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