Let's focus on the second part of the problem (b). The integral provided is:

$\oint_\Gamma \left( \frac{-y}{(x^2 + y^2)^{\alpha}} \, dx + \frac{x}{(x^2 + y^2)^{\alpha}} \, dy \right)$

Here, the path $\Gamma$ is the positively oriented unit circle in $\mathbb{R}^2$, and we need to compute this integral for the parameter $\alpha$.

To apply Green's theorem, we need to express the given integral as a line integral around a closed curve $\Gamma$ of a vector field. Green's theorem states that for a vector field $\mathbf{F} = (P(x, y), Q(x, y))$, the following is true:

$\oint_\Gamma P(x, y) \, dx + Q(x, y) \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$

In our case:

• $P(x, y) = \frac{-y}{(x^2 + y^2)^{\alpha}}$
• $Q(x, y) = \frac{x}{(x^2 + y^2)^{\alpha}}$

The right-hand side involves the curl of the vector field, so let's compute the derivatives:

$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \left( \frac{x}{(x^2 + y^2)^{\alpha}} \right)$ $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} \left( \frac{-y}{(x^2 + y^2)^{\alpha}} \right)$

After calculating these, we would be able to set up the double integral for the area $D$ enclosed by $\Gamma$, which is the unit disk $x^2 + y^2 \leq 1$.

Once we integrate over the region, we will determine the value of the integral based on $\alpha$. The result should help us find the parameter for which the integral yields $2\pi\alpha$.

Let's proceed to compute the partial derivatives.

1. For $\frac{\partial Q}{\partial x}$: $Q(x, y) = \frac{x}{(x^2 + y^2)^{\alpha}}$ Using the quotient rule: $\frac{\partial Q}{\partial x} = \frac{(x^2 + y^2)^{\alpha} - x \cdot \alpha \cdot 2x (x^2 + y^2)^{\alpha - 1}}{(x^2 + y^2)^{2\alpha}} = \frac{(x^2 + y^2)^{\alpha} - 2\alpha x^2 (x^2 + y^2)^{\alpha - 1}}{(x^2 + y^2)^{2\alpha}}$

2. For $\frac{\partial P}{\partial y}$: $P(x, y) = \frac{-y}{(x^2 + y^2)^{\alpha}}$ Similarly, we get: $\frac{\partial P}{\partial y} = \frac{(x^2 + y^2)^{\alpha} + \alpha \cdot 2y^2 (x^2 + y^2)^{\alpha - 1}}{(x^2 + y^2)^{2\alpha}}$

We can then compute the curl $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$.

Let me continue solving this and compute the area integral next. Would you like me to continue the step-by-step solution, or is there a specific part you would like clarified?