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Diffential elements and analysis dimensional

  1. Mar 30, 2014 #1
    The definition for volume element is simples, is ##dV=dxdydz##, ok. But, if you integrate this you'll have problems, because ##\int dV = \int dxdydx## no make sense in the right side of equation and, on the other hand, ##\iiint dV =\iiint dxdydx## no make sense in the left side of equation... so, this problem is eliminated if you define the volume element like ##d^3V = dxdydz##, now the tiple integral make sense: [tex]\iiint d^3V = \iiint dxdydz[/tex] However, to think if a quantity physical, in infinitesimal size, have simple, double, triple, ..., differential is non-intuitive, is a concept very analytical. But, ignore this information is metematically wrong.

    So, which is correct form for deal with this?
     
  2. jcsd
  3. Mar 30, 2014 #2

    Mark44

    Staff: Mentor

    Not really, if you work with definite integrals. As a triple integral, the one on the left would be ##\int_R dV##, where R is the three-dimensional region over which integration takes place. A triple integral can be rewritten as three iterated integrals, such as you show on the right, with suitable limits of integration.

    I'm not sure that your integral with dV makes any sense as an indefinite integral. What would you have as the antiderivative?
     
    Last edited: Apr 1, 2014
  4. Mar 30, 2014 #3
    I omitted the limits...

    BTW, which is the correct definition for volume element, d³V or dV ?
     
  5. Mar 31, 2014 #4

    Mark44

    Staff: Mentor

    It's dV, a differential volume element.
     
  6. Apr 1, 2014 #5
    Are you saing that [tex]\int\limits_{V} = \iiint\limits_{x\;y\;z}[/tex] ?
     
    Last edited by a moderator: Apr 1, 2014
  7. Apr 1, 2014 #6

    Mark44

    Staff: Mentor

    Yes, but the three iterated integrals on the right side don't have to be in that particular order. They can appear in any of six ways.

    What I said was that a triple integral (##\int_R f(x, y, z) dV##) can be written as three iterated integrals - one possible order is ##\int \int \int f(x, y, z) dx~dy~dz##.
     
  8. Apr 2, 2014 #7
    If this notation proceeds: [tex]\iiint\limits_{x\;y\;z}f(x,y,z)dxdydz[/tex] (attention for the argument of the function f)

    So this too proceeds: [tex]\int\limits_{V}f(V)dV[/tex] What is a function of a tridimensional argument?
     
  9. Apr 2, 2014 #8

    Mark44

    Staff: Mentor

    Based on the order of the differentials, it would be written as
    $$\iiint\limits_{z\;y\;x}f(x,y,z)dxdydz $$
    The innermost integration is with respect to x, so the inner limits of integration are x values. Next, the integration is with respect to y. Finally, the outer integral is with respect to z, so the limits of integration are z values.
    No, it doesn't. We started with a function of three variables. You can't just switch it be be a function of one variable.

    For example, if f(x, y, z) = 2xy + e3z, how does f(V) make any sense at all?

    See just above for an example of a function with three arguments.
     
    Last edited: Apr 2, 2014
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