Change of variable integral problem

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Homework Help Overview

The discussion revolves around a change of variable integral problem involving the Jacobian determinant in the context of double integrals. Participants are exploring the relationship between the Jacobian and the transformations of variables in integration.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses confusion regarding the inversion of the Jacobian during integration. Some participants clarify the relationship between the Jacobian and the differential area elements in the context of variable transformations.

Discussion Status

Participants are actively discussing the properties of the Jacobian and its role in changing variables for integrals. Some guidance has been provided regarding the correct formulation of the Jacobian and its implications for integration, though no consensus has been reached.

Contextual Notes

The original poster references previous tutorial problems but does not recall similar applications of the Jacobian, indicating a potential gap in understanding or experience with this concept.

NewtonianAlch
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Homework Statement


http://img716.imageshack.us/img716/7453/28042782.jpg

The Attempt at a Solution



What I do not understand is how the Jacobian suddenly gets inverted when doing the integral, I have looked over my other tutorial problems for similar solutions and I do not recall doing something like this.

http://img140.imageshack.us/img140/7459/45340834.jpg
http://img220.imageshack.us/img220/7548/63185150.jpg
 
Last edited by a moderator:
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The Jacobian calculated in the solution is
$$J = \begin{vmatrix}
\partial u/\partial x & \partial u/\partial y \\
\partial v/\partial x & \partial v/\partial y
\end{vmatrix}$$ so you have that ##du\,dv = J\,dx\,dy##. Because you want to find ##\int dx\,dy##, you need to integrate ##\int (1/J)\,du\,dv##. If you used the inverse transformations and found instead
$$J' = \begin{vmatrix}
\partial x/\partial u & \partial x/\partial v \\
\partial y/\partial u & \partial y/\partial v
\end{vmatrix},$$ then you'd have ##dx\,dy = J'\,du\,dv##.
 
Yes, it looks like velas response is correct, but let me add a notation that might help you remember which way you are going. There is a visual believability about

dudv=|\frac{\partial(u,v)}{\partial(x,y)}|dxdy

and so also for

dxdy=|\frac{\partial(x,y)}{\partial(u,v)}|dudv

See? It sort of looks like things cancel correctly.
 
Last edited:
algebrat said:
Yes, it looks like velas response is correct, ...

vela has a history of giving correct responses!
 
Thanks for the responses, yea I sort of see it now. I guess a bit more practice with it is required.
 

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