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Thread: Symmetry in Multiple Integration

  1. #1
    Senior Member AllanCuz's Avatar
    Apr 2010

    Symmetry in Multiple Integration

    I can do multiple integration easily enough, but I am absolutely horrible at detecting symmetry. In fact, even if I know something is symmetrical I can't always discern what this means. In particular, I'm talking about double/triple integrals where some part of the equation simplifies to 0 due to symmetry.

    For example, if we have

    $\displaystyle \iiint (xy + z^2)dV$

    Over the domain defined by

    $\displaystyle 0 \le z \le 1 - |x| - |y| $

    XY will cancel via symmetry. I understand if we have just X, or just Y and they are bounded by a domain that goes from negative to posative bounds that are equal, then yeah...they will be zero.

    But how come XY = 0?

    Similarly, if we have

    $\displaystyle \iiint (3+2xy)dV$

    over the hemisphere D given by

    $\displaystyle x^2 + y^2 + z^2 \le 4$

    $\displaystyle z\ge0$

    This simplifies to

    $\displaystyle \iiint (3)dV$

    I simply do not see how. I know we are symmetric about the X and Y plane, sure, but what does this actually mean?

    Last edited by AllanCuz; Apr 12th 2010 at 08:39 PM.
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  2. #2
    Apr 2010
    Just do it without. Symmetry for Calc III problems are harder to detect than regular Calc I or II integrals. It makes things so much easier.

    To prove symmetry you need to make sure that the domain and the integrand is symmetric. If it is not, then you are screwed and just have to do it the long ways anyways.

    So to save you time, especially on timed tests and quizzes don't worry about it and just do the problem.
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