Stoke's theorem rewritten, not in book, but am I right?

[flyingpig]

Homework Statement

This is Stoke's theorem in my textbook and I am trying to write it in a new form

\iint_S curl \mathbf{F} \cdot d\mathbf{S}

The Attempt at a Solution

\iint_S curl \mathbf{F} \cdot d\mathbf{S} = \iint_S curl \mathbf{F} \cdot \hat{n} dS = \iint_S curl \mathbf{F} \cdot \frac{(\mathbf{r_u} \times \mathbf{r_v})}{|\mathbf{r_u} \times \mathbf{r_v}|} |\mathbf{r_u} \times \mathbf{r_v}| dA = \iint_S curl \mathbf{F} \cdot (\mathbf{r_u} \times \mathbf{r_v}) dA

 

[Pengwuino]

... what is \bf{r}_u and \bf{r}_v?

 

[flyingpig]

The partial derivatives of my parametric surface which I forgot to define...

So r(u,v)

 

[HallsofIvy]

Homework Statement

This is Stoke's theorem in my textbook and I am trying to write it in a new form

\iint_S curl \mathbf{F} \cdot d\mathbf{S}

No, that is not Stoke's theorem.

The Attempt at a Solution

\iint_S curl \mathbf{F} \cdot d\mathbf{S} = \iint_S curl \mathbf{F} \cdot \hat{n} dS = \iint_S curl \mathbf{F} \cdot \frac{(\mathbf{r_u} \times \mathbf{r_v})}{|\mathbf{r_u} \times \mathbf{r_v}|} |\mathbf{r_u} \times \mathbf{r_v}| dA = \iint_S curl \mathbf{F} \cdot (\mathbf{r_u} \times \mathbf{r_v}) dA

 

[flyingpig]

Sorry I meant that

\iint_S curl \mathbf{F} \cdot d\mathbf{S} = \oint \mathbf{F} \cdot d\mathbf{r}

 

[Dick]

The idea is correct. Except that n might be either +/-(r_u x r_v)/|r_u x r_v| depending on the direction of the normal implied by the problem.