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Interpreting a double integral as volume

  1. Mar 11, 2008 #1
    The function F(x,y) = 4x^2y^3 over the disk x^2 + y^2 =1 is supposed to be zero over the disk. I'm wondering how you can see it?

    I cannot see this or imagine it in 3D. Is it because the function is odd in terms of y?

    F(x,-y) = -F(x,y) ? independent of wheher x is positive or negative?

    Because functions like G(x,y) = y^3 is obviously odd and so the double integral would be zero over the same disk, but I have an extra x^2 term. How is one supposed to think this?
  2. jcsd
  3. Mar 11, 2008 #2
    The fact that it's even with respect to the x variable just means that it is symetrical about x = 0 along any line parallel to the x-axis, but it is odd with respect to the y variable so the canceling out of the area occurs along the y direction only...
  4. Mar 11, 2008 #3
    So the integral (volume) is not zero?
  5. Mar 11, 2008 #4
    It is zero. Have you calculated double integrals yet? If you integrate over any line parallel to the y axis using symmetric limits, so x stays constant, you see that the integral is zero:

    [tex] \int_{-1}^{1}4x^2 y^3 dy[/tex]
    (Note that x is treated as a constant here, since it doesn't vary along a line parallel to the y-axis)

    [tex] 4x^2\int_{-1}^{1}y^3 dy[/tex]

    [tex] 4x^2 \left[y^4/4\right]_{-1}^{1} [/tex]

    [tex] 4x^2 [1/4-1/4] = 4x^2 . 0 = 0 [/tex]

    i.e. once the function is odd with respect to one variable that is enough for the whole thing to be zero (once your limits symmetric) because once that intergral is zero, the whole double integral (or triple integral or whatever) must be zero. Does this makes sense?
  6. Mar 11, 2008 #5
    Yup. thx
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