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Green’s Theorem
In this section we are going to investigate the relationship between certain kinds of line integrals
(on closed paths) and double integrals
...
Here is a sketch of such a curve and region
...
Also
notice that a direction has been put on the curve
...
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
...
Green’s Theorem
Let C be a positively oriented, piecewise smooth, simple, closed curve and let D be the region
enclosed by the curve
...
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,
∫ Pdx + Qdy
C
or
∫ Pdx + Qdy
C
Both of these notations do assume that C satisfies the conditions of Green’s Theorem so be
careful in using them
...
Let’s work a couple of examples
...
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
...
0 ≤ x ≤1
0 ≤ y ≤ 2x
We can identify P and Q from the line integral
...
P = xy
Q=x y
2
3
So, using Green’s Theorem the line integral becomes,
∫ xy dx + x
C
2
y dy = ∫∫ 2 xy − x dA
3
3
D
1
3
⌠
= ∫ 2 xy − x dy dx
⌡0 0
2x
1
⌠ ⎛1 4
⎞
= ⎮ ⎜ xy − xy ⎟ dx
⎠0
⌡0 ⎝ 2
2x
1
= ∫ 8 x 5 − 2 x 2 dx
0
1
⎛4 6 2 3⎞
=⎜ x − x ⎟
3 ⎠0
⎝3
2
=
3
Example 2 Evaluate
∫
y 3 dx − x 3 dy where C is the positively oriented circle of radius 2
C
centered at the origin
...
Let’s first identify P and Q from the line integral
...
Since D is a disk it seems like the best way to do this integral is to use polar coordinates
...
∫y
C
3
dx − x dy = −3∫∫ ( x + y ) dA
3
2
2
D
2π
= −3⌠
⌡0
∫
2
0
r dr dθ
2π
3
⌠ 1 4
= −3⎮
r dθ
⌡0 4 0
= −3∫
2π
0
2
4 dθ
= −24π
So, Green’s theorem, as stated, will not work on regions that have holes in them
...
So, let’s see how we can deal with those kinds of regions
...
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 boundary of D1 is C1 ∪ C3 while the boundary of D2 is
C2 ∪ ( −C3 ) and notice that both of these boundaries are positively oriented
...
Finally, also note that we can think
of the whole boundary, C, as,
C = ( C1 ∪ C3 ) ∪ ( C2 ∪ ( −C3 ) ) = C1 ∪ C2
since both C3 and −C3 will “cancel” each other out
...
∫∫ ( Q
x
− Py ) dA =
∫∫ ( Q
x
D1 ∪ D2
D
− Py ) dA = ∫∫ ( Qx − Py ) dA + ∫∫ ( Qx − Py ) dA
D1
D2
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
...
Using this fact we get,
∫∫ ( Q
x
− Py ) dA =
D
∫ Pdx + Qdy + ∫ Pdx + Qdy + ∫ Pdx + Qdy − ∫ Pdx + Qdy
C1
=
C3
C2
C3
∫ Pdx + Qdy + ∫ Pdx + Qdy
C1
C2
Finally, put the line integrals back together and we get,
∫∫ ( Q
x
− Py ) dA =
D
∫ Pdx + Qdy + ∫ Pdx + Qdy
C1
C2
∫
=
Pdx + Qdy
C1 ∪C2
=
∫ Pdx + Qdy
C
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
...
This idea will help us in dealing with regions that have holes
in them
...
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
...
We originally said that a curve had a positive
orientation if it was traversed in a counter-clockwise direction
...
For the boundary of the hole this definition won’t work and we
need to resort to the second definition that we gave above
...
However, if we cut the disk in half and rename all
the various portions of the curves we get the following sketch
...
Also notice that we can use Green’s
Theorem on each of these new regions since they don’t have any holes in them
...
Also
recall from the work above that boundaries that have the same curve, but opposite direction will
cancel
...
This will be true in general for regions that have holes in
them
...
Example 3 Evaluate
∫
y 3 dx − x 3 dy where C are the two circles of radius 2 and radius 1
C
centered at the origin with positive orientation
...
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
...
∫
C
y 3 dx − x3 dy = −3∫∫ ( x 2 + y 2 ) dA
D
2π
= −3⌠
⌡0
∫
2
1
2π
r 3 dr dθ
⌠ 1 4
= −3⎮
r dθ
⌡0 4 1
2
2π
15
⌠
= −3⎮
dθ
⌡0 4
45π
=−
2
We will close out this section with an interesting application of Green’s Theorem
...
A = ∫∫ dA
D
Let’s think of this double integral as the result of using Green’s Theorem
...
There are many functions that will satisfy this
...
P = 0 P = −y
Q= x Q=0
y
P=−
2
x
Q=
2
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
...
Let’s take a quick look at an example of this
...
Solution
We can use either of the integrals above, but the third one is probably the easiest
...
So, to do this we’ll need a parameterization of C