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Does the orientation you evaluate line integrals matter?

  1. Dec 6, 2013 #1
    rTf1iaC.png

    If instead of evaluating the above line integral in counter-clockwise direction, I evaluate it via the clockwise direction, would that change the answer? What if I evaluate ##C_1## and ##C_3## in the counter-clockwise direction, but I evaluate ##C_2## in the clockwise direction?
     
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  3. Dec 6, 2013 #2

    SteamKing

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    It sure will. Can you say how it will affect the answer? Think Green's Theorem.
     
  4. Dec 6, 2013 #3
    It will be the negative.

    If the direction does matter, in which direction would I evaluate the below line integral?

    cbxIR.png
     
  5. Dec 6, 2013 #4

    AlephZero

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    You would get nonsense.

    Actually, it doesn't even make sense to talk about integrating part of a closed path "counter-clockwise" or "clockwise". Where is the center of the "clock face", if you don't have a complete closed path?
     
  6. Dec 6, 2013 #5
    I just asked Dr. Martín Argerami. Here is his response:

    Is this right or wrong? It seems to me like the line integral always changes sign when I evaluate in another direction.
     
  7. Dec 6, 2013 #6

    SteamKing

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  8. Dec 6, 2013 #7

    Office_Shredder

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    ainster, it does always change sign when you evaluate in the opposite direction. For intuition think about one dimensional integrals, and
    [tex] \int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx, [/tex]

    which is an example of a line integral over a function which is reversed in direction.
     
  9. Dec 7, 2013 #8

    vanhees71

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    It's neither right nor wrong but imprecise ;-)).

    Of course, the direction of the path (or the orientation of a closed path) in a line integral matters, because the integral flips its sign when you change the direction/orientiation.

    In Stokes's Theorem (or in Green's Theorem in the two-dimensional case) the correct relative orientation of the area and the path matters. For Stokes's Theorem in [itex]\mathbb{R}^3[/itex] you can chose the orientation of the surface arbitrary, i.e., you make an aribtrary choice of the direction of your surface-normal vector field (i.e., you make it point to the one or the other side of the surface). Then for Stokes's Theorem to hold in its standard form, the closed boundary curve of the surface must be oriented in the sense of the right-hand rule, i.e., pointing with the thumb of the right hand in direction of the surface-normal vectors, your fingers curl in the direction of the boundary curve's orientation.

    For Green's Theorem in [itex]\mathbb{R}^2[/itex] the orientation of the boundary curve is such that if you walk along the curve you always have the area to your left.

    Of course, Green's Theorem can be seen as a special case of Stokes's Theorem of a vector field in [itex]\mathbb{R}^3[/itex] with only field components in [itex]x[/itex]- and [itex]y[/itex]-direction and depending only on the coordinates [itex](x,y)[/itex] (expressed in terms of Cartesian Coordinates).
     
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