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Double integral with a circle connecting the two

  1. May 8, 2014 #1
    I'm trying to figure out what this one symbol was I saw. I also have a guess that I would like to see if is correct. I saw a double integral with a circle connecting the two. What does this mean? Here is my guess. Is it used when dealing with Stoke's Theorem? Since ∫F°dS =∫∫ curl(F)°dS (Both F and S are vectors, just don't know how to make the arrow) can you write the integrals in ∫∫curl(F)°dS with a circle connecting the two if the first integral ∫F°dS is simple, closed, has continuous first order partial derivatives and is positively oriented? Is that what I saw? Just someone connecting the double integral in Stoke's Theorem with a circle to show the curve is positively oriented and meets all the other required criteria? Thanks.
     
  2. jcsd
  3. May 8, 2014 #2

    adjacent

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    attachment.php?attachmentid=35830&stc=1&d=1306196087.png
    Are you talking about this?
     
  4. May 8, 2014 #3
    Yes that's the one! Was I correct at all?
     
  5. May 8, 2014 #4

    jtbell

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    I think that symbol indicates a surface integral over a closed surface. Similarly, a single integral sign with a circle over it indicates a line integral along a closed path. It's common to see the latter used to represent both kinds of integrals, however.
     
  6. May 8, 2014 #5
    Okay got ya. My only question now is that the only closed surfaces I have dealt with are when we use the Divergence Theorem to work with surfaces like a sphere. The Divergence Theorem is a triple integral though. Is it correct to use the symbol in question on the double integral in Stokes Theorem? I'm hesitant to assume that because I thought Stoke's Theorem simply applied to closed curves that don't lye on the xy-plane, not closed surfaces. Some help would be great. Thanks.
     
  7. May 8, 2014 #6

    jtbell

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    You're right, it wouldn't be appropriate to use it for the surface integral in Stokes's Theorem, if it really does mean a closed surface. Maybe it's just a generic surface integral, then. Or a typo.

    I have a vague memory of seeing that symbol, but it's been a long time. How old is the book that you're looking at?

    Aha, I've seen it more recently... it's available in my MathType software, along with a similar symbol for a triple integral.

    I've also found a Wikipedia page:

    http://en.wikipedia.org/wiki/Integral_symbol

    which lists it simply as "closed surface integral", and the triple-integral version as "closed volume integral." But then that raises another question: what's the difference between a "closed volume integral" and an "open volume integral"? :confused:
     
  8. May 8, 2014 #7

    pasmith

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    There is none in [itex]\mathbb{R}^3[/itex], but there will be in [itex]\mathbb{R}^4[/itex] or higher.
     
  9. May 8, 2014 #8
    Alright, starting to narrow it down now. Just to make sure, I can use the closed surface integral (the double integral with the hoop) whenever i'm working with closed, positively oriented surfaces, and I can use the triple integral with the hoop when doing something like Stoke's Theorem. Is this correct? Also, I can only use these symbols when dealing with vector calculus correct?

    Pasmith, are you saying that there are no closed volume integrals in R^3? What about something like a sphere? Or did you mean it the other way around?
    Thanks.
     
  10. May 8, 2014 #9

    pasmith

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    A closed surface is a 2-dimensional compact manifold without boundary. It turns out that the boundary of a compact 3-dimensional manifold with boundary is a closed surface.

    By analogy, a closed volume would be a compact 3-dimensional manifold without boundary, and would be the boundary of a compact 4-dimensional manifold with boundary.

    It turns out that if a volume in [itex]\mathbb{R}^3[/itex] is compact then it has a boundary. Thus the concept isn't useful in [itex]\mathbb{R}^3[/itex].

    However closed volumes do exist in [itex]\mathbb{R}^4[/itex], for example the subset [itex]\{ (x_1,x_2,x_3,x_4) : \sum_{i=1}^4 x_i^2 = 1\}[/itex].
     
  11. May 9, 2014 #10
    Do you think you could expand on this a bit? In vector calculus, i'm at the stage where I mostly know how to solve problems and don't have a strong grasp on deep understanding. You know how math is, you learn how to solve problems first because of the pressure of tests and then as time progresses you understand the true meaning of what you are doing. I love vector calculus so much though that i'm trying to do as much of both at the same time as I can.
     
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