Stokes theorem states that
intS curl F . dS = intC F . dr
If we rewrite the left hand side in terms of a parametrization
r(u,v) from a region D in the u-v plane, we get
intD curl(F) . (ru x rv) du dv
If we rewrite the right hand side in terms of this parametrization, we get,
letting B be the boundary of D
intB (F . ru, F . rv) . dr
(Check this! it amounts to the chain rule). By Green's theorem (applied
in the u-v plane) this is
intD d/du (F . rv) - d/dv (F . ru) du dv
Thus if we can show that
curl(F) . (ru x rv) = d/du (F . rv) - d/dv (F . ru)
we'll have shown that Stokes theorem follows from Green's theorem.
This can be done using vector triple product identities as follows:
first the vectors ru and rv are
constant in u and v, so that only derivatives of F matter: by identities which can
be found in 9.4 of the book and neglecting further derivatives of ru
because ru and rv are constant:
(nabla x F) . (ru x rv) = ((nabla x F) x ru) . rv
= ((ru . nabla) F - nabla (F . ru)) . rv
= (ru . nabla)(F . rv) - (rv . nabla)(F . ru)
= d/du (F . rv) - d/dv (F . ru) (*)
= dF/du . rv - dF/dv . ru (**)
Now lets stop pretending ru and rv are constant.
We'd really like the original expression (*), but with ru and rv
not constant, so let's try expanding (*) via the chain rule. We get:
dF/du . rv + F . rvu - dF/dv . ru - F . ruv.
Thankfully, two terms cancel by Clairaut's theorem, and the two expressions
(*) and (**) are equal, even if ru and rv are allow to vary.
If P(u,v) = F . ru and Q(u,v) = F . rv,,
we see that the line integral around the boundary, using the parametrization, is precisely
the line integral with respect to the vector field (P,Q). Stokes Theorem is reduced to