An integration in the z-direction produces
∫∫∫
V
∂F 3
∂z dzdx dy =
∫∫
R
F 3 (x, y, z )
z 2 (x,y)
z 1 (x,y)
dx dy
=
∫∫
R
F 3 (x, y, z 2 (x, y )) dx dy −
∫∫
R
F 3 (x, y, z 1 (x, y )) dx dy.
The element of surface area on the upper and lower surfaces can be represented by
On the surface S 2 , dS 2 =
dx dy
ˆe 3 ·ˆen 2
On the surface S 1 , dS 1 = dx dy
−ˆe 3 ·ˆen 1
so that the above integral can be expressed as
∫∫∫
V
∂F 3
∂z
dV =
∫∫
S 2
F 3 ˆe 3 ·ˆen 2 dS 2 +
∫∫
S 1
F 3 ˆe 3 ·ˆen 1 dS 1 =
∫∫
S
F 3 eˆ 3 ·ˆendS,
which establishes the desired result.
Similarly by dividing the surface into appropriate sections and projecting the
surface elements of these sections onto appropriated planes, the remaining integrals
may be verified.
Example 8-3. Verify the divergence theorem for the vector field
F(x, y, z ) = 2 xˆe 1 − 3 yˆe 2 + 4 zˆe 3
over the region in the first octant bounded by the surfaces
z=x^2 +y^2 , z = 4, x = 0, y = 0
Solution The given surfaces define a closed region over which the integrations are
to be performed. This region is illustrated in figure 8-4. The divergence of the given
vector field is given by
div F=∇·F =∂F∂x^1 +∂F∂y^2 +∂F∂z^3 = 2 −3 + 4 = 3 ,
and thus the volume integral part of the Gauss divergence theorem can be deter-
mined by summing the element of volume dV =dx dy dz first in the z−direction from
the surface z=x^2 +y^2 to the plane z= 4. The resulting parallelepiped is then summed
