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Line integral: how can it be > 0?

  1. Apr 2, 2014 #1
    Using my understanding of calculus, I don't understand why line integrals in 3-d space
    can give a result > 0. You are following a line and integrating under that line. The line
    has some length. But according to my understanding of calculus, it does not have a width.
    What is this arbitrary width, and where does it come from?

    Thank you
     
  2. jcsd
  3. Apr 2, 2014 #2

    jbunniii

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    Imagine a line segment ##[a,b]##. Its length is ##b-a##. If we embed this line segment in 3-d space, then its length is still ##b-a##. Of course, its area and volume are zero. The line integral is measuring length (perhaps with a nonuniform weighting applied), not area or volume.
     
  4. Apr 2, 2014 #3
    Yes, that's where I get confused. It measures length with a non-uniform weight applied. Let's say that weight is the height. We have a length and a height. But the area is length*width*height, integrate to solve the integral. But width is zero, so the integral should be zero. What am I missing?
     
  5. Apr 2, 2014 #4

    jbunniii

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    Area is a 2-dimensional measure: length*height. If you calculate length*width*height, you are measuring volume, not area.

    Think about a very thin sheet of paper suspended in 3-dimensional space. To measure its area, I multiply length*height to get, say, 100 square inches. To measure its volume, I multiply length*height*width to get zero. Both calculations are legitimate, depending on what it is you are trying to measure.
     
  6. Apr 2, 2014 #5

    pasmith

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    That would be a volume. Area is always a product of two lengths.

    A line integral is best thought of as
    [tex]\sum (\mbox{quantity per unit length})(\mbox{small length along a curve}).[/tex]
    It's true that we can find the signed area bounded by the curve [itex]y = f(x)[/itex], the line [itex]y = 0[/itex] and the lines [itex]x = a[/itex] and [itex]x = b[/itex] by
    [tex]
    \sum (\mbox{perpendicular distance of $(x,f(x))$ from $x = 0$})(\mbox{small length along $x$ axis})
    [/tex]
    to get [itex]\int_a^b f(x)\,dx[/itex], and that's what motivates the formal definition of the Riemann integral*, but from a physical point of view one has really
    [tex]
    \sum (\mbox{small area on the $(x,y)$ plane})
    [/tex]
    which is a special case of a surface integral, which can be thought of as
    [tex]
    \sum (\mbox{quantity per unit area})(\mbox{small area on a surface}).
    [/tex]

    Similarly one can have a volume integral, which may be thought of as
    [tex]
    \sum (\mbox{quantity per unit volume})(\mbox{small volume}).
    [/tex]

    *One could equally well motivate the Riemann integral as
    [tex]
    (\mbox{position}) = \sum (\mbox{instantaneous velocity})(\mbox{short time})
    [/tex]
     
  7. Apr 2, 2014 #6
    Thanks
     
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