To evaluate the contour integral $\oint_C \mathbf{F} \cdot d\mathbf{r}$ for $\mathbf{F} = \langle 0, -3z, -2y \rangle$, we will use Stokes' Theorem. Stokes' Theorem relates a surface integral over a surface $S$ to a line integral over the boundary curve $C$ of the surface. Specifically, Stokes' Theorem states:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$

where $\nabla \times \mathbf{F}$ is the curl of $\mathbf{F}$.

First, we need to compute the curl of $\mathbf{F}$. The vector field $\mathbf{F}$ is given by:

$\mathbf{F} = \langle 0, -3z, -2y \rangle$

The curl of $\mathbf{F}$ is given by:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & -3z & -2y \end{vmatrix}$$ Evaluating this determinant, we get: $$\nabla \times \mathbf{F} = \left( \frac{\partial (-2y)}{\partial y} - \frac{\partial (-3z)}{\partial z} \right) \mathbf{i} - \left( \frac{\partial (0)}{\partial z} - \frac{\partial (0)}{\partial x} \right) \mathbf{j} + \left( \frac{\partial (-3z)}{\partial x} - \frac{\partial (0)}{\partial y} \right) \mathbf{k}$$ Simplifying each term, we find: $$\nabla \times \mathbf{F} = (-2 - (-3)) \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} = \mathbf{i}$$ So, $$\nabla \times \mathbf{F} = \langle 1, 0, 0 \rangle$$ Next, we need to compute the surface integral of $$\nabla \times \mathbf{F}$$ over a surface $$S$$ bounded by the curve $$C$$. The surface $$S$$ can be taken as the disk in the plane $$x + y + z = 0$$. The unit normal vector to the surface $$S$$ (since the plane is $$x + y + z = 0$$) is $$\mathbf{n} = \langle 1, 1, 1 \rangle / \sqrt{3}$$. The differential surface element $$d\mathbf{S}$$ is: $$d\mathbf{S} = \mathbf{n} \, dS = \frac{1}{\sqrt{3}} \langle 1, 1, 1 \rangle \, dS$$ Then, the surface integral becomes: $$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S \langle 1, 0, 0 \rangle \cdot \frac{1}{\sqrt{3}} \langle 1, 1, 1 \rangle \, dS$$ This simplifies to: $$\iint_S \frac{1}{\sqrt{3}} \, dS = \frac{1}{\sqrt{3}} \iint_S dS$$ The area of the disk $$S$$ with radius 5 is $$\pi (5)^2 = 25\pi$$. Therefore, $$\iint_S dS = 25\pi$$ Thus, $$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \frac{1}{\sqrt{3}} \times 25\pi = \frac{25\pi}{\sqrt{3}}$$ Therefore, $$\oint_C \mathbf{F} \cdot d\mathbf{r} = \frac{25\pi}{\sqrt{3}}$$ ### Dependence on the Radius and Location of the Center 1. **Dependence on the radius of the circle**: Yes, the circulation depends on the radius of the circle since the integral over the surface $$S$$ involves the area of the disk, which is directly proportional to the square of the radius. 2. **Dependence on the location of the center of the circle**: No, the circulation does not depend on the location of the center of the circle as long as the surface lies within the same plane and is bounded by the curve $$C$$. Do you have any further questions or need more details? ### Questions to Explore: 1. What is the divergence theorem, and how is it applied? 2. How do you parameterize a surface for surface integrals? 3. What are some common applications of Stokes' Theorem in physics? 4. How do you calculate the curl of a vector field in spherical coordinates? 5. What is the physical interpretation of the curl of a vector field? ### Tip: When applying Stokes' Theorem, always ensure that the orientation of the curve and the surface are consistent with the right-hand rule to avoid sign errors.