Let's go through each part of the question step-by-step:

Given Data:

The vector field is given by: $\vec{F} = \langle P, Q \rangle = \left\langle \frac{-y}{(x - y)^2}, \frac{x}{(x - y)^2} \right\rangle$ The domain is: $\mathcal{D} = \{(x, y) \in \mathbb{R}^2 \ | \ y < x\}$

Solution:

(i) Picture of the Set $D$ and Simple Connectedness

The set $D$ consists of points $(x, y)$ where $y < x$. This is the region below the line $y = x$ in the $xy$-plane. A sketch would show the half-plane under this line.

Now, to determine if $D$ is simply connected:

• A region is simply connected if it has no holes, meaning any closed loop within the region can be continuously contracted to a point without leaving the region.
• Since $y < x$ creates a half-plane with no holes, $D$ is indeed simply connected.

(ii) Verifying $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

1. Calculate $\frac{\partial P}{\partial y}$: $P = \frac{-y}{(x - y)^2}$ Differentiate $P$ with respect to $y$: $\frac{\partial P}{\partial y} = \frac{-1 \cdot (x - y)^2 - (-y)(2)(x - y)(-1)}{(x - y)^4} = \frac{- (x - y)^2 - 2y(x - y)}{(x - y)^3} = \frac{x}{(x - y)^3}$

2. Calculate $\frac{\partial Q}{\partial x}$: $Q = \frac{x}{(x - y)^2}$ Differentiate $Q$ with respect to $x$: [ \frac{\partial Q}{\partial x} = \frac{1 \cdot (x - y)^2 - x \cdot 2(x - y)}{(x - y)^4} = \frac{(x - y) - 2x(x - y)}