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Help understanding closed line integrals

  1. Mar 2, 2013 #1
    Hi
    I'm currently studying Electromagnetism, and we keep coming across this symbol:
    [itex]\oint[/itex]
    A closed line integral, something I have never really been able to understand.

    If a normal integral works like this:

    http://imageshack.us/a/img109/3732/standardintegral.png [Broken]

    where f(x) is the "height" of the function, and dx is an infinitesimally small "width", then the area equals

    f(x1)*d(x1) + f(x2)*d(x2) + f(x3)*d(x3) + f(x4)*d(x4) ...

    dx= x2-x1 and dx is infinitesimally small.

    So, can someone please explain closed line integrals in a similar way?
    In an attempt to answer this myself I created this picture, but it didn't help. Can someone please say what I did wrong?

    http://imageshack.us/a/img825/4049/loopintegral.png [Broken]
    Thank you for reading
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 2, 2013 #2
    I have just been told that y is just a function of x. and r should just be treated as a constant in each example. But the question still stands, how can you explain closed line integrals in terms of a series of multiplications and additions.
    Thanks!
     
  4. Mar 2, 2013 #3

    SteamKing

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    It's a tricky topic for a forum to teach. Have you studied vector calculus yet?
     
  5. Mar 3, 2013 #4
    Yes, I have covered multivariable functions, partial DE's, grad, curl, and div. All we have been taught about closed line integrals is that it gives the area enclosed by a function which comes back on itself (is a closed loop). We have never been taught how the maths works. Is it possible just to tell me what I did wrong in the area of a circle example?
     
    Last edited: Mar 3, 2013
  6. Mar 3, 2013 #5

    pasmith

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    There is no difference in the definitions of a line integral over a closed curve and a line integral over a non-closed curve. Indicating that the curve is closed is largely done to call attention to the fact that it is the boundary of a surface.
     
  7. Mar 4, 2013 #6
    Forget about area as the basic concept of integration. Calculating areas is one application of integration, but there are others and it is hard to usefully generalize it to understanding line integrals.

    Think if an integral like [itex] \int_a^b f(x)\, dx [/itex] as "summing up" the values of the function over that interval (integration: the act of combining or adding parts to make a unified whole). The best physical example I can think of is when f(x) represents the mass density at point x of a length of straight wire. Then the integral of f(x) represents the total mass of the wire between points a and b.

    There are two principal types of line integrals. One is an integral with respect to arclength:
    [itex] \int_\gamma f\, ds.[/itex]

    The meaning of this is the same as what I just described. If f is a function that tells you the mass density of a wire (which takes the shape of the curve gamma), then the integral of f over gamma gives you the mass.

    The other type looks like:
    [itex] \int_\gamma F \cdot d\vec{r} [/itex]
    This is the one you would come across the most in electromagnetism. F is a vector field such as the magnetic or electric field. Basically r(t) represents a curve (which might be a geometrical curve or it could actually be a trajectory of a particle depending on context). As you go around the curve, you are calculating the component of the vector field in the direction tangent to the curve. This is a physically meaningful quantity in many contexts. For example, if F represents an electric field and r(t) represents the trajectory of a charged particle, then [itex] F \cdot r'(t) [/itex] represents the rate of change of the electric potential of the particle at time t. So when you sum up all those values, you get the accumulated change in electric potential from the starting point to the ending point of the trajectory.

    The technical points of explicitly calculating these integrals given formulas are covered in a calculus textbook.

    As a previous poster noted, the fact that the curves are closed plays no special significance in the definition of the integral. However, it does have physical consequences in particular cases. In my electric potential example, if your ending point is the same as your starting point, then the accumulated change in potential would be 0 (since the electric potential is a function of position).
     
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