Different forms of Stokes' theorem

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In summary, Stokes' theorem encompasses various forms that relate surface integrals and line integrals across different contexts in vector calculus. The most familiar form connects the surface integral of the curl of a vector field over a surface to the line integral of the vector field along the boundary of that surface. Other variations include the generalized Stokes' theorem, which applies to manifolds and differential forms, and specific cases like Green's theorem and the divergence theorem, which further illustrate the relationships between fields and integrals in two and three dimensions. Each form highlights the fundamental connection between local properties of vector fields and their global behavior over surfaces and curves.
  • #1
MatinSAR
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Homework Statement
Find the different forms using ##\vec V=\vec a \phi## and ##\vec V=\vec a \times \vec P## for constant ##\vec a##.
Relevant Equations
Stokes' theorem.
What am I trying to do for ##\vec V=\vec a \phi## :
##R.H.S= \oint \vec V \cdot d \vec \lambda=\oint \vec a \phi \cdot d \vec \lambda=\vec a \cdot \oint \phi d \vec \lambda ##

##L.H.S= \iint_S \vec \nabla \times \vec V \cdot \vec d \sigma=\iint_S \vec \nabla \times (\vec a \phi) \cdot \vec d \sigma=\iint_S (\phi \vec \nabla \times \vec a + (\vec \nabla \phi) \times \vec a) \cdot \vec d \sigma= ?##
I think ##\phi \vec \nabla \times \vec a + (\vec \nabla \phi) \times \vec a## should be 0. why is this wrong?

##\phi \vec \nabla \times \vec a## is 0 because ##\vec a## is a constant vector.
##(\vec \nabla \phi) \times \vec a## is 0 because ##\vec \nabla## acts on both ##\phi## and ##\vec a## so it should be zero.

Edit:
Now I think ##(\vec \nabla \phi) \times \vec a## is not 0 because ##\vec \nabla## acts only on ##\phi## so we can rewrite it as ##- \vec a \times (\vec \nabla \phi).##
 
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  • #2
MatinSAR said:
Now I think ##(\vec \nabla \phi) \times \vec a## is not 0 because ##\vec \nabla## acts only on ##\phi## so we can rewrite it as ##\vec a \times (\vec \nabla \phi).##
Right.
 
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  • #3
haruspex said:
Right.
Thanks for the reply @haruspex .
##L.H.S= \iint_S \vec \nabla \times \vec V \cdot d \vec \sigma=\iint_S \vec \nabla \times (\vec a \phi) \cdot d \vec \sigma=\iint_S (\phi \vec \nabla \times \vec a + (\vec \nabla \phi) \times \vec a) \cdot d \vec \sigma=##
##- \iint_S \vec a \times (\vec \nabla \phi) \cdot d \vec \sigma=- \iint_S \vec a \cdot (\vec \nabla \phi) \times d \vec \sigma=\vec a \cdot \iint_S d \vec \sigma \times (\vec \nabla \phi) ##
L.H.S = R.H.S
##\vec a \cdot \iint_S d \vec \sigma \times (\vec \nabla \phi) = \vec a \cdot \oint \phi d \vec \lambda ##
##\iint_S d \vec \sigma \times (\vec \nabla \phi) = \oint \phi d \vec \lambda##

I hope I won't have problem with other part. (##\vec V=\vec a \times \vec P##)
 
  • #4
I've managed to prove 2nd part using what I've learnt here : https://www.physicsforums.com/threads/vector-operators-grad-div-and-curl.1057533/

But I'm not sure if my proof is mathematically true or it is nonsense. Picture of my work:
2023_12_23 3_15 PM Office Lens.jpg


I would be grateful if someone could point out the problem with my proof.
 

FAQ: Different forms of Stokes' theorem

What is Stokes' theorem in vector calculus?

Stokes' theorem in vector calculus relates a surface integral over a surface \( S \) to a line integral over its boundary \( \partial S \). Mathematically, it is expressed as \(\int_{S} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r}\), where \( \mathbf{F} \) is a vector field, \( \nabla \times \mathbf{F} \) is its curl, and \( d\mathbf{S} \) is the surface element vector.

How is the Divergence theorem related to Stokes' theorem?

The Divergence theorem, also known as Gauss's theorem, relates the flux of a vector field through a closed surface to the volume integral of the divergence of the field inside the surface. It is expressed as \(\oint_{S} \mathbf{F} \cdot d\mathbf{S} = \int_{V} (\nabla \cdot \mathbf{F}) \, dV\). While Stokes' theorem deals with the curl of a field and a surface integral, the Divergence theorem deals with the divergence of a field and a volume integral. Both theorems are special cases of the more general Stokes' theorem in differential geometry.

What is the general form of Stokes' theorem in differential geometry?

The general form of Stokes' theorem in differential geometry states that for a manifold \( M \) with boundary \( \partial M \), and a differential form \( \omega \) of degree \( n-1 \), the integral of the exterior derivative \( d\omega \) over \( M \) is equal to the integral of \( \omega \) over \( \partial M \). Mathematically, it is expressed as \(\int_{M} d\omega = \int_{\partial M} \omega\). This general form encompasses various classical theorems, including Stokes' theorem in vector calculus, the Divergence theorem, and Green's theorem.

How does Green's theorem relate to Stokes' theorem?

Green's theorem is a special case of Stokes' theorem applied to a region in the plane. It relates a double integral over a region \( D \) to a line integral over its boundary \( \partial D \). Mathematically, it can be written in the form \(\int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \oint_{\partial D} (P \, dx + Q \

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