Finding a Line Integral over a Closed Curve

\(\mathbf{F}=\mathbf{F}(x,y)=\frac{-y\mathbf{i}+x\mathbf{j}}{x^{2}+y^{2}}\) is the gradient of \(f(x,y) =\tan ^{-1}\frac{y}{x},\) \(x\neq 0,\) since \[\begin{equation*} \nabla\! \ f=\frac{\partial }{\partial x}\tan ^{-1}\frac{y}{x}\mathbf{i}+\frac{\partial }{\partial y}\tan ^{-1}\frac{y}{x}\mathbf{j}=\frac{ \frac{-y}{x^{2}}}{1+\frac{y^{2}}{x^2}}\mathbf{i}+\frac{\frac{1}{x}}{1+\frac{y^{2}}{x^2}}\mathbf{j}=\frac{-y\mathbf{i}+x\mathbf{j}}{x^{2}+y^{2}} \end{equation*}\]

So, \(\mathbf{F}\) is a conservative vector field on any region \(R\) that contains no points on the \(y\)-axis (\(x=0\)). Then by the corollary \[\begin{equation*} \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=0 \end{equation*}\]

along any closed, piecewise-smooth curve \(C\) that does not cross or touch the \(y\)-axis.