Page 115 - Vector Analysis

§4.3 The Surface Integrals 111

Remark 4.37. Write ψ : V Ñ Σ as
()

ψ(u, v) = x(u, v), y(u, v), z(u, v) .

Then if q = (u0, v0),

 xu(u0, v0) xv(u0, v0) 
yu(u0, v0) yv(u0, v0)
[] =   = [ , v0)]...[ψ,2(u0, v0 ] .
(Dψ)(q) [ψ,1(u0 )]

zu(u0, v0) zv(u0, v0)

The injectivity of Dψ(q) is then translated to that the two vectors
()

ψ,1 (u0, v0) ” ψu(u0, v0) = xu(u0, v0), yu(u0, v0), zu(u0, v0)
()

ψ,2 (u0, v0) ” ψv(u0, v0) = xv(u0, v0), yv(u0, v0), zv(u0, v0)

are linearly independent. Therefore, the range of Dψ(q) is the span of the two vectors ψ,1 (q)

and ψ,2 (q) and is indeed a plane for all q P V.

Let p P Σ and q = ψ´1(p). Since Dψ(q) is injective, each v P TpΣ corresponds a

u[vn]iq=u[eDvψec(tqo)r][(aa,,bb]T) ,PaRnd2 such that v = aψ,1 (q) + bψ,2 (q). This vector (a, b) P R2 satisfies
can be computed by

[ a ] ([Dψ(q)]T[Dψ(q)])´1[Dψ(q)]T[v]
b
= .

Example 4.38. Let S2 = ␣(x, y, z) P R3 ˇ x2 + y2 + z2 = 1( be the unit sphere in R3.
ˇ

If p = (x0, y0, z0) P S2, then either x0, y0 or z0 is non-zero. Suppose that z0 ‰ 0. Let

r = 1 ´ ax20 + y02 ą 0. Define

# (x, y, a1 ´ x2 ´ y2) if z0 ą 0 ,
ψ(x, y) = (x, y, ´a1 ´ x2 ´ y2) if z0 ă 0 ,

()
V = B (x0, y0), r , and U = ψ(V). Then ψ : V Ñ U is a bijection. Let φ = ψ´1. Then

tU , φu is a coordinate chart at p; thus S2 is a surface.

There exists another coordinate chart. Let U1 = S2z(0, 0, ´1) and U2 = S2z(0, 0, 1).

Define the map φ1 : U1 Ñ R2 by that φ1(p) is the unique point on R2 such that (0, 0, ´1),

φ1(p) and (x, y, 0) are on the same straight line. Similarly, define φ2 : U2 Ñ R2 by that φ2(p)

is the unique point on R2 such that (0, 0, 1), φ2(p) and (x, y, 0) are on the same straight

line. It is easy to check that if p P S2, then either t U1, φ1u or t U2, φ2u is a coordinate chart

at p.