Let $\mathbf{F}(x,y)=\frac{-y\mathbf{i}+x\mathbf{j}}{x^2+y^2}$ and let $C$ be the circle $\mathbf{r}(t)=(\cos t)\mathbf{i}+(\sin t)\mathbf{j},0\le t\le 2\pi$.
A. Compute $\frac{\partial Q}{\partial x}$

Note: Your answer should be an expression of $x$ and $y$; e.g. " $3xy-y"$
B. Compute $\frac{\partial P}{\partial y}$

Note: Your answer should be an expression of $x$ and $y$; e.g. " $3xy-y"$
C. Compute ${\int }_C\mathbf{F}\cdot d\mathbf{r}$

Note: Your answer should be a number
D. Is $\mathrm{F}$ conservative? Type $\mathrm{Y}$ if yes, type $\mathrm{N}$ if no.

Step 1 ：The vector field $\mathbf{F}(x,y)=\frac{-y\mathbf{i}+x\mathbf{j}}{x^2+y^2}$ can be written as $\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}$, where $P(x,y)=-\frac{y}{x^2+y^2}$ and $Q(x,y)=\frac{x}{x^2+y^2}$.

Step 2 ：Compute $\frac{\partial Q}{\partial x}=\frac{\partial }{\partial x}\left(\frac{x}{x^2+y^2}\right)$. Using the quotient rule for differentiation, we get $\frac{\partial Q}{\partial x}=\frac{(x^2+y^2)\cdot 1-2x\cdot x}{(x^2+y^2{)}^2}=\frac{y^2}{(x^2+y^2{)}^2}$.

Step 3 ：Compute $\frac{\partial P}{\partial y}=\frac{\partial }{\partial y}(-\frac{y}{x^2+y^2})$. Using the quotient rule for differentiation, we get $\frac{\partial P}{\partial y}=\frac{(x^2+y^2)\cdot (-1)-2y\cdot (-y)}{(x^2+y^2{)}^2}=\frac{x^2}{(x^2+y^2{)}^2}$.

Step 4 ：The line integral of $\mathbf{F}$ over the curve $C$ is given by ${\int }_C\mathbf{F}\cdot d\mathbf{r}={\int }_0^{2\pi }\mathbf{F}(r(t))\cdot r^{\prime }(t)dt$. Since $r(t)=(\cos t)\mathbf{i}+(\sin t)\mathbf{j}$, we have $r^{\prime }(t)=(-\sin t)\mathbf{i}+(\cos t)\mathbf{j}$. Substituting $r(t)$ into $\mathbf{F}$, we get $\mathbf{F}(r(t))=\frac{-\sin t\mathbf{i}+\cos t\mathbf{j}}{{\cos }^{2}t+{\sin }^{2}t}=-\sin t\mathbf{i}+\cos t\mathbf{j}$. Then, $\mathbf{F}(r(t))\cdot r^{\prime }(t)=(-\sin t\mathbf{i}+\cos t\mathbf{j})\cdot ((-\sin t)\mathbf{i}+(\cos t)\mathbf{j})={\sin }^{2}t+{\cos }^{2}t=1$. So, ${\int }_C\mathbf{F}\cdot d\mathbf{r}={\int }_0^{2\pi }1dt=2\pi$.

Step 5 ：A vector field $\mathbf{F}$ is conservative if and only if $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$. From previous steps, we have $\frac{\partial P}{\partial y}=\frac{x^2}{(x^2+y^2{)}^2}$ and $\frac{\partial Q}{\partial x}=\frac{y^2}{(x^2+y^2{)}^2}$. Since $\frac{\partial P}{\partial y}\ne \frac{\partial Q}{\partial x}$, $\mathbf{F}$ is not conservative. So, the answer is $\boxed{{\displaystyle N}}$.