The question is:(adsbygoogle = window.adsbygoogle || []).push({});

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 :)

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Question and answer on 'not exact' forms

**Physics Forums | Science Articles, Homework Help, Discussion**