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

$$\oint\:dv=0$$

where $v$ is Volume.

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

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    I had never heard of cyclic integrals either, so I used google and came up with this: http://www.sciforums.com/showthread.php?23985-Cyclic-Integrals&s=0eaf4a9feab589b64ce6a27fdc2b79c12012-08-15
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    http://www.sciforums.com/archive/index.php/t-23985.html2012-08-15
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    You've seen [this](http://books.google.com/books?hl=en&id=my8hPO-JFPoC&pg=PA19)? (Personally, I think this is more physics than math.)2012-08-15
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    Are you sure that it was a cyclic integral over *volume*? Normally this notation is only used for closed lines/surfaces.2012-08-15
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    @celtschk, thermodynamics defines integration differently from math, and it could theoretically be both volume, lines etc..2012-08-15
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    @picakhu: OK, I overlooked the word "thermodynamics" (maybe because in my thermodynamics course IIRC we've never calculated integrals).2012-08-15
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    Seems likely that the integral refers to the change in volume of some gas (or something) over the course of some thermodynamical process, which is represented by a curve through the parameter space.2012-08-15

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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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    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.2012-08-15
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    Integration of an exact differential, such as $dv$, will indeed give $0$ around a closed curve.2012-08-15
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    (This follows from Stokes' theorem. http://en.wikipedia.org/wiki/Stokes_theorem)2012-08-15