Green's theorem gives a relationship between the line integral of a two-dimensional vector field over a closed path in the plane and the double integral over the region it encloses. The fact that the integral of a (two-dimensional) conservative field over a closed path is zero is a special case of Green's theorem.
Green's theorem is itself a special case of the much more general Stokes' theorem. The statement in Green's theorem that two different types of integrals are equal can be used to compute either type: sometimes Green's theorem is used to transform a line integral into a double integral, and sometimes it is used to transform a double integral into a line integral.
Let be a positively oriented, piecewise smooth, simple closed curve in a plane, and let be the region bounded by . If and are functions of defined on an open region containing and have continuous partial derivatives there, then where the path integral is traversed counterclockwise.
Evaluate the integral where is the boundary of the upper half of the unit disk, traversed counterclockwise.
The boundary is defined piecewise, so this integral would be tedious to compute directly. Green's theorem gives where is the upper half disk. This integral can be computed easily as (The integral could also be computed using polar coordinates.)
Let be a continuous function of two variables with continuous partial derivatives, and let be the gradient of defined by Show that for any closed curve
This is a straightforward application of Green's theorem: because the mixed partial derivatives and are equal.
Green's theorem can be used "in reverse" to compute certain double integrals as well. It is necessary that the integrand be expressible in the form given on the right side of Green's theorem. Here is a very useful example.
Let be a plane region enclosed by a simple closed curve Show that the area of equals any of the following integrals (where the path is traversed counterclockwise):
As an application, compute the area of an ellipse with semi-major axes and
The proof of the first statement is immediate: Green's theorem applied to any of the three integrals above shows that they all equal which is the area of
To compute the area of an ellipse, use the parametrization to get (The integral of is a standard trigonometric integral, left to the reader.)
This method is especially useful for regions bounded by parametric curves.
A cardioid is a curve traced by a fixed point on the perimeter of a circle of radius which is rolling around another circle of radius It is parameterized by the equations where What is the area inside the cardioid?
Here so The first two integrals are straightforward applications of the identity They are equal to and respectively. The third integral is simplified via the identity and equals So the final answer is
One of the fundamental results in the theory of contour integration from complex analysis is Cauchy's theorem:
Let be a holomorphic function and let be a simple closed curve in the complex plane. Then
The proof reduces the problem to Green's theorem.
Write and Then the integral is These integrals can be evaluated by Green's theorem:
But holomorphic functions satisfy the Cauchy-Riemann equations and Therefore, both integrals are 0 and the result follows.
Green's theorem is a special case of the three-dimensional version of Stokes' theorem, which states that for a vector field where is the normal vector to the region and is the curl of When and is a region in the -plane, the setting of Green's theorem, is the unit vector and the third component of is so the theorem becomes and the left side is just as desired.
This proof is the reversed version of another proof; watch it here.
Let be a piecewise smooth, simple closed curve in the plane. Let this smooth curve be enclosed in the region , and assume that and and their first partial derivatives are continuous at each point in the region containing . We need to prove that
Let us say that the curve is made up of two curves and such that
Now, under the following conditions, integrating with respect to between and yields
Integrating the resulting integrand over the interval , we obtain
Thus, we arrive at the first half of the required expression
Similarly, we can arrive at the other half of the proof. Let us say the curve is made up of two curves and such that
Now, integrating with respect to between and yields
Integrating the resulting integrand over the interval we obtain
Thus, we arrive at the second half of the required expression.
Summing both the results finishes the proof of Green's theorem: