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·dS
∆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)