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Double Integral volumes: Triangular base

  1. Jan 3, 2012 #1

    ∫∫R 5x2 + 2y2

    where R is triangle (1,1) (2,0) (2,2)

    I see the lines bounding the triangle are y=x y=2-x and x=2, and have tried many attempts at setting up the correct limits.

    Would it be correct to split this into 2 triangles, or are the limits y=xy=2-x for y and 21 for x, correct
  2. jcsd
  3. Jan 3, 2012 #2
  4. Jan 3, 2012 #3
    Thanks for the help.... but

    when I then integrate WRT y I get

    2∫1 [5x2y+(2/3) y3] which I have to substitute in my limits of y=x and y=2-x

    but then this leaves me with a (2/3) (2-x)3 term, which does not simplify well, and does not lead me to the correct answer.

    Any ideas where I have gone wrong, thanks!
  5. Jan 3, 2012 #4
    Well I would just leave that term in the form (2/3) (2-x)^3 and then integrate it with respect to x without factoring it out, that term should just become (-1/6) (2-x)^4, by the power rule. Beyond that I can't see where you went wrong without seeing your work.
  6. Jan 3, 2012 #5
    Right, here goes....

    Integrating WRT y

    2∫1 [5x2y+(2/3)y3 ]

    then entering limits

    2∫1 [5x3+(2/3)x3] - [5x2(2-x) + (2/3)(2-x)3]

    simplifying to....

    2∫1 (32/3)x3 - 10x2 - (2/3)(2-x)3

    applying the second integral WRT x

    [ (32/12) x4 - (10/3) x3 + (1/6)(2-x)4 ]

    then entering limits of 2 and 1 leaves me with 14, where I am meant to find 33/2.

    Again thanks in advance! I can't help thinking my original set up of the limits in the problem is wrong
  7. Jan 3, 2012 #6
    hmm, I keep getting 75/2, I'm quite confident the limits of integration are correct, I'm not sure what is going wrong..
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