Example 9-3. Multiply both sides of Newton’s second law F
=m
a =md

(^2)
r
dt^2

by

d
r

dt

and then integrate from P 0 (x 0 , y 0 , z 0 )to P(x, y, z),to obtain

∫P

P 0

F
 ·d
r =

∫P

P 0

F
·d
r

dt

dt =

∫P

P 0

md

(^2)
r
dt^2
·d
r
dt
dt

∫P
P 0
m
2
d
dt
(
d
r
dt
) 2
dt =
∫P
P 0
m
2
d
dt
(
V^2
)
dt

mV^2
2
P
P 0

mV^2
(^2) P−
mV^2
(^2) P 0
(9 .59)

which states that the work done in moving from P 0 to P 1 equals the change in kinetic

energy. Now if F
is derivable from a potential function φsuch that F
=−∇ φ, then

∫P

P 0

F
·d
r =

∫P

P 0

−∇ φ d
r =

∫P

P 0

−dφ =−φ

P

P 0

=φ(x 0 , y 0 , z 0 )−φ(x, y, z) (9 .60)

Equating the results from equations (9.59) and (9.60) and rearranging terms shows

that

φ(x 0 , y 0 , z 0 ) +

m

2 V

(^2) (x 0 , y 0 , z 0 ) = φ(x, y, z ) + m
2 V
(^2) (x, y, z ) (9 .61)

This equation states that the sum of the kinetic energy and the potential energy has

a constant value. A result which is known as the principal of conservation of energy.

As a result, any force fields which are derivable from potential functions are called

conservative force fields.

Example 9-4. If F
is a solenoidal vector field, then div F
= 0 and one can write

F
= curl V
for some vector potential V .
Consider an arbitrary region enclosed by a

surface Sand then select a simple closed curve Con this surface which divides the

surface into two regions, call these regions S 1 and S 2. The flux through this volume

is given by

∫∫

S

F
·d
S=

∫∫

S 1

F
·dS
 1 +

∫∫

S 2

F
·dS
 2 =

∫∫∫

V

div F dV
 = 0

which implies that ∫∫

S 1

F
·dS
 1 =−

∫∫

S 2

F
·dS
 2