where f 1 , f 2 , f 3 are arbitrary functions which have been held constant during the
partial differentiation process. Now add the first and second equations, add the first
and third equation and add the second and third equations to obtain
2 φ=2 xy^2 +xz +yz^2 +f 1 +f 2
2 φ=xy^2 + 2xz +yz^2 +f 1 +f 3
2 φ=xy^2 +xz + 2 yz^2 +f 2 +f 3
In order that these three equations be the same, require that
f 1 +f 2 =xz +yz^2 , f 1 +f 3 =xy^2 +yz^2 , f 2 +f 3 =xy^2 +xz (9 .49)
Now solve the equations (9.49) for f 1 , f 2 and f 3 to show that
f 1 =z^2 y, f 2 =xz, f 3 =xy^2 ,
The potential function can then be expressed as
φ=xy^2 +xz +yz^2.
Observe that any constant can be added to this potential function to obtain a more
general result, since the derivative of a constant is zero.
Solenoidal Fields
A vector field which is solenoidal satisfies the property that the divergence of the
vector field is zero. An alternate definition of a solenoidal vector field is obtained
from Gauss’ divergence theorem
∫∫∫
V
∇·F dV =
∫∫
S
F·dS. (9 .50)
If ∇·F = 0,then
∫∫
S
F·dS = 0 which implies that the total flux, through the simple
closed surface surrounding the volume V, is zero.
It has been shown that an irrotational vector field is derivable from a potential
function. An analogous result holds for solenoidal vector fields. That is, if F is a
solenoidal vector field which is continuous and differentiable , then there exists a vector
potential V such that F =∇× V = curl V . However, this vector potential is not unique,
for if V is a vector satisfying F = curl V , then the vector potential V∗=V+∇ψ, where
