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Homework Help: I need help with double integrals

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data
    calculate the integral of f(x,y) = xy over the area that is defined from the equations below

    x^2+y^2 <= 1, x>=0, y>=1/2


    3. The attempt at a solution

    first i tried to figure out the graph

    [PLAIN]http://img571.imageshack.us/img571/9039/33928873.jpg [Broken]

    so i want to find the value from the black surface and z = xy right?

    so i say that it is equals to

    [URL]http://latex.codecogs.com/gif.latex?\int_{1/2}^{sqrt(1-x^2)}\int_{0}^{1}(xy)dxdy[/URL]

    what am i doing wrong?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 8, 2010 #2

    LCKurtz

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    Several things. For one thing your outer integral must have constant limits. If you wish to integrate in the x direction first, x must go from x = 0 to x on the circle, which will be a function of y. Then the y limits will be constants and you can tell from your picture, y doesn't go from 0 to 1. What are the correct y limits?
     
    Last edited by a moderator: May 4, 2017
  4. May 8, 2010 #3
    from 0 to 1 is for x variable

    i have dx first then dy

    do you mean something else?

    thanks in advance

    Edit: yea you're right

    but again if I put the second integral outside the first one inside and change dxdy to dydx i get the wrong answers
     
    Last edited: May 8, 2010
  5. May 8, 2010 #4
    i find 1/16 which is 0.0625 but the result is 9/128 which is 0,07

    how i find these results:

    first i calculate the integral of xy from 1/2 to sqrt(1-x^2) dy

    i find (3*x)/8-x^3/2

    then i calculate the integral of (3*x)/8-x^3/2 from 0 to 1 dx and i find 1/16
     
  6. May 8, 2010 #5

    LCKurtz

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    It would be easier to follow what you are doing if you would first write out the double integral, with limits, that you are doing. If you integrate x second it does not go clear to 1 as you can see from your picture. It is probably easier to set it up integrating x first.
     
  7. May 8, 2010 #6
    this is the integral

    [PLAIN]http://img687.imageshack.us/img687/4043/qwe.gif [Broken]
    [STRIKE]
    why doesnt x go from 0 to 1? I want the black part of the shape only, where x goes from 0 to 1[/STRIKE]

    edit: oh...

    you're right, so there is the problem... but how can I find x? hm..

    Edit2:

    ok found x,

    y = 1/2
    x^2+y^2 = 1

    so x = +- sqrt(3)/2

    let me check now

    Edit3:

    YEA!! It works now :)

    the final integral

    [PLAIN]http://img202.imageshack.us/img202/9528/asdfs.gif [Broken]

    Thanks a lot for your help LCKurtz :)
     
    Last edited by a moderator: May 4, 2017
  8. May 8, 2010 #7

    LCKurtz

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    You're welcome. Don 't forget to change that upper limit on your first integral when you write it up; it isn't 1 either.
     
    Last edited by a moderator: May 4, 2017
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