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Finding the appropriate transformation to apply to an integral

  • Thread starter Pr0grammer
  • Start date
  • #1

Homework Statement


Evaluate [tex]\int_0^1 \! \int_0^{1-x} \! \sqrt{x+y} \left(y-2x\right)^{2} \, \, \mathrm{d}y \, \mathrm{d}x.[/tex] by applying the appropriate transformation.


Homework Equations


N/A


The Attempt at a Solution


So far, the best I can come up with is u=1-x, v=x+y, which gives me [tex]\int_0^1 \! \int_0^u \! \sqrt{v} \left(v-3-3u\right)^{2} \, \, \mathrm{d}v \, \mathrm{d}u.[/tex].

I know how to evaluate it after applying the transformation. As far as I can tell, one of the transformations should be 1-x to deal with the limit, and the other should be x+y to deal with the square root. I was wondering if there was a better way to do the transformation, though, since it still seems like it might be somewhat painful to integrate.
 
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Answers and Replies

  • #2
22,097
3,277
That was actually a very good transformation. The integral you end up with is not that hard, is it?
 
  • #3
Wolfram suggests that the correct answer should be 2/9: http://www.wolframalpha.com/input/?i=int+(sqrt(x%2By)(y-2x)^2)+dy+dx,+x%3D0+to+1,+y%3D0+to+1-x+

However, after evaluating my transformation, I got 1892/315, which Wolfram agreed with: http://www.wolframalpha.com/input/?i=int+(sqrt(v)(v-3u-3)^2)+dv+du,+v%3D0+to+u,+u%3D0+to+1

Any idea where the mistake might be? (Also, I got -1 for the Jacobian from x=1-u and y=u+v-1. Is that right?)
 

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