Page 150 - Vector Analysis

146 CHAPTER 4. Vector Calculus

Proof of the Stokes theorem. Using (4.23) and then applying the divergence theorem on
surfaces with boundary (Theorem 4.83), we find that

żż ż ż

curlu ¨ N dS = divΣ(u ˆ N) dS = (u ˆ N) ¨ (T ˆ N) ds = (u ¨ T) ds

ΣΣ BΣ BΣ

in which the identity (u ˆ N) ¨ (T ˆ N) = u ¨ T is used. ˝

Example 4.88. Let Σ be the surface given in Example 4.51, and F : R3 Ñ R3 be a vector-
valued function given by F(x, y, z) = (y, ´x, 0). Then by the definition of line integral,

¿ żπ ( )
2 t dt
(sin2 t, ´ cos t sin t, 0) ¨ cos2 t ´ sin2 t, 2 sin t cos t, ´ sin
F ¨ dr =

C ´ π
2

żπ ( sin2 t cos2 t ´ sin4 t ´ 2 sin2 t cos2 )
2 t dt

=

´ π
2
(t )ˇ
π ż π 1 ´ cos 2t ´ sin 2t π π
´ dt 2 =´ .
ż2 sin2 2
=´
tdt = = ´ ˇ

´ π 2´ π 2 4 ˇ´ π 2
2 2 2

while by the fact that curlF = (0, 0, ´2), the Stokes theorem implies that

¿ż ż ż π ż π´ϕ
2
F ¨ dr = (0, 0, ´2) ¨ N dS = ´2 cos ϕ sin ϕ d(θ, ϕ) = ´2 sin ϕ cos ϕ dθdϕ

CΣ ψ´1(Σ) 0ϕ

ż π (π 1 )ˇϕ= π
2 2ϕ ˇ 2
sin 2ϕ cos cos sin
= ´ (π ´ 2ϕ) dϕ = 2ϕ ´ ϕ 2ϕ +
0 2 2 ˇϕ=0
πππ π
=´ ´ + =´ .
222 2

Example 4.89. Let C be a smooth curve parameterized by t P [0, 2π] .
()

r(t) = cos(sin t) sin t, sin(sin t) sin t, cos t ,

Then the curve C is a closed curve on S2, and divide S2 into two parts. Let Σ denote the

part with smaller area. z

xy

2. Plot C1 and C2 on the θφ-plane. The curve C divides the unit sphere into two parts, and let Σ
be the part containing the point (0, 1, 0). Identify the corresponding region of Σ on θφ-plane.

3. Find the surface area of Σ.