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Orientation of Surface and edge of surface

  1. Jan 3, 2012 #1
    Consider a function Q, from some set D in R2 to some set in R3. Suppose Q parametrizes a surface S.

    Orientation of the edge of the surface must be compatible with orientation of the surface. I read that if the parametrization of the Boundary of the domain D give it anticlockwise direction, then the parametrization of the surface S will always induce the proper orientation on the edge of S. Why is this so?

    thanks
     
  2. jcsd
  3. Jan 3, 2012 #2

    chiro

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    Hey Castilla.

    It might help if you visualize the right hand rule with respect to the surface. Think about how that relates to the other properties of orientation.
     
  4. Jan 4, 2012 #3
    Hello, Chiro. Up to my limited knowledge, a coordinate system follos the RHR when:
    - the thumb points in direction of positive y,
    - second finger points in direction of positive z and
    - third finger points in direction of positive x.

    But I can't see how this would help me to see that, if my parametrization of the boundary of the flat domain D gives that boundary a counterclockwise direction, then the parametrized surface and its edge have compatible orientations. Please give me more hints. Thank you.
     
  5. Jan 4, 2012 #4

    HallsofIvy

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    The association of the orientation of a boundary of a surface to the orientation of the surface itself is simply a convention. The standard is to use the "right hand rule" as chiro says. With the thumb of your right hand pointing in the direction of a normal vector, your fingers curl in the direction of the orientation of the boundary.

    But we could as easily have chose a "left hand orientation" and, as long as we are consistent, we will get the same results. As I said, the choice of "right hand orientation" is just a convention.
     
  6. Jan 4, 2012 #5
    I understand that this conventions about orientation of surface and edge have only one reason, that of treating equivalently either side of the surface.

    But I found this in a webpage:

    It turns out, that if we parametrize a surface S starting from a domain D in R2 whose boundary
    curve C is traveled in the anti-clockwise sense, then the parametrization will automatically
    induce the proper orientation on the edge of S.

    that was the reason of my question.
     
  7. Jan 4, 2012 #6

    LCKurtz

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    Sometimes Calculus texts state problems in the form where the surface is given to be oriented upwards or downwards, as in finding the flux through ##x^2+y^2=4## in the xy plane, oriented downwards. Then you don't have choice of clockwise or counterclockwise. You must choose the parameterization of the curve to be compatible with the given orientation. In this case it would be clockwise as viewed from above.
     
  8. Jan 5, 2012 #7
    Hallsoftivy, I am afraid I am getting a bit confused.

    Classical Stokes theorem states an equation with a surface integral in one side and a line integral in the other side. Line integrals depend on the orientation of the curve. Let's say that function α parametrizes that curve in one direction and function β parametrizes it in the other direction. Obviously it can't be only a matter of convention to decide which of both line integrals will go in the Stokes statement: the surface integral can't be equal to both line integrals!
     
  9. Jan 9, 2012 #8
    Let's suppose we have a function r that parametrizes a surface S in R3, taking as domain a flat region R in R2.

    I see in many books that in the statement of Stokes Theorem they say that the parametrizacion of the boundary of R must give counter-clockwise direction to said boundary. But I suppose that the function r can change that direction or can preserve it, and this would also apply if that boundary had clockwise direction.

    So way they deem necesary for the Theorem that the parametrizacion of the boundary of R must be counter-clockwise??

    Thanks
     
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