If A and B are two points in an open region D in space, the work

done in moving a particle from A to B by a field F defined on D usually depends on the path taken. There are some special fields which allow the integral's value to stay the same for all paths from A to B.

DEFINITION:

Let F be a field defined on an open region D in space, and suppose that for any two points A and B, the work

done in moving from A to B is the same over all paths from A to B. Then the integral

is path independent in D and the field F is conservative on D.

An alternate definition states that under conditions normally met in practice, a field F is conservative if and only if it is the gradient field of a scalar function f, or the field F is conservative if and only if F = Ñ f for some f.

DEFINITION:

If F is a field defined on D and F = Ñ f for some scalar function f on D, then f is called a potential function for F.

For all the examples that we will be doing in this course, we will assume that all the curves are piecewise smooth, i.e. made up of finitely many smooth pieces connected end to end. Some other assumptions that we will make are the following. First, the components of F will have continuous first partial derivatives, which will make the mixed partial equal. Second, D is to be an open region in space, and finally D is connected.

THEOREM:

THE FUNDAMENTAL THEOREM OF LINE INTEGRALS

1.

Let F = M i + N j + P k be a vector field whose components are continuous throughout an open connected region D in space. Then there exists a differentiable function f such that

if and only if for all points A and B in D the value

is independent of the path joining A to B in D.

2.

If the integral is independent of the path from A to B, its value is

THEOREM:

The following statements are equivalent:

1.

2.

The field F is conservative on D.

figure 1

figure 2

How do we make a closed loop with two paths from point A to point B? We can reverse the direction of one of the paths. (See figure 1) If A and B lie on a loop, we can reverse part of the loop to make two paths from A to B. (See figure 2)

Here is a summary of what theorem 1 and 2 tells us.

We now have two questions left. (1) How do we know when a given field F is conservative? (2) If F is in fact conservative, how do we find a potential function f (so that F = Ñ f)? Here are the answers.

THE COMPONENT TEST FOR CONSERVATIVE FIELDS

Let F = M (x, y, z) i + N (x, y, z) j + P (x, y, z) k be a field whose component functions have continuous first partial derivatives. Then F is conservative if and only if

EXAMPLE 1:

Is F = yz i + xz j + xy k conservative, and if it is, find its potential function f.

SOLUTION: To determine if F is conservative, we must determine if

is true. Let M = yz, N = xz, and P = xy.

F is conservative. Now to determine the potential function f.

Therefore, the potential function f (x, y, z) = xyz + C.

EXAMPLE 2: 

Is F = (y sin z) i + (x sin z) j + (xy cos z) k conservative, and if it is find its potential function f.

SOLUTION: Let M = y sin z, N = x sin z, and P - xy cos z.

F is conservative. Now to determine the potential function f.

The potential function for this vector field is f (x, y, z) = xy sin z + C.

EXAMPLE 3:

Determine if F = y i + (x + z) j - y k is conservative or not.

SOLUTION: Let M = y, N = x + z, and P = -y.

Since it fails the first condition, we can conclude that F is not conservative.

EXAMPLE 4:

Find the potential function f for the field F = e ^{y + 2z} i + x e ^{y + 2z} j + 2xe^y+2zk.

The potential function is f (x, y, z) = e ^{y + 2z} + C.

It is often convenient to express work and circulation integrals in the "differential" form.

DEFINITION:

The form M (x, y, z) dx + N (x, y, z) dy + P (x, y, z) dz is called the differential form. A differential form is exact on a domain D in space if

for some scalar function f throughout D.

The differential form M dx + N dy + P dz is exact if and only if

This statement is equivalent to saying that the field F = M i + N j + P k is conservative.

EXAMPLE 5:

Show that the differential form in the integral

is exact. Then evaluate the integral.

SOLUTION: Let M = 2x, N = 2y, and P = 2z.

Therefore, the differential form is exact, and df = 2x dx + 2y dy + 2z dz. Now to find the potential function.

Now use the potential function and the fundamental theorem of line integrals to evaluate the given integral.

EXAMPLE 6:

Show that the differential form in the integral

is exact. Then evaluate the integral.

Therefore, the differential form is exact, and df = (2x ln y - yz) dx + (x ²/y - xz) dy - xy dz.

f (x, y, z) = x ² ln y - xyz + C