Page 137 - Vector Analysis

§4.6 The Divergence Theorem 133

Example 4.72. Find the flux integral of the vector field F(x, y, z) = (x, y2, z) upward
through the first octant part Σ of the cylindrical surface x2 + z2 = a2, 0 ă y ă b.

z
a

ab
xy

Figure 4.4: The surface Σ
Fist, we parameterize Σ by

?
ψ(u, v) = (u, v, a2 ´ u2), (u, v) P V = (0, a) ˆ (0, b) .

Since the first fundamental form g associated with tV, ψu is g = }ψ,1 ˆψ,2 }2R3 = a2 ,
and the upward-pointing unit normal is N(x, y, z) = ( x , 0, z ), we have a2 ´ u2

aa

ż F ¨ N dS = ż 1 (u2 + a2 ´ u2) ? a u2 d(u, v) = a2 ż 1 u2 d(u, v)
a a2 ´ ?
Σ V V
a2 ´

żbża 1 u ˇu=a πa2b
ˇ
= a2 ? dudv = a2b arcsin = .
0 0 a2 ´ u2 a ˇu=0 2

4.6.2 Measurements of the flux - the divergence operator

Let Ω Ď R3 be an open set, and u : Ω Ñ R3 be a C 1 vector field. Suppose that O is a
bounded open set of class C 1 such that O Ď Ω with outward-pointing unit normal vector
field N. Then the flux integral of u over B O in the direction N is

ż

u ¨ N dS .

BO

Consider a special case that O = B(a, r) for some ball in R3 centered at a with radius r.
ż

We first compute u3N3 dS. Consider

BB(a,r)

ψ+(x1, x2) = ( x2, a3 + ar2 ´ (x1 ´ a)2 ´ (x2 ´ ) , (x1, x2) P D(a, r) ,
x1, a2)2 (x1, x2) P D(a, r) ,
( )
ψ´(x2, x2) = x1, x2, a3 ´ ar2 ´ (x1 ´ a)2 ´ (x2 ´ a2)2 ,