Mathematical Physics Vol 1

Chapter 2. Vector analysis

50

Figure 2.6: Oriented curve division.

Consider now a bounded function ϕ ( r )= ϕ ( X )= ϕ ( x , y , z ) , where r is the position vector of thepoint X = X ( x , y , z ) . The function ϕ can be either a scalar or a vector function. Let us partition the oriented curve l = AB by points T 0 , T 1 ,..., T n into intervals ∆ T i =( T i − 1 , T i ) , i = 1 ,..., n . Let the point X i , whose position vector ρ i , belong to the interval ∆ T i , i.e. X i ∈ ∆ T i . Note that in this case i denotes the i –th point, rather than a vector component.

Figure 2.7: Oriented arc AB .

Consider the product

ϕ ( ρ i ) ◦ ( r i − r i

ϕ ( ρ i ) ◦ ∆ r i = ϕ ( X i ) ◦ ∆ r i ,

(2.69)

− 1 )=

where r i = r ( t i ) is the position vector of point T i . This product can be either a vector or a scalar, depending on the nature of the function ϕ and the meaning of the ◦ symbol for the product (scalar or vector product). Consider now the sum ( integral sum)

n ∑ i = 1

ϕ ( X i ) ◦ ∆ r i .

I =

(2.70)

Definition If there exists a limit value of the sum (2.70), when the greatest absolute value | ∆ r i | tends to zero, then this value is called the line integral of the function f ( X ) along the curve l = AB and denoted by

ϕ ( X i ) ◦ ∆ r i = Z l

n ∑ i = 1

ϕ ( r ) ◦ d r .

lim max | ∆ r i |→ 0

(2.71)