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Homework Help: Finding Moments

  1. Jul 3, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm not sure how to go about finding the following moments:

    M_{x}= \int \int\ [/tex] y dx dy
    M_{y}= \int \int\ [/tex] x dx dy

    Where the region is bounded by the ellipse:
    [tex]\frac{(x-2)^2}{16}} + \frac{(y-4)^2}{36}}[/tex] = 1

    2. Relevant equations
    Listed above...

    3. The attempt at a solution

    I drew the ellipse and found the bounds to be
    -2 [tex]\leq y[/tex] [tex]\leq10[/tex]
    -2 [tex]\leq x[/tex] [tex]\leq 6[/tex]

    Then I tried integrating with those bounds, but I can't seem to get the right answers. Any help/hints would be greatly appreciated.
  2. jcsd
  3. Jul 3, 2008 #2


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    Those boundaries describe a rectangle, not an ellipse. Do you see why?
  4. Jul 3, 2008 #3
    Ooohh yes, you're right.

    So does this mean that I have to find the bounds of my integral as functions?
    Would I have to find the equation of the ellipse in terms of y, and make that my bounds for the y-integral. and then keep the x integral as numbers?

    like D = {(x,y)| [tex]a \leq x \leq b, g_{1}\leq y \leq g_{2} [/tex]}
  5. Jul 3, 2008 #4


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    Yes that's what you would do. This problem may be easier in polar coordinates.
  6. Jul 5, 2008 #5
    Actually I just found out that I might have to use polar coordinates.
    But since the region is not a circle, but an ellipse, how would I be able to write my x and y in polar coordinate form?
    Normally we have x=rcos(theta), and y=rsin(theta) for a circle.

    would it be...
    x=4rcos(theta) + 2
    y=6rsin(theta) + 4
    Last edited: Jul 5, 2008
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