To solve this problem, we are asked to evaluate the line integral of the vector field $\mathbf{F}(x, y) = y\mathbf{i} - x\mathbf{j}$ along the curve $C$, which is a circle of radius 2 centered at the origin, traversed counterclockwise starting at $(2, 0)$.

Solution:

1. Check if the field is conservative: A vector field $\mathbf{F} = P(x, y)\mathbf{i} + Q(x, y)\mathbf{j}$ is conservative if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}.$ Here, $P(x, y) = y$ and $Q(x, y) = -x$. Compute the partial derivatives:

   \quad \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(-x) = -1.$$ Since $$\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$$, the vector field is **not conservative**.

2. Use Green's Theorem: Since $C$ is a closed curve, we can apply Green's Theorem: $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA.$ Substituting $P(x, y) = y$ and $Q(x, y) = -x$, we have: $\frac{\partial Q}{\partial x} = -1, \quad \frac{\partial P}{\partial y} = 1.$ Thus, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -1 - 1 = -2.$

3. Evaluate the double integral: The region $R$ is the circle of radius 2 centered at the origin. In polar coordinates, the area element $dA = r \, dr \, d\theta$. The bounds are: $0 \leq r \leq 2, \quad 0 \leq \theta \leq 2\pi.$ The integral becomes: $\iint_R (-2) \, dA = \int_0^{2\pi} \int_0^2 (-2) r \, dr \, d\theta.$

   First, integrate with respect to $r$: $\int_0^2 (-2)r \, dr = -2 \int_0^2 r \, dr = -2 \left[ \frac{r^2}{2} \right]_0^2 = -2 \cdot \frac{4}{2} = -4.$

   Next, integrate with respect to $\theta$: $\int_0^{2\pi} (-4) \, d\theta = -4 \int_0^{2\pi} 1 \, d\theta = -4 [\theta]_0^{2\pi} = -4(2\pi - 0) = -8\pi.$

Final Answer:

The value of the line integral is: $\oint_C \mathbf{F} \cdot d\mathbf{r} = -8\pi.$

Would you like further clarification on any step? Here are some related questions:

1. How does Green's Theorem relate to line integrals and flux?
2. Why was it necessary to verify whether the field is conservative?
3. Can this problem be solved without Green’s Theorem?
4. What happens if the circle's radius is changed or oriented differently?
5. How would the solution change if the curve was traversed clockwise?

Tip: Always check if a vector field is conservative before applying Green's