Conservative Vector Field: Understanding Circulation and Potential Function

[Damidami]

Let f(x,y) = ( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}) with f : D \subset \mathbb{R}^2 \to \mathbb{R}^2

I know if I take D = D_1 = \mathbb{R}^2 - \{ (0,0) \} 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 D = D_2 = \{(x,y) \in \mathbb{R}^2 : (x,y) \neq (0,a), a \geq 0\}, that is \mathbb{R}^2 without the ray starting at the origin and going in the positive-y axis.

With D_2 one has f 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

\phi'_x = \frac{-y}{x^2 + y^2}
\phi'_y = \frac{x}{x^2 + y^2}
so
\phi \approx - \arctan(\frac{x}{y}) + c(y)
\phi \approx \arctan(\frac{y}{x}) + c(x)

And can't see how to get the expression for the potential function.
Is it \phi(x,y) = \arctan(\frac{y}{x}) - \arctan(\frac{x}{y}) + c ?Any idea what is happening here?
Thanks!

 

[identity1]

Do you remember the trig identity \arctan(x)+\arctan(\frac{1}{x})=\frac{\pi}{2}? Your expressions are in fact consistent :P The potential function \arctan(x)+C works fine.

 

[identity1]

* I mean the potential function \phi(x, y)=\arctan(\frac{y}{x})+C.

 

[Damidami]

Hi identity1,
Thanks for your reply.
Some things still I don't understand.
That equation \arctan(x) + \arctan(\frac{1}{x}) = \frac{\pi}{2} seems to work only for positive real values of x, since for example
\arctan(-2) + \arctan(\frac{-1}{2}) = \frac{-\pi}{2}

Say, for example, I want to calculate the circulation of f with domain D_2 from a curve that starts at A = (-1,1) and ends at B=(1,1)

If I use \phi_1(x,y) = -\arctan(\frac{x}{y}) + C
\phi_1(B) - \phi_1(A) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}

Instead if I use \phi_2(x,y) = \arctan(\frac{y}{x}) + C
\phi_2(B) - \phi_2(A) = -\frac{\pi}{4} - \frac{\pi}{4} = \frac{-\pi}{2}

So it changes the sign for the same curve oriented in the same direction from A to B.

Something seems to be wrong, the sign shouldn't change, doesn't it?
Thanks.