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First I was asked to

  1. find the line integral $$ I=\int_{C} \dfrac{-y}{x^{2}+y^{2}} dx + \dfrac{x}{x^{2}+y^{2}} dy, $$ where $C$ is the closed curve $$ C=\{ (x,y) \in \Bbb R^{2} \mid (x-2)^{2}+4y^{2}=1 \} $$ and $C$ is oriented counterclockwise.

It seems that we have all of the conditions necessary for Green's Theorem, thus $$ I=\iint_A (\dfrac{y^2-x^2}{x^{2}+y^{2}}-\dfrac{y^2-x^2}{x^{2}+y^{2}})dxdy=0. $$ Is it correct?

Next, the follow-up question seems to be an application of the previous one but I don't know how they can be related.

  1. Let $U=\{ (x,y) \in \Bbb R^{2} \mid x>0 \}$. Find a $C^{\infty}$-function $\varphi$ on $U$ satisfying the following $$ \frac{\partial \varphi}{\partial x}(x,y)=\frac{-y}{x^{2}+y^{2}},\ \ \frac{\partial \varphi}{\partial y}(x,y)=\frac{x}{x^{2}+y^{2}},\ \ \varphi(1,0)=1. $$

Any help would be much appreciated.

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On the half-plane $x>0$ the field is conservative and hence you can just choose any path in this half-plane and integrate from $(1,0)$ to $(x,y)$ to get $$\varphi(x,y)=1+\int_{(1,0)\to (x,y)}\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ For example, going along the axes, you have $$\varphi(x,y)=1+\int_1^x\frac{-0}{x^2+0^2}dx+\int_0^y\frac{x}{x^2+y^2}dy$$ $$=1+x\int_0^y\frac{dy}{x^2+y^2}$$ $$=1+\arctan\frac{y}{x}$$

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  • $\begingroup$ So my computation on the first question is correct? And how do you prove that your obtained function is smooth? $\endgroup$ – user Oct 9 '16 at 8:29
  • $\begingroup$ The first derivatives give the field. And I think it is easy to see that your field is $C^{\infty}$ when $x>0$. $\endgroup$ – velut luna Oct 9 '16 at 8:35
  • $\begingroup$ Maybe we can check directly that your $\varphi(x,y)$ is smooth. Because $x^2+y^2>0$ whenever $x>0$. $\endgroup$ – user Oct 9 '16 at 8:55
  • $\begingroup$ Yes. If you are comfortable with it. $\endgroup$ – velut luna Oct 9 '16 at 8:59

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