Can anyone explain what a cyclic integral is? My professor used it in his Thermodynamics lecture. One of the equations was


where $v$ is Volume.

Isn't the integral of $dv$ equal to $v$? Can anyone explain in simple terms?

  • $\begingroup$ I had never heard of cyclic integrals either, so I used google and came up with this: sciforums.com/… $\endgroup$ – Sam Jones Aug 15 '12 at 12:13
  • $\begingroup$ sciforums.com/archive/index.php/t-23985.html $\endgroup$ – picakhu Aug 15 '12 at 12:13
  • $\begingroup$ You've seen this? (Personally, I think this is more physics than math.) $\endgroup$ – J. M. is a poor mathematician Aug 15 '12 at 12:14
  • $\begingroup$ Are you sure that it was a cyclic integral over volume? Normally this notation is only used for closed lines/surfaces. $\endgroup$ – celtschk Aug 15 '12 at 12:15
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    $\begingroup$ @celtschk, thermodynamics defines integration differently from math, and it could theoretically be both volume, lines etc.. $\endgroup$ – picakhu Aug 15 '12 at 12:16

The circle indicates that the (line) integral is taken around a closed curve. The integral of the differential $dv$ (whatever $v$ is) will always just be the net change in $v$. Around a closed curve, there is 0 net change, because you end up where you started. This article seems to explain in greater detail: http://en.wikipedia.org/wiki/Line_integral

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    $\begingroup$ It is not true that integration around a closed curve will always give $0$ (otherwise that concept would be quite useless). A counterexample is Ampere's law. $\endgroup$ – celtschk Aug 15 '12 at 12:18
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    $\begingroup$ Integration of an exact differential, such as $dv$, will indeed give $0$ around a closed curve. $\endgroup$ – Sean Eberhard Aug 15 '12 at 12:29
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    $\begingroup$ (This follows from Stokes' theorem. en.wikipedia.org/wiki/Stokes_theorem) $\endgroup$ – Sean Eberhard Aug 15 '12 at 12:35

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