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Stokes' Theorem

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Hey!!! :eek:

green.png

$\overrightarrow{F}=M \hat{i}+ N \hat{j}+ P \hat{k}$

To prove the Stokes' Theorem we apply Green's Theroem at $ABE$, $BCE$, $CDE$.

$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}=\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$
$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F} }d \overrightarrow{R}=\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}+\int_B^E{\overrightarrow{F}}d \overrightarrow{R}+\int_E^B{\overrightarrow{F}}d \overrightarrow{R}+\int_C^E{\overrightarrow{F}}d \overrightarrow{R}+\int_E^C{\overrightarrow{F}}d \overrightarrow{R}=\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}$

Isn't the Green's Theorem:
$\oint_S{\overrightarrow{F} d \overrightarrow{R}}=\iint_R{\nabla \times \overrightarrow{F} dA}$?
Is this the same as $\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hey!!! :eek:

View attachment 2092

$\overrightarrow{F}=M \hat{i}+ N \hat{j}+ P \hat{k}$

To prove the Stokes' Theorem we apply Green's Theroem at $ABE$, $BCE$, $CDE$.

$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}=\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$
$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F} }d \overrightarrow{R}=\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}+\int_B^E{\overrightarrow{F}}d \overrightarrow{R}+\int_E^B{\overrightarrow{F}}d \overrightarrow{R}+\int_C^E{\overrightarrow{F}}d \overrightarrow{R}+\int_E^C{\overrightarrow{F}}d \overrightarrow{R}=\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}$

Isn't the Green's Theorem:
$\oint_S{\overrightarrow{F} d \overrightarrow{R}}=\iint_R{\nabla \times \overrightarrow{F} dA}$?
Is this the same as $\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$?
Hi!! (Blush)

They look the same, but they are not.

Green's Theorem only applies to closed curves in a 2D plane (usually the XY plane).
Stokes' Theorem is more general and applies to any 3D curve that is the boundary of a surface.

Your proof seems to split up a 3D curve into 3 2D curves.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Hi!! (Blush)

They look the same, but they are not.

Green's Theorem only applies to closed curves in a 2D plane (usually the XY plane).
Stokes' Theorem is more general and applies to any 3D curve that is the boundary of a surface.

Your proof seems to split up a 3D curve into 3 2D curves.
Is $\hat{n} d \sigma = dA$?

So can we replace $\iint_{ABCDE} \nabla \times \overrightarrow{F} \cdot \hat{n} d \sigma$ with $\iint_{ABCDE} \nabla \times \overrightarrow{F} \cdot dA$ ?? (Wondering)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Is $\hat{n} d \sigma = dA$.

So can we replace $\iint_{ABCDE} \nabla \times \overrightarrow{F} \cdot \hat{n} d \sigma$ with $\iint_{ABCDE} \nabla \times \overrightarrow{F} \cdot dA$ ?? (Wondering)
Yes and no.
The symbol $dA$ can indeed mean $\hat n d\sigma$.
But in this particular case I believe $dA$ is supposed to be an infinitesimal surface element in the XY plane, because that is its meaning in Green's theorem.
And $\hat n d\sigma$ is intended to be an infinitesimal surface element in 3D, which is how it is used in Stokes' theorem. (Wasntme)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
Yes and no.
The symbol $dA$ can indeed mean $\hat n d\sigma$.
But in this particular case I believe $dA$ is supposed to be an infinitesimal surface element in the XY plane, because that is its meaning in Green's theorem.
And $\hat n d\sigma$ is intended to be an infinitesimal surface element in 3D, which is how it is used in Stokes' theorem. (Wasntme)
Applying the Green's theorem at $ABE, BCE, CDE$ we get:
$$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}=\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$$

Since it is the Green's theorem, why don't we use $dA$? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Applying the Green's theorem at $ABE, BCE, CDE$ we get:
$$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}=\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$$

Since it is the Green's theorem, why don't we use $dA$? (Wondering)
I don't know.
It's just a symbol and either can be used anyway. (Wasntme)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,047
$\overrightarrow{F}=M \hat{i}+ N \hat{j}+ P \hat{k}$

To prove the Stokes' Theorem we apply Green's Theroem at $ABE$, $BCE$, $CDE$.

$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}=\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$
$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F} }d \overrightarrow{R}=\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}+\int_B^E{\overrightarrow{F}}d \overrightarrow{R}+\int_E^B{\overrightarrow{F}}d \overrightarrow{R}+\int_C^E{\overrightarrow{F}}d \overrightarrow{R}+\int_E^C{\overrightarrow{F}}d \overrightarrow{R}=\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}$
Why do we take at the integral $\displaystyle{\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma}$ the limit $ABCDE$, and at the integral $\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}$ we take the same limit??(Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
$\overrightarrow{F}=M \hat{i}+ N \hat{j}+ P \hat{k}$

To prove the Stokes' Theorem we apply Green's Theroem at $ABE$, $BCE$, $CDE$.

$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}=\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$
$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F} }d \overrightarrow{R}=\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}+\int_B^E{\overrightarrow{F}}d \overrightarrow{R}+\int_E^B{\overrightarrow{F}}d \overrightarrow{R}+\int_C^E{\overrightarrow{F}}d \overrightarrow{R}+\int_E^C{\overrightarrow{F}}d \overrightarrow{R}=\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}$
Why do we take at the integral $\displaystyle{\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma}$ the limit $ABCDE$, and at the integral $\oint_{ABCDE}{\overrightarrow{F}}d \overrightarrow{R}$ we take the same limit??(Wondering)
In the first line we have from Green's theorem that:
$$\oint_{ABE} \overrightarrow{F}d \overrightarrow{R} = \iint_{ABE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$$
So:
$$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}
=\iint_{ABE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma
+\iint_{BCE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma
+ \iint_{CDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$$
Those 3 surface integrals on the right hand side can be bundled to yield:
$$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}
=\iint_{ABCDE}{ \nabla \times \overrightarrow{F} \cdot \hat{n}}d \sigma$$


In the second line:
$$(\oint_{ABE}+\oint_{BCE}+\oint_{CDE}){ \overrightarrow{F}}d \overrightarrow{R}$$
we follow the path of the line integrals.
In this path for instance BE is first traversed in one direction and than back in the other direction.
We can split this path up in the path ABCDE and the traversals of each of BE and CE back and forth.


As a result we can conclude that the right hand side of the first line must be equal to the right hand side of the second line. (Mmm)