green's theorem


A short example of Green's theorem.

Green's Theorem says: for C a simple closed curve in the xy-plane and D the region it encloses, if F = P(x,yi + Q(x,yj, then
int_C F.dr = int int_D dQ/dx - dP/dy dA
where C is taken to have positive orientation (it is traversed in a counter-clockwise direction). Note that Green's Theorem applies to regions in the xy-plane.
quarter circle with radius 2 in the first quadrant
figure 1: the region of integration for the example

An Example
Consider F = 3xy i + 2y2 j and the curve C given by the quarter circle of radius 2 shown to the right. We are taking C to have positive orientation: that is, we are traversing it in the counter-clockwise direction.

We could evaluate the line integral of F.dr along C directly, but it is almost always easier to use Green's theorem. Thus we have

int_C F.dr = int int_D dQ/dx - dP/dy dA = int int_D -3x dA

In this case the domain D is most easily described in polar coordinates, for which 0 <= r <= 2 and 0 <= theta <= pi/2, so that the integral becomes

int_C F.dr =int int_D -3x dA =int_0^pi/2 int_0^2 -3r^2 cos(t) dt dr = -8

Some Related Strands

Another Example
Green's Theorem only works for simple, closed curves. Further, we assume a positive orientation. The following illustrates what happens when one or both of these are not the case.

unclosed quarter circle in the first quadrant
figure 2: a non-closed simple curve C1
angle bracket in the first quadrant
figure 3: C2

Consider the same vector field we used above, F = 3xy i + 2y2 j, and the curve C1 shown in figure 2, which is the quarter circle starting at the point (0,2) and ending at (2,0). To find the line integral of F on C1 we can't apply Green's Theorem directly, but can do it indirectly.

First, note that the integral along C1 will be the negative of the line integral in the opposite direction. Thus by reversing signs we can calculate the integrals in the "positive" direction and get the integral we want.

Second, note that, calling C2 the curve shown in figure 3, we can write the integral we did in the first example in terms of the integrals on C1 and C2:

-int_C F.dr = int_C1 F.dr + int_C2 F.dr

(be sure you see where the negative sign on the left-hand side of this equation comes from!) so that to find only the line integral on C1 we can subtract:

int_C1 F.dr = -int_C F.dr - int_C2 F.dr

We already know the first term on the right-hand side from above (it's 8), so that the line integral we want is equal to 8 minus the line integral on C2, which is easy to find. On the section along the x-axis, y is zero, and so F is zero and there is no contribution to the line integral. On the section along the y-axis, x is zero, and F = <0, 2y2>. We can parameterize the curve with r(t) = <0, 2-2t> (with 0 <= t <= 1), so

int_C2 F.dr = int_0^1 2(2-2t)^2 (-2 dt) = -16/3

And we have

int_C1 F.dr = 8 + 16/3 = 40/3


green's theorem
Last modified: Wed Mar 31 16:07:16 EST 2004
©2004 Gavin LaRose, UM Math Dept.