Example 9-7. The field lines of the vector field F=F(x, y, z ) = yˆe 1 +xˆe 2 +zˆe 3
are determined from the differential system
dx
y
=dy
x
=dz
z
.
By trial and error one can construct the functions
V 1 =^1
z
V 2 =^1
z
V 3 =−(x+y)
z^2
so that V·F= 0.One can then construct the exact differential equation
V ·dr =^1
zdx +
1
zdy −
(x+y)
z^2 dz = grad μ^1 ·dr =dμ^1 = 0
from which to determine
μ 1 =x+y
z
=c 1
Similarly, by using trial and error, one can show that the functions
W 1 =x W 2 =−y W 3 = 0
are such that W ·F = 0.This produces the exact differential equation
W ·dr =x dx −y dy = grad μ 2 ·dr =dμ 2 = 0
which is easily integrated. One finds that
μ 2 =
x^2
2 −
y^2
2 =c^2.
Note also that the trial and error method might produce all kinds of results. For
example, let
P 1 =
1
2 z P^2 =−
1
2 z P^3 =
1
2 (x−y),
then one can show P·F = 0.Consequently,
P·dr =^1
2 z dx −
1
2 z dy +
1
2 (x−y)dz = grad μ^3 ·dr =dμ^3 = 0 (9 .85)
is an exact differential which can be integrated. The equation (9.85) implies that
∂μ 3
∂x
=^1
2
z, ∂μ^3
∂y
=−^1
2
z, ∂μ^3
∂z
=^1
2
(x−y)
