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In cartesian coordinates, $dA = dx\,dy$ . As $x=\rho \cos\phi\ $ and $y=\rho \sin \phi \ $,

then: $dx= \cos\phi\, d\rho - \rho \sin\phi\, d\phi$ and $dy = \sin\phi\, d\rho + \rho \cos\phi \, d \phi $.

So $dx\, dy = \rho\, d\rho\, d\phi(\cos^2\phi-\sin^2\phi) $, (ignoring second order terms in $d\rho$ and $d\phi$)

Why is this not equal to the correct result of $dA=\rho\, d\rho\, d\phi$?

Bernard
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Vishal Jain
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1 Answers1

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Because $d\rho\, d\phi=-d\phi\, d\rho$

Bernard
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Fareed Abi Farraj
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  • I did not know the multiplication of differentials was not commutative, is there any reason for this being so? – Vishal Jain Dec 29 '19 at 14:38
  • It is actually an exterior product (i.e. an alternating product) $;\mathrm d\rho\wedge\mathrm d\phi=-\mathrm d\phi\wedge\mathrm d\rho$, $\mathrm d\rho\wedge\mathrm \rho=0$, &c. – Bernard Dec 29 '19 at 14:51
  • Yes it is from the defintion of the exterior product, and for the same reason $d\rho \wedge d\rho =-d\rho \wedge d\rho=0$ – Fareed Abi Farraj Dec 29 '19 at 15:01