, 3t \rangle \cdot \langle -4, 3, 0 \rangle \, dt = \int_0^1 -12t \, dt = -6 \)**Segment 3:** - \( \frac{d\vec{r}_3}{dt} = \langle 0, -3, 2 \rangle \) - \( \vec{F}(\vec{r}_3(t)) = \langle 3 - 3t, -2t, 3 - 3t \rangle \) - Integral: \( \int_0^1 \langle 3 - 3t, -2t, 3 - 3t \rangle \cdot \langle 0, -3, 2 \rangle \, dt = \int_0^1 (6t - 6) \, dt = -3 \)Adding the integrals from all segments, \(\oint_{C} \vec{F} \cdot d\vec{r} = 0 - 6 - 3 = -9 \).

The curl of \( \vec{F} = \langle y, -z, y \rangle \) is given by:\[\text{curl } \vec{F} = abla \times \vec{F} = \left(\frac{\partial y}{\partial y} - \frac{\partial (-z)}{\partial z}, \frac{\partial y}{\partial x} - \frac{\partial y}{\partial z}, \frac{\partial (-z)}{\partial x} - \frac{\partial y}{\partial y}\right)\]Calculating each component:1. \( \frac{\partial y}{\partial y} - \frac{\partial (-z)}{\partial z} = 1 + 1 = 2 \)2. \( \frac{\partial y}{\partial x} - \frac{\partial y}{\partial z} = 0 - 0 = 0 \)3. \( \frac{\partial (-z)}{\partial x} - \frac{\partial y}{\partial y} = 0 - 1 = -1 \)Thus, \(\text{curl } \vec{F} = \langle 2, 0, -1 \rangle \).

The surface \( S \) is given by the plane equation \( z = 2 - \frac{x}{2} - \frac{2y}{3} \), which can be rewritten as \( x + \frac{4}{3}y + 2z = 4 \). The normal vector \( \vec{n} \) to this plane is \( \langle 1, \frac{4}{3}, 2 \rangle \).

Compute the dot product \( (\text{curl } \vec{F}) \cdot \vec{n} \):\[\langle 2, 0, -1 \rangle \cdot \langle 1, \frac{4}{3}, 2 \rangle = 2 \cdot 1 + 0 \cdot \frac{4}{3} + (-1) \cdot 2 = 2 - 2 = 0\]Since the dot product is zero, the surface integral becomes:\[\iint_{S} 0 \, dS = 0\]Hence, \( \iint_{S} (\text{curl } \vec{F}) \cdot \vec{n} \, dS = 0 \).

From previous computations, we found that:- The line integral \( \oint_{C} \vec{F} \cdot d\vec{r} = -9 \).- The surface integral \( \iint_{S} (\text{curl } \vec{F}) \cdot \vec{n} \, dS = 0 \).These are not equal, indicating a mistake in our setup or calculations. Recheck parameterizations, curl calculation, or integral evaluations.