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Double integral (6x^2 -40y)dA

  1. Jul 17, 2006 #1
    double integral (6x^2 -40y)dA where it is a trianglewith vertices (0,3) , (1,1) and (5,3)

    may i know how to divide the region according to this triangle??
  2. jcsd
  3. Jul 17, 2006 #2


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    The fisrt step is to draw a picture. It is then easier to see what needs to be done.

    Please provide us with a complete problem statement. What you have posted is not clear.
  4. Jul 17, 2006 #3
    ∫∫ (for D) (6x^2-40y)dA
    ,D is the triangle with vertices (0,3), (1,1) and (5,3).

    i have drawn the picture
  5. Jul 17, 2006 #4
    but i don't know how to divide the region
    pls help
  6. Jul 17, 2006 #5
    I would divide it at the point (1,1) perpendicularly to the x-axis.
    Then the two regions are bounded by a constant and a straight line .
  7. Jul 17, 2006 #6


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    I wouldn't divide it. I would integrate with respect to y. As y varies from 1 to 3, the left side is the the line from (0,3) to (1,1) and the right boundary is the line from (1,1) to (5,3). What are the equations of those two lines, written as x= ay+ b?
  8. Jul 17, 2006 #7
    Even better !
  9. Jul 18, 2006 #8
    [What are the equations of those two lines, written as x= ay+ b?]

    do u mean we have to formulate another eqn ??or just integrate with respect to the axis coordinate using the eqn given??
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