Identity proof using Stoke's Theorem

In summary, the conversation discussed the use of Stoke's Theorem to prove that an open surface with boundary C, a space curve, and a scalar field f(\vec r) can be evaluated using the vector function \textbf{F}\equiv f\textbf{c}, where \textbf{c} is a constant vector. After some calculation and clarification of the use of constants in the integral, it was determined that the constant vector can be pulled out of the integrals, leading to the conclusion that the integral must equal zero.
  • #1
pretzsp
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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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  • #2
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:
 
  • #3
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)
 
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  • #4
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?
 
  • #5
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.
 

FAQ: Identity proof using Stoke's Theorem

1. What is Stoke's Theorem?

Stoke's Theorem is a mathematical concept that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field along the closed boundary curve of the surface.

2. How is Stoke's Theorem used to prove identity?

Stoke's Theorem is used in multivariable calculus to prove identities involving vector fields. By relating the surface integral to the line integral, it allows for the evaluation of one integral in terms of the other, which can then be simplified and equated to prove an identity.

3. What is the significance of using Stoke's Theorem for identity proof?

Stoke's Theorem provides a powerful tool for proving identities involving vector fields. It allows for the evaluation of complex integrals through simpler ones, and can also be used to convert between different types of integrals, such as surface and line integrals.

4. What types of identities can be proven using Stoke's Theorem?

Stoke's Theorem can be used to prove a wide range of identities involving vector fields, including the fundamental theorem of calculus, Green's theorem, and the Divergence theorem. It is also a key component in the proof of the Helmholtz decomposition theorem.

5. Are there any limitations to using Stoke's Theorem for identity proof?

While Stoke's Theorem is a powerful tool, it does have some limitations. It can only be used for identities involving vector fields, and the surfaces and boundaries must be well-defined and smooth. It also requires a good understanding of multivariable calculus and vector calculus concepts.

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