question about this line integral whether it is exact

this form $\omega=-\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy$ ,which is not defined at origin .it is not exact on the whole x-y plane .

but when y is not zero , $\omega=d(-\arctan(x/y))$,

my question is

when consider a circle whose center is on x-axis and the circle doesn't contain origin, for example $(x-2)^2+y^2=1$ , then integral of ω along this circle seems to be zero.(I have not calculated, but by symmetry it looks the same as the integral along $x^2+(y-2)^2=1$,which is zero). so is this always zero for all the closed curves that cross the x axis and not contain origin ? If it is true, how to use Fundamental Theorem for Line Integrals to calculate $\int_{C}\omega =\arctan(B)-\arctan(A)$ while point B and A are on $x$-axis where $\arctan$ is not defined.

Solution 1:

$\DeclareMathOperator{\arccot}{arccot}$If $$ \omega = \frac{-y\, dx + x\, dy}{x^{2} + y^{2}},\quad (x, y) \neq (0, 0), $$ and if $\theta$ is a continuous choice of polar angle in an open set $G$ not containing the origin, we have $\omega = d\theta$ in $G$.

We do have $\omega = d(-\arctan(x/y))$ when $y \neq 0$, but the domain of the "potential function" $-\arctan(x/y)$ is the union of two disjoint open half-planes, and $-\arctan(x/y)$ has no continuous extension to a connected domain.

The "principal branch" of polar angle $\theta$, defined off the non-negative horizontal axis, is instead given by $$ \theta(x, y) = \begin{cases} \arctan(y/x) & 0 < x, \\ \arccot(x/y) & 0 < y, \\ \arccot(x/y) - \pi & y < 0. \end{cases} $$ (One can, and should, check that any two of these formulas agree where their domains overlap. One conceptual advantage of giving overlapping domains is that smoothness is apparent.)

If you run your argument with this (or an equivalent) definition of polar angle, order is restored to the mathematical universe.