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Orientation and stokes

Determining the proper orientation of a boundary given the orientation of the normal vector. Created by Sal Khan.

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  • spunky sam blue style avatar for user Ethan Dlugie
    Why is Sal drawing 3D bodies? I thought Stokes theorem deals with 2D surfaces in 3D space.
    (12 votes)
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  • blobby green style avatar for user ben14A
    for the end of the video, I've learned in physics and in a course of mechanics whats called the rule of the right hand (or whatever it's called in english) for the momentum . anyway, your thumb is the direction of the normal vector and the four fingers (when you twist your wrist) show the way you "walk" to. can i apply this rule here or is it just a coincidence that the two rules match?
    (8 votes)
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  • leafers ultimate style avatar for user stephen
    Can we not just use the right hand rule?
    (6 votes)
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  • blobby green style avatar for user Adriaan Knox
    Am I becoming delirious from studying for my final or is this little man really walking the wrong way on this video to be keeping the surface to his left?
    (2 votes)
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    • blobby green style avatar for user verheyen.thierry
      Try your keyboard and a pencil. First walk along with your pencil around your keyboard(the side with your keys) with the keyboard on the left of your pencil (make sure you know what you define as the left side of your pencil). Now take the backside of your keyboard ( the side with no keys and do the same). Counterclockwise in the first case, it becomes clockwise in the second.

      Visualizing it is cumbersome, so I just use the right hand rule. :)
      (2 votes)
  • blobby green style avatar for user mohzakiyah1997
    Can't we use the right-hand rule? It seems it works perfectly.
    (2 votes)
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  • blobby green style avatar for user sunilkapdi2014
    plz tell me about volume integrals ....to find volume integrals,..
    (1 vote)
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    • piceratops tree style avatar for user Alex Cole
      A volume integral is simply a triple integral. For example, think of some mass density, p, which is a function of x, y and z - let p = xyz.
      Now, say you want to find the total amount of mass in a cubic region bounded by the origin, and a point at (2,4,6). You can draw this as a cuboid in the xyz plane.
      You can set up a volume integral of p with respect to x, y and z, with 3 limits for x (0 to 2), y (0 to 4) and z (0 to 6).
      Now, you solve the integral in any order - for simplicity integrate with respect to (wrt) x first (ie. (x^2)/2), then you can factor out that x term leaving only yz inside the integral - you can calculate the limits of the x term outside of the integral. Then solve for y, factor out, and finally solve for z.

      If you'd like to give the above example a try, let p = 5xyz, and use the same cuboid (0,0,0) to (2,4,6), and you should find that the volume integral solves to 1440kg.
      (2 votes)
  • starky ultimate style avatar for user bwohanwang
    The little guy's head should point to the normal of where he is. The first example, where the normal is pointing up, by the time the point moves to where the little guy is, it should be twisted to pointing outside, i.e. downward. Same for the second example, the normal is pointing inside.
    The bottle cap example, are we twisting the bottle or the cap? If we hold the cap stationary and twist the bottle instead, we get the other result. After applying the right-hand rule, we come to the conclusion that the bottle stays and cap turns.
    Thanks for the video!
    (1 vote)
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  • leafers ultimate style avatar for user Simon
    there is also the right hand trick:
    if you imagine that you grab the surface with your four fingers oriented in the direction of the rotation, your thumb indicate the normal vector orientation! ;P
    (1 vote)
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  • blobby green style avatar for user BILLY SACHS
    Enjoyed the bottle analogy. Lefty loosy righty tighty
    (1 vote)
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  • piceratops ultimate style avatar for user kingpattycake
    How does orientation relate to physics ?
    (1 vote)
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Video transcript

I've rewritten Stokes' theorem right over here. What I want to focus on in this video is the question of orientation because there are two different orientations for our boundary curve. We could go in that direction like that, or we could go in the opposite direction. We could be going like that. And there are also two different orientations for this normal vector. The normal vector might pop out like that, or it could actually go into the surface like that. So we have to make sure that our orientations are consistent, and what I want to do is give you two different ways of thinking about it. And you might think of others, but these are the ones that work for me. In order for Stokes' theorem to hold, we have to make sure that we're not actually picking the negative of one or the other orientations. And so the easiest way for me to remember-- it is if our normal vector is-- let's say, it goes in that direction. And if you have some hypothetical person traversing the boundary of our surface, and their direction is pointed. And their head is pointed in the same direction as the normal vector-- so this is the normal vector. So their head is pointed in the exact same direction as the normal vector-- or you could say maybe their body or, really, their head-- So that's them. Then the direction that you would have to actually traverse the boundary is the direction that would allow this person to keep the surface to their left. So over here, he would have to go in this direction in order to keep the surface to his left. So he would have to go just like that. If we oriented the surface differently-- so let me redraw the surface right over here and draw similar surface. So if we had a surface-- so this surface looks very similar. This is a very similar looking surface that I'm drawing right over here just to give a idea of some of the contours. But if we said that the normal vector for this surface, if we orient it in the opposite way-- so if we said that the normal vector here was actually pointing downward like that, then we would have to, in order for Stokes' theorem to hold, we would have to traverse the boundary in a different direction because, once again, if I draw my little character right over here, his head is pointed in the direction of the normal vector. He is now upside down. So let me draw him. So this is him running right over here. I could draw a better job. This is him running right over here. Now, in order to keep-- and from his point of view, this would kind of look like a some type of a pool or a ditch of some kind. It would actually go down. Here, it looks like a hill to him. But since he's upside down, in order for him to keep the boundary to his left, he would have to now go in the other direction. So depending on the orientation of your normal vector, which is really the orientation of your actual surface, will dictate how you need to traverse the path. Now, another way to think about it-- and this idea was introduced by one of the viewers on YouTube, but it's a valid way of thinking about it-- is to imagine that the surface is a bottle cap. And so let me draw some type of a bottle over here. So I'll draw. Let me draw a bottle. You could imagine some type of a glass soda bottle. So what we really care about is the cap of the bottle-- so make it feel like it's glass. So there, that's our bottle, And let me draw its cap. Let me draw the cap of the bottle because that's what we care about. We can kind of imagine that being the surface. So this is the cap of our bottle, and you just need to think about, well, which way would I have to twist the cap in order to make the cap move up, in order to take the cap off. And you could think of the normal vector as the direction that the cap would move, and the twisting is the direction that you would have to traverse the path. So you would have to twist the bottle that way, or you could think about the other way. If you twisted the bottle the other way, then the cap would move down. So the normal vector is the direction that the cap would move, and the direction that you would traverse the boundary is how you would actually twist it. So either of these are ways of thinking about it, but they're important to keep in mind, especially once the shapes start getting a little bit more convoluted and oriented in strange ways.