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

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

Note: Your answer should be an expression of $x$ and $y$; e.g. " $3 x y-y "$
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 \(\frac{\partial Q}{\partial x}=\frac{(x^{2}+y^{2})\cdot1-2x\cdot x}{(x^{2}+y^{2})^{2}}=\frac{y^{2}}{(x^{2}+y^{2})^{2}}\).

Step 3 ：Compute \(\frac{\partial P}{\partial y}=\frac{\partial}{\partial y}\left(-\frac{y}{x^{2}+y^{2}}\right)\). 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 \(\int_{C} \mathbf{F} \cdot d \mathbf{r}=\int_{0}^{2\pi} \mathbf{F}(r(t)) \cdot r'(t) dt\). Since \(r(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}\), we have \(r'(t)=(-\sin t) \mathbf{i}+(\cos t) \mathbf{j}\). 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, \(\mathbf{F}(r(t)) \cdot r'(t)=(-\sin t \mathbf{i}+\cos t \mathbf{j})\cdot ((-\sin t) \mathbf{i}+(\cos t) \mathbf{j})=\sin^{2} t+\cos^{2} t=1\). So, \(\int_{C} \mathbf{F} \cdot d \mathbf{r}=\int_{0}^{2\pi} 1 dt=2\pi\).

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 \(\frac{\partial P}{\partial y}\neq\frac{\partial Q}{\partial x}\), \(\mathbf{F}\) is not conservative. So, the answer is \(\boxed{N}\).