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Multiple integral with substitution

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data
    This is going to be confusing to read, as I don't know how to make this look right. The first integral is from 0 to L-2d, the second from x1+d to L-d, and the third from x2 to L. (F(x)=1)

    1.) 0[tex]\int[/tex]L-2d,x1+d[tex]\int[/tex]L-d,x2+d[tex]\int[/tex]L dx3dx2dx1


    2.) 0[tex]\int[/tex]L-2d,x1+d[tex]\int[/tex]L-d (L-x2-d)dx2dx1

    3.) substituting y2 for L-x2-d

    4.) 0[tex]\int[/tex]L-2d,0[tex]\int[/tex]L-X1-2d y2dy2dx1

    2. Relevant equations



    3. The attempt at a solution
    The answer to this isn't that important, as I already have the solution. What I don't understand is why after substituting y2, the upper and lower bounds of the corresponding integral aren't the other way around, with L-x1-2d the lower bound, and 0 the upper. Is this because dy2=-dx2? Probably, but I just want to make sure I understand what's going on.
     
    Last edited: Nov 21, 2009
  2. jcsd
  3. Nov 22, 2009 #2

    HallsofIvy

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    So this is
    [tex]\int_0^{L-2d}\int_{x_1+d}^{L-d}\int_{x_2+d}^L dx1dx2dx3[/tex]?
    (You say above "the third from x2 to L" but have "x2+ d" in the integral)
    Click on the LaTex to see the code.


    Okay, good.

    If y2= L- x2- d, then when x2= L-d, the upper limit, y2= L-(L-d)-d= L-L+d-d= 0. When x2= x1+ d, the lower limit, y2= L- (x1+d)-d= L-x1-d-d= L- x1- 2d.

    However, just as you say, dy2= -dx2 and :
    [tex]\int_a^b f(x)dx= -\int_b^a f(x)dx[/tex]

    Taking that "-" out reverses the limits of integration.
     
  4. Nov 22, 2009 #3

    tiny-tim

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    eugh! :yuck:

    Either use the X2 and X2 tags just above the Reply box …

    0L-2d

    or type \int_{}^{} in LaTeX …

    [tex]\int_{0}^{L-2d}[/tex] :smile:

    Hi mlarson9000! :smile:
    Yes, it's because dy2 = -dx2

    but I personally think that's badly and confusingly written :frown:, and the author should have kept the limits the same way up, and put a minus in front of the integral.
     
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