15.4 Green's Theorem

This theorem states that the value of a double integral over a simply connected region  R is determined by the value of a line integral about the region.  What it means to be connected was in the last section. What it means to be simply connected is this: A plane region is simply connected if it is enclosed by one  simple closed curve. The drawing here should help:

The purple region is simply connected.

The orange is not connected.

The cyan is enclosed by three curves, so not simply connected.

The blue region the curve crosses itself so it is not simple.

Theorem 15.8 Green's Theorem

Let R be a simply connected region with a piecewise smooth boundary C, oriented counterclockwise.  (This means the curve is only traversed once and the interior of the region is always on the left of the curve. Like the purple region above. ) If M and N have continuous partial derivatives in an open region containing R, then

You should read the proof in the text.

Ex 1 Use Green's Theorem to evaluate the line integral:

along the path from (0,0) to (1,1) along the graph y = x³ and back from (1,1) to (0,0) along the graph of y = x.

By Green's theorem, we know

The region R below dictates the limits of integration

Ex 1 (continued) do the above example without using Green's theorem.

We need to add up both line integrals. C1 from (0,0) to (1,1) along the path y = x³ and then  add the line integral along C2 the path from (1,1) back to (0,0) along y = x. Then we must add the results.

The curve C1 is described simply by  . Then the integral becomes

The curve C2 is described by  

 Note that the sum of the integrals is 1/4 as expected. This was not nearly as nice, though!!

Ex 2 While subject to the force  , a particle travels once around the circle of radius 3 centered at the origin. Use Green's Theorem to find the work done by F.

From Ex 1 we know that  so the integral will be

Switching to polar coordinates brings us to this point:

Ex 3 Evaluate the line integral using Green's theorem:

where C is the curve r = 1+cos(q).

We must (obviously?) convert to polar coordinates for the double integral.

Note that the field is NOT conservative (you can't find a potential function f(x,y) such that F = gradient of f ) though the closed path integral still comes out to be zero.

Theorem 15.9 Line Integrals for Calculating Area

If R is a plane region bounded by a piecewise smooth curve C, oriented counterclockwise, then the area of R is given by 

Note that this comes about since

Therefore we have Green's Theorem stating

The expression on the right, as you know, is the area of R.

Ex 4 Show that the area of an ellipse is pab for a and b the lengths of the major and minor axes.

The ellipse can be viewed as

Parametric equations for the counterclockwise curve are

Alternate Forms for Green's Theorem

Alt Form 1

If F is a vector field in R², then we can write the vector valued function as if it was in R³: 

Then we can write curl of F as

If we then evaluate the dot product of curl of F and the unit vector k, we get

Therefore in Green's Theorem with given good conditions, we can write

If we extend Green's Theorem to space, we get Stoke's Theorem discussed in Unit V lesson 8.

Alt Form 2

If we use arc length s as a parameter for C then we have  so a unit tangent vector T tangent to C is   the outward unit normal vector to C would be  as can be seen by this sketch:

Hence for the vector field  we can apply Green's Theorem to obtain the following:

Ultimately we have the formula

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Assignment pg 1095 #3, 9,15,19,21,25