Page 144 - Vector Analysis

140 CHAPTER 4. Vector Calculus

Theorem 4.83. Let Σ Ď R3 be an oriented C 1-surface with C 1-boundary B Σ, N : Σ Ñ S2
be a continuous unit normal vector field on Σ, and T : B Σ Ñ S2 be tangent vector on B Σ
such that T is compatible with N (which means T ˆ N points away from Σ). Then

żż @ u P TΣ X C 1(Σ; R3) X C (Σ; R3) ,

u ¨ (T ˆ N) ds = divΣu dS

BΣ Σ

where divΣ is the surface divergence operator.

Proof. Let tVm, ψmuKm=1 denote a collection of local parametrization of Σ such that ψm(Vm)X
B Σ = H for 1 ď m ď J, and ψm(Vm) X B Σ is non-empty and connected for J + 1 ď m ď K.
W.L.O.G., we can assume that Vm = Bm ” B(0, rm) for some rm ą 0. Write Um = ψm(Vm),
and let tgmuKm=1 be the associated metric tensor, as well as the associated first fundamental
form gm = det(gm). Let tζmuKm=1 be a partition-of-unity of Σ subordinate to t UmuKm=1.
Then

ż Kż

divΣu dS = ÿ ζmdivΣu dS

Σ m=1 UmXΣ

= J 2ż ˝ ψm ) B [?gm gmαβ ( ˝ ψm) ¨ B ψm )]
ÿ B yα (u dy
ÿ B yβ
(ζm
m=1
α,β=1 Bm

+ K 2 ż ˝ ψm) B [?gmgmαβ ( ˝ ψm) ¨ B ψm )] .
(ζm B yα (u dy
ÿ ÿ B yβ
Bm+
m=J +1 α,β=1

Let n denote the outward-pointing unit normal on either BBm for 1 ď m ď J or BBm+ for
J + 1 ď m ď K. Since ζm ˝ ϑm = 0 on BB(0, rm) for 1 ď m ď J, and ζm ˝ ϑm = 0 on
ty2 ą 0u X BB(0, rm) for J + 1 ď m ď K, the divergence theorem (on R2) implies that

ż divΣu dS = K 2ż [?gmgmαβ ( ˝ ψm) ¨ B ψm )] B (ζm ˝ ψm) dy
(u B yβ B yα
Σ ÿÿ

´

m=1 α,β=1 ψm´1(UmXΣ)

+ K 2 ż ˝ ψm )nα[?gmgmαβ ( ˝ ψm) ¨ B ψm )] dy1
(u
ÿ ÿ (ζm B yβ

m=J +1 α,β=1 BmXty2=0u

Kż (u ¨ ∇Σζm) ˝ ψm?gm dy

ÿ
=´

m=1 ψm´1(UmXΣ)

K ż [ 2 nα?gmgmαβ B ψm ]
B yβ
+ ÿ (ζm ˝ ψm)(u ˝ ψm) ¨ ÿ dy1 .

m=J +1 BmXty2=0u α,β=1