Begin2.DVI

where the derivatives are to be evaluated at = 0 .The volume of the sphere of radius

centered at the point (x 0 , y 0 , z 0 )is given by^43 π^3 and consequently the limit of the

ratio of VolumeFlux as tends toward zero produces the scalar relation

div F
= lim∆V→ 0

∆S→ 0

∫∫

R

F
·d
S

∆V

=∂F^1

∂x

+∂F^2

∂y

+∂F^3

∂z =0

(8 .8)

Recalling the definition of the operator ∇,the mathematical expression of the diver-

gence may be represented

div F
=∇·F
=

(

∂

∂x

ˆe 1 + ∂

∂y

ˆe 2 + ∂

∂z

ˆe 3

)

·(F 1 ˆe 1 +F 2 ˆe 2 +F 3 ˆe 3 )

=

∂F 1

∂x +

∂F 2

∂y +

∂F 3

∂z

(8 .9)

Example 8-2. Find the divergence of the vector field

F
(x, y, z) = x^2 yeˆ 1 + (x^2 +yz^2 )ˆe 2 +xyz ˆe 3

Solution: By using the result from equation (8.9), the divergence can be expressed

div F
=∇·F
=

∂(x^2 y)

∂x +

∂(x^2 +yz^2 )

∂y +

∂(xyz)

∂z = 2xy +z

(^2) +xy = 3xy +z 2

The Gauss Divergence Theorem

A relation known as the Gauss divergence theorem exists between the flux and

divergence of a vector field. Let F
(x, y, z )denote a vector field which is continuous

with continuous derivatives. For an arbitrary closed sectionally continuous surface S

which encloses a volume V, the Gauss’ divergence theorem states

∫∫∫

V

div F dV
=

∫∫∫

∇·F dV
 =

∫∫

S

F
·dS
=

∫∫

S

F
·eˆndS. (8 .10)

which states that the surface integral of the normal component of F
summed over

a closed surface equals the integral of the divergence of F
summed over the volume

enclosed by S. This theorem can also be represented in the expanded form as

∫∫∫

V

(

∂F 1

∂x

+∂F^2

∂y

+∂F^3

∂z

)

dx dy dz =

∫∫

S

(F 1 ˆe 1 +F 2 ˆe 2 +F 3 ˆe 3 )·ˆendS, (8 .11)

where ˆenis the exterior or positive normal to the closed surface.