Why is the Jacobian for polar coordinates sometimes negative?

[laser]

Homework Statement

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Relevant Equations

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Proving this geometrically [1] gives $J = r.$

Why is the $-r$ one wrong? Why is $(x, y) \rightarrow (\theta, r)$ is different from $(x, y) \rightarrow (r, \theta)$? Edit: In Paul's Notes [2] it seems like $J$ is always positive, but online says it can be negative...

[1] The first answer on https://math.stackexchange.com/questions/1656814/how-to-prove-dxdy-r-dr-d-theta
[2] https://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx

 

[laser]

I don't think I can delete this post - but I think I've figured out the issue. $J$ can be positive or negative but in the integration formula we take the absolute value of $J$. My bad for reading it wrong!

 

[Mark44]

If you switch a row or column of a determinant, the sign of the determinant changes. Regarding the text in the image you posted, in both cases the transformations are from polar to Cartesian rather than from Cartesian to polar as written in that text. The relevant equations are $x = r\cos(\theta)$ and $y = r\sin(\theta)$.

The equations for converting from Cartesian to polar are different.

 

[Mark44]

$J$ can be positive or negative but in the integration formula we take the absolute value of $J$. My bad for reading it wrong!

How is your post related to integration? You didn't mention anything of the sort in your earlier post.

 

[pasmith]

How is your post related to integration? You didn't mention anything of the sort in your earlier post.

Aside from in the title, which contains "dA = rdrdtheta".

@laser: The change of varaible formula in area integrals is derived from <br /> \mathbf{n}\,dS = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\,du\,dv so that <br /> dS = \left\| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right\|\,du\,dv. Swapping the order of u and v changes the direction of \mathbf{n} but does not change dS. We treat 2D as being in the plane z = 0, so that the cross-product reduces to <br /> \left(\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} <br /> \frac{\partial y}{\partial u} \right)\mathbf{e}_z and taking the norm of this gives <br /> \left| \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} <br /> \frac{\partial y}{\partial u} \right|.

 

[WWGD]

Just to point out that this change of coordinates is only valid locally, i.e., there's no global change of coordinates in this case.