Jim Lambers
MAT 280
Spring Semester 2009-10
Lecture 19 Notes
These notes correspond to Section 13.4 in Stewart and Section 8.1 in Marsden and Tromba.
Green’s Theorem
We have learned that if a vector field is conservative, then its line integral over a closed curve 𝐶 is
equal to zero. However, if this is not the case, then evaluation of a line integral using the formula
∫
∫
F ⋅ 𝑑r =
𝐶
𝑏
F(r(𝑡)) ⋅ r′ (𝑡) 𝑑𝑡,
𝑎
where r(𝑡) is a parameterization of 𝐶, can be very difficult, even if 𝐶 is a relatively simple plane
curve. Fortunately, in this case, there is an alternative approach, using a result known as Green’s
Theorem.
We assume that F = ⟨𝑃, 𝑄⟩, and consider the case where 𝐶 encloses a region 𝐷 that can be
viewed as a region of either type I or type II. That is, 𝐷 has the definitions
𝐷 = {(𝑥, 𝑦) ∣ 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑔1 (𝑥) ≤ 𝑦 ≤ 𝑔2 (𝑥)}
and
𝐷 = {(𝑥, 𝑦) ∣ 𝑐 ≤ 𝑦 ≤ 𝑑, ℎ1 (𝑦) ≤ 𝑥 ≤ ℎ2 (𝑦)}.
Using the first definition, we have 𝐶 = 𝐶1 ∪ 𝐶2 ∪ (−𝐶3 ) ∪ (−𝐶4 ), where:
∙ 𝐶1 is the curve with parameterization 𝑥 = 𝑡, 𝑦 = 𝑔1 (𝑡), for 𝑎 ≤ 𝑡 ≤ 𝑏
∙ 𝐶2 is the vertical line segment with parameterization 𝑥 = 𝑏, 𝑦 = 𝑡, for 𝑔1 (𝑏) ≤ 𝑡 ≤ 𝑔2 (𝑏)
∙ 𝐶3 is the curve with parameterization 𝑥 = 𝑡, 𝑦 = 𝑔2 (𝑡), for 𝑎 ≤ 𝑡 ≤ 𝑏
∙ 𝐶4 is the vertical line segment with parameterization 𝑥 = 𝑎, 𝑦 = 𝑡, for 𝑔1 (𝑎) ≤ 𝑡 ≤ 𝑔2 (𝑎)
We use positive orientation to describe the curve 𝐶, which means that the curve is traversed
counterclockwise. This means that as the curve is traversed, the region 𝐷 is “on the left”.
In view of
∫
∫
F ⋅ 𝑑r = ⟨𝑃 𝑑𝑥 + 𝑄 𝑑𝑦,
𝐶
𝐶
1
we have
∫
∫
∫
∫
∫
𝑃 𝑑𝑥
𝑃 𝑑𝑥 +
𝑃 𝑑𝑥 +
𝑃 𝑑𝑥 +
𝑃 𝑑𝑥 =
−𝐶4
−𝐶3
𝐶2
𝐶1
𝐶
∫
∫
∫
∫
𝑃 𝑑𝑥
𝑃 𝑑𝑥 −
𝑃 𝑑𝑥 −
𝑃 𝑑𝑥 +
=
𝐶1
𝑏
∫
=
𝑃 (𝑥(𝑡), 𝑦(𝑡))𝑥′ (𝑡) 𝑑𝑡 +
∫
𝑃 (𝑥(𝑡), 𝑦(𝑡))𝑥′ (𝑡) 𝑑𝑡 −
∫
𝑎
𝑏
∫
𝑎
𝑏
∫
𝐶4
𝐶3
𝐶2
∫
𝑔1 (𝑏)
𝑔2 (𝑎)
𝑃 (𝑥(𝑡), 𝑦(𝑡))𝑥′ (𝑡) 𝑑𝑡 −
𝑃 (𝑥(𝑡), 𝑦(𝑡))𝑥′ (𝑡) 𝑑𝑡
𝑔1 (𝑎)
𝑔2 (𝑏)
∫
∫
𝑎
𝑔1 (𝑏)
𝑎
𝑏
𝑔2 (𝑎)
𝑃 (𝑡, 𝑔2 (𝑡))(1) 𝑑𝑡 −
𝑃 (𝑏, 𝑡)(0) 𝑑𝑡 −
𝑃 (𝑡, 𝑔1 (𝑡))(1) 𝑑𝑡 +
=
𝑔2 (𝑏)
𝑃 (𝑎, 𝑏)(0) 𝑑𝑡
𝑔1 (𝑎)
𝑏
∫
[𝑃 (𝑡, 𝑔1 (𝑡)) − 𝑃 (𝑡, 𝑔2 (𝑡))] 𝑑𝑡
=
𝑎
∫ 𝑏∫
𝑔2 (𝑡)
= −
𝑃𝑦 (𝑡, 𝑦) 𝑑𝑦 𝑑𝑡
𝑎
𝑔1 (𝑡)
∫ ∫
= −
𝐷
∂𝑃
𝑑𝐴.
∂𝑦
Using a similar approach in which 𝐷 is viewed as a region of type II, we obtain
∫
∫ ∫
∂𝑄
𝑄 𝑑𝑦 =
𝑑𝐴.
𝐶
𝐷 ∂𝑥
Putting these results together, we obtain Green’s Theorem, which states that if 𝐶 is a positively
oriented, piecewise smooth, simple (that is, not self-intersecting) closed curve that encloses a region
𝐷, and 𝑃 and 𝑄 are functions that have continuous first partial derivatives on 𝐷, then
)
∫
∫ ∫ (
∂𝑄 ∂𝑃
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 =
−
𝑑𝐴.
∂𝑥
∂𝑦
𝐶
𝐷
Another common statement of the theorem is
)
∫ ∫ (
∫
∂𝑄 ∂𝑃
−
𝑑𝐴 =
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦,
∂𝑥
∂𝑦
𝐷
∂𝐷
where ∂𝐷 denotes the positively oriented boundary of 𝐷.
This theorem can be used to find a simpler approach to evaluating a line integral of the vector
field ⟨𝑃, 𝑄⟩ over 𝐶 by converting the integral to a double integral over 𝐷, or it can be used to find
a simpler approach to evaluating a double integral over a region 𝐷 by converting it into an integral
over its boundary.
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To show that Green’s Theorem applies for more general regions than those that are of both
type I and type II, we consider a region 𝐷 that is the union of two regions 𝐷1 and 𝐷2 that are of
both type I and type II. Let 𝐶 be the positively oriented boundary of 𝐷, let 𝐷1 have positively
oriented boundary 𝐶1 ∪ 𝐶3 , and let 𝐷2 have positively oriented boundary 𝐶2 ∪ (−𝐶3 ), where 𝐶3
is the boundary between 𝐷1 and 𝐷2 . Then, 𝐶 = 𝐶1 ∪ 𝐶2 . It follows that for functions 𝑃 and 𝑄
that satisfy the assumptions of Green’s Theorem on 𝐷, we can apply the theorem to 𝐷1 and 𝐷2
individually to obtain
)
)
)
∫ ∫ (
∫ ∫ (
∫ ∫ (
∂𝑄 ∂𝑃
∂𝑄 ∂𝑃
∂𝑄 ∂𝑃
𝑑𝐴 =
𝑑𝐴 +
𝑑𝐴
−
−
−
∂𝑥
∂𝑦
∂𝑥
∂𝑦
∂𝑥
∂𝑦
𝐷
𝐷1
𝐷2
∫
∫
=
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦
𝐶1 ∪𝐶3
𝐶2 ∪(−𝐶3 )
∫
∫
=
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
𝐶1
𝐶3
∫
∫
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦
𝐶2
−𝐶3
∫
∫
=
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
𝐶1
𝐶2
∫
∫
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 −
𝐶3
𝐶3
∫
∫
=
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦
𝐶1
𝐶2
∫
=
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦
∫𝐶1 ∪𝐶2
=
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦.
𝐶
We conclude that Green’s Theorem holds on 𝐷1 ∪ 𝐷2 . The same argument can be used to easily
show that Green’s Theorem applies on any finite union of simple regions, which are regions of both
type I and type II.
Green’s Theorem can also be applied to regions with “holes”, that is, regions that are not simply
connected. To see this, let 𝐷 be a region enclosed by two curves 𝐶1 and 𝐶2 that are both positively
oriented with respect to 𝐷 (that is, 𝐷 is on the left as either 𝐶1 or 𝐶2 is traversed). Let 𝐶2 be
contained within the region enclosed by 𝐶1 ; that is, let 𝐶2 be the boundary of the “hole” in 𝐷.
Then, we can decompose 𝐷 into two simply connected regions 𝐷′ and 𝐷′′ by connecting 𝐶2 to 𝐶1
along two separate curves that lie within 𝐷. Applying Green’s Theorem to 𝐷′ and 𝐷′′ individually,
we find that the line integrals along the common boundaries of 𝐷′ and 𝐷′′ cancel, because they
3
have opposite orientations with respect to these regions. Therefore, we have
)
)
)
∫ ∫ (
∫ ∫ (
∫ ∫ (
∂𝑄 ∂𝑃
∂𝑄 ∂𝑃
∂𝑄 ∂𝑃
−
𝑑𝐴 =
−
𝑑𝐴 +
−
𝑑𝐴
∂𝑥
∂𝑦
∂𝑥
∂𝑦
∂𝑥
∂𝑦
′
𝐷
𝐷′′
∫
∫ 𝐷
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
=
𝐶2
𝐶1
∫
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦.
=
𝐶1 ∪𝐶2
Therefore, Green’s Theorem applies to 𝐷 as well.
Example The vector field
〈
F(𝑥, 𝑦) = ⟨𝑃 (𝑥, 𝑦), 𝑄(𝑥, 𝑦)⟩ =
𝑦
𝑥
− 2
, 2
2
𝑥 + 𝑦 𝑥 + 𝑦2
〉
is conservative on all of ℝ2 except at the origin, because it is not defined there. Specifically, F = ∇𝑓
where
𝑦
𝑓 (𝑥, 𝑦) = tan−1 .
𝑥
Now, consider a region 𝐷 that is enclosed by a positively oriented, piecewise smooth, simple closed
curve 𝐶, and also has a “hole” that is a disk of radius 𝑎, centered at the origin, and contained
entirely within 𝐶. Let 𝐶 ′ be the positively oriented boundary of this disk. Then, the boundary of
𝐷 is 𝐶 ∪ (−𝐶 ′ ), because, as a portion of the boundary of 𝐷, rather than the disk, it is necessary
for 𝐶 ′ to switch orientation. Applying Green’s Theorem to compute the line integral of F over the
boundary of 𝐷 yields
)
∫
∫
∫ ∫ (
∂𝑄 ∂𝑃
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 +
𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 =
−
𝑑𝐴 = 0,
∂𝑥
∂𝑦
𝐶
−𝐶 ′
𝐷
since F is conservative on 𝐷. It follows that
∫
∫
F ⋅ 𝑑r = −
𝐶
∫
F ⋅ 𝑑r =
−𝐶 ′
F ⋅ 𝑑r,
𝐶′
so we can compute the line integral of F over 𝐶, which we have not specified, by computing the line
integral over the circle 𝐶 ′ , which can be parameterized by 𝑥 = 𝑎 cos 𝑡, 𝑦 = 𝑎 sin 𝑡, for 0 ≤ 𝑡 ≤ 2𝜋.
This yields
∫
∫ 2𝜋
F ⋅ 𝑑r =
𝑃 (𝑥(𝑡), 𝑦(𝑡))𝑥′ (𝑡) 𝑑𝑡 + 𝑄(𝑥(𝑡), 𝑦(𝑡))𝑦 ′ (𝑡) 𝑑𝑡
𝐶′
0
)
(
)
∫ 2𝜋 (
𝑎 sin 𝑡
𝑎 cos 𝑡
=
−
(−𝑎 sin 𝑡) 𝑑𝑡 +
(𝑎 cos 𝑡) 𝑑𝑡
(𝑎 cos 𝑡)2 + (𝑎 sin 𝑡)2
(𝑎 cos 𝑡)2 + (𝑎 sin 𝑡)2
0
4
∫
2𝜋
=
0
∫
𝑎2 sin2 𝑡
𝑎2 cos2 𝑡
𝑑𝑡
+
𝑑𝑡
𝑎2 cos2 +𝑎2 sin2 𝑡
𝑎2 cos2 𝑡 + 𝑎2 sin2 𝑡
2𝜋
=
1 𝑑𝑡
0
= 2𝜋.
We conclude that the line integral of F over any positively oriented, piecewise smooth, simple closed
curve that encloses the origin is equal to 2𝜋. □
Example Consider a 𝑛-sided polygon 𝑃 with vertices (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ), . . ., (𝑥𝑛 , 𝑦𝑛 ). The area 𝐴
of the polygon is given by the double integral
∫ ∫
1 𝑑𝐴.
𝐴=
𝑃
Let 𝑃 (𝑥, 𝑦) = −𝑦/2 and 𝑄(𝑥, 𝑦) = 𝑥/2. Then
(
) (
( ))
∂𝑄 ∂𝑃
1
1
−
=
− −
= 1.
∂𝑥
∂𝑦
2
2
It follows from Green’s Theorem that if ∂𝑃 is positively oriented, then
∫
∫
1
𝐴=
𝑄 𝑑𝑦 + 𝑃 𝑑𝑥 =
𝑥 𝑑𝑦 − 𝑦 𝑑𝑥.
2 ∂𝑃
∂𝑃
To evaluate this line integral, we consider each edge of 𝑃 individually. Let 𝐶 be the line segment
from (𝑥1 , 𝑦1 ) to (𝑥2 , 𝑦2 ), and assume, for convenience, that 𝐶 is not vertical. Then 𝐶 can be
parameterized by 𝑥 = 𝑡, 𝑦 = 𝑚𝑥 + 𝑏, for 𝑥1 ≤ 𝑥 ≤ 𝑥2 , where
𝑦2 − 𝑦1
, 𝑏 = 𝑦1 − 𝑚𝑥1 .
𝑚=
𝑥2 − 𝑥1
We then have
∫
∫ 𝑥2
𝑥 𝑑𝑦 − 𝑦 𝑑𝑥 =
𝑚𝑡 𝑑𝑡 − (𝑚𝑡 + 𝑏) 𝑑𝑡
𝐶
𝑥1
∫ 𝑥2
= −
𝑏 𝑑𝑡
𝑥1
= 𝑏(𝑥1 − 𝑥2 )
= 𝑦1 (𝑥1 − 𝑥2 ) − 𝑚𝑥1 (𝑥1 − 𝑥2 )
= 𝑦1 (𝑥1 − 𝑥2 ) + (𝑦2 − 𝑦1 )𝑥1
= 𝑥 1 𝑦2 − 𝑥 2 𝑦1 .
We conclude that
1
𝐴 = [(𝑥1 𝑦2 − 𝑥2 𝑦1 ) + (𝑥2 𝑦3 − 𝑥3 𝑦2 ) + ⋅ ⋅ ⋅ + (𝑥𝑛−1 𝑦𝑛 − 𝑥𝑛 𝑦𝑛−1 ) + (𝑥𝑛 𝑦1 − 𝑥1 𝑦𝑛 )] .
2
□
5
Practice Problems
Practice problems from the recommended textbooks are:
∙ Stewart: Section 13.4, Exercises 1, 3, 7-17 odd
∙ Marsden/Tromba: Section 8.1, Exercises 1-9 odd
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