Identity proof using Stoke's Theorem

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Homework Help Overview

The discussion revolves around applying Stokes' Theorem to demonstrate a relationship involving a scalar field and an open surface with boundary. The participants are exploring the implications of the theorem in the context of vector calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to evaluate the expression involving the curl of the gradient of a scalar field but encounters difficulties. Some participants suggest using a specific vector function related to the scalar field, while others question the validity of manipulating the dot product with a constant vector.

Discussion Status

The discussion is active, with participants providing insights and clarifications regarding the application of Stokes' Theorem. There is an exchange of ideas about the mathematical steps involved, and some guidance has been offered regarding the treatment of constant vectors in integrals.

Contextual Notes

Participants are navigating the complexities of vector calculus and the assumptions inherent in the application of Stokes' Theorem. There is an emphasis on understanding the manipulation of vector quantities within the context of the theorem.

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Homework Statement



Show using Stoke's Theorem that
gif.latex?\bg_white%20\iint_S%20(\nabla%20f)\times%20d\vec%20S%20=%20-\oint_C%20fd%20\vec%20r.gif


S is an open surface with boundary C (a space curve). [tex]f(\vec r)[/tex] is a scalar field.

Homework Equations



Stoke's theorem [tex]\iint_S (\nabla\times \vec F) \cdot d\vec S = \int_C F \cdot d\vec r[/tex]

The Attempt at a Solution



Thus far I've tried evaluating [tex](\nabla f)\times d\vec S[/tex] by taking components in various ways and then trying to force it to become something that I can take the curl of... with no success at all. I'm completely stumped now.
 
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Try applying Stokes' theorem to the vector function [itex]\textbf{F}\equiv f\textbf{c}[/itex], where [itex]\textbf{c}[/itex] is any constant vector.:wink:
 
Well, what I get here is

[tex]\iint_S \vec c\cdot\nabla f \times d\vec S = -\oint \vec c \cdot fd\vec r[/tex]

And I'm not entirely sure whether I can just "cancel" the c dot product...

(should be a vector c there, by the way)
 
Last edited:
Well, since [itex]\textbf{c}[/itex] is a constant vector, you can certainly pull it out of the integrals and say

[tex]\vec c\cdot\left(\iint_S \nabla f \times d\vec S\right) = \vec c\cdot\left(-\oint fd\vec r\right)[/tex]

[tex]\implies \vec c\cdot \left(\iint_S \nabla f \times d\vec S+\oint fd\vec r\right)=0[/tex]

And since this is true for arbitrary, constant [itex]\vec{c}[/itex], what must you conclude?
 
Yeah, those last steps were fine for me, I just wasn't sure if that even with c a constant vector that it was "allowed" to just pull it out like so. Thanks for the help.
 

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