An example computation with a form that is closed but not exact I'm working on the following question in munkres:

Let $A = \mathbb{R}^2-0$; let $$\omega = (-y\,dx + x\, dy)/(x^2+y^2)$$ in $A$. Show $\omega$ is closed but not exact in $A$. 

This is a multi-part question, so I've posted an image of all the parts (I wrote the question so it could be searched).

My questions correspond to the parts of problem:
b. This clearly "smells" like polar coordinates. Indeed, if $\phi(x,y) = \tan^{-1}(y/x)$ the formulas are verified "almost everywhere" on B. However, the fact that the formula above doesn't work everywhere (namely $x=0$) gives me pause. The way he asks the question makes me think there's a way of showing uniqueness without producing an explicit formula for $\phi$, but I cannot see the argument. Thoughts?
c. My original thought was to consider $f_1(x,y,t) = (x^2+y^2)^{1/2}\cos t - x$ and $f_2(x,y,t) = (x^2+y^2)^{1/2}\cos t - y$. Then use implicit fxn theorem to conclude there exists $g(x,y) = t$ -- this gets me regularity of $g$ too. But there are some periodic problems. For $f_1$: $\frac{\partial f_1}{\partial t} = 0$ when $t = k\pi$ so this $g(x,y)$ isn't defined everywhere. How would one proceed showing this then? How is the hint used?
d. Do b/c provide an explicit formula then? How is the hint used? 
e. So in d, we showed $\omega$ is exact. The given formula shows $\omega$ is a one form. Then by this part, we conclude $\phi$ is constant on $B$?
f. I don't understand what the hint is suggesting I do.
Generically:
I think the point of the problem is to show that the domain of the form matters -- is this correct?
 A: Yes, it's just polar coordinates. Define $\phi:B\to \mathbb R$ as follows:
If $x>0, y\ge0,$ then $\phi(x,y)=\tan^{-1}(y/x).$ 
If $x\le0,y>0$ then $\phi(x,y)=-\tan^{-1}(x/y)+\frac{\pi}{2}.$ 
If $x<0,y\le0$ then $\phi(x,y)=\tan^{-1}(y/x)+\pi.$
If $x>0,y\le 0$ then $\phi(x,y)=-\tan^{-1}(x/y)+\frac{3\pi}{2}.$
Then, $\phi$ is smooth on $B$. Now, fix a point $p=(x,y):x>0,y\ge0.$  Then,  $d\phi: T_pB\to T_{\phi(p)}\mathbb R$ is given by 
$(d\phi)_p=(\partial_x)_pdx+(\partial_y)_pdy=\frac{-y}{x^2 + y^2}dx+\frac{x}{x^2 + y^2}dy=\omega.$
The same formula for $\omega$ is obtained in the other quadrants of $\mathbb R^2\cap B$.
If $g$ is a closed $0$-form, then $dg=0$. Now,  in local coordinates, $dg=\partial_xgdx+\partial_ygdy.$ As $dg$ is $identically$ zero in $B$, we have  $dg(\frac{\partial}{\partial x})_p=(\partial_xg)_p=0$ for all $p\in B.$ Similarly, $(\partial_yg)_p=0.$ Now you can either use the hint or just observe that since both partial derivatives of $g$ vanish on the connected set $B$, in fact $g$ must be constant there.
If $\omega=df$ on $A$ then in particular $df-d\phi=\omega-\omega=0$ on $B$ so $f-\phi=c,$ some constant, on $B$. Using the hint, $\lim_{y\to 0^-}(f(1,y)-\phi(1,y))=f(1,0)+2\pi=c$ and $\lim_{y\to 0^+}(f(1,y)-\phi(1,y))=f(1,0)=c$, which implies that $2\pi=0,$ a contradiction.
But it's easier just to integrate both sides around the unit circle. That is, if $\omega=df$ then $\int \omega=2\pi$ by direct calculation, whereas $\int df=0$ by the FTC. Or you can argue that if $\omega=df$ then, $df =\frac{-y}{x^2 + y^2}dx+\frac{x}{x^2 + y^2}dy$ with $\partial f_x=\frac{-y}{x^2 + y^2}$ and $\partial f_y=\frac{x}{x^2 + y^2}$. But then, you get a contradiction because the mixed partials are not equal. 
