Treating the vector function F as a function of , one can expand F in a Taylor’s
series about = 0,to obtain
F=F(x 0 , y 0 , z 0 ) + dF
d +
^2
2!
d^2 F
d^2 +
^3
3!
d^3 F
d^3 +··· (8 .7)
where all the derivatives are to be evaluated at = 0 .Substituting the expressions
for the unit normal and the Taylor’s series into the flux integral produces
φ=^2 μ 0 +^3 μ 1 +^4 μ 2 +···,
where
μ 0 =
∫π
0
∫ 2 π
0
F(x 0 , y 0 , z 0 )·ˆensin θ dφ dθ
μ 1 =
∫π
0
∫ 2 π
0
dF
d =0 ·ˆensin θ dφ dθ
μ 2 =
∫π
0
∫ 2 π
0
d^2 F
d^2 =0
·ˆensin θ dφ dθ,
plus higher order terms in . The vector F(x 0 , y 0 , z 0 )is a constant and an evaluation
of the integral defining μ 0 produces μ 0 = 0 .To calculate the second integral defining
μ 1 observe that the chain rule for functions of more than one variable produces the
result
dF
d
=∂
F
∂x
∂x
∂
+∂
F
∂y
∂y
∂
+∂
F
∂z
∂z
∂
dF
d =
∂F
∂x sin θcos φ+
∂F
∂y sin θsin φ+
∂F
∂z cos θ
This result can be expressed in the component form
dF
d =0 =
(
∂F 1
∂x sin θcos φ+
∂F 1
∂y sin θsin φ+
∂F 1
∂z cos θ
)
ˆe 1
+
(
∂F 2
∂x
sin θcos φ+∂F^2
∂y
sin θsin φ+∂F^2
∂z
cos θ
)
eˆ 2
+
(
∂F 3
∂x sin θcos φ+
∂F 3
∂y sin θsin φ+
∂F 3
∂z cos θ
)
eˆ 3
=0
where all the derivatives are to be evaluated at = 0.Evaluating the integrals defining
μ 1 ,produces
μ 1 =
4
3 π
(
∂F 1
∂x +
∂F 2
∂y +
∂F 3
∂z
)
=0
The flux integral then has the form
φ=
4
3 π
3
(
∂F 1
∂x +
∂F 2
∂y +
∂F 3
∂z
)
+^4 μ 2 +^5 μ 3 +···
