Discontinuity in the value of arctan(y/x)

[GwtBc]

Homework Statement

Let $\omega = \frac{xdy -ydx}{x^2+y^2}$ and for $p, q > 0$ define the curve $C_{p,q,x,y}$:
$$ (-p,-q) \rightarrow (x,-q) \rightarrow (x,y) $$
Let $$f(x,y) = \int_{C_{p,q,x,y}}\omega$$
Show that $f(x,y)$ is discontinuous at $x=0$ if $y \geq 0$ is fixed, but not if $y < 0$. What is the significance of the value of this discontinuity?

Relevant Equations

Found the discontinuity to be $2\pi$ if $y> 0$ and $\pi$ if $y=0$. However, I don't really know what the significance of this value is in this context.

after two simple line integrals we find that

$$ f(x,y) = \arctan{p/q} + \arctan{y/x} + \pi/2$$
if $x > 0$ and
$$f(x,y) = \arctan{p/q} + \arctan{y/x} - \pi/2$$
if $x < 0$.
And then we can just take the limits to find the value of the discontinuity (as given above). But what is the significance of the value? I know it's a vague question but is there something very fundamental?

Perhaps something that's related is that earlier in the question there is discussion of how with $\theta = \arctan(y/x)$, $d\theta$ is closed (as a differential form), but not exact.

 

[PeroK]

The most obvious thing that comes to mind is that at $y = 0$ you are jumping from one branch of the $\arctan$ graph to the next.