# Homework Help: Identity proof using Stoke's Theorem

1. May 16, 2010

### pretzsp

1. The problem statement, all variables and given/known data

Show using Stoke's Theorem that

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

2. Relevant equations

Stoke's theorem $$\iint_S (\nabla\times \vec F) \cdot d\vec S = \int_C F \cdot d\vec r$$

3. The attempt at a solution

Thus far I've tried evaluating $$(\nabla f)\times d\vec S$$ 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.

2. May 16, 2010

### gabbagabbahey

Try applying Stokes' theorem to the vector function $\textbf{F}\equiv f\textbf{c}$, where $\textbf{c}$ is any constant vector.

3. May 16, 2010

### pretzsp

Well, what I get here is

$$\iint_S \vec c\cdot\nabla f \times d\vec S = -\oint \vec c \cdot fd\vec r$$

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: May 16, 2010
4. May 16, 2010

### gabbagabbahey

Well, since $\textbf{c}$ is a constant vector, you can certainly pull it out of the integrals and say

$$\vec c\cdot\left(\iint_S \nabla f \times d\vec S\right) = \vec c\cdot\left(-\oint fd\vec r\right)$$

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

And since this is true for arbitrary, constant $\vec{c}$, what must you conclude?

5. May 16, 2010

### pretzsp

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.