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6797695 No.6797695 [Reply] [Original]

Can someone explain in layman terms the generalized stokes theorem, the notation and the intuition on why it works?

I learnt green's theorem and kelvin-stokes theorem on calc 3, then I heard about generalized stokes, from which we can derive a plethora of theorems.

If you're using notation not used up to calc 3, state it beforehand. I just learned that <span class="math">\partial x[/spoiler] can denote a boundary instead of a partial operator and I guess there are more of these weird notations.

>> No.6797715

Bump

>> No.6797735

>>6797695

The boundary of a finite bounded n-surface is always compact. The behaviour of a field on this n-surface is completely determined by the field on the boundary, because of continuity and other properties it has to satisfy.

Most simple example is a curve (I might fuck this formula up, beware):

<span class="math">\Integral^b_af(x)dx = F(b) - F(a)[/spoiler]

where the boundary of the space you integrate over (the curve) is a set of two distinct points a and b. By the above logic, the solution is completely determined by the addition of the value of the field at these two points. The only thing you have to always remember here is that you have to be able to assign an orientation to the boundary. In this case, the orientation of the point b is opposite that of point a, giving a minus sign. Convention is to assign it to the starting point of a curve.

Similarly, the boundary of a cylinder is two circles, with opposite orientations.

Etc. etc.. It's really beautiful.

>> No.6797736

>>6797735

> Unknown countrol sequence

Ah, fuck this crap.

Integral over f(x)dx from a to b = F(b) - F(a).

>> No.6797743

>>6797735
Why do they have opposite orientations? Also, what does my pic related mean if we don't collapse it to FTC and generalise after that?

>> No.6797744

>>6797735
Also, why is the surface almost defined by its boundary?

>> No.6797762

>>6797743

Not sure what FTC means.

Opposite orientations so that you can define a notion of "inside" the boundary. Think of a cylinder bounded by two circles: o=o
Say, the left circle is oriented such that its normal points to the surface of the cylinder. Then, if the opposite circle is oriented in the exact opposite way, its normal will also point towards the surface of the cylinder. That's what makes the "field is determined by values on the boundary" notion consistent.

>>6797744

I did not say that. If I did, it's wrong. What I meant to say is the behaviour, meaning the differential of a field is determined by the values of the field on the boundary of that finite surface.

>> No.6797779

The most concentrated idea is this:

If you start at a set, and then sum all the changes in this set through a region, you will end up at a set which is equal to your first set plus the changes.

This isn't rigurous, but it's a nice way to picture it.

For instance, if you have a point of a function in position (a,x) and you add all the changes in the point (the derivative of the function) in the range ]a,b[ then you end with the point (b,y). The difference between these two points is the sum of changes in the path.

>> No.6797782

>>6797779
If this idea reads confusing, consider this: The fundamental theorem of calculus is the first special case of stoke's theorem you learn. Picture the similarity between it, green's, then divergence, then curl stoke's, and you'll get it.

I'm the faggot who always says "stoke's theorem" when someone asks for favorite equation and shit.

>> No.6797838

>>6797736
if the boundary is {a,b} and you want to "integrate" F on that boundary, why isn't it F(a)+F(b)? Why is it F(b)-F(a)?