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Question about orientation and surface integrals

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm a bit confused as to how to determine which component must be positive or negative if the question gives you a surface and says the normal vector is pointing outward or inward. Some surfaces have it so that the z component is positive if n is pointing outward and others have it as x or y. How do you figure out which component it should be?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 23, 2012 #2

    sharks

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    Let's consider the simplest example. If you have a surface z=0, then if the downward component is considered outward, you can directly deduce that the unit normal vector is [itex](0,0,-1)^T[/itex]. If the upward component is outward, then [itex]\hat n[/itex] is [itex](0,0,1)^T[/itex].

    In the case of of the x-axis, if the leftward component (in the 3D Cartesian coordinate system) is considered outward, then [itex]\hat n[/itex] is [itex](1,0,0)^T[/itex]. Now, you can deduce the rest. Simply follow the positive or negative directions along the axis relative to which the outward component is parallel to.

    If you're dealing with a surface, then you need to project it onto the xy, xz or yz planes, and you can then easily see the general direction of the outward component. But in this case, you can only directly deduce the sign of [itex]\hat n[/itex] but you'll have to calculate its value using: [itex]\frac{∇ \phi }{\left | ∇ \phi \right |}[/itex], where [itex]\phi (x,y,z)[/itex] is a function of the surface.
     
    Last edited: Apr 23, 2012
  4. Apr 23, 2012 #3

    tiny-tim

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    Hi Kuma! :smile:

    If the question mentions the direction of the normal, then that's the positive direction (for calculating flux etc).

    And if the question doesn't mention it, why does it matter? :confused:

    (I don't understand the x y z part of your question … can you give an example?)
     
  5. Apr 23, 2012 #4
    Thanks for the replies.

    See his first and second example here: http://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx

    On the first example he uses the y-component and I'm kind of confused about trying to figure out what outward and inward means first off, and which direction they should point. On the second example he uses the z-component.
     
  6. Apr 23, 2012 #5

    tiny-tim

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    In Example 1, it points out the general rule …
    "we’ve been told that the surface has a positive orientation and by convention this means that all the unit normal vectors will need to point outwards"​

    "outward" means what it says! :biggrin:

    it's a closed surface, so it has an inside and an outside, and the outward direction goes outside
    In Example 2, again it tells you to use positive orientation.

    The y and z components have nothing to do with it. :confused:
     
  7. Apr 23, 2012 #6
    Wait what...? :(

    Well what does this mean then

    Also note that in order for unit normal vectors on the paraboloid to point away from the region they will all need to point generally in the negative y direction. On the other hand, unit normal vectors on the disk will need to point in the positive y direction in order to point away from the region.

    I figured that meant that the y-component should be negative. I'm asking about components cause the problems I'm working on, the solutions mention component signs. For example here is an excerpt from one of the solutions

    "Since n is pointing outward, the z–component must be positive, therefore (u, v) is orientation reversing."

    My main problem here is trying to figure out the normal vector. It messes me up because in that problem, I had to multiply it by (-1), but I didn't. I'm just having trouble figuring out orientation and what it means for the normal vector.
     
  8. Apr 23, 2012 #7

    tiny-tim

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    yes, it's pointing out that there's a paraboloid on the left, capped by a disc on the right

    so the outward direction on the paraboloid is left, but on the disc is right

    (and left is y negative, while right is y positive)
     
  9. Apr 23, 2012 #8

    HallsofIvy

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    "upward" or "downward" normals refer to normals with, respectively, positive and negative z- components. "Right" and "left" typically refer to positive and negative x-components. You should be able to see that from looking at a graph.

    To talk about "inward" and "outward" normals, you have to have a closed surface so you have inward and outward normals.
     
  10. Apr 23, 2012 #9
    Oh okay. So here's another question. If you have a sphere for example then outward vectors could be considered as the x,y, and z axes. But if you have a plane, it has to specify either upward or downward then?

    Also this brings me back to my confusion of each component of the normal when it is pointing in or out. This is one question I was working on:
    F(x,y,z) = (x,y,z)
    S is that part of the ellipsoid x^2/4 + y^2 + z^2 = 1 which lies in the first octant and n points outward.

    now the solution says this about n

    "Since n is pointing outward, the z–component must be positive, therefore (u, v) is orientation reversing."

    Why is it that the z-component has to be positive and not the x or y?
     
  11. Apr 23, 2012 #10

    sharks

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    This should help. I think that my reply has gone unnoticed :confused:

    If you project that section of the sphere onto the x-y plane, then you will end up with a plane, and the outward normal to that plane is definitely in the direction of the positive z axis.
     
  12. Apr 23, 2012 #11
    I saw your first reply :)

    But why project it to the xy plane as opposed to any other (yz, xz)? How would I know which plane to project it to?
     
  13. Apr 24, 2012 #12

    tiny-tim

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    Kuma, forget for the moment your worries about x y z …

    are you now happy with what is meant by "positive orientation" and "outward direction" for a general closed surface?​

    (for an open surface, ie a surface with a boundary, those terms are of course meaningless)
     
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