Different forms of Stokes' theorem

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The discussion revolves around the application of Stokes' theorem to the vector field defined as ##\vec V=\vec a \phi##. The right-hand side is evaluated as ##\vec a \cdot \oint \phi d \vec \lambda##, while the left-hand side involves the curl of the vector field, leading to the expression ##\phi \vec \nabla \times \vec a + (\vec \nabla \phi) \times \vec a##. Initially, it was thought that both terms could be zero, but it was clarified that ##\phi \vec \nabla \times \vec a## is zero due to ##\vec a## being constant, while ##(\vec \nabla \phi) \times \vec a## is not zero and can be rewritten as ##\vec a \times (\vec \nabla \phi)##. The conclusion reached is that the left-hand side equals the right-hand side, confirming the validity of the application of Stokes' theorem in this context.
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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.
 
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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