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The differential 1-form

$$w = \frac{x\,dy-y\,dx}{x^2+y^2}$$

is equal to the exterior derivative

$$d\theta=d(\arctan\frac{y}{x})$$

I understand by chain rule, how multivariable derivative of $\arctan(\frac{y}{x})=\frac{x-y}{x^2+y^2}$, but I have no idea how to present with $dx$ and $dy$ in the result. I do not have a good concept of what an exterior derivative is.

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  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$
    – user
    Jan 24, 2018 at 22:05

1 Answer 1

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Note that in polar coordinates

$$\arctan\bigg( \frac{y}{x} \bigg) =\arctan\bigg( \frac{\rho \sin \theta}{\rho \cos \theta}\bigg)= \arctan (\tan\theta)=\theta$$

thus

$$d\theta=d(\arctan\frac{y}{x})$$

and

$$d\theta=\frac{\partial \theta}{\partial x}\,dx+\frac{\partial \theta}{\partial y}\,dy=\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy=\frac{x\,dy-y\,dx}{x^2+y^2}$$

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