Stokes' Theorem formula question

[aimee3]

I was wondering, how you break down dS to something with dA? I know that dS is equal to ndS. The n is equal to grad f / (magnitude of grad f) and the dS is equal to the same as the magnitude of grad f right? So is the formula the same as double integral of region D (curl F * grad f) dA?

 

[grey_earl]

I don't know what f is (as opposed to F), but yes, that should be it.
(A comes from area, while S comes from surface - it is basically the same)

 

[LCKurtz]

I was wondering, how you break down dS to something with dA? I know that dS is equal to ndS. The n is equal to grad f / (magnitude of grad f) and the dS is equal to the same as the magnitude of grad f right? So is the formula the same as double integral of region D (curl F * grad f) dA?

Let's say you parameterize your surface S in terms of u and v as

\vec R = \vec R(u,v)

Now since

dS = |\vec R_u\times \vec R_v|dudv
and
\hat n =\pm\frac{\vec R_u\times\vec R_v}{|\vec R_u\times\vec R_v|}
with the sign chosen to agree with the orientation of the surface, you have
d\vec S =\hat n dS =\pm\frac{\vec R_u\times\vec R_v}{|\vec R_u\times\vec R_v|}<br /> |\vec R_u\times\vec R_v|dudv=\pm\vec R_u\times \vec R_vdudv
This allows you to express everything in terms of u and v with appropriate uv limits.