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.
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 counterclockwise 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 on D
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.
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,

Example 2 Evaluate
where C
is the positively oriented circle of radius 2 centered at the origin.
Solution
Okay, a circle will satisfy the conditions of Green’s
Theorem since it is closed and simple and so there really isn’t a reason to
sketch it.
Let’s first identify P
and Q from the line integral.
Be careful with the minus sign on Q!
Now, using Green’s theorem on the line integral gives,
where D is a
disk of radius 2 centered at the origin.
Since D is a
disk it seems like the best way to do this integral is to use polar
coordinates. Here is the evaluation of
the integral.

So, Green’s theorem, as stated, will not work on regions
that have holes in them. However, many
regions do have holes in them. So, let’s
see how we can deal with those kinds of regions.
Let’s start with the following region. Even though this region doesn’t have any
holes in it the arguments that we’re going to go through will be similar to
those that we’d need for regions with holes in them, except it will be a little
easier to deal with and write down.
The region D will
be and recall that the symbol is called the union and means that D consists of both D_{1} and D_{2}. The boundary of D_{1} is while the boundary of D_{2} is and notice that both of these boundaries are
positively oriented. As we traverse each
boundary the corresponding region is always on the left. Finally, also note that we can think of the
whole boundary, C, as,
since both and will “cancel” each other out.
Now, let’s start with the following double integral and use
a basic property of double integrals to break it up.
Next, use Green’s theorem on each of these and again use the
fact that we can break up line integrals into separate line integrals for each
portion of the boundary.
Next, we’ll use the fact that,
Recall that changing the orientation of a curve with line
integrals with respect to x and/or y will simply change the sign on the
integral. Using this fact we get,
Finally, put the line integrals back together and we get,
So, what did we learn from this? If you think about it this was just a lot of
work and all we got out of it was the result from Green’s Theorem which we
already knew to be true. What this
exercise has shown us is that if we break a region up as we did above then the
portion of the line integral on the pieces of the curve that are in the middle
of the region (each of which are in the opposite direction) will cancel
out. This idea will help us in dealing
with regions that have holes in them.
To see this let’s look at a ring.
Notice that both of the curves are oriented positively since
the region D is on the left side as
we traverse the curve in the indicated direction. Note as well that the curve C_{2} seems to violate the
original definition of positive orientation.
We originally said that a curve had a positive orientation if it was
traversed in a counterclockwise direction.
However, this was only for regions that do not have holes. For the boundary of the hole this definition
won’t work and we need to resort to the second definition that we gave above.
Now, since this region has a hole in it we will apparently
not be able to use Green’s Theorem on any line integral with the curve . However, if we cut the disk in half and
rename all the various portions of the curves we get the following sketch.
The boundary of the upper portion (D_{1})of the disk is and the boundary on the lower portion (D_{2})of the disk is . Also notice that we can use Green’s Theorem
on each of these new regions since they don’t have any holes in them. This means that we can do the following,
Now, we can break up the line integrals into line integrals
on each piece of the boundary. Also
recall from the work above that boundaries that have the same curve, but
opposite direction will cancel. Doing
this gives,
But at this point we can add the line integrals back up as
follows,
The end result of all of this is that we could have just
used Green’s Theorem on the disk from the start even though there is a hole in
it. This will be true in general for
regions that have holes in them.
Let’s take a look at an example.
Example 3 Evaluate
where C
are the two circles of radius 2 and radius 1 centered at the origin with positive orientation.
Solution
Notice that this is the same line integral as we looked at
in the second example and only the curve has changed. In this case the region D will now be the region between these
two circles and that will only change the limits in the double integral so
we’ll not put in some of the details here.
Here is the work for this integral.

We will close out this section with an interesting
application of Green’s Theorem. Recall
that we can determine the area of a region D
with the following double integral.
Let’s think of this double integral as the result of using
Green’s Theorem. In other words, let’s
assume that
and see if we can get some functions P and Q that will satisfy
this.
There are many functions that will satisfy this. Here are some of the more common functions.
Then, if we use Green’s Theorem in reverse we see that the
area of the region D can also be
computed by evaluating any of the following line integrals.
where C is the
boundary of the region D.
Let’s take a quick look at an example of this.
Example 4 Use
Green’s Theorem to find the area of a disk of radius a.
Solution
We can use either of the integrals above, but the third
one is probably the easiest. So,
where C is the
circle of radius a. So, to do this we’ll need a
parameterization of C. This is,
The area is then,
