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Homework Help: Change of variables

  1. Feb 3, 2006 #1
    Hi, I would like some help with the following question.

    Q. Let f be continuous on [0,1] and let R be the triangular region with vertices (0,0), (1,0) and (0,1). Show that:

    [tex]
    \int\limits_{}^{} {\int\limits_R^{} {f\left( {x + y} \right)} dA = \int\limits_0^1 {uf\left( u \right)} } du
    [/tex]


    By making the substitutions u = x+y and v = y, I got it down to:

    [tex]
    \int\limits_{}^{} {\int\limits_R^{} {f\left( {x + y} \right)} dA = } \int\limits_0^1 {\int\limits_0^u {f\left( u \right)} dvdu}
    [/tex]

    The above leads to the given result. However, I was stuck on trying to get bounds for the integrals so I'm not sure if I've justified those limits of integration properly.

    From the boundary line y = 1 - x, the substitution u = x+y gives u = 1. From the boundary line x = 0, the substitutions yield u = 0 + y = v so that v = u. The boundary line y = 0 yields v = 0. The lower limit for u is what I am lacking in. The three boundaries that I've just obtained completely describe the region R anyway so I decided to say that at the origin x,y = 0 which gives u = 0. I'm not sure whether I should've done something else to obtain the lower u limit.

    Any help would be good.
     
    Last edited: Feb 3, 2006
  2. jcsd
  3. Feb 3, 2006 #2

    benorin

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    Homework Helper

    If the original bounds are 0<=y<=1 and 0<=x<=1-y

    then for u=x+y and v=y we have

    0<=x<=1-y ==> y<=x+y<=1 ==> v<=u<=1

    and, of chourse 0<=v<=1
     
  4. Feb 3, 2006 #3

    benorin

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    A better way: (take all inequalities as inclusive)

    Let L1: x+y=1, 0<y<1 ==> L1': u=1, 0<v<1
    Let L2: y=0, 0<x<1 ==> L2': v=0, 0<u-v<1, but v=0, so 0<u-0<1 or just 0<u<1
    Let L3: x=0, 0<y<1 ==> L3': u-v=0 i.e. u=v, 0<v<1
     
  5. Feb 3, 2006 #4

    benorin

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    So the transformed region is /| = a right triangle in the uv-plane formed by cutting the unit square along u=v, lower triangle
     
  6. Feb 3, 2006 #5
    Thanks for the help Benorin. I just thought of another way to justify setting u = 0 as a lower bound for the u integral. I've already established 3boundary lines which already account for the 'shape' of the original region in the xy plane. From a quick sketch I can see that adding the line y = - x as a boundary line still gives the same region so y = - x => y + x = 0 = u. But that's kind of a fudge method, your method is the right one.
     
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