Page 138 - Vector Analysis

134 CHAPTER 4. Vector Calculus

where D((a, r) is the disk in R2 given by ␣(x1, x2) P R2 ˇ (x1 ´ a1 )2 + (x2 ´ a2)2 ď r2(. Since
ˇ

B B(a, r)z ψ+(D(a, r)) Y ψ´(D(a, r)) is the equator of the sphere B B(a, r) which has zero

area, we must have

żż ż

u3N3 dS = u3N3 dS + u3N3 dS .

BB(a,r) ψ+(D(a,r)) ψ´(D(a,r))

Note that (N ˝ ψ˘)(x1, x2) = 1 ( x2 ) ´ ) In view of Example 4.49, we have
r ψ˘(x1, a.

ż
u3N3 dS

ψ+(D(a,r))

= ż u3(ψ+(x1, x2)) ar2 ´ (x1 ´ a1)2 ´ (x2 ´ a2)2 r dA
r a1)2
D(a,r) ar2 ´ (x1 ´ ´ (x2 ´ a2)2

ż
= u3(ψ+(x1, x2)) dA .

D(a,r)

and similarly, żż
Therefore,
u3N3 dS = ´ u3(ψ´(x1, x2)) dA .

ψ+(D(a,r)) D(a,r)

ż ż [u3 ]
u3N3 dS = x2))
(ψ+(x1, x2 )) ´ u3(ψ´(x1, dA

BB(a,r) D(a,r) ( ż ?a3+ r2´(x1´a1)2´(x2´a2)2
?
= ż B u3 (x1, x2, x3) ) A
a3´ r2´(x1´a1)2´(x2´a2)2 B x3 dx3 d
D(a,r)

ż B u3 dx .
= B(a,r) B x3

Similarly,

ż u1N1 dS = ż B u1 dx and ż u2N2 dS = ż B u2 dx ;
B x1 B x2
BB(a,r) B(a,r) BB(a,r) B(a,r)

thus we conclude that 3

ż ż ÿ B ui
u ¨ N dS =
B(a,r) i=1 B xi dx .

BB(a,r)

The computation above motivates the following

Definition 4.73 (The divergence operator). Let u : Ω Ď Rn Ñ Rn be a vector field. The

divergence of u is a scalar function defined by

divu = n Bui .

ÿ

i=1 B xi