Find $\oint_{\!\!\!\!C}\left( \dfrac{-y}{x^{2}+y^{2}}\,dx+\dfrac{x}{x^{2}+y^{2}} \,dy\right)$ if:

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Solution For $P=\dfrac{-y}{x^{2}+y^{2}}$ and $Q=\dfrac{x}{ x^{2}+y^{2}}$, we have $$\dfrac{\partial P}{\partial y}=\dfrac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}\qquad \hbox{and} \qquad \dfrac{\partial Q}{\partial x}=\dfrac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}$$

Notice that, except at the point $(0,0),$ the partial derivatives $\dfrac{ \partial P}{\partial y}$ and $\dfrac{\partial Q}{\partial x}$ are continuous.

(a) It is difficult to find the line integral along $C$. Moreover, because $x^{2/3}+y^{2/3}=1$ contains the point $(0,0)$ in its interior, we cannot use Green’s Theorem as we did in Example 2 to change the line integral to a double integral. However, since $\dfrac{ \partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ throughout $R,$ we can use a different curve that leads to an easier line integral. For example, suppose we replace $x^{2/3}+y^{2/3}=1$ by the unit circle $C_{1}$: $x^{2}+y^{2}=1,$ noting that the conditions required for $\int_{C}( P\,dx+Q\,dy)$ to equal $\int_{C_{1}}( P\,dx+Q\,dy)$ are met. See Figure 38.

Figure 38

On $C_{1},$ $x=\cos \theta$, $y=\sin \theta$, $0\leq \theta \leq 2\pi$. Then $dx=-\sin \theta d\theta$, $dy=\cos \theta d\theta$, and $$\begin{eqnarray*} \oint_{C}\left( \dfrac{-y}{x^{2}+y^{2}}\,dx+\dfrac{x}{x^{2}+y^{2}}\,dy\right) &=&\oint_{C_{1}}\left( \dfrac{-y}{x^{2}+y^{2}}\,dx+\dfrac{x}{x^{2}+y^{2}} \,dy\right)\\ &=&\int_{0}^{2\pi }(\sin ^{2}\theta +\cos ^{2}\theta )\,d\theta =2\pi \end{eqnarray*}$$

(b) In this case, the interior of the curve $C$ does not contain the origin. Since $$\begin{equation*} \dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x} \end{equation*}$$

in the simply connected region $R$ enclosed by $C$, it follows that $\mathbf{ F}=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}$ is a conservative vector field. Since $C$ is closed, $$\oint_{C}\left( \dfrac{-y}{x^{2}+y^{2}}\,dx+\dfrac{x}{x^{2}+y^{2}}\,dy\right)= \oint_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r} =0$$