# Path Independent IntegralsPotential FunctionsExact Differential Forms

In Lecture 11 we introduced potential functions and conservative fields, and we stated several important theorems. In this lecture, we will continue to prove the theorems and we will give examples of potential functions and exact differential forms.

### Theorems:

F=f <==> F is conservative
if F=f, the result (xf)=0 ==> xF = 0
xF = 0 ==>
F is conservative <==> (any closed path on D)

• D is simply connected.
• F has continuous second derivatives on D.
• C is piecewise smooth.

### Proofs

Prove xF ==>

Apply Stoke's Theorem

Why must D be simply connected?
Consider F = (-y / r^2) i + (x / r^2) j + z k where r^2 = x^2 + y^2

To make F have continuous second derivatives, exclude z axis from D.

If C surrounds the origin, try to apply the theorem.
xF=0 ==>
From the line integral over the unit circle, it could be shown that the integral should equal 2*(pi). We get the wrong answer with the theorem because C surrounds a part of D that is not simply connected.

Prove that F conservative ==>

We can prove the reverse of this statement easily by working through the proof backwards.

### Summary

If xF=0, F is conservative and there exists f such that F=f and =f(B)-f(A)

### Example

Evaluate

Recall F (dot) dr = (Mi + Nj + Pk) (dot) (dxi + dyj + dzk
=Mdx + Ndy + Pdz

F = yzi + xzj + xyk
Test xF=0 ?
= i(x-x) - j(y-y) + k(z-z) = 0

So, F is conservative. Find f:
df/dx = M = yz
Integrate with respect to x:
f = yzx + g(y,z)
Now differentiate this with respect to y.
df/dy = N = xz = d(xyz +g(y,z))/dy = zx + dg/dy
dg/dy=0 ==> g=g(z)
Now differentiate f with respect to z.
df/dz = P = xy = d(xyz + g(z))/dz = xy + dg/dz
dg/dz=0 ==> g=constant
f = xyz + constant
So, the integral equals f(3,5,0)-f(1,1,2) = 0 + const - 2 - const = -2

### Exact Differential Forms

Mdx + Ndy + Pdz is called an exact differential form if it equals:
(df/dx)dx + (df/dy)dy + (df/dz)dz = df

To test for exactness - Does xF=0 ?
Or, does dP/dy = dN/dz and dM/dz = dP/dx and dN/dx = dM/dy ?

#### Example

Evaluate
Is it exact?
M = 2 x cosy ==> dM/dy = -2 x siny
N = - x^2 siny ==> dN/dx = -2 x siny
Yes, it is exact. Find f.
df/dx = M = 2 x cosy
f = x^2 cosy + g(y)
df/dy = N = -x^2 siny = d(x^2cosy + g(y))/dy = -x^2 siny + dg/dy
dg/dy = 0 ==> g=constant
f = x^2 cosy + constant
The integral thus evaluates to: 0cos1 - 1cos0 = -1

Lecture written by Aric Shafran on Thursday, October 6, 1995
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