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Course: Multivariable calculus > Unit 5
Lesson 5: Stokes' theorem- 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
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Conditions for stokes theorem
Understanding when you can use Stokes. Piecewise-smooth lines and surfaces. Created by Sal Khan.
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In R^3 the same boundary will be shared by many surfaces. For instance a circular loop can be the boundary of a circle, a hemisphere, a paraboloid, etc. Does Stoke's theorem imply that the flux over all of these surfaces is the same?(8 votes)
Yes, the amazing outcome of the theorem is that it doesn't matter which surface you have as long as it has the same boundary.(14 votes)
Can I use stokes to solve flux problems? How do you know which theorem out of stokes,divergence,green to use?(1 vote)
As Maclaurin is to Taylor, Green is to Stokes, that is Green (2 dimensions) is a special case of Stokes (3 dimensions).
Stokes' Theorem equates the single integral of a function f along the boundary of a surface with the double integral of some kind of derivative of f along the surface itself.
Gauss's Theorem (a.k.a. the Divergence Theorem) equates the double integral of a function along a closed surface which is the boundary of a three-dimensional region with the triple integral of some kind of derivative of f along the region itself.
Thus the situation in Gauss's Theorem is "one dimension up" from the situation in Stokes's Theorem, so it should be easy to figure out which of these results applies. If you see a three dimensional region bounded by a closed surface, or if you see a triple integral, it must be Gauss's Theorem that you want. Conversely, if you see a two dimensional region bounded by a closed curve, or if you see a single integral (really a line integral), then it must be Stokes' Theorem that you want.
This is only two theorems: what about Green's Theorem? Green's Theorem is in fact the special case of Stokes's Theorem in which the surface lies entirely in the plane. Thus when you are applying Green's Theorem you are technically applying Stokes's Theorem as well, however in a case which leads to some simplifications in the formulas. Especially, when you have a vector field in the plane, the curl of the vector field is always a purely vertical vector, so it makes sense to identify this with a scalar quantity, and this scalar quantity is precisely the "derivative" appearing in the double integral in Green's Theorem.
All are used for flux calculations.(8 votes)
Is it also a condition for F to be well defined and continuous differentiable at S (class C1)?(3 votes)
It is a condition for F to be continuously differentiable at S.(3 votes)
Wikipedia says that the divergence theorem is a more generalized form of Stokes' theorem. Is this true?(2 votes)
Perhaps, you are referring to the generalised Stokes' theorem which is the generalised version of all 3 of these theorems.(1 vote)
At3:57, I think Sal means they're not smooth at the corners, not edges.(2 votes)- Can you give an example of a curve that is simple & closed, but not piece-wise smooth? Thanks!(2 votes)
- Hi bkmurthy99, here's an example. Keep in mind we need to get a little beyond what most multivariable courses cover in terms of rigorous definitions. Recall smooth means differentiable in this context, and piecewise-smooth means we can at least split up our function into finitely many pieces that are themselves smooth. To violate this condition, we'd need a function everywhere undifferentiable. There's a pathological (meaning it's wild!) function called the Weierstrass function, basically a fractal curve, that is continuous everywhere but differentiable nowhere. Since we can't break it up into finitely many smooth pieces (or any smooth pieces actually), it's not piecewise-smooth. Normally it's a function from R to R, but to make it a simple, closed curve, we just have to connect up two ends of the graph to make a jagged circle-ish boundary. It's simple because it doesn't intersect itself (which I haven't rigorously shown, but is probably true), and closed because the Weierstrass function is symmetric, so we can definitely join up two ends of it into that jagged circle. See this image for a visual of the function. Imagine picking up the graph between x = -1 and x = 1, and forming a circle with that strand of graph. https://i.stack.imgur.com/RgN7d.png
Hope this example clears up how it's theoretically possible to have a simple, closed, not piecewise-smooth curve. In practice, since the real world is constrained by the Planck-length and quantum mechanical stuff, every actual curve is piecewise smooth in real life. Hope this helps!(1 vote)
So could a normal surface, with continuous derivatives everywhere (gradually changing slope), also be called a piecewise smooth surface? Even though it's not really "piecewise" per se? because it's not necessarily smooth in pieces...
I'm asking because I was asked this recently on a test..(1 vote)- Hi Karun, yes, even perfectly smooth surfaces can be called piecewise smooth as well. The reason comes down to the adjective piecewise smooth being more general than just smooth. The definition of piecewise smooth is that it's possible to break up the surface into a finite number of sections, each of which is itself smooth. For your regular, smooth surface, you can "break" it into just one piece, namely itself, to satisfy the definition. Similarly, we could say a circle is piecewise smooth, even though it's also perfectly smooth on its own. Hope this helps :)(1 vote)
- why we use green's theorem,stoke,s theorem and divergence theorem?reply must(1 vote)
We use those theorems to turn very complicated line integrals into very simple double (or triple, if you are talking about the 3-d divergence theorem) integrals.(1 vote)
- I still don't see why the integration of the boundary is the same as the integration of the total surface area of the boundary. This is counter-intuitive because it's assumed in our everyday thinking that the surface area of an objection is larger than the mere boundary line of that object. ? I wish this part could be made clear (unless I'm understanding the theorem's conclusion wrong)?(1 vote)
I believe you also need orientability, right?(1 vote)
Yes, you need orientability. Otherwise, there would not be any well-defined surface to integrate over.(1 vote)
3:57
Video transcript
Now that we've explored Stokes' Theorem a little bit, I want to talk about the situations in wich we can use it. You'll see that this is pretty general theorem. But we do have to thing about what type of surfaces and what type of boundary are those surfaces we are actually dealing with and the case of Stokes', we need surfaces that are piecewise... piecewise-smooth piecewise-smooth surfaces so this surface right over here it is actually smooth not just even piecewise-smooth. Sounds like a very fancy term but all the smooth part means that you have just continuous derivatives and since we are talking about surfaces we're going to have continuous partial derivatives regardless of which direction you pick. So this is continuous derivatives, and another way to think about that conceptually is if you pick a direction on the surface if you say that we go in that direction, the slope in that direction changes gradually, doesn't jump around. If you pick this direction right over here, the slope is changing gradually. So we have a continuous derivative. And you're like "what does the 'piecewise' means?" Well, the piecewise actually allow us to use Stokes' Theorem with more surfaces. Because if you have a surface that looks like... Let's say a surface that looks like this. Let's say looks like a cup. So this is the opening of the top of the cup let's say that has no opening on top so we can see the backside of the cup and this is the side of the cup and this right over here is the bottom of the cup and if it was transparent we could actually see through it. So surfaces like this is not entirely smooth because it has edges. There are points right over here. So this edge right over here... If we pick this... let's say if we pick this direction to go and if we go this direction along the bottom, then right we get to the edge, all of the sudden the slope changes dramatically jumps. So the slope is not continuous at that edge. The slope jumps and we start going straight up. And so this entire surface is not smooth. But the piecewise actually give us an out. This tell us that it's okay as long as we can break the surfaces up into pieces that are smooth. And so this cup we can break it up but we were doing this wen tackling surfaces integrals we can break it up into the bottom part, which is a smooth surface, it has continuous derivative, and the sides which kind of wraps around is also is also a smooth surface so most things you'll encounter in a traditional calculus course actually do, especially surfaces, do fit piecewise-smooth. And the thing is though actually very hard to visualize. I imagine this all outer pointy fractely looking things where it's hard to break it up into pieces that are actually smooth. That's for surface part but we also have to care about the boundary, in order to apply Stokes' Theorem. And that is that right over there. The boundary needs to be a simple, which means that doesn't cross itself, a simple closed piecewise-smooth boundary. So once again: simple and closed that just means so this is not a simple boundary. If it is really crossing itself or intersecting itself, although you can break it up into to tow simple boundaries. But something like this is a simple boundary. So that is a simple boundary right over there. It also have to be closed wich really means that just loops in on itself. You just have something like that. It actually has to close and actually has to loops in on itself. In order to use Stokes' Theorem and once again it has to be piecewise-smooth but now we are talking about a path or a line or curve like this and a piecewise-smooth just means that you can break it up into sections were derivatives are continuous. The way I've drawm this one, this one and this one, the slope is changing gradually. So over there the slope is like that. It is changing gradually as we go around this path. Something that is not smooth, a path that is not smooth might look something like this. Might look something like that. And the places that this aren't smooth are at the edges: not smooth there, not smooth there and not smooth there. But we have to be simple-closed and this is simple and closed. And it's not smooth but it is piecewise-smooth. We can break it up into this section of the path. Which is that line right over there is smooth, that line over there is smooth, that line is smooth, and that line is smooth. And we've done that when evaluating in line integrals. We broke it up into smooth segments that we can then use to actually compute line integral. So if you find... if you have a boundary where the... if you have a surface that is piecewise-smooth and its boundary is a simple-closed piecewise-smooth boundary, you're good to go.