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I am trying to demonstrate that $dx*dy$ (in cartesian coordinates) is equal to $r*dr*dφ$ (polar coordinates). I know the image, but I want to follow an other way:

$$x=r*cosφ$$ $$y=r*sinφ$$

$$d(r*cosφ)*d(r*sin(φ)) \rightarrow ?$$

Aron Wolf
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  • Try to better formulate your question, please. Take a look at the example 2 at this wiki page

    https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

     – Gustavo Mar 18 '16 at 05:27
  • You need to use the wedge product of 1-forms, which in turn computes the determinant of the Jacobian matrix. – Ted Shifrin Mar 18 '16 at 05:48
  • The $d\cdots$ notation is misleading. See the question and my answer in http://math.stackexchange.com/questions/667091/is-line-element-mathematically-rigorous/667126#667126. – Martín-Blas Pérez Pinilla Mar 18 '16 at 09:53

2 Answers2

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I suggest you use Jacobian matrix since this is not very rigorous. $$\begin{align*} |d(r \cos \phi)| |d(r \sin \phi)| &= |dr \cos \phi + r\,d\cos\phi| |dr \sin \phi + r\,d\sin \phi| \\ &= |\cos \phi\,dr - r \sin \phi\,d\phi| |\sin\phi\,dr + r \cos\phi\,d\phi| \\ &= r(\sin\phi)^2\,dr\,d\phi + r(\cos\phi)^2\,dr\,d\phi \end{align*}$$ The $(dr)^2$ terms and $(d\phi)^2$ terms are negligible.

Henricus V.
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  • ok, but it is -r(sinϕ)^2drdϕ+r(cosϕ)^2drdϕ = drdϕ( (cosϕ)^2 - (sinϕ)^2) and the problem is that (cosϕ)^2 - (sinϕ)^2 is not equal to 1 – Aron Wolf Mar 18 '16 at 05:40
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DiffrlArea

In the $x,y$ co-ordinate system element of area or differential of area is $ dx\cdot dy $ (red rectangle).

In the $r,\varphi$ co-ordinate system element of area or differential of area is $ dr\cdot r d \varphi $ ( gray sector segment ).

The infinitesimal differential areas ( or volumes) can be computed treating them as if the sides are finite. It is directly geometrical.

The same result is obtained using Jacobian conversion. Recently I also read that Leibnitz used such product evaluations when Newton treated more using time as differential.I used to think it was some sort of engineering shortcut!

$$ dx\cdot dy = dr \cdot r d \varphi $$

Narasimham
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