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5:27 AM
@robjohn hi
sir I have some question on stokes theorem?
like divergence, we know as scalar quantity in curl basically we are talking about a loop through which field pass assuming it as rotational field
I cannot understand what is the basic difference between gauss divergence theorem and stokes theorem?
 
5:56 AM
@robjohn I have difficulty in physically analyzing stokes theorem>
 
6:16 AM
@JackRod What do you mean "physically analyzing"?
You might want to look at this
In vector calculus and differential geometry the generalized Stokes theorem (sometimes with apostrophe as Stokes' theorem or Stokes's theorem), also called the Stokes–Cartan theorem, is a statement about the integration of differential forms on manifolds, which both simplifies and generalizes several theorems from vector calculus. It is a generalization of Isaac Newton's fundamental theorem of calculus that relates two-dimensional line integrals to three-dimensional surface integrals.Stokes' theorem says that the integral of a differential form ω over the boundary of some orientable manifold Ω...
It says that Stokes' Theorem, Green's Theorem, and the Divergence Theorem are special cases of the same more general theorem.
 
6:46 AM
@robjohn sir actually I wanted to say how it is different from divergence theorem
except the understanding of curl and $.\nabla$
we have been doing electrodynamics where we considering field as a electric field
in general I can understand divergence of field E in closed-loop is a always zero
which is what the gauss says $\int\nabla.E.da=0$
but how do i define stocks in the same context
 
I am not sure what you are asking.
 
ok sir can u help me in understanding proof of this theorem?
 
which theorem?
 
please sir
stokes
 
$\int_\Sigma(\nabla\times f)\cdot n\,\mathrm{d}S=\int_{\partial\Sigma}t\cdot f\,\mathrm{d}s$?
where $t$ is the unit tangent
is that the form of Stokes you are using?
 
7:03 AM
yes
 
have you looked at Stokes Theorem on Wikipedia?
 
yes
but there was no proof
 
the amount of writing and diagramming will take a lot of time. It is probably best to find a proof in a book where that has already been done. Just making the diagrams and writing the formulas is very involved.
 
can you do screen video call if u have diagram
based proof
or anything better
 
7:31 AM
@robjohn
do u have good proof link?
 
No, i would have to search on google just as you could
 
leave it sir I would try to understand from the book I have
I will get back to u if i have a problem?
 
I am writing up something, but it will take a long time. read your book and see if you can understand from there.
 
ok sir
 
8:06 AM
$$
\begin{align}
\int_\Sigma\nabla\times F\cdot\mathrm{d}S
&=\int_\Sigma\left[
\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\mathrm{d}y\,\mathrm{d}z
+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\mathrm{d}z\,\mathrm{d}x
+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathrm{d}x\,\mathrm{d}y
\right]\\
&=\int_\Sigma\left[
\left(\frac{\partial P}{\partial y}\,\mathrm{d}y+\frac{\partial P}{\partial z}\,\mathrm{d}z\right)\mathrm{d}x
where $F=(P,Q,R)$
$\int_\Sigma \mathrm{d}P\,\mathrm{d}x$ is the integral from the negatively oriented edge of $\partial\Sigma$ to the positively oriented edge of $\partial\Sigma$ along a strip with constant $x$ then integrated in $x$
This gives $\int_{\partial\Sigma}P\,\mathrm{d}x$
Similarly with the other terms
You have to remember that the orientation when integrating $\mathrm{d}x\,\mathrm{d}y$ is opposite of the orientation when integrating $\mathrm{d}y\,\mathrm{d}x$. This is the basis of differential forms.
@JackRod: here is what I have for now.
 

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