The line integral (9.45) can be expressed as the sum of the line integrals along the
straight line paths P 0 P 1 , P 1 P 2 , P 2 P illustrated in figure 9-3, where
Along P 0 P 1 ,one finds dy =dz = 0, y =y 0 , z =z 0
Along P 1 P 2 ,there exists the conditions dx =dz = 0, z =z 0 , x held constant
Along P 2 P , use dx =dy = 0, x and yboth held constant.
This produces the integral
∫PP 0F·dr =∫ xx 0F 1 (x, y 0 , z 0 )dx +∫yy 0F 2 (x, y, z 0 )dy +∫zz 0F 3 (x, y, z )dz=φ(x, y, z )−φ(x 0 , y 0 , z 0 ).(9 .46)If F is irrotational, then ∇× F = 0 which implies that
∂F 2
∂z=∂F^3
∂y, ∂F^1
∂z=∂F^3
∂x, ∂F^1
∂y=∂F^2
∂x. (9 .47)
This hypothesis leads to the result F = grad φ=∇φor its equivalence
∂φ
∂x=F 1 , ∂φ
∂y=F 2 , ∂φ
∂z=F 3.To demonstrate this take the partial derivatives of both sides of the equation (9.46)
and show
∂φ
∂x=F 1 (x, y 0 , z 0 ) +∫yy 0∂F 2 (x, y, z 0 )
∂xdy +∫zz 0∂F 3 (x, y, z )
∂xdz∂φ
∂y =F^2 (x, y, z^0 ) +∫zz 0∂F 3 (x, y, z)
∂y dz
∂φ
∂z=F 3 (x, y, z ).(9 .48)Use the results from equation (9.47), to simplify the first set of integrals and find
∂φ
∂x =F^1 (x, y^0 , z^0 ) +∫yy 0∂F 1 (x, y, z 0 )
∂y dy +∫zz 0∂F 1 (x, y, z )
∂z dz
=F 1 (x, y 0 , z 0 ) + F 1 (x, y, z 0 )y
y 0+F 1 (x, y, z )z
z 0
=F 1 (x, y 0 , z 0 ) + F 1 (x, y, z 0 )−F 1 (x, y 0 , z 0 ) + F 1 (x, y, z)−F 1 (x, y, z 0 )
=F 1 (x, y, z).