Let $A = \mathbb{R}^2 \backslash \{ (0, 0) \}$ and let $f : A \longrightarrow \mathbb{R}^2$ be a vector field such that $f(x,y) = (-y,x)$.
Show that there exists no polar-angle-function of $f$ on $A$.
(Hint: integrate $f$ over the unit circle)
I am struggling to solve this exercise. I do not understand what integrating over the unit circle tells me about the existence of a polar-angle-function. The integral over the unit circle is $2\pi$. Does anyone have a hint?
EDIT 1:
A polar-angle-function (name translated from German) is a function $\theta : U \longrightarrow \mathbb{R}$ such that a curve $\alpha : U \longrightarrow \mathbb{R}^2 \backslash \{ (0,0) \}$ can be represented as
$$ \alpha(t) \enspace = \enspace |\alpha(t)| \cdot \Big( \cos \theta(t) \, , \; \sin \theta(t) \Big)^T$$
EDIT 2:
I calculated the integral of $f$ over the unit circle as follows: Let the unit circle be parametrised by $\gamma : [0,2\pi) \longrightarrow \mathbb{R}^2 : t \longrightarrow \big( \cos t, \sin t \big)^T$. Then the integral is:
\begin{align} \oint_{S^1} f(x,y) \, d\gamma \enspace &= \enspace \int_0^{2\pi} f \big( \gamma_x(t), \gamma_y(t) \big) \cdot \dot{\gamma}(t) \, dt \\ &= \enspace \int_{0}^{2\pi} \Big( - \sin t, \cos t \Big) \cdot \Big( - \sin t, \cos t \Big) \, dt \\ &= \enspace \int_0^{2\pi} \underbrace{\sin^2 t + \cos^2 t}_{= \, 1} \, dt \\ &= \enspace 2 \pi \end{align}
EDIT 3:
Lemma: Let $\Omega \subset \mathbb{R}^2$ be star-shaped with respect to $x_0 \in \Omega$. Let $g: \Omega \longrightarrow \mathbb{R}^2 \backslash \{(0,0)\}$ be a continuous vector field. Then there exists a continuous map $\vartheta: \Omega \longrightarrow \mathbb{R}$, such that
$$g(x) \enspace = \enspace |g(x)| \cdot \Big( \cos \vartheta(x), \sin \vartheta(x) \Big)$$
for all $x \in \Omega$. The function $\vartheta$ is called the polar-angle-function and is unique up to addition of a konstant function $x \longmapsto 2\pi j$ for some $j \in \mathbb{Z}$.