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Finding positive orientation for a surface

  1. Dec 7, 2012 #1
    If I parameterized a surface by the vector function r(a,b) I would then proceed to find the normal to the surface by crossing ra and rb. But how would I determine which normal raXrb or rbXra has the positive orientation?

    Im assuming you would first need to know if the concave side of the surface is facing away or towards the origin? But from there I dont know.
     
  2. jcsd
  3. Dec 8, 2012 #2

    haruspex

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    I don't see how it could be related to second order derivatives. Consider e.g. just moving around in the XY plane. Let's say the cross product is defined with a right-hand corkscrew rule, so (1,0,0)x(0,1,0) = (0,0,1). If increasing a moves r in the +ve x-direction and increasing b moves r in the +ve y-direction then raxrb is in the +ve z direction. But if you swap the sign of ra it will be in the negative z-direction.
     
  4. Dec 9, 2012 #3

    HallsofIvy

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    The distinction between "positive" and "negative" orientation for a surface is purely arbitrary. A surface can have either of two "orientations", the normal vector pointing in either of two directions out of the surface. But to designate one of those as the "postive orientation" and the other as "negative orientation" requires some other information.

    Could you give us the exact wording of an example. Usually such "surface orientation" probles will say "oriented by postive z component" or in the case of a closed surface oriented by "outward" or "inward" normals.
     
  5. Dec 9, 2012 #4

    I parameterized the surface
    x = x
    y = x^2+ z^2
    z=z

    Wrong picture look at the picture in post # 5
    Now to find the orientation towards the origin do I do rxXrz or rzXrx?
     

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    Last edited: Dec 9, 2012
  6. Dec 9, 2012 #5
    Correct problem
     

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