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When I did further maths at college, we spent a couple of hours on a particular kind of integration, where the function was integrated with respect to the length of the path along the function, typically starting at the point x = 0, y = f(0), and typically calling the path length variable s. I remember that this was curious for all sorts of reasons, but not any specific reason.

It may have been called implicit integration, but googling for that phrase seems to suggest I am remembering it wrong.

I get puzzled looks every time I describe this to people who have studied maths for years for some reason. Does this sound familiar and if so, what's the common name for this?

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    How about ["line integral"](http://en.wikipedia.org/wiki/Line_integral) ?2012-02-14
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    Also called the Contour Integral.2012-02-14
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    @TheChaz In that course, the function to be integrated was not a field; more like f(x) = sin(x). Which brings the question of what exactly was integrated; I believe it was the Y of the point.2012-02-14
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    How about "finding arc length with integration"?2012-02-14
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    @TheChaz Not that. On further thought, I don't think it was the Y that got integrated; that doesn't make much sense. Perhaps the tangent angle relative to that at the starting point?...2012-02-14
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    When you say "the length of the path along the function", do you mean along the _graph_ of the function?2012-02-14
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    @MichaelHardy I do, yes. I'm starting to doubt that the function was expressed in cartesian coordinates, but I don't see how polar would change anything... Perhaps the graph was described by a function of the length along the graph, rather than y = f(x), does this ring a bell? Sorry, not my best question :)2012-02-14

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I've found it! It was called an intrinsic equation, and presumably we integrated it to obtain a cartesian representation.

Perhaps there was no special term for the integration; maybe it was something like "Integrating an intrinsic curve".

There are obviously a few things I got wrong in the question, which made answering it a bit of a guesswork.

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    Great then. Can you accept your own answer? (If your doubts have been cleared, obviously)2012-02-23