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What is the Geometrical Interpretation of line integral?

  1. Sep 30, 2012 #1
    So I'm able to calculate them no problem. But the problem is, I don't really understand what they mean. For example, W=∫Fdr

    I understand that a vector field is something that defines a vector at every point. Then if we pick a curve in the field, we are integrating along that curve. So does it just give the area under that curve? And wouldn't the area under that curve be the same regardless of if the vector field is there or not? Would appreciate someone clearing this up for me :)
     
  2. jcsd
  3. Sep 30, 2012 #2
    This is not the area under the curve, especially for the line in 3D space.
     
  4. Sep 30, 2012 #3
    Taking the line integral in a scalar field f gives you the area under a curve along the surface f, just like you said.
    It's a bit more complicated in a vector field. You can sort of imagine the line integral is the normal integral of a second equation whose value at a point is equal to the dot product of the displacement vector of the curve at that point and the vector field at that point.

    I don't think I did a very good job of explaining it, wikipedia has a really good animation on their 'line integral' page describing it.
     
  5. Sep 30, 2012 #4
    Here is a simple explanation, without full mathematical rigour.
    It should help understand the principle of a line integration (integration along a path or line).
    Once that is agreed you can proceed to vector integration which needs more explanation.

    First you should realise that the ordinary integral you have met is a line integral for a function of one variable. The line in this case is the length along the x axis that we are using.

    This is noted in the first sketch where the area under the curve is divided into small strips of width δx and height equal to the value of the function at the midpoint of the strip.

    So the area of one strip is zδx and the area under the curve is the area of all the strips added up. This sum becomes an integral as the strip width tends to zero.

    You have probably seen something like this before.

    No if we move on to fig 2 and let z now be a function of two variables, our line representing z = f(x,y) can wiggle all over (or rather above) the x-y plane.

    I have shown a particular line and erected a 'wall' under this line between the line and the xy plane.
    As before, the wall is divided into small strips.
    The area of each strip is the strip length δs times the strip height f(x,y) = zδs
    As before we add up all the strips, take the limit and obtain an integral equal to the area of the 'wall' under the curve.
    This is again the line integral.

    Please note that δs has components δx and δy, which is how we actually calculate the line integral.

    Does this help?
     

    Attached Files:

  6. Sep 30, 2012 #5
    An integral is a summation. A line integral is a summation of all the values along a path or curve with an appropriate scaling for the distance traveled.
     
  7. Oct 1, 2012 #6
    Thank you all for the help. My question now is, once you have the area under that "wall" why does the answer change when it is in a vector field. Shouldn't the area be same regardless?
    The line integral equation I'm learning is

    ∫Fdr

    where F is some vector function
     
  8. Oct 1, 2012 #7
    Not quite.

    The integral is ∫F.dr

    ie the dot product where both F and r are vectors.
     
  9. Oct 1, 2012 #8
    Right, so F is some vector field and we are integrating along some curve in the field. In other words, every point on that curve will correspond to a different F vector and r vector. And if you take the sum of all Fdr along the curve, that is the line integral?
     
  10. Oct 1, 2012 #9
    They do not necessarily have to be different.
    That's correct.
     
  11. Oct 1, 2012 #10
    Mikey and Kashishi are correct we need to move from looking at integration as an area to looking at it as a sum.

    Look again later as there is more to this for vectors, I will post later, when I have time.
     
  12. Oct 1, 2012 #11
    Ok, thank you for all the help :)
     
  13. Oct 1, 2012 #12
    "if you take the sum of all Fdr along the curve, that is the line integral?"...yes


    Nabeel: try reading here, carefully,
    and then ask questions if you need.

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

    "In qualitative terms, a line integral in vector calculus can be thought of as a measure of the total effect of a given field along a given curve."

    This means the portion of the field along the curve.....as in the dot product mention in prior posts...so the line integral along a curve normal to the field would be what?? If you can answer that, you have the idea!!!
     
    Last edited: Oct 1, 2012
  14. Oct 2, 2012 #13
    OK to proceed to part 2 vector line integrals.

    In order to do this we should review what a vector is. Or rather, consider the question
    "Where do vectors live?"

    We shall see that they do not, in general inhabit the same space as the coordinate system, but live in a space of their own.

    Figs 1 to 3 show the x-y plane.

    In Fig1 we have our line, pq or y=h.
    We can draw geometric vectors pq and pr as shown in fig 2 and take the cross product D1 x D2 = D1D2 sin (90) = area under the line pq.

    However we could put a different pair of geometric vectors at p, as in Fig3 and the cross product area would not equal the area under the curve or the line integral.

    Now all of these vectors live completely in the x-y plane (space) and are represented by lines from points p to q p to r etc in that plane.

    If we now introduce another type of vector at p such as a force F in Fig4,

    Whilst the application point of F is in the x-y space (plane) the value is not it is in the force space.

    Fig 5 illustrates this with numbers and shows that if F has x and y components of 1 and 1 it does not mean that it can be represented by geometric vectors PR and PQ in the x-y space.

    It is represented by force components in its own (force) space.

    So back to our line integral along our original Fig1 line, pq.

    In Fig 6 I have shown a distributed series of forces, such as you might find in a uniformly distributed load on a beam, and consider the line integrals along pq.

    Now a line integral is the summation of the effect of the some function along the path pq.

    That is a line∫ = ∫(some function)dx

    There are many such possible functions we could calculate, so I have chosen the dot product of our vector and the vector x that is a line drawn drawn along the x axis to the position on pq of interest.

    I have done this because this is the perpendicular distance from the y axis of any point.
    The dot product gives us the moment of any vertical force acting at this point about the y axis.

    The line integral from p to q sums all these moments and gives the total moment about the y axis.

    This has shown that line integrals may well not yield the area under the curve or indeed any area at all. However many have useful physical interpretations and, of course, these are the ones we choose to select.

    Does this clear things up at all?
     

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