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Orienting boundary with surface

Determining the proper orientation of the boundary given the orientation of the surface. Created by Sal Khan.

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  • starky ultimate style avatar for user Paul van Sommeren
    Isn't it easier to use a right-hand rule? Curling four fingers across the direction of the curve with your thumb pointing to the positive normal vector direction? Or are there cases where this does not apply?
    (20 votes)
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    • blobby green style avatar for user JV Rodrigues
      You can do that for every contour. The right-hand rule its just a more intuitive way to know how a cross product will look like. Positive it your thumb its on the positive direction of z or x or y, depending on which axis you're working and negative if it points to the negative direction.Some call it the Screwdriver rule, if the screwdriver is turning a screw and it would take it out of the surface or point or whatever the resultant vector is positive and vice-versa for the negative situation.

      Did i made it more clear to you?
      (3 votes)
  • blobby green style avatar for user anupbwalvekar
    Hello,
    I am struggling with following question for so many days now. It would be great if you could help.
    If I have a Normal at a point on a surface on the surface boundary and a tangent at the same point on the curve, how do I check whether the curve is in the positive orientation with surface? In short, how can I verify that the surface is really on the left or not from the curve tangent and surface normal information?
    In case of planar surfaces, it is quite easier to solve because surface normal is constant all over the surface. On a non planar surface it is more difficult because the normal changes. e.g. in case of cylinder which is cut by two circular curves (one at z=0 and one at z = h) the curve orientations are required to be opposite of each other.

    Thank you very much in advance.

    Regards,
    Anup
    (4 votes)
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    • blobby green style avatar for user Cody  Hugie
      If the concern is making sure the curve is positively orientated I believe you could find the "curl" with a line integral. If the line integral turned out positive the surface path would be counterclockwise which is positive by convention and tell you the surface was on your left. If the value was negative then the path of the line integral is clockwise making the surface on your right.
      (2 votes)
  • duskpin seedling style avatar for user Martin Velasquez
    So, it's just the right hand rule? Using the thumb as N and wrapping fingers is where the circulation is going? N positive = counterclockwise. N negative = clockwise?
    (3 votes)
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    • male robot hal style avatar for user matthias.estner
      For the simple case of a "normal" surface it's just the right hand rule right.
      Image for example a cylinder with open end on both ends.

      So now you can't you the right hand rule. The normal vector of the sides of the cylinder is pointing radial outwards. On you have to orientate the TWO boundary curves (because in this case you have 2 boundary Curves) according to the rule Sal explained. And the two curved will have opposite orientation. Hope this makes sense to you.
      (2 votes)
  • winston baby style avatar for user Andrew
    This should be in the Tips & Thanks section, but since people rarely view that section I will leave this in the Questions.

    No need to watch this video
    Just watch the video after this :
    https://www.khanacademy.org/math/multivariable-calculus/greens-theorem-and-stokes-theorem/stokes-theorem/v/orientation-and-stokes

    It's basically the identical video but it has an added insight at the end (using the "corkscrew" method to visualize direction). Personally, I prefer the right hand rule but any of these methods work
    (2 votes)
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  • leafers sapling style avatar for user Jack
    When he says that if you orient your head in the direction of the normal vector and you were to walk along the path, the surface would be on your left (-), does that imply that you are outside of the surface?
    (1 vote)
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  • blobby green style avatar for user gurungdebendra82
    a question about direction. in a 3d cordinate system, if we have a curve, what is the normal at a point going to be like. is it a plane perpendicular to the tangent.
    well i know for a surface we can find a normal at a point p, and than its tangent plane by simple formula <p-x>.<n> = 0.
    (1 vote)
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  • blobby green style avatar for user gurungdebendra82
    this may be in the context of geometry, does integrating ∯1.n̂ds just gives the area enclosed by the curve.
    (1 vote)
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Video transcript

I've restated Stokes' theorem. And what I want to do in this video is make sure that we get our orientation right. Because when we think about a normal vector to a surface there are actually two normal vectors. There-- based on the way I've drawn it right over here, there could be the one that might pop outward, like this. Or there might be the one that pops inward, just like that. Both of those would be normal to the surface right there. And also, when we think about a path that goes around the boundary of a surface, there's two ways to think about that path. We could be going-- based on how I've oriented it right now, we could go in a counterclockwise orientation, or direction. Or we could go in a clockwise orientation, or direction. So in order to make sure we're using Stokes' theorem correctly, we need to make sure we understand which each convention it is using. And the way we think about it is, whatever the normal direction we pick-- and so let's say we pick this normal direction right over here, the one I am drawing in yellow. So if we pick this as our normal vector. So we're essentially saying maybe that's the top, one way of doing it is, that's the top of our surface, then the positive orientation that we need to traverse the path in is the one that if your head was pointed in the direction of the normal vector, and you were to walk along that path, the inside, or the surface itself would be to your left. And so, if my head is pointed in the direction of the normal vector-- so this is me right over here-- my head is pointed in the direction of the normal vector-- I'm wearing a big arrow hat right over there-- and if I'm walking around the boundary, the actual surface needs to be to my left. So I need to be-- this is me walking right over here-- I need to be walking in the counterclockwise direction just like that. Then that's the convention that we use when we're thinking about Stokes' theorem. If oriented this thing differently, or if we said that no, no, no, no, no, this is not the normal vector. This is not the, essentially, the top that we want to pick. If we wanted to pick it the other way, if we wanted to go in that direction, If we wanted that to be our normal vector, in order to be consistent, we would have to now do the opposite. I would now have to have my head going in that direction. And then I would have to walk, once again, and this might be a little bit harder to visualize. I would have to walk in the direction that the surface is to my left. And now, in this situation, instead of the surface looking like a hill to me, the surface would look like some type of a bowl, or some type of a valley or something like that to me. And the way that I would have to do it now, and it's a little bit hard to visualize the upside down Sal, but the upside down Sal would have to walk in this direction in order for the bowl, or the dip, to be to my left. So that's just important to keep this in mind in order for this to be consistent with this right over here. Put your head in the direction of the normal vector. Or you can kind of view that as the top of the direction that the top of the surface is going in. And then the contour, or the direction that you would have to traverse the boundary in order for this to be true, is the direction with which the surface is to your left.