Let [itex] f(x,y) = ( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}) [/itex] with [itex] f : D \subset \mathbb{R}^2 \to \mathbb{R}^2 [/itex](adsbygoogle = window.adsbygoogle || []).push({});

I know if I take [itex] D = D_1 = \mathbb{R}^2 - \{ (0,0) \} [/itex] the vector field is not conservative, for the circulation over a circunference centered at the origin does not equal zero.

But let's see what happens if I take [itex] D = D_2 = \{(x,y) \in \mathbb{R}^2 : (x,y) \neq (0,a), a \geq 0\} [/itex], that is [itex] \mathbb{R}^2 [/itex] without the ray starting at the origin and going in the positive-y axis.

With [itex] D_2 [/itex] one has [itex] f [/itex] has to be conservative, because the domain is simply connected, and it satisfices the necesary condition (jacobian continuous and symmetric). So together they satisfty a sufficient condition.

The problem arises when I want to calculate the potential function. I get

[itex] \phi'_x = \frac{-y}{x^2 + y^2}[/itex]

[itex] \phi'_y = \frac{x}{x^2 + y^2}[/itex]

so

[itex] \phi \approx - \arctan(\frac{x}{y}) + c(y) [/itex]

[itex] \phi \approx \arctan(\frac{y}{x}) + c(x) [/itex]

And can't see how to get the expression for the potential function.

Is it [itex] \phi(x,y) = \arctan(\frac{y}{x}) - \arctan(\frac{x}{y}) + c [/itex] ?

Any idea what is happening here?

Thanks!

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# Conservative vector field?

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