[alright1234 (OP)]

Stokes' theorem is equating a line integral with a surface integral. The result of the line integral is a length which is not equivalent to the the surface area of the surface integral; units do not seem to match


m =/ m^2.............................................................equ 1

[Bored chemist]

Three points.
Mathematical theorems are true.
You have forgotten to provide any reference, so it's impossible to know what you are talking about
It really would be better if you tried to learn a bit about a subject before  getting all shouty about it.

[chiralSPO]

This would be a good place to start: https://en.wikipedia.org/wiki/Stokes'_theorem

It does seem somewhat counterintuitive that the path integral of a closed loop would give any info about what is inside... but recall that it only applies in specific cases, ie. when looking at "smooth" and "orientable" surfaces (manifolds), which means that it is already limited to cases in which information about one specific point also has some bearing on its immediate neighbors (and their neighbors, and their neighbors, etc.)--thus information is somewhat delocalized.

The theorem proves that this type of delocalization ultimately allows limited (no pun intended) information about an n-dimensional manifold to be deduced by information gained by analyzing the n-1 dimensional boundary of that manifold.

It's been quite a while since I studied or used multivariate calculus, but I recall that some of the derivations of the theorem were pretty straight forward (at least in specific cases--I wouldn't even know where to start with the general form!)

[alright1234 (OP)]

Three points.
Mathematical theorems are true.

Until they are proven false.

The result of the Stoke's line integral is a length which is not equivalent to the the surface area of the surface integral.

[alancalverd]

Oh dear. Any integral has a higher dimensionality than the integrand. That was lesson 1 at the Open University access course for those who had not studied physical sciences. It's probably on the shelves at Harvard, covered in dust and unread by philosophers..

[chiralSPO]

The result of the Stoke's line integral is a length which is not equivalent to the the surface area of the surface integral.

The integral of a function over a line is NOT the length of the line (HINT: it must relate to the function as well). Stokes' theorem does NOT equate a surface area with a length--which is what you appear to be claiming.

Please seek a better understanding of this topic before knocking it.

[Bored chemist]

Until they are proven false.

No.
Once again, you are showing that you don't know what you are talking about.
An idea in maths is only called a theorem after it has been proved to be true.
And that's irrevocable.
2+2 won't stop being 4

[alright1234 (OP)]

Is a length equal to an area?

[Bored chemist]

Is a length equal to an area?

No, but, as has been pointed out, an integral along a path is not a length.

[alright1234 (OP)]

Is a length equal to an area?

No, but, as has been pointed out, an integral along a path is not a length.

Your statement is patently incorrect.

[Bored chemist]

Is a length equal to an area?

No, but, as has been pointed out, an integral along a path is not a length.

Your statement is patently incorrect.

Among the things I can calculate as a path integral is the energy stored in a compressed gas.
Are you saying that the energy in a compressed gas is a line?

Again, it really would be better if you learned more before getting shouty

[alright1234 (OP)]

Mathematically, Is a length equal to an area?

[Bored chemist]

Mathematically, Is a length equal to an area?

No.
Integrals are often areas.

[alright1234 (OP)]

Stokes' theorem is equating a line integral with a surface integral. The result of the line integral is a length which is not equivalent to the the surface area of the surface integral which proves Stokes' theorem is mathematically invalid.

[chiralSPO]

Stokes' theorem is equating a line integral with a surface integral. The result of the line integral is a length which is not equivalent to the the surface area of the surface integral which proves Stokes' theorem is mathematically invalid.

Sorry. This only proves that you don't know what you're talking about. Multiple people have provided explanations or links to detailed explanations in an effort to educate you, while you have only repeated the same ridiculous straw man argument over and over. Please try to understand, or at least realize that you don't understand. This is the last post I will make in this thread.

[alright1234 (OP)]

Please indicate in your own words how Stokes' theorem justifies equating a line integral with a surface integral. The links you prove are incomprehensible to me. Please use simple words and examples that I can understand. A couple of example would be good since if it is possible It would be monumental proof.

[Bored chemist]

Please indicate in your own words how Stokes' theorem justifies equating a line integral with a surface integral.

You need to show that it does equate them.

[alright1234 (OP)]

∫ F ⋅dr =∬ curl F ⋅ dS

http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx

[Bored chemist]

Curl is a derivative operator.
So the units work fine.

[alright1234 (OP)]

Curl is a derivative operator.
So the units work fine.

The surface integral has precedence before the derivative of the curl.