Green's Theorem
Green's Theorem
Suppose is a simply connected region i.e. no holes in and the boundary of is a closed, piecewise smooth curve oriented counterclockwise. The following famous theorem states the connection between the line integral over and the double integral over :
Theorem (Green's theorem) Let be a 2D vector field such that both and are continuous and have continuous first order partial derivatives on some open region containing . Then
(Note: The symbol means is a closed curve.)
Proof: By definition, .
To proof Green's theorem, it suffices to prove the following:
(a)
(b)
Consider (a):
We regard as a type I region such that it is bounded from below by curve , which is the graph of and from above by , which is the graph of for . Then their parametrizations are and for .
(FTC)
Consider (b):
We regard as a type II region and use the similar argument, it can be shown that
Combining (a) and (b), Green's theorem is proved.
Remark: If , then is conservative and .
Example: We revisit the earlier example - evaluate , where is the loop along the triangle from to to and back to . Here we use Green's theorem to find the line integral.
Let and . Then we have
By Green's theorem,
, where is the triangular region with vertices and .
Regarding as a type I region, we have
Remark: The double integral is easier to compute than the sum of three line integrals over three sides of the triangle.
Applications
Finding area of a region
We can use Green's theorem to find the area of a simply connected region using the vector field . Suppose we are given the parametrization of the boundary of (oriented counterclockwise). Then
Remark: Other vector fields like also works. The mechanism of a useful tool called plainmeter, which is used for measuring the area of an arbitrary shape by tracing its boundary, is based on Green's theorem.
Example: Let be an ellipse parametrized by for . Use Green's theorem to find the area of the region enclosed by .
Answer:
is obviously a simply connected region. Then by Green's theorem, we have
Therefore, the area of the ellipse is .
Exercise: Use Green's theorem to find the area enclosed by the astroid , where . The applet below shows the astroid. (Hint: You can use the parametrization for .)
Multiply connected regions
Although Green's theorem can only be applied to simply connected regions, we can use the following trick to deal with regions that have "holes":
Let be an annulus, as shown in the image below
The boundary of the annulus consists of two circles - the outer one is oriented counterclockwise and the inner one is oriented clockwise. Then we make two "cuts" at a point and divide into two simply connected regions and , enclosed by the curves as shown above. By Green's theorem, we have
Notice that is the outer boundary of (counterclockwise) and is the boundary of the "hole" of (clockwise).
This trick can also be applied to any multiply-connected region i.e. a connected region with holes. Suppose is the outer boundary of (counterclockwise) and is the boundary of the hole (clockwise), . Make the cuts as shown in the diagram below. Then the Green's theorem for is as follows:
Example: Let for . Evaluate , where is a piecewise smooth, counterclockwise simple closed curve if
(a) does not enclose .
(b) encloses .
Answer:
First, notice that .
(a) Suppose does not enclose . is defined on the region enclosed by and is simply connected. Therefore, by Green's theorem, we have
.
(b) Since is not defined at i.e. is not continuous on the region enclosed by , Green's theorem cannot be directly applied. We consider the multiply-connected region bounded by and , which is the circle centered at with radius (for small enough ) and oriented clockwise, as shown in the applet below. Using Green's theorem for multiply-connected regions, we have
Consider the parametrization of : for . We have
Hence, for any simple closed curve that encloses .