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Mathematics
Differential Geometry
Infinitesimal area element in polar coordinate
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[QUOTE="Orodruin, post: 5462023, member: 510075"] Because this is not what an area element means. You need to relate the displacement vectors corresponding to a change in each coordinate to the spanned area. This is done through the Jacobian determinant, not by multiplying infinitesimals together. You [I]can[/I] multiply the infinitesimals if you use an anti-symmetric tensor product which defines an area form. For example: \begin{align} dx \wedge dy &= [\cos(\theta) dr - r \sin(\theta) d\theta]\wedge [\sin(\theta) dr + r \cos(\theta) d\theta] \nonumber \\ &= \cos(\theta)\sin(\theta) (dr \wedge dr - r^2 d\theta \wedge d\theta) - r \sin^2(\theta) d\theta \wedge dr + r \cos^2(\theta) dr \wedge d\theta \nonumber \\ &= r\, dr \wedge d\theta. \nonumber \end{align} Edit: Updated LaTeX display for the anti-symmetric tensor product. [/QUOTE]
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Mathematics
Differential Geometry
Infinitesimal area element in polar coordinate
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