Solution (a) Let $P(x,y) =x+y$ and $Q(x,y) =y-x.$ Then $$\dfrac{\partial P}{\partial y}=\dfrac{\partial }{\partial y}(x+y) =1\qquad\hbox{and}\qquad \dfrac{\partial Q}{\partial x}=\dfrac{\partial }{\partial x}(y-x) =-1$$

Since $\dfrac{\partial P}{\partial y}\neq \dfrac{\partial Q}{\partial x}$, the force field $\mathbf{F}$ is not a conservative vector field.

(b) To show that the work $W$ done by $\mathbf{F}$ is dependent on the path, we choose two paths beginning at the origin $(0,0)$ and ending at the point $(1,1)$.

For Path 1, we let $C$ be defined by $y=x$, where $0\leq x\leq 1.$ That is, $x(t) =t,$ $y(t) =t,$ $0\leq t\leq 1.$ Then $\mathbf{F}( x(t),y(t)) =2t\,\mathbf{i}$, $\mathbf{r}(t) =t\mathbf{i}+t\mathbf{j}$, and $d\mathbf{r=\,}\left( \mathbf{i}+\mathbf{j}\right) dt.$ So, $$\begin{equation*} \mathbf{F}\,{\cdot}\, d\mathbf{r}=2t\,dt \end{equation*}$$

The work $W$ done by $\mathbf{F}$ along $C$ is $$W=\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{0}^{1}2t\,dt=\big[ t^{2}\big] _{0}^{1}=1$$

For Path 2, we let $C$ be the piecewise-smooth curve made up of:

For $C_{1},$ $\mathbf{F}( x(t),y(t)) =t\,\mathbf{i}-t\mathbf{j}$, $\mathbf{r}(t) =t\mathbf{i}$, and $d\mathbf{r}=\mathbf{i}\,dt$. So, $$\mathbf{F}\,{\cdot}\, d\mathbf{r}=t\,dt$$

For $C_{2},$ $\mathbf{F}( x(t),y(t)) =( 1+t) \,\mathbf{i}+( t-1) \mathbf{j}$, $\mathbf{r}(t) =\mathbf{i}+t \mathbf{j}$, and $d\mathbf{r}=\mathbf{j}\,dt$. So, $$\begin{equation*} \mathbf{F}\,{\cdot}\, d\mathbf{r}=( t-1)\, dt \end{equation*}$$

The work $W$ done by $\mathbf{F}$ along Path 2 is $$\begin{equation*} W=\int_{C_{1}}\mathbf{F}\,{\cdot}\, d\mathbf{r}\,+\int_{C_{2}}\mathbf{F}\,{\cdot}\, d \mathbf{r}=\int_{0}^{1}t\,dt +\int_{0}^{1}( t-1)\, dt=0 \end{equation*}$$

Since the work done along Path 1 and Path 2 are not equal, the work $W$ done by the force $\mathbf{F}$ is dependent on the path.