Different forms of Stokes' theorem

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SUMMARY

This discussion focuses on the application of Stokes' theorem to the vector field defined as ##\vec V = \vec a \phi##. The participants analyze the right-hand side (R.H.S) and left-hand side (L.H.S) of the theorem, concluding that ##\phi \vec \nabla \times \vec a + (\vec \nabla \phi) \times \vec a## simplifies to ##- \vec a \times (\vec \nabla \phi)##. They confirm that both sides of the equation equate, validating the use of Stokes' theorem in this context. The discussion also touches on the implications for the second part of the theorem involving ##\vec V = \vec a \times \vec P##.

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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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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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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##)
 
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.
 

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