# Stokes' Theorem formula question

1. Jun 19, 2011

### 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?

2. Jun 19, 2011

### 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)

3. Jun 19, 2011

### LCKurtz

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
$$d\vec S =\hat n dS =\pm\frac{\vec R_u\times\vec R_v}{|\vec R_u\times\vec R_v|} |\vec R_u\times\vec R_v|dudv=\pm\vec R_u\times \vec R_vdudv$$
This allows you to express everything in terms of u and v with appropriate uv limits.