Professor Dave here, let’s check out Green’s Theorem.
Now that we know about line integrals and double integrals, we are ready to learn an
interesting theorem. Recall that with line integrals, we can integrate along a curve
C. And if we have a vector field F = then the form of the line integral is the
integral along C of F dot dr, which becomes the integral along C of P dx + Q dy. This
is the case for all curves C. However, a special theorem exists for “closed curves”, that
is, curves that form a complete loop and enclose an area which we will label D. For these cases,
the line integral over C of P dx + Q dy ends up being equal to the double integral over
the domain D of (∂Q/dx - ∂P/dy)dxdy. This is called Green’s Theorem. This is a very
useful theorem, as line integrals are typically harder or more complicated to calculate than
ordinary integrals, so this gives us a more convenient way to calculate them for closed
curves. Before we move on to an example, there’s an important distinction we must make regarding
the direction of the curve. Green's Theorem applies to closed curves that run counterclockwise.
Another way to think of it is if you walked along the path in the direction of the arrows,
the enclosed domain will always be on your left. This is referred to as “positive orientation”.
For curves that go in a clockwise direction, or “negative orientation”, Green’s theorem
can still be applied, but with a sign change. So the line integral can still be expressed
as a double integral, just with a negative sign right here out front. It will be important
to remember this detail to get correct answers. Now for an example, let’s say we have the
curve shown here made of four line segments that make a square, from (0,0) to (1,0) to
(1,1) to (0,1) and back to the origin. We also have a vector field F = and
we want to find the line integral over C of F dot dr. We could split the curve into the
four line segments and do each line integral individually, but using Green’s theorem
can give us a much faster and simpler calculation. Our integration domain D ends up simply being
a square where x and y both run from 0 to 1. So we have a double integral over this
region of (∂Q/∂x - ∂P/∂y)dxdy. Our Q here is x^2 so ∂Q/∂x is 2x. Our P is
xy so ∂P/∂y is x. Plugging these in we get the double integral of (2x – x) dxdy,
or just x dxdy. Integrating x first, we get (x^2)/2 evaluated from 0 to 1, which gives
(1^2)/2 minus 0, or just 1/2. We still have the y integral to evaluate but there’s no
y dependence left, we just have a constant. So integrating 1/2 dy gives y/2 evaluated
from 0 to 1, which just gives 1/2 minus 0, or 1/2. This is the value of our line integral.
As you can see, Green’s theorem gives us a much simpler method for evaluating line
integrals on closed curves. Of course not every boundary will be so simple to express.
We will have to use the same kind of thought process we discussed in the tutorial on double
integrals to properly write the bounds of integration. For a quick look at this, consider
the curve C given here, where three line segments make a triangle. The line segments are a straight
line from (0,0) to (1,0), from (1,0) to (1,1), and finally along the line y = x back to the
origin. We would have do these three lines individually to calculate the line integral,
but if we set up the integration for the contained region correctly we can use Green’s theorem.
If we want to integrate y first in our double integral, y will run from 0 to the curve y
= x, so x is the upper bound. Meanwhile x will simply run from 0 to 1. Thus the line
integral over C of P dx + Q dy will be the integral from 0 to 1 and then from 0 to x
of (∂Q/∂x - ∂P/∂y) dydx. And with that, we’ve laid down the essentials for
using Green’s theorem, so the rest just comes down to setting up the bounds of integration.
Let’s check comprehension.