Math 294
Lecture 6
Given by Prof. Alan Zehnder on 9/13/95 at 10:10 in Kaufmann Auditorium
Green's Theorem
In Lecture 5 we introduced divergence
and curl and stated Green's Theorem. In this lecture, we will
give examples of how to use Green's Theorem and then sketch a
proof of the theorem.
Green's Theorem
F=Mi+Nj. R is the region enclosed by C.
Example
Let F = 2xi - 3yj
C: r(t) = a cos t i + a sin t j, 0<=t<=2*pi
/||<b>v</b>||<br>
<b>n</b> = (a cost<b>i</b> + a sint<b>j</b>)/||<b>v</b>||<br></font></tt>
We would then dot n with <b>F</b> and integrate over C.
<img src=)
Using Green's theorem is much easier:
Now, to calculate the circulation without Green's Theorem,
we find T, dot it with F, and integrate over C.
Again, Green's Theorem is much simpler:
Thus, Green's Theorem is a very good alternative to the line integral.
Another Example
One might notice that in the previous example, divF was constant,
making the integral very easy. This is not always true.
Suppose F=2x^2i-3y^2j
divF=4x-6y
One More Example
Given an integral of the form
there are two ways to solve using Green's Theorem.
- Let M=x^2 and N=x-y and plug into the Flux equation.
- Let M=x-y and N=x^2 and plug into the Circulation equation.
Proof of Green's Theorem
We will not do a complete proof - we will just give a sketch of one way to
prove it.
First, we will assume:
- F and its first partial derivatives are continuous on R.
- C consists of a finite number of smooth curves.
- C is a simple curve.
Special Case of C - paths where horizontal and vertical lines cross C only
twice.
We must prove that:
We can prove for the dN/dx term in a similar way, though we will not show
it here.
To extend this proof to other shapes:
- Rectangular Shapes. This is not very difficult, but we will not show
it here.
- Note that in the second drawing, the two additional line segments cover
the same distance
and point in opposite directions. They cancel out so that the same result
is reached from the
two regions in the second drawing as from the entire region in the first
drawing.
- Again we can break the region into many acceptable shapes and sum to
get the same result.
