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Double integral over triangular region

  1. Jul 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Integrate f(u,v)= v - sqrt(u) over the triangular region cut from the first quadrant by the line u+v=64 in the uv plane.

    2. Relevant equations
    I am assuming u is the equivalent of the x axis in the xy plane and v the equivalent of y in the xy plane.
    I am taking the triangle as a Type I region.

    3. The attempt at a solution
    limits of integration:

    0≤u≤65, 0≤v≤64-u

    ∫ ∫ v-sqrt(u) dvdu

    ∫ (1/2)v2 - sqrt(u)v evaluated from v=0 to v=64-u

    ∫ 2048 - 64u + (1/2)u2 - 64*u1/2 + u3/2

    2048u - 32u2 + (1/6)u3 - (2/3)*64*u3/2+(2/5)*u5/2 evaluated from u=0 to u=64

    I get a really narley fraction, namely, 4638576/30, which is of course wrong.
     
  2. jcsd
  3. Jul 27, 2013 #2
    Everything looks good, I double checked your calculations and came to a different result. Double check the final evaluation. I came up with ~ 8,000 (the exact value is for you to figure out).
     
  4. Jul 28, 2013 #3

    haruspex

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    I get more like 32000.
     
  5. Jul 28, 2013 #4
    Yes, the exact answer is 524288/15 ~ 35000.

    This was a terrible problem not because it was difficult but because the numbers were so ugly.
     
  6. Jul 28, 2013 #5

    haruspex

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    It's not so bad if you leave as much as possible in powers of 2.
    211u - 25u2 + (1/6)u3 - (2/3)*26*u3/2+(2/5)*u5/2 where u = 26:
    217 - 217 + (1/6)218 - (2/3)*26*29+(2/5)*215 = 216{(2/3) - (1/3)+(1/5)} = 219/15
     
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