Page 141 - Vector Analysis

§4.6 The Divergence Theorem 137

Corollary 4.76. Let Ω Ď Rn be a bounded Lipschitz domain, and f P C 1(Ω) X C (Ω). Then

ż Bf dx = ż f Ni dS ,
B xi
Ω BΩ

where Ni is the i-th component of the outward-pointing unit normal N of Ω.

Letting v be the product of a scalar function and a vector-valued function in Theorem
4.75, we conclude the following

Corollary 4.77. Let Ω Ď Rn be a bounded Lipschitz domain, and v P C 1(Ω; Rn) X C (Ω; Rn)
be a vector-valued function and φ P C 1(Ω) X C (Ω) be a scalar function. Then

żż ż

φ divv dx = (v ¨ N)φ dS ´ v ¨ ∇φ dx , (4.18)

Ω BΩ Ω

where N is the outward-pointing unit normal on B Ω.

Example 4.78. Let Ω be the the first octant part bounded by the cylindrical surface
x2 + z2 = a2 and the plane y = b, and F : Ω Ñ R3 be a vector-valued function defined by
F(x, y, z) = (x, y2, z).

z
a

ab
xy

Figure 4.5: The domain Ω and its five pieces of boundaries

With N denoting the outward-pointing unit normal of B Ω,

??
ż ż a ż b ż a2´x2 ż a ż a2´x2

divF d(x, y, z) = (2 + 2y) dzdydx = (b2 + 2b) dzdx

Ω 0 00 00

πa2(b2 + 2b)
=.

4

On the other hand, we note that the boundary of Ω has five parts: Σ as given in Example
4.72, two rectangles R1 = tx = 0u ˆ [0, b] ˆ [0, a], R2 = [0, a] ˆ [0, b] ˆ tz = 0u, and two