Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change of variables problem

  1. Nov 18, 2005 #1
    let f be continuous on [0,1] and R be a triangular region with vertices (0,0), (1,0) and (0,1). Show:

    the double integral over the region R of f(x+y)dxdy = the integral from 0 to 1 over u f(u)du

    I recognize it is a change of variables problem but I'll be damned if I can create a set of functions u and v that yield the right jacobian. If anyone can help, its much appreciated.
  2. jcsd
  3. Nov 18, 2005 #2
    I think it's obvious that u=x+y, v=x for example for v...this leads to |det(Jacob)|=1..then byconstruction of u and v, f(x+y)=f(u) which is independent of v, hence you can integrate over v (take care of the limit of integration, i was not able at first glance to be convinced of the u limit, but for v you get integral from 0 to 1 dv, which gives 1, and hence "disappears")...which hence lead an integral of the form integral from 0 to 1 over f(u) du...
  4. Nov 19, 2005 #3
    yeah that certainly would be a nice result if that were the answer to the problem but theree is a u out front and so a jacobian of 1 will not get the right results, but thanks anyway.
  5. Nov 20, 2005 #4
    Oh, ok, then u could do something like : int{0,1}du[u f(u)]=F(1)-int{0,1}du[ F(u)] .where F(u)=int{0,u}ds[ f(s)]+C...by integration by parts,you have to adjust the integration constant c so that only the double integral remains....then use the first computation backward and get x+y as argument.........does it work like this better...?
    Last edited: Nov 20, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook