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Integration in real world.

  1. Nov 16, 2013 #1
    What is the difference between ∫X.dY and ∫Y.dX in the physical world? I know what the difference is in pure mathematics. ∫X.dY represents the are bounded by the curve and the Y axis while ∫Y.dX represents the area bounded by the curve and the X axis. But I am unable to translate this into physical situations.

    For example when we are doing ∫F.dX, we are calculating force multiplied by displacement for very small displacements and then adding them all up. Similarly ∫X.dF should mean that we are multiplying the displacement of the particle for a very small change in F with that small change. Should they not mean the same thing and give the same result? Yet only the first is called the work done. One other example that I can think of is that of pressure-volume work. ∫p.dv is the work done. I can't even think of what ∫v.dp means. What does it mean?
  2. jcsd
  3. Nov 16, 2013 #2


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    X and Y are arbitrary variables. If you do not define them, the integrals are meaningless.

    Integrate the volume for a pressure change. If volume is a function of pressure, this is possible (I am not sure if there is a useful physical interpretation for that).
  4. Nov 16, 2013 #3


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    No. we are multiplying "the displacement of the particle", not "the displacement of the particle for a very small change in F". In other words, "the displacement from origin", not "the small change in displacement".

    If you write down a mathematical formula "at random", usually it doesn't mean anything physically.
  5. Nov 16, 2013 #4
    That makes sense. But ∫v.dp should mean something since dH=dU+pdV+Vdp. Here dU is the change in internal energy and pdV is the work done. What is Vdp?
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