Calculus- Early Transcendentals, 2021a

16.4. Green’s Theorem 547
The proof of Green’s Theorem will follow a discussion and several examples.
∫
To indicate that an integral is being done over a closed curve in the counter-clockwise direction, we
∮
C
usually write . We also use the notation ∂D to mean the boundary of D oriented in the counterclockwise
C
∮ ∫
direction. With this notation, = .
C
∂D
We already know one case, not ∮ particularly interesting, in which this theorem is true: If f is conservative,
we know that the integral f · dr = 0, because any integral of a conservative vector field around
C
a closed curve is zero. We also know in this case that ∂ f 1 /∂y = ∂ f 2 /∂x, so the double integral in the
theorem is simply the integral of the zero function, namely, 0. So in the case that f is conservative, the
theorem says simply that 0 = 0.
Example 16.13
We illustrate the theorem by computing both sides of
∫
∫∫
x 3 dx+ xy 2 dy =
∂D
D
y − 0dA,
where D is the triangular region with corners (0,0), (1,0), (0,−1).
Solution. Starting with the double integral:
∫∫
D
y 2 − 0dA =
∫ 1 ∫ −1+x
0 0
y 2 dydx =
∫ 1
0
(−1 + x) 3
dx =
3
(−1 + x)4
12
∣
1
0
= − 1
12 .
There is no single formula to describe the boundary of D, so to compute the left side directly we
need to compute three separate integrals corresponding to the three sides of the triangle, and each of these
integrals we break into two integrals, the “dx” partandthe“dy” part. The three sides are described by
y = 0, y = −1 + x, andx = 0. The integrals are then
∫
∫ 1 ∫ 0 ∫ 0 ∫ −1
∫ 0 ∫ −1
x 3 dx+ xy 2 dy = x 3 dx+ 0dy+ x 3 dx+ (y + 1)y 2 dy+ 0dx+ 0dy
∂D
0
0 1
0
0 0
= 1 4 + 0 − 1 4 + − 1
12 + 0 + 0 = − 1
12 .
Alternately, we could describe the three sides in vector form as 〈t,0〉 for t from 0 to 1, 〈t + 1,t〉 for t
from 0 to −1, and 〈0,−1 + t〉 for t from 0 to 1. Note that in each case, as t ranges from lower to upper
bound, we follow the corresponding side in the correct direction. Now
∫
∫ 1
∫ −1
∫ 1
x 3 dx+ xy 2 dy = t 3 +t · 0 2 dt + (t + 1) 3 +(t + 1)t 2 dt + 0 + 0(−1 +t)dt
∂D
=
0
∫ 1
0
0
t 3 dt +
∫ −1
0
(t + 1) 3 +(t + 1)t 2 dt = − 1
12 . ♣
0