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Can't solve double integral

  1. Jun 25, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫∫D√(9x2+4y2) dx dy


    D is the region: x2/4+y2/9=1

    My understanding is that i have to integrate the function of a density to calculate the mass of plate which is ellipse. Problem is i cant and shouldnt be able to integrate this integral at my level, so am i missing some way of simplification?
     
    Last edited: Jun 25, 2014
  2. jcsd
  3. Jun 25, 2014 #2

    Orodruin

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    Have you considered a change of variables (or two)?
     
  4. Jun 25, 2014 #3
    to polar cordinates?
     
  5. Jun 25, 2014 #4

    Orodruin

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    Try getting the coordinates to a form where making the change to polar coordinates make more sense first. Hint: An ellipse does not give the simplest boundary condition in polar coordinates.
     
  6. Jun 25, 2014 #5

    Zondrina

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    Why not let ##x = 2u## and ##y = 3v##.

    Then switch to polar.
     
  7. Jun 25, 2014 #6
    i do not understand how this will help.

    can please someone solve the problem for me im desperate
     
  8. Jun 25, 2014 #7

    Orodruin

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    Why don't you try doing it? It will be much more instructive than if we solve it for you.
     
  9. Jun 25, 2014 #8

    Zondrina

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    There's a theorem you should be familiar with, it should be along the lines of:

    ##\int_R \int f(x,y) dxdy = \int_{R'} \int f(u,v) |J| dudv##.

    Where ##|J|## is the Jacobian matrix of the transformation, ##R## and ##R'## are the regions. You have an integral, which looks like the left side of the above theorem. The theorem tells you that you can change the region and co-ordinates of a double integral, and get the same answer.

    So making the substitution I noted in an earlier post, what happens to the region ##D##? It gets mapped to some region ##D'## through an invertible map. What is the result of plugging the substitution into the integral on the right?
     
  10. Jun 25, 2014 #9
    ∫∫R'√(36u^2+12v^2) J du dv

    ?
     
  11. Jun 25, 2014 #10

    Zondrina

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    No, the integral becomes ##36 \int \int_{D'} \sqrt{u^2 + v^2} |J| dudv##. Where of course ##dx = 2du## and ##dy = 3dv##.

    The region ##D → D' := u^2 + v^2 = 1##. Notice this is the equation of a circle with radius 1.

    The Jacobian is the matrix of partial derivatives of the transformation. Can you compute it?

    For reference: http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
     
  12. Jun 25, 2014 #11

    Orodruin

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    What is ##4\times 3^2## and what is the expression for ##J##?

    If you are not familiar with jacobians, try the following instead: what is ##du## if ##u = x/2##? (It is ok to use this type of reasoning as long as you are making parameter transformations of the type considered here, i.e., u = f(x) and v = g(y))
     
  13. Jun 25, 2014 #12

    Orodruin

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    ##\sqrt{36} = 6## ...
     
  14. Jun 25, 2014 #13
    ok, i understand a little now.

    J=6 ?
     
  15. Jun 25, 2014 #14
    is the final solution 36pi?
     
  16. Jun 25, 2014 #15

    Orodruin

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    Yes, J=6 (or rather |J|=6) in the first variable change.

    What is the area element ##du\, dv## expressed in polar coordinates?
     
  17. Jun 25, 2014 #16
    24pi
     
  18. Jun 25, 2014 #17

    Zondrina

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    Don't forget ##dx## and ##dy##.
     
  19. Jun 25, 2014 #18

    Orodruin

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    I would have agreed if you did not still have the jacobian in your expression.
     
  20. Jun 25, 2014 #19

    Orodruin

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    What are your integration limits in polar coordinates? (The answer is the same as I get by doing it in my head, which I do not trust at 1am, but I just want to make sure you've got it right)
     
  21. Jun 25, 2014 #20
    r=(0,1)
    fi=(0,2pi)
     
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