Finding a potential function - vector fields Let F be the vector function - $\overrightarrow{F} = (\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$
We know that F is conservative on every path that doesnt encircle the point $(0,0)$
Calculating it's potential function gave me - $\phi = arctan(\frac{x}{y})$
the problem is with the points $(x,0)$,
what can we say about those points? the potential is not defined there.
How can I, for example, calculate the work on a linear route from (0,1) to (1,0)? I would like to say $\phi(0,1) - \phi(1,0)$ but $\phi(1,0)$ is not defined.
thanks!
 A: You can't define the potential function everywhere, only on a region that doesn't wind around the pole at $ ( 0 , 0 ) $.  You can use $ \arctan ( y / x ) $ on the region where $ x > 0 $ and separately on the region where $ x < 0 $; you can also use $ \operatorname { arccot } ( x / y ) = \pi / 2 - \arctan ( x / y ) $ on the region where $ y > 0 $ and separately on the region where $ y < 0 $.  You can also add a constant to any of these.  Altogether, these cover the entire region except $ ( 0 , 0 ) $, and where they overlap, they differ by constants (multiples of $ \pi $, in fact).  So if you want to cross the boundary of one of them, then you can switch to another one, adding the appropriate constant.  In this way, you can patch together a single potential function to use on your entire path, as long as it doesn't wind around the origin.
In particular, since $ \arctan ( y / x ) $ agrees with $ \operatorname { arccot } ( x / y ) $ in the first quadrant, you can use those function (with no need to even add a constant) to cover your entire line segment.  To be explicit about the combined function, it's $$ \phi ( x , y ) : = \cases { \arctan ( y / x ) & for \( x > 0 \), \\ \operatorname { arccot } ( x / y ) & for \( y > 0 \). } $$  (And you can extend this into the 4th quadrant, but not all the way around.)  In particular, $ \phi ( 1 , 0 ) = \arctan ( 0 / 1 ) = 0 $ and $ \phi ( 0 , 1 ) = \operatorname { arccot } ( 0 / 1 ) = \pi / 2 $, so the integral is $ \pi / 2 - 0 = \pi / 2 $ (which I think you knew it had to be).
