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Limits of Integration of a Triangle

  1. Feb 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose you have a Triangle with the vertices, (0,0) (1,1) and (0,1). Integrating along that path.

    I have some differential function dZ where Z = Z(x,y)

    2. Relevant equations

    3. The attempt at a solution

    If I need to integrate, then I need to find the limits of integration. Am I correct with the following.

    0 < x < y (x is between x and y)
    0 < y < 1 (y is between 0 and 1).

    I have attached my awful MS Paint drawing to demonstrate the triangle.
    http://[url=https://ibb.co/dgJzMF][PLAIN]https://image.ibb.co/bP8Naa/trainglethemrmo.png [Broken][/url][/PLAIN]
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Feb 9, 2017 #2


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    This is a double integral, where the ends of the integrals need to change so that the integrals only cover the triangle. Start with a double integral with unknown ends and fill in the blanks:
    ??(∫??Z(x,y) dx)dy

    I don't think that I should say more on a homework problem. Give it a try.
  4. Feb 10, 2017 #3
    How to I integrate if the integral is a sum such as dz = y dx + (x+2y)dy ?

    And are my limits of integration correct?
  5. Feb 10, 2017 #4


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    What are
    I may have misunderstood. Are you integrating Z or dZ?
    What are your limits? I don't see them.
  6. Feb 10, 2017 #5
    You didn't misunderstood, I mistyped, my apologies. It is dZ.

    My limits I assumed are
    0<x< y
  7. Feb 10, 2017 #6


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    You need to put the integrand into the formula of the integrals with correct limits on the integrals and keep track of which integration has dx≡0 or dy≡0.
  8. Feb 10, 2017 #7
    That makes sense. Thank you! :)
  9. Feb 10, 2017 #8


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    I'm having second thoughts. It doesn't seem right that terms with dx≡0 or dy≡0 would immediately disappear from the calculation. I would need to rethink this. You can try it with dx≡0 or dy≡0 and also with them constant and see if one approach makes more sense. Maybe someone more familiar with this can clarify.
  10. Feb 11, 2017 #9
    Is this an integral over the complex plane? Do you have ##\mathop\int\limits_{T} f(z)dz##. If so, you can break it up into three integrals ##\mathop\int\limits_{T}=\mathop\int\limits_{T_1}+\mathop\int\limits_{T_2}+\mathop\int\limits_{T_3}## and for starters, if you let ##z=x+iy## over the complex plane then for example, you would have along the real axis: ##\mathop\int\limits_{T_1}f(x+iy)(dx+idy)=\mathop\int\limits_{a}^{b} f(x+iy)dx##. Right?
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