198
IX – Multivariate Differential and Integral Calculus
for any form of degree 2 and any 2-dimensional path σ : I
2
−→ X of
class C
1 .
Denoting by σ
i (s, t) the coordinates of σ(s, t) with respect to a basis for
E, if
= p ij dx
i
∧ dx
j ,
with respect to this basis, then the expression integrated can be calculated
by replacing the dx
i by the dσ
i = D 1 σ
i (s, t)ds + D 2 σ
i (s, t)dt and so dx
i
∧ dx
j
by
J
ij (s, t)ds ∧ dt where J
ij = D 1 σ
i .D 2 σ
j
− D 1 σ
j .D 2 σ
i
is the Jacobian of the functions σ
i and σ
j with respect to s, t. So the right
hand side of (21’) reduces to the classical double integral
p ij [σ(s, t)] J
ij (s, t).dsdt =
r(s, t)dsdt .
(9.22)
The computation is immediate as long as the mechanism of exterior products
and inverse images has been understood.
Version (20) of the result obtained suggests a far quicker proof, which
amount to applying Gauss’ formula (9) to the square I
2 = K. Indeed, the
left hand side is by definition a curvilinear integral: the integral over I of
the inverse image ω ◦ σ . But denoting by ∂K the path with initial point 0
consisting in following the border of K counterclockwise, clearly
∂σ = σ ◦ ∂K ,
and so ω ◦ ∂σ = ω ◦ (σ ◦ ∂K) = (ω ◦ σ) ◦ ∂K by (8.19) . The left hand side
of (20) is, therefore, the integral of ω ◦ σ along the path ∂K. On the other
hand, by (8.21),
dω ◦ σ = d(ω ◦ σ) .
Setting ω ◦ σ = θ to be a form of degree 1 on K, by (21’), relation (20) means
that
∂K
θ =
K
dθ ,
which reduces to (9) as expected.
The preceding calculation can be generalized. Consider two Cartesian
spaces E and F , two open sets U ⊂ E and V ⊂ F , a map f : U −→ V and
a form of degree 2 on V . Let σ : I
2
−→ U be a 2-dimensional path in U .
This gives a path f ◦ σ : I
2
−→ V in V . Having said this,
σ
◦ f =
f ◦σ
.
(9.23)
IX – Multivariate Differential and Integral Calculus
for any form of degree 2 and any 2-dimensional path σ : I
2
−→ X of
class C
1 .
Denoting by σ
i (s, t) the coordinates of σ(s, t) with respect to a basis for
E, if
= p ij dx
i
∧ dx
j ,
with respect to this basis, then the expression integrated can be calculated
by replacing the dx
i by the dσ
i = D 1 σ
i (s, t)ds + D 2 σ
i (s, t)dt and so dx
i
∧ dx
j
by
J
ij (s, t)ds ∧ dt where J
ij = D 1 σ
i .D 2 σ
j
− D 1 σ
j .D 2 σ
i
is the Jacobian of the functions σ
i and σ
j with respect to s, t. So the right
hand side of (21’) reduces to the classical double integral
p ij [σ(s, t)] J
ij (s, t).dsdt =
r(s, t)dsdt .
(9.22)
The computation is immediate as long as the mechanism of exterior products
and inverse images has been understood.
Version (20) of the result obtained suggests a far quicker proof, which
amount to applying Gauss’ formula (9) to the square I
2 = K. Indeed, the
left hand side is by definition a curvilinear integral: the integral over I of
the inverse image ω ◦ σ . But denoting by ∂K the path with initial point 0
consisting in following the border of K counterclockwise, clearly
∂σ = σ ◦ ∂K ,
and so ω ◦ ∂σ = ω ◦ (σ ◦ ∂K) = (ω ◦ σ) ◦ ∂K by (8.19) . The left hand side
of (20) is, therefore, the integral of ω ◦ σ along the path ∂K. On the other
hand, by (8.21),
dω ◦ σ = d(ω ◦ σ) .
Setting ω ◦ σ = θ to be a form of degree 1 on K, by (21’), relation (20) means
that
∂K
θ =
K
dθ ,
which reduces to (9) as expected.
The preceding calculation can be generalized. Consider two Cartesian
spaces E and F , two open sets U ⊂ E and V ⊂ F , a map f : U −→ V and
a form of degree 2 on V . Let σ : I
2
−→ U be a 2-dimensional path in U .
This gives a path f ◦ σ : I
2
−→ V in V . Having said this,
σ
◦ f =
f ◦σ
.
(9.23)