Begin2.DVI

or to solve for one quantity φ=φ(x, y, z ),then it should be obvious that it would be

easier to solve for the one quantity φand then calculate the components F 1 , F 2 , F 3 by

calculating the gradient grad φ. The function φ, which defines the scalar field from

which F is derivable is called the potential function associated with the irrotational

vector field F .

In a simply-connected^5 region R, let F define an irrotational vector field which

is continuous with derivatives which are also continuous. The following statements

are then equivalent.

1. ∇× F = curl F= 0 and the vector field F is irrotational.

2. F=∇φ= grad φand F is derivable from a scalar potential function φ=φ(x, y, z )

by taking the gradient of this function.

3. The dot product F·dr =dφ , where dφ is an exact differential.

4. The line integral W=

∫P 2

P 1 F·dr is the work done in moving through the vector

field F between two points P 1 and P 2 ,and this work done is independent of the

curve selected for connecting the points P 1 and P 2.

5. The line integral

∫

C

©F·dr = 0, which implies that the work done in moving

around a simple closed path is zero.

If a vector field F=F(x, y, z ) = F 1 (x, y, z)ê 1 +F 2 (x, y, z)ê 2 +F 3 (x, y, z)ê 3 is derivable

from a scalar function φ=φ(x, y, z ) such that F = grad φ= ∇φ (sometimes F is

defined as the negative of the gradient due to a particular application that requires

a negative sign), then F is called a conservative vector field , and φ is called the

potential function from which the field is derivable. Set F = grad φ, and equate the

like components of these vectors and obtain the scalar equations

F 1 (x, y, z ) = ∂φ ∂x

, F 2 (x, y, z) = ∂φ ∂y

, F 3 (x, y, z ) = ∂φ ∂z

.

These equations imply that

F·dr =∇φ·dr =∂φ ∂x

dx +∂φ ∂y

dy +∂φ ∂z

dz =dφ (9 .40)

is an exact differential. Consequently the statement 2 implies the statement 3.

If F = grad φ, then the line integral