The problem asks us to analyze the vector field $F: \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}^2$ defined by $F(x, y) := \left( \frac{y}{x^2 + y^2}, \frac{-x}{x^2 + y^2} \right).$ Specifically, part (d) involves checking whether the restrictions $F|_{U_1}$ and $F|_{U_2}$ are gradient fields for the sets $U_1 := \{ (x, y) \in \mathbb{R}^2 : x > 0 \}$ and $U_2 := \{ (x, y) \in \mathbb{R}^2 : y > 0 \}$. If they are gradient fields, we should determine the potential function $\Phi_j$ for $F|_{U_j}$ with $j = 1, 2$.

Solution Outline

1. Check if $F$ is a gradient field: For $F$ to be a gradient field in each region $U_1$ and $U_2$, it must satisfy the condition $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$ where $F = (F_1, F_2)$.

2. Potential Function: If $F$ is a gradient field in each region, we would find a scalar potential function $\Phi$ such that $F = \nabla \Phi$.

Step-by-Step Solution

1. Compute the Partial Derivatives:

   • For $F(x, y) = \left( \frac{y}{x^2 + y^2}, \frac{-x}{x^2 + y^2} \right)$:
     • $F_1(x, y) = \frac{y}{x^2 + y^2}$
     • $F_2(x, y) = \frac{-x}{x^2 + y^2}$
   • Calculate $\frac{\partial F_1}{\partial y}$ and $\frac{\partial F_2}{\partial x}$ to determine if they are equal.

2. Determine Potential Function: If $F|_{U_1}$ and $F|_{U_2}$ are gradient fields, find potential functions $\Phi_1$ and $\Phi_2$ for these regions.

Would you like the full calculations and steps for each part?

Further Exploration

1. What properties make a vector field conservative or a gradient field?
2. How can you determine if a vector field is path-independent?
3. Why are the regions $U_1$ and $U_2$ necessary to consider separately?
4. How does excluding the origin $(0,0)$ impact the vector field?
5. Can the concept of potential functions apply to fields in more than two dimensions?

Tip: Always check if a vector field is conservative by verifying if it satisfies the equality of mixed partial derivatives (e.g., $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$).