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Does ##g(x,y,z)## (the equation of the surface) need positive z or

  1. Dec 9, 2013 #1
    Does ##g(x,y,z)## (the equation of the surface) need positive z or negative z when doing a surface integral?

    CmTxw.png

    The author specifies that ##g## either has ##z## or ##-z## depending on the orientation of the surface but doesn't specify which belongs to which orientation.
     
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  3. Dec 9, 2013 #2

    Office_Shredder

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    If you use z-f(x,y) then the gradient of g will be (df/dx,df/dy,1) and the unit normal will always be pointing upwards, if you use f(x,y)-z similarly you can see that the unit normal is always pointing downwards (the z-component of the gradient will be -1). Whether you want the unit normal pointing upwards or downwards depends on the context of the problem. For example if you are applying Stoke's theorem to calculate a line integral around the boundary of a surface, then whether you want it pointing up or down depends on which direction the integral is around the boundary.
     
  4. Dec 9, 2013 #3

    HallsofIvy

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    At every point on any differentiable surface there are an infinite number of "normal" vectors, two of which are "unit normal" vectors. There exist a unit vector in each of the two directions, each of which is -1 times the other. That is why you need to specify a direction. How you do that depends upon how you are "given" the surface. "up" or "down" ("positive z" and "negative z") may make sense if there is a coordinate system associated to the surface. "Inner" and "outer" will always make sense for a sphere or a paraboloid but obviously wouldn't for a plane.
     
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