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Show that on R^2\{0} (without zero),

let w=(xdy-ydx)/(x^2+y^2) and show (a) closed (b) not exact.

(a) is straightforward,

and for (b), the following is the solution lecturer provided.

Firstly convert to polar coordinates letting x=rcos(p) y=rsin(p) where p is supposed to be angle, and get w=dp.

And then! (from here on I don't understand) suppose w=df where f is smooth on R^2\{0}. On R dp=df implies d(p-f) is constant. This contradicts that f smooth as in any neighbourhood of (1,0) there exists p1,p2 in R^2\{0} such that f(p1)-f(p2)>pi (as in pi=3.14...)so f is not continuous.

Could someone 'explain' what he's doing in this solution? I don't understand how exterior derivatives got to do with these functions being constant etc.

I appreciate your help :)

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# Question and answer on 'not exact' forms

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