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Functions having the same integral are equal

  1. Apr 30, 2013 #1
    Suppose that for any solid region D, it is true that
    [tex] \int\int\int_{D}f(x,y,z)dV = \int\int\int_{D}g(x,y,z)dV [/tex]

    Then is it the case that f(x,y,z) is g(x,y,z). I am not sure if it's true but I seem to need it to equate the integral and differential forms of Gauss's law.

    Any thoughts?

    BiP
     
  2. jcsd
  3. Apr 30, 2013 #2

    pwsnafu

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    Hint: consider f-g.
     
  4. Apr 30, 2013 #3

    CompuChip

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    To be more precise: it is true almost everywhere.

    If you have learned measure theory, then the words "almost everywhere" have a very precisely defined meaning. Otherwise, you can just take it very loosely, and think of examples like
    $$f(x) = x^2; g(x) = \begin{cases} x^2 & \text{if } x \neq 0 \\ -1 & \text{ if } x = 0 \end{cases}$$
     
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