Let's go through each part of the problem step-by-step.

Given

We have a vector field: $\vec{F} = \langle P, Q \rangle = \left\langle \frac{-y}{(x - y)^2}, \frac{x}{(x - y)^2} \right\rangle$ with domain $\mathcal{D} = \{ (x, y) \in \mathbb{R}^2 \mid y < x \}$.

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

The domain $D$ consists of all points in the plane where $y < x$. This is the half-plane below the line $y = x$. To determine if $D$ is simply connected:

• A region is simply connected if it has no holes and if any closed loop within it can be continuously contracted to a point.
• Since $D$ is the half-plane below $y = x$ and does not contain any holes, it is simply connected.

(ii) Show that $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ on $D$

To check if $\vec{F}$ might be conservative, we calculate the partial derivatives of $P$ and $Q$ with respect to $y$ and $x$, respectively.

For $P = \frac{-y}{(x - y)^2}$: $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} \left( \frac{-y}{(x - y)^2} \right) = \frac{- (x - y)^2 - (-y)(-2)(x - y)}{(x - y)^4} = \frac{- (x - y)^2 + 2y (x - y)}{(x - y)^3}$

For