Begin2.DVI

evaluated. The geometry suggests a change to cylindrical coordinates. In cylindrical

coordinates the following relations are satisfied:

x=rcos θ, y =rsin θ, z =x^2 +y^2 =r^2


r =xˆe 1 +yˆe 2 +zˆe 3 =rcos θˆe 1 +rsin θˆe 2 +r^2 ˆe 3

E= 1 + 4 r^2 , F = 0, G =r^2

ˆen=^2 rcos θ

ˆe 1 + 2 rsin θˆe 2 −ˆe 3

√

1 + 4r^2

F
·ˆen=^4 r

(^2) cos (^2) θ− 6 r (^2) sin (^2) θ− 4 r 2
√
1 + 4r^2
, dS =r
√
1 + 4 r^2 drdθ.

The integral over the surface S 4 can then be expressed in the form

∫∫

S 4

F
·dS
=

∫ π 2

0

∫ 2

0

(4 cos^2 θ−6 sin^2 θ−4)r^3 drdθ =− 10 π.

Physical Interpretation of Divergence

The divergence of a vector field is a scalar field which is interpreted as represent-

ing the flux per unit volume diverging from a small neighborhood of a point. In the

limit as the volume of the neighborhood tends toward zero, the limit of the ratio of

flux divided by volume is called the instantaneous flux per unit volume at a point or

the instantaneous flux density at a point.

If F
(x, y, z) defines a vector field which is continuous with continuous

derivatives in a region Rand if at some point P 0 of R, one finds that

div F >
0 ,then a source is said to exist at point P 0.

div F <
0 ,then a sink is said to exist at point P 0.

div F
= 0,then F
is called solenoidal and no sources or sinks exist.

The Gauss divergence theorem states that if div F
= 0,then the flux φ=

∫∫

S

F
·dS

over the closed surface vanishes. When the flux vanishes the vector field is called

solenoidal, and in this case, the flux of the vector field F
into a volume exactly equals

the flux of the field F
out of the volume. Consider the field lines discussed earlier

and visualize a bundle of these field lines forming a tube. Cut the tube by two plane

areas S 1 and S 2 normal to the field lines as in figure 8-5.