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Tuesday, 26 March 2013

line integrals part 6


In this section we are going to investigate the relationship between certain kinds of line integrals (on closed paths) and double integrals.

Let’s start off with a simple (recall that this means that it doesn’t cross itself) closed curve C and let D be the region enclosed by the curve.  Here is a sketch of such a curve and region.
GreensThm_G1
First, notice that because the curve is simple and closed there are no holes in the region D.  Also notice that a direction has been put on the curve.  We will use the convention here that the curve C has a positive orientation if it is traced out in a counter-clockwise direction.  Another way to think of a positive orientation (that will cover much more general curves as well see later) is that as we traverse the path following the positive orientation the region D must always be on the left.

Given curves/regions such as this we have the following theorem.

Green’s Theorem
Let C be a positively oriented, piecewise smooth, simple, closed curve and let D be the region enclosed by the curve.  If P and Q have continuous first order partial derivatives onD then,
                                              

Before working some examples there are some alternate notations that we need to acknowledge.  When working with a line integral in which the path satisfies the condition of Green’s Theorem we will often denote the line integral as,



Both of these notations do assume that C satisfies the conditions of Green’s Theorem so be careful in using them.

Also, sometimes the curve C is not thought of as a separate curve but instead as the boundary of some region D and in these cases you may see C denoted as .

Let’s work a couple of examples.

Example 1  Use Green’s Theorem to evaluate  where C is the triangle with vertices  with positive orientation.

Solution
Let’s first sketch C and D for this case to make sure that the conditions of Green’s Theorem are met for C and will need the sketch of D to evaluate the double integral.
GreensThm_Ex1_G1

So, the curve does satisfy the conditions of Green’s Theorem and we can see that the following inequalities will define the region enclosed.

                                              

We can identify P and Q from the line integral.  Here they are.
                                                    

So, using Green’s Theorem the line integral becomes,
                                        
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