194
IX – Multivariate Differential and Integral Calculus
In particular,
dω(x; h, k) =
d
2
dtds
I(x; sh, tk) for s = t = 0 .
Having done this, we can “ retrace ” our calculations; as the right hand side
of (3) is clearly zero for t = 0, the FT and (4) show that
d
ds
I(x; sh, tk) =
t
0
dω(x + sh + vk; h, k).dv ;
and as I(x; sh, tk) = 0 for s = 0, it also follows that
I(x; sh, tk) =
s
0
du
t
0
dω(x + uh + vk; h, k)dv .
(9.5)
Hence for s = t = 1,
∂P (x;h,k)
ω =
I 2
dω(x + uh + vk; h, k)dudv ,
(9.6)
which is an extended double integral over the square I
2 = I × I in the plane;
this supposes that h and k are sufficiently small so that the surface of the
parallelogram P (x; h, k) can be contained in the open subset on which ω is
defined.
The simplest case can be obtained by assuming that ω = pdx + qdy is a
form on R
2 , and so
dω = (D 1 q − D 2 p)dx ∧ dy = p 12 dx ∧ dy .
(9.7)
(4) can then be written
∂P
pdx + qdy =
I 2
p 12 (x + uh + vk)
h
1 k
2
− h
2 k
1
dudv ,
(9.8)
where P is a parallelogram with initial point x generated by the vectors h
and k. In particular, if x = 0 and if h and k are the unit vectors of the
coordinate axes, then we get the Green-Riemann formula, unless it be the
Gauss formula,
∂I 2
pdx + qdy =
I 2
(D 1 q − D 2 p) dxdy
(9.9)
for the square I
2 , provided the boundary ∂I
2 of I
2 is given the usual positive
orientation. Cauchy proved it directly:
I 2
D 1 qdxdy =
1
0
dy
1
0
D 1 q(x, y)dx =
1
0
[q(1, y) − q(0, y)] dy =
=
∂I 2
qdy