Begin2.DVI

(Ben Green) #1
where

μ 0 =

∫ 2 π

0

F 0 ·dξ, F 0 =F(x 0 , y 0 , z 0 )

μ 1 =

∫ 2 π

0

dF
d ·d
ξ

μ 2 =

∫ 2 π

0

1
2!

d^2 F
d^2

·dξ

···

where all the derivatives are evaluated at = 0 and dξ= (−sin θˆe 1 + cos θˆe 2 )dθ. The

vector F 0 is a constant and the integral μ 0 is easily shown to be zero. The vector

dF
d

evaluated at = 0 , when expanded is given by

dF
d =

∂F
∂x cos θ+

∂F
∂y sin θ=

(
∂F 1
∂x cos θ+

∂F 1
∂y sin θ

)
ˆe 1

+

(
∂F 2
∂x cos θ+

∂F 2
∂y sin θ

)
ˆe 2

+

(
∂F 3
∂x

cos θ+∂F^3
∂y

sin θ

)
ˆe 3 ,

where the partial derivatives are all evaluated at = 0.It is readily verified that the

integral μ 1 reduces to

μ 1 =π

(
∂F 2
∂x −

∂F 1
∂y

)
.

The area of the circle surrounding P 0 is π^2 ,and consequently the ratio of the circu-

lation divided by the area in the limit as tends toward zero produces

( curl F)·ˆe 3 =∂F^2
∂x

−∂F^1
∂y

. (8 .35)


Similarly, by considering other planes through the point P 0 which are parallel to the

xz and yz planes, arguments similar to those above produce the relations

( curl F)·ˆe 2 =

∂F 1
∂z −

∂F 3

∂x and ( curl

F)·ˆe 1 =∂F^3
∂y −

∂F 2
∂z. (8 .36)

Adding these components gives the mathematical expression for curl F. One finds

the curl F can be written as

curl F =

(
∂F 3
∂y

−∂F^2
∂z

)
ˆe 1 +

(
∂F 1
∂z

−∂F^3
∂x

)
ˆe 2 +

(
∂F 2
∂x

−∂F^1
∂y

)
ˆe 3. (8 .37)
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