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Supposedly simple double integral

  1. Nov 3, 2007 #1
    double integral of xy dA
    in the triangular region of (0,0), (3,0), (0,1).
    my problem that I am having is finding the limits I am suposed to find dx or dy in. I figure I should use 0 to 3 for dx, but then i do dy from 0 to what? Help appreciated.
     
  2. jcsd
  3. Nov 3, 2007 #2

    Avodyne

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    Try drawing a picture of the region. Then, for a given value of x, what values of y lie within the region? This gives you the limits of integration for y, given x. (Of course, you must then do the y integral before you do the x integral.)
     
  4. Nov 3, 2007 #3
    so then the parameters for y would be: 0 to x/3?
     
  5. Nov 4, 2007 #4

    HallsofIvy

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    Yes, because the upper boundary is the line y= x/3.

    It is a very good exercise to "swap" the limits of integration. Suppose you wanted to integrate with respect to x first and then y? Clearly to cover the entire triangle, you must take y going from 0 to 1. For each y, then, x must go from the left boundary, x= 0, to the "right" boundary which is still that line y= x/3. That is, x must go from x= 0 to x= what? Do the integral of xy both ways and see if you get the same thing.
     
  6. Nov 4, 2007 #5
    still not getting the right answer.

    I used the parameters dy= 0 to 1 and dx= 0 to -3y+3 and got 2.375.
    the answer was wrong.
    I did it the other way with dy=0 to x/3+1 and dx= 0 to 3 and got another wrong answer.
    what am I doing wrong?
     
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