Page 127 - Vector Analysis

§4.3 The Surface Integrals 123

Since JΨ = det ([ ]) the identity above implies that |JΨ|(?g2 ˝ Ψ) = ?g1, so we conclude
DΨ ,

that

ż (f ˝ ψ1)?g1 dA = ż (f ˝ ψ2)?g2 dA . (4.13)

V1 V2

Therefore, the surface integral of f over Σ is independent of the choice of parameterizations

of Σ. In particular, the surface area of a regular C 1-surface which can be parameterized by

a global parametrization is also independent of the choice of parameterizations.
As noticed in Remark 4.45, the first fundamental form ?g associated with the parametriza-

tion tV, ψu can be viewed as the Jacobian of the map ψ. Therefore, we arrive at the con-
clusion that dS ‘‘ = ”?g dA. dS is called the surface element. Moreover, similar to the

reason provided in Remark 4.22, the surface integral of a positive continuous function f

over Σ, where f is considered as the mass density of the surface given by

f (x) = lim the mass of ψ(∆)
diam(∆)Ñ0 the surface area of ψ(∆)

ψ´1 (x)P∆

is the total mass of the surface.
Next, we study the surface area of general regular surfaces that cannot be parameterized

using a single pair tV, ψu. Let Σ Ď R3 be a regular surface, and tVi, ψiuiPI be a collection
of local parameterizations satisfying that for each p P Σ there exists i P I such that tVi, ψiu
is a local parametrization of Σ at p. If there exists a countable collection of non-negative
functions tζjujPJ defined on Σ such that

1. For each j P J, spt(ζj) ” the closure of ␣x P Σ ˇ ζj (x) ‰ 0( Ď Vi for some i P I;
ˇ

2. ř ζj(x) = 1 for all x P Σ,

jPJ

then intuitively we can compute the surface area by

żż

ÿ (4.14)

dS = ζj dS ,

Σ jPJ Σ

where the surface integral of ζj over Σ is defined by (4.12) since spt(ζj) Ď ψ(Vi) and ζj = 0

outside spt(ζj). In other words, each term on the right-hand side of (4.14) can be evaluated

by

ż ż (ζj ˝ ψi)?gi dS .
ζj dS =

Σ Vi