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I'm trying to derive the area element $dA$ in polar coordinates using total differentials (just for the sake of trying: I know the existence of the Jacobian), but I can't get the correct result.


Here's how:

$$ \left\{ \begin{array}{c} x\left(\rho,\theta\right)=\rho\cos\theta\\ y\left(\rho,\theta\right)=\rho\sin\theta \end{array}\right. $$

The total differentials of $x\left(\rho,\theta\right)$ and $y\left(\rho,\theta\right)$ are given by

$$ \begin{array}{c} dx=\cos\theta\, d\rho-\rho\sin\theta\, d\theta\\ dy=\sin\theta\, d\rho+\rho\cos\theta\, d\theta \end{array} $$

The area element $dA$ can therefore be written as

$$ dA=dx\, dy=\cos\theta\,\sin\theta\,\left(d\rho\right)^{2}-\rho^{2}\,\cos\theta\,\sin\theta\,\left(d\theta\right)^{2}+\left(\cos^{2}\theta-\sin^{2}\theta\right)\rho\, d\rho\, d\theta $$

By neglecting second order terms

$$ dA\approx\left(\cos^{2}\theta-\sin^{2}\theta\right)\rho\, d\rho\, d\theta $$


Where am I wrong with this? Thanks in advance.

Davide La Vardera
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1 Answers1

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You're going wrong because $dx$ and $dy$ are vectors. This is something the Jacobean formalises and encapsulates, so you can just churn out the formula instead of thinking about it each time.

In this case:

$ dx = \cos \theta ~d\rho ~\hat{\rho} - \rho \sin \theta ~d\theta ~\hat{\theta}$ and

$ dy = \sin \theta ~d\rho ~\hat{\rho} + \rho \cos \theta ~d\theta ~\hat{\theta}$

$dA = dx \times dy = \rho \cos^2\theta ~d\rho~d\theta~(\hat{\rho} \times\hat{\theta}) - \rho \sin^2\theta ~d\rho~d\theta~(\hat{\theta}\times \hat{\rho} )$

Finally, $\hat{\rho} \times \hat{\theta} = \hat{z}$, to get your area element.

stochasticboy321
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