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example if we have cylindrical coordinates : $x = r \ cos(\theta)$ $y = r \ sin(\theta)$ if we treated x as function of $r$ and $\theta$ then:

$dx = \partial x / \partial r\ *\ dr\ + \partial x/ \partial \theta\ * \ d\theta$

$dy = \partial y / \partial r\ *\ dr\ + \partial y/ \partial \theta\ * \ d\theta$

why we cannot use multiplication as $dxdy = (\partial y / \partial r\ *\ dr\ + \partial y/ \partial \theta\ * \ d\theta)*(\partial x / \partial r\ *\ dr\ + \partial x/ \partial \theta\ * \ d\theta)$

Mahmoud Emadeldeen
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  • The short answer is that such "differentials" are not real numbers, so they do not obey the same arithmetic which real numbers follow. Therefore, since they aren't real numbers, there is actually no reason to even expect such "naive multiplication" to give the right answer. – peek-a-boo Jul 12 '19 at 04:10
  • I'm pretty sure this has been asked before, but I can't find it. Anyway, you have to use the wedge product (a.k.a. the exterior product). See here, for instance: https://math.stackexchange.com/a/2296042/1242 – Hans Lundmark Jul 12 '19 at 04:16
  • sorry sir but what is exterior product belong to , i mean what is the math branch the contain or study this type of product – Mahmoud Emadeldeen Jul 12 '19 at 17:57

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