Stokes theorem states that
int_{S} curl F . dS = int_{C} F . dr
If we rewrite the left hand side in terms of a parametrization
r(u,v) from a region D in the uv 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
int_{B} (F . r_{u}, F . r_{v}) . dr
(Check this! it amounts to the chain rule). By Green's theorem (applied
in the uv plane) this is
int_{D} d/du (F . r_{v})  d/dv (F . r_{u}) du dv
Thus if we can show that
curl(F) . (r_{u} x r_{v}) = d/du (F . r_{v})  d/dv (F . r_{u})
we'll have shown that Stokes theorem follows from Green's theorem.
This can be done using vector triple product identities as follows:
Lets pretend
first the vectors r_{u} and r_{v} 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 r_{u}
because r_{u} and r_{v} are constant:
(nabla x F) . (r_{u} x r_{v}) = ((nabla x F) x r_{u}) . r_{v}
= ((r_{u} . nabla) F  nabla (F . r_{u})) . r_{v}
= (r_{u} . nabla)(F . r_{v})  (r_{v} . nabla)(F . r_{u})
= d/du (F . r_{v})  d/dv (F . r_{u}) (*)
= dF/du . r_{v}  dF/dv . r_{u} (**)
Now lets stop pretending r_{u} and r_{v} are constant.
We'd really like the original expression (*), but with r_{u} and r_{v}
not constant, so let's try expanding (*) via the chain rule. We get:
dF/du . r_{v} + F . r_{vu}  dF/dv . r_{u}  F . r_{uv}.
Thankfully, two terms cancel by Clairaut's theorem, and the two expressions
(*) and (**) are equal, even if r_{u} and r_{v} are allow to vary.
If P(u,v) = F . r_{u} and Q(u,v) = F . r_{v},,
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
Green's Theorem.
