Using Green’s theorem to find area
Typically we use Green’s theorem to calculate as an alternative to calculate some
line integral ∫
CF ⋅ ds. If, for example, we are in two dimension, C is a
simple closed curve, and F(x,y) is defined everywhere inside C, we can
use Green’s theorem to convert the line integral into to double integral.
Instead of calculating line integral ∫
CF ⋅ ds directly, we calculate the double
integral
 D dA | | |
Can we use Green’s theorem to go the other direction? If we are given a double
integral, can we use Green’s theorem to convert the double integral into a
line integral and calculate the line integral? If we are given the double
integral
 Df(x,y)dA, | | |
we can use Green’s theorem only if there happens to be a vector field
F(
x,y) so
that
f(x,y) =  - . | | |
However, we haven’t learned any method to find such a vector field
F. So, we
aren’t likely to use Green’s theorem in this direction very often.
There is one important exception to this rule, however, and that is when we are
using a double integral to calculate the area of a region D. Recall that
the area of a region D is equal to the double integral of f(x,y) = 1 over
D:
Area of D =  DdA =  D1dA. | | |
If
f(
x,y) = 1, it is easy to find a vector field
F so that
 - = f(x,y) = 1. | | |
There are many such vector field
F, but we’ll pick the vector field
F(
x,y) = (
-y∕2
,x∕2). You can confirm that indeed
-
= 1.
In summary, if C is a counterclockwise oriented simple closed curve that bounds a
region where you can apply Green’s theorem, the area of the region D bounded by
C = ∂D is
Area of D = ∫
CF ⋅ ds =  ∫
Cxdy - ydx, | | |
where
F(
x,y) = (
-y∕2
,x∕2).
Example
Use Green’s Theorem to calculate the area of the disk D defined by x2 + y2 ≤ 4.
Solution: Since we know the area of the disk of radius 2 is 4π, we better get 4π
for our answer.
The boundary of D is the circle of radius 4. We can parametrized it in a
counterclockwise orientation using
| c(t) = (2 cos t, 2 sin t), 0 ≤ t ≤ 2π. | | |
Then
| c'(t) = (-2 sin t, 2 cos t), | | |
and, by Green’s theorem,
| area of D | =  dA | |
|
| =  ∫
Cxdy - ydx | |
|
| =  ∫
02π[(2 cos t)(2 cos t) - (2 sin t)(-2 sin t)]dt | |
|
| =  ∫
02π4(sin 2t + cos 2t)dt = 2 ∫
02πdt = 4π. | | |
Thankfully, our answer agrees with what we know it should be.
