# Change of variable integral problem

1. Jun 7, 2012

### NewtonianAlch

1. The problem statement, all variables and given/known data
http://img716.imageshack.us/img716/7453/28042782.jpg [Broken]

3. 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 [Broken]
http://img220.imageshack.us/img220/7548/63185150.jpg [Broken]

Last edited by a moderator: May 6, 2017
2. Jun 7, 2012

### vela

Staff Emeritus
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 $dx\,dy = J'\,du\,dv$.

3. Jun 7, 2012

### algebrat

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: Jun 7, 2012
4. Jun 7, 2012

### SammyS

Staff Emeritus
vela has a history of giving correct responses!

5. Jun 8, 2012

### NewtonianAlch

Thanks for the responses, yea I sort of see it now. I guess a bit more practice with it is required.