Page 126 - Vector Analysis

122 CHAPTER 4. Vector Calculus

thus making a change of variable y = tan2 θ we conclude that

the surface area of Σ = 2 żπ arcsin tan θ d(tan2 θ) = 2 żπ θ ( tan2 θ)
4 sec θ 4 d
0 0
[ ]
2θ ˇθ= π ż π θdθ
θˇ 4 4
tan2 tan2
= ˇθ=0 ´

[π 0 ] [π ]
2 dθ 2
ż π ˇθ= π
4 θ)ˇ 4
(sec2 (tan
= 4 ´ θ ´ 1) = 4 ´ θ ´ ˇθ=0
π
[ (0 π )]

= 2 ´ 1´ = π´2.
44

4.3.3 The surface element and the surface integral

Let Σ Ď R3 be a regular surface, and tV, ψu be a parametrization of Σ such that ψ(V) = Σ.

If f : Σ Ñ R is a bounded continuous function, the surface integral of f over Σ, denoted by

ż

f dS, is defined by

Σ ż ż
=
f dS (f ˝ ψ)?g dA . (4.12)

ΣV

żż

In particular, if f ” 1, the number dS ” 1 dS is the surface area of Σ.

ΣΣ

Since the surface integrals defined by (4.12) seems to depend on a given parametrization,

before proceeding we show that the surface integral is indeed independent of the choice of the

parameterizations. Suppose that tV1, ψ1u and tV2, ψ2u are two local C 1-parameterizations
of a regular surface Σ at p, g1, g2 denote the metric tensors associated with the parameter-
izations tV1, ψ1u, tV2, ψ2u, respectively, and g1 = det(g1), g2 = det(g2) are corresponding
first fundamental forms. Let Ψ = ψ2´1 ˝ ψ1. Then the change of variables formula (Theorem
3.31) implies that

ż (f ˝ ψ2)?g2 dA = ż Ψ)(?g2 ) ż ψ1)(?g2 )
Ψ |JΨ| Ψ |JΨ|
(f ˝ ψ2 ˝ ˝ dA = (f ˝ ˝ dA ,

V2 V1 V1

where JΨ is the Jacobian of the map Ψ. Using (4.7), we find that

[DΨ]T[(Dψ2) ˝ Ψ]T[(Dψ2) ˝ ][ ] = [ ]T[Dψ1] ;
Ψ DΨ Dψ1

thus by the fact that g1 = det ( ]T[Dψ1]) and g2 = det ([Dψ2]T[Dψ2 ) we obtain that
[Dψ1 ],

det ([DΨ])2(g2 ˝ Ψ) = g1 .