MTH 234.018
Green’s Theorem
Section 16.4
As as an example of Green’s Theorem, let’s compute the area of a triangle with base b and heigh h. To
simplify things, let’s assume the base is along the x-axis, so we can draw it on the x, y plane as:
Here the closed boundary curve C (oriented counter-clockwise) is made up of the three segments C1 , C2 ,
1
and C3 , as in the picture. We already know the area is bh, but let’s see Green’s Theorem in action.
2
Ordinarily, we would compute area by integrating the constant function 1 over the region:
ZZ
Area =
dA
D
In class we noticed something about the vector field F~ = hP, Qi =
−y x
,
: it has the property that
2 2
∂Q ∂P
−
= 1. So by Green’s Theorem, we have:
∂x
∂y
ZZ
ZZ I
I
∂Q ∂P
1
Green’s
~
Area =
1 dA =
−
dA =
F · d~r =
xdy − ydx
∂x
∂y
2 C
C
D
D
The integral over C will be the sum of the integrals over the segments C1 ,C2 , and C3 . So we will do them
separately, and add them up.
1
First, let’s do C1 . We can parameterize it as
~r(t) = ht, 0i
for 0 ≤ t ≤ b
Then the derivative (velocity) is
~r 0 (t) = h1, 0i
So the integral over C1 is
Z
1
F~ · d~r =
2
C1
Z
0
b
1
h0, ti · h1, 0i dt =
2
Z
b
(0 − 0) dt = 0
0
Next, let’s do C3 . It is the segment from (a, h) to (0, 0). So it can be parameterized as
~r(t) = (1 − t) ha, hi
for 0 ≤ t ≤ 1
Then the derivative is
~r 0 (t) = h−a, −hi
So the integral over C3 is
Z
Z
Z
1 1
1 1
F~ · d~r =
(1 − t) h−h, ai · h−a, −hi dt =
(1 − t)(ah − ah)dt = 0
2 0
2 0
C3
Finally, the last segment is C2 . It goes from (b, 0) to (a, h). So it can be parameterized as
~r(t) = (1 − t) hb, 0i + t ha, hi = hb + (a − b)t, hti
for 0 ≤ t ≤ 1
The derivative is
~r 0 (t) = ha − b, hi
So the integral of F~ =
Z
C2
1
F~ · d~r =
2
Z
1
2
h−y, xi over C2 is
1
h−ht, b + (a − b)ti · ha − b, hi dt =
0
1
2
Z
1
(h(b − a)t + h(b + (a − b)t)) dt =
0
So all together, we have
Z
Area =
C1
F~ · d~r +
Z
F~ · d~r +
C2
Z
C3
2
1
F~ · d~r = 0 + bh + 0
2
1
2
Z
1
bh dt =
0
1
bh
2