Understanding Vector Calculus Proof: Divergence Theorem and Scalar Field

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SUMMARY

The discussion centers on the proof of the vector calculus identity involving the divergence theorem and a scalar field. The equation presented is ∫Vf(∇·A)dV=∫SfA·dS-∫VA·(∇f)dV, where f represents a scalar field and the surface integral is over a closed surface. Participants suggest rewriting the expression as ∫V f(∇·A) + A·(∇f)dV = ∫S fA·dS to facilitate the proof. This approach effectively utilizes the divergence theorem to simplify the proof process.

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physiks
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The following is used as part of a proof I'm trying to understand:

Vf(.A)dV=∫SfA.dS-∫VA.(f)dV
where f is a scalar field, and the surface integral is taken over a closed surface (which presumably encloses the volume).

I'm not sure how to go about proving this. I can see the divergence theorem will come into play at some stage, but the scalar field seems to be in the way to start with. This is probably really simple, I'm a little rusty with my vector calculus.

Clues would be helpful, thanks :)
 
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Hi physiks,
Rewrite your expression like $$\int_V f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)\, dV = \int_S f\mathbf{A} \cdot \mathbf{dS}$$ and work with the left hand side to show the right hand side.
 
CAF123 said:
Hi physiks,
Rewrite your expression like $$\int_V f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)\, dV = \int_S f\mathbf{A} \cdot \mathbf{dS}$$ and work with the left hand side to show the right hand side.

Got it, thanks!
 

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