# How do you show that $d\theta = \frac{x dy - y dx }{x^2 + y^2}$?

If $(r, \theta)$ are polar coordinates on $\mathbb{R}^2\backslash \left\{ (0,0)\right\}$, then how do I show/prove that $d\theta =\dfrac{x dy - y dx}{x^2 + y^2}$?

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What set are you working on? I don't think this makes much sense on R^2 as a whole. Perhaps on R^2 - {0} this might work... –  Jesse Madnick Oct 5 '10 at 5:59
Well, since $\theta=\arctan\left(\frac{y}{x}\right)$, you can try implicit differentiation. Think of $\theta$, x, and y as a function of a new parameter, call it t, differentiate the whole mess with respect to t, and "cancel out" the $\mathrm{d}t$ term afterward. –  Ｊ. Ｍ. Oct 5 '10 at 6:06
Yes, I meant R^2-{0} –  JimJones Oct 5 '10 at 6:06
@Jyotirmoy: I have restored the [differential-geometry] tag because the d here may mean differential form. @Jim please clarify if it is the case. –  KennyTM Oct 5 '10 at 9:13

$$x=r\cos\theta$$ $$y=r\sin\theta$$

So, $$dx=\cos\theta\thinspace dr -r\sin\theta\thinspace d\theta$$ $$dy=\sin\theta\thinspace dr +r\cos\theta\thinspace d\theta$$

Solving for $d\theta$ and using $\sin^2\theta+\cos^2\theta=1$ $$d\theta={{dy\thinspace\cos\theta-dx\thinspace\sin\theta}\over {r}}$$ Multiplying and dividing by $r$ $$={{dy\thinspace (r\cos\theta)-dx\thinspace (r\sin\theta)}\over {r^2}}$$ Using the definition of $x$ and $y$ in terms of polar coordinates $$={{x\thinspace dy-y\thinspace dx}\over {x^2+y^2}}$$

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Thank you so much, this was very helpful –  JimJones Oct 5 '10 at 17:11

On $R^2\setminus\{0\}$ we have $\theta=\Im(\ln(x+iy))$ (here $\Im$ stands for "the imaginary part of"). Therefore $$d\theta = \Im\left(\frac1{x+iy}\right)dx+\Im\left(\frac{i}{x+iy}\right)dy=\Im\left(\frac{x-iy}{x^2+y^2}\right)dx+\Im\left(\frac{i(x-iy)}{x^2+y^2}\right)dy=$$$$\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy=\frac{xdy-ydx}{x^2+y^2}.$$

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First you need a definition of $\theta$ (and/or $d \theta$) to compare with the form involving $dx$ and $dy$. In coordinates any definition is only "local" and limited to regions of $(x,y)$ that exclude loops around $0$ (which would modify $\theta$ by $2\pi$). In practice any local definition will be as a function of $x/y$ or of $y/x$ according to which line or half-line from $0$ is excluded as a "branch cut". e.g., $\theta = \arctan(y/x)$.

If $\theta$ is defined using trigonometric functions one then needs a definition of those functions not using angles, or the whole exercise is logically vacuous, defining $\theta$ in terms of $\theta$ (using polar coordinates is an example of this). An angle-free definition is possible but it is ad hoc if there is no clear relation to geometry, such as defining inverse sine or tangent functions as integrals. A more "correct" definition is to use arc length on the projection of $(x,y)$ to the unit circle. Here if $u(x,y)=(x,y)/\sqrt{x^2+y^2}$ is the projection one would have $d\theta = |u(x+dx,y+dy) - u(x,y)|$ up to second order terms in $dx$ and $dy$.

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