- Stokes' theorem intuition
- Green's and Stokes' theorem relationship
- Orienting boundary with surface
- Orientation and stokes
- Orientations and boundaries
- Conditions for stokes theorem
- Stokes example part 1
- Stokes example part 2
- Stokes example part 3
- Stokes example part 4
- Stokes' theorem
- Evaluating line integral directly - part 1
- Evaluating line integral directly - part 2
Stokes' theorem intuition
Conceptual understanding of why the curl of a vector field along a surface would relate to the line integral around the surface's boundary. Created by Sal Khan.
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- I don't understand why we have to dot the curl of F with the normal vector. Since we are integrating over the surface aren't we integrating F on the surface anyway?(15 votes)
- Stokes theorem says that ∫F·dr = ∬curl(F)·n ds.
If you think about fluid in 3D space, it could be swirling in any direction, the curl(F) is a vector that points in the direction of the AXIS OF ROTATION of the swirling fluid. curl(F)·n picks out the curl who's axis of rotation is normal/perpendicular to the surface. So if the axis of rotation is perpendicular to the surface, then the swirling is happening parallel to (and right on) the surface.(37 votes)
- I still don't really understand Stokes theorem. Can someone help explain it to me?(11 votes)
- I think this is the best way to teach it.
It is a speccial case of the Holographic principle...The Holographic Principle means that information inside a reigon is completely encoded onto its boundary.
So, in Stokes' theoremn, the flux through a reigon (the surface $S$ ) is exactly the same as the work through its boundary (the curve $C$).(25 votes)
- What would happen if we had a circular surface where at each point on the boundary the vector was perpendicular. Would that have curl or would it be equivalent to the 2nd example Sal gives where it all cancels out? (Or is this in fact not at all related?)(6 votes)
- If the all of the vectors on the surface were perpendicular (or normal) to the surface, then the curl of the vector field would always be zero along the surface, so the surface integral of the curl is 0. Also, for the line integral, the dot product is 0 because they are perpendicular, so the line integral is 0.
So yes, both the surface integral of the curl and the the line integral would be zero.(8 votes)
- 11:09How is n ds the same as d s ? Also ds is supposed to be a differential of area?(8 votes)
- Which surface are we talking of exactly? There are tons of surfaces which can have the boundary C right? Are we talking of any of those surfaces?(6 votes)
- In principle, given a boundary C, Stokes theorem also holds for any 3-D surface with boundary C. Roughly speaking, this means that with any closed curve C, along with any surface S with boundary C, the line integral of F around C is equal to the sum of the "curls" of F on the surface S. Sal started of with simple surfaces in this video to help the viewer develop an intuitive understanding of what Stokes theorem means physically.(2 votes)
- At10:47; by dotting the vector field with the normal wouldn't you get the projection of the vector field perpendicular to the surface? I understand that we need the projection of the vectorfield ON the surface, but I don't see how dotting with the normal vector accomplishes that?(4 votes)
- Stokes theorem says that ∫F·dr = ∬curl(F)·n ds. We don't dot the field F with the normal vector, we dot the curl(F) with the normal vector.
If you think about fluid in 3D space, it could be swirling in any direction, the curl(F) is a vector that points in the direction of the AXIS OF ROTATION of the swirling fluid. curl(F)·n picks out the curl who's axis of rotation is normal/perpendicular to the surface.(5 votes)
- is there a video where curl is explained?(2 votes)
- who was stokes?(3 votes)
- Sir George Gabriel Stokes was an Irish physicist and mathematician (13 August 1819 – 1 February 1903) who made great contributions to fluid dynamics and optics.
If you watch The Walking Dead and have watched season 5 or 6, you will know that there is a pastor named Gabriel Stokes.
Sir George Stokes, the creator of Stokes theorem, had a father who's name was Gabriel Stokes who also happened to be a pastor.
For all you Walking Dead fans, I'd like you to take a moment to appreciate this coincidence(5 votes)
- Around11:47, does it matter whether we say that we calculate a double integral over a "surface" (symbol S) over over a "region" (symbol R)?(3 votes)
- They both mean the same thing in the current context. They both basically mean summing up an infinite number of infinitesimally small area chunks over the entire surface/region.(4 votes)
- What is the logical contrast between line integral, surface integral and volume integral(2 votes)
- Line integral finds length by summing points, surface integral finds area by summing points to get lines the summing lines, volume finds, well, volume by summing points, then the lines, then the surfaces.
That's a somewhat pithy explanation, but it's how I tend to intuitively understand it physically. If it's not what you were looking for I'm happy to try explaining it another way!(4 votes)
So I've drawn multiple versions of the exact same surface S, five copies of that exact same surface. And what I want to do is think about the value of the line integral-- let me write this down-- the value of the line integral of F dot dr, where F is the vector field that I've drawn in magenta in each of these diagrams. And obviously, it's different in each of these diagrams. And the only part of the vector field that I've drawn is a part that's along the surface. I could have drawn the part of the vector field that's off the surface, but we're only going to be concerned with what's going on on the surface. So the vector field could be defined in this entire three-dimensional space in here as well. And these are obviously different vector fields, and we can see that visually based on how we drew it. And the contour that we care about-- remember, we're going to take a line integral, so the path matters. The path that we care about is the counterclockwise boundary of our surface, so it's going to be this right over here. The counterclockwise boundary of our surface is what we're going to be taking F dot dr along. So this right over here, and let me draw the orientation. It's going to be counterclockwise. And we're going to do it in every one of these situations, for every one of these surfaces and every one of these F's. And what I really want to think about is how the value of F dot dr over that contour, how it might change from example to example. And obviously, the only difference between each of these is what the vector field F is doing. So first let's think about this example right over here. At this part of the contour, this bottom part right over here, our vector field is going in the exact same direction as our line, as our contour. So we're going to get positive values of F dot dr down here, and we're going to keep summing them up. We're taking an integral. Then as we go up the curve or as we go kind of uphill right over here, we see that our vector field is going essentially orthogonal. It's going perpendicular. It's going perpendicular to our contour. So our F dot dr, we're not going to get any value from that. F dot dr and all of these, this part of the contour, it's all going to be 0. So we're not going to get anything right over there. Actually, let me just write nothing. So we get nothing right over there. Maybe I'll write 0. I'll write we'll get 0 right over there. And then up here, when we're at this part of the contour, our vector field is going in the exact opposite direction as our path. Up here our path is going, I guess, from the right to the left, while our vector field is going from the left to the right. And so we're actually going to get negative values. We are going to get negative values up here, and they're going to sum up to a reasonable negative value. And if the vector field is constant, and I kind of drew it like it is, and if this length is equal to this length, then these two values are going to cancel out. When you add this positive sum to this negative sum, they're going to be 0. And then once again, when you go downhill again, the vector field is perpendicular to our actual path, so we're going to get 0. So based on the way I've described it, your line integral of F dot dr for this version of F in this example right over here, it might all cancel out. You will get something, if we make the assumptions that I made, this might be equal to 0. So in this example, F dot dr could be equal to 0. Now let's think about what's going on in this situation. In this one, just like the last one, as we go along the bottom, the vector field is going in the exact same direction as our contour, so we're going to get positive values. And once we go up the hill, the vector field is going perpendicular to our path, so it's not going to add anything really to it. So we're just going to get 0 along this part. But then up here, our vector field has switched directions. And once again, it's going in the exact same direction as our path, so we're going to get more positive values right over there. And then as we go down here, it's not going to add anything because our vector field is perpendicular to our path. So we're going to get 0. But notice, now these two ends don't cancel out with each other. We're going to get a positive value. We are going to get a positive value. And what was the difference between this version of F, this vector field, and this vector field right over here? Well, this vector field switched directions so that the top part didn't cancel out with the bottom part. Or another way to think about it is, it had some curl. There's a little bit of spinning going on. If this was describing the velocity of a fluid, and if you were to put a stick right over there on the surface, the stick would spin. It has some spin or some curl, however you want to describe it. This right over here has no curl. If you put a stick right over here, it would just flow with the fluid, but the stick itself would not spin. So we got a positive value for the line integral in this situation. And we also have, it looks like, a positive curl. Now let's think about this one. In this situation, as we go along this part of our contour, our vector field F is going in the exact same direction, so we're going to get positive values. Now as we go uphill, our vector field F, it's also kind of turned in that direction, and so we'll get more positive values. And now as we go in that direction, our vector field F is still going in the direction of our contour, we're going to get more positive values. And as we go down, once again the vector field F is going in the direction of our contour, so we'll get even more positive values. So in this situation, the value of our line integral of F dot dr is even more positive. And we see that the actual vector field along the surface-- and remember, the vector field might be doing all sorts of crazy things off the surface. Actually, let me draw that in that same magenta color. It might be doing all sorts of crazy things off the surface, but what we really care about is what's happening on the surface. And because this vector field is, I guess you could say curling, or it's spinning along the surface, it allows it to go with the boundary along all the points, and we get a very positive value for this line integral. So we have a higher curl. So more curl, it looks like, is leading to a more positive line integral. Now let's think about what's happening in this situation right over here. This situation down here, our vector field is going in the same direction as our path, so we're going to get positive values. Just like the first situation, as we go up the hill like this or up the surface like that, our vector field is perpendicular to our surface, so it's really not going to add anything to our line integral. And then as we go along this top part, this first part of the top part right over here, the vector field is going against us. So it's negative right over here. We're going in the exact opposite direction of our path. And then, right as we get to the end, the vector field flips direction, and we get a little bit of positive right over here because it's a little bit of it going in the same direction. And then we go back downhill. When we go back downhill, it adds nothing because our vector field is going perpendicular to our path. So the big difference between this case and the case up here is the case up here-- well, actually, I could compare between these two or these two. But the difference between this one and this one is that at least this part of the vector field has switched directions. So we get a little bit of positive value. And one way to think about it, this one is going to be less positive than that, if we take the line integral, but more positive than that. And one other way to think about it is, we have a little bit of curl going on right over here. Our vector field switched directions right around there, or I guess you could say that it's spinning right around there. So if you put a stick, if that was in the water, it would start spinning. But everywhere else, there isn't a lot of curl. So you have some curl, but it's over a little small region of the surface. While over here, you had curl going on over a larger portion of our surface. And so up here, you had a more positive curl, more positive line integral. Here you have a curl over less of the surface, and you're going to have a less positive line integral. Now let's think about this one over here. This vector field, along the surface there is some curl. There's some curl going on right over there. If you put a stick in the water, if you view that as the velocity of water, the stick would spin. So you have some curl. But then it switches direction again, so you have curl there as well, and it's actually curl in the opposite direction. So to some degree, if you were to sum all of this up, maybe it would cancel out. And it makes sense. It makes sense that it would cancel out because when you take the line integral around the whole thing, just like this first situation, it looks like it will add up to 0. Because even though you have some curl, the curls cancel out each other. And so when you go to this top part of the surface, the vector field is going in the exact same direction as this bottom part of the surface. So if you were to take your line integral, the same one that we care about, just like the first situation it would be positive down here, 0 as we go up the curve. And then as we go down here, the vector field has switched directions twice, so it's still going against the path of our contour, just like this first situation. So it would be negative up there. And then as we go down, it would be 0. So this thing right over here also looks just like the first one because the curls essentially cancel out. We switched direction twice. So over here, our line integral might also be 0. Now, the whole reason why I went through this exercise is to give you an intuition of why it might make sense that if we have more curl happening over more of this surface, why that might make the value of this line integral be larger. And so, hopefully, it starts to give you an intuition that maybe, just maybe, the value of this line integral, the value of F dot dr over that contour, over this contour that's going in the counterclockwise direction-- we'll talk more about orientation in future videos-- maybe this is equal to the sum of the curls over the surface. And so let's think about this. It could be a surface integral. So we're going to go over the surface, and what we care about is the curl of F. We care about the curl of F. But we don't just care about the curl of F generally because F might be spinning in a direction, in a way, that's off the surface. We care about how much it's curling on the surface. So what we would want to do is we'd want to take the curl of F and dot it with the normal vector at any point to the surface and then multiply that times the surface itself. And this is just to say that the more surface where we have more curling going on, the more that the line integral, the value of the line integral, might be. And we saw that when we compared these three examples. And another way of writing all of this is the surface integral-- let me write the surface in that same brown color-- the surface integral of the curl of F, which would just be another vector that tells us how much we are spinning generally, but we want to care how much we're spinning along the surface. So we're dotting it with the normal vector. Or another way to write this whole thing is to say dot ds. So if you take essentially the sum across the entire surface of how much we're curling, how much we're spinning along that surface, then maybe, just maybe, this will be equal to the value of the line integral as we go around the boundary of the surface. And it actually turns out that this is the case. And obviously, I haven't proven it to you here, but hopefully you have some intuition why this makes sense. And this idea that this is equal to this is called Stokes' theorem, and we'll explore it more in the next few videos.