Maths Notation: What Does \oint Mean?

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Homework Help Overview

The discussion revolves around the mathematical notation \oint, specifically its meaning in the context of integrals. Participants are exploring whether it denotes a line integral over a closed curve or a surface integral over a closed surface, and how it is used in different contexts, particularly in textbooks.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the specific meaning of the \oint notation and its application in various integral contexts. Some are providing examples from textbooks to illustrate their points, while others are seeking clarification on whether it can apply to non-closed curves or surfaces.

Discussion Status

The conversation is active, with participants sharing examples from textbooks and discussing the implications of the notation. There is no explicit consensus on its use in non-closed contexts, but several interpretations are being explored.

Contextual Notes

Participants reference specific examples from textbooks, noting that the context may influence the interpretation of the \oint notation. There is an ongoing discussion about whether the notation must always indicate closed paths or surfaces.

Defennder
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This is probably a dumb question but what does the circle through the integral sign mean?
\oint

I was thinking perhaps it could denote either the line integral over a closed curve, or the surface integral over a closed surface, depending on the context. But it seems the textbook I use employ that notation even when the line/surface integral is not over a closed curve/surface.
 
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Yes, that symbol means the integral over a closed curve. I've never seen it used for the surface integral over a closed surface but I guess it could be. I would be very surprised if a textbook used that to indicate an integral over a curve that was NOT closed. Could you give an example of that?
 
Ok this is from the 7th Edn of Engineering Electromagnetics by William Hayt and John Buck, pg 212 in the chapter on time-invariant magnetic fields:

It follows that only the integral form of the Biot-Savart law can be verified experimentally,
\textbf{H} = \oint \frac{Id\textbf{L} \times \textbf{a}_R}{4\pi R^2}

I suppose that it might be explained that the current I would be flowing in a closed circuit and hence a closed path. But what is the interpretation of \oint as applied to surface integrals? Must it be a closed surface?
 
I've seen that in Gauss's Law:

\oint E\! dA = Q / \epsilon_o

And, yes, to my understanding it must be a closed surface.
 

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