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Unformatted text preview: 16.4: GREEN’S THEOREM KIAM HEONG KWA 1. GREEN’S THEOREM Let C be a simple closed curve in the sense that if r(t), a S t S b, is
a parametrization of C', then r(t0) = r(t1), a S to < t1 3 b, if and only
to = a and t1 = b. If the curve C' is given the positive orientation,
i.e., a single counterclockwise traversal of C, then it is said to be
positively oriented. Equivalently, the region D enclosed by the curve
C is always on the left as the point r(t) traverses C if the latter is
positively oriented. Convention 1. We always assume that D contains also all the points
on 0. Likewise, if C is given the negative orientation, i.e., a single clockwise
traversal of C, then C is said to be negatively oriented. The region D bounded by C is always on the right as the point r(t) traverses C in
this case. We are now ready to state: Theorem 1 (Green’s Theorem). Let C be a positively oriented, piece
wise smooth, simple closed curve that bounds a plane region D C 1R2.
If P(x,y) and Q($, y) are functions whose partial derivatives are con
tinuous on an open region that contains D, then (1.1) £Pdw+Qdy=//D (—g—g—g—f) dA, where we have written 390 for f0 to indicate that the curve C’ is given
the positive orientation. Convention 2. We also write faD for f0, where 8D is the boundary
curve 0' being positively oriented. Date: November 11, 2010. 2 KIAM HEONG KWA More generally, if 01 is a negatively oriented, piecewise smooth, sim
ple closed curve that lies in the interior of a positively oriented, piece
wise smooth, simple closed curve 02, then (1.2) /01de+Qdy+/02Pdm+Qdy=//D dA, where D is the region between the curves Cl and 02. Example 1 (Problem 2 in the text). We shall compute f0 my dit+as2 dy ( a) directly and { b ) using Green’s theorem, where C is the rectangle with
vertices (0,0), (3,0), (3,1), and (0,1). The rectangle C consists of four line segments, each of which is
parametrized as follows: 012x=x, y=0, 03x53;
C22x=3, y=y, OSySI;
C3zm=3t, y=1, OStS3;
C4za:=0, y=1—t, Ogtgl. It is readily seen that fck my doc + x2 dy = 0 for k = 1 and k = 4. So
faydx+m2dy=/ mydm+a§2dy+/ xydx+x2dy
0 02 CB = 1 32dy+ 3(3—t)(1><—1>dt
9 To apply Green’s theorem, one sets P(:t,y) = my and Q(a:,y) = 112,
so that 62,, — Py = :L‘. 30 by (1.1), one has fxydzc+as2dy= mdA
C D
1 3
= / / xda: dy
0 0
_ 21
_ 2 ,
where D is the plane rectangular region bounded by 0'. Example 2 (Problem 5 in the text). Let D be the plane triangular
region with uertices (0,0), (2,2), and (2,4) and let 0 = 8D. Then by 16.4: GREEN’S THEOREM 3 Green’s theorem, 0 a
‘2d +2x2d=// [—223 ———a:2 dA
éwrr yy L)%<m WM) =// 2$ydA D
2 29: =/ / Zmydydm
0 a: = 12. Example 3 (Problem 10 in the text). Let D be the plane elliptical
region bounded by the ellipse C : 3:2 + my + y2 = 1. Then by Green’s
theorem, . 8 8 .
£S1nydx+xcosydy—//D [6:(xcosy)—a—y(smy) dA
=// OdA
D =0. In fact, it can~ be shown likewise that for any piecewise smooth, simple
closed curve C, ﬁsinydm+xcosydy = 0 C
because a a
ah? cosy) — 8—y(smy) = 0. A veriﬁcation of this statement without Green’s theorem will have been
diﬁ‘icult! The Area of a Region Bounded by a Simple Closed Curve.
Let D be a plane region bounded by a piecewise smooth, simple closed
curve 0. Recall that the area A(D) of D can be computed as the double
integral ffD 1 dA. On the other hand, by choosing the functions P and
Q in (1.1) such that
8Q (9P 1.3 —— — — = 1
( ) ('93; By ’
we see that A(D) is also equal to the line integral f0 P dz: + Q dx. To
ﬁnd some suitable functions P and Q, observe that (1.3) is equivalent
to (m) £W—M=%W+me 4 KIAM HEONG KWA for any oz 6 R. If, in addition to (1.4), we stipulate that Q depends
only on :10 and P depends only on y, then (1.4) holds only if d d
an — aw] — d—y[P+ (1 — a>y1— ﬂ
for some constant B E R, from which it follows that Q(:z:) = (a + mm and P(y) = (a + ,8 — 1)y. Using the arbitrariness of a, we can rewrite the last equation as 62(90) = 723 and P(y) = ("r — 1)y for an arbitrary 'y E R. Clearly, the functions P and Q satisfy (1.3).
Hence Green’s theorem implies that for any 7 E R. Note that equation (5) on p. 1058 of the text is ob
tained by setting 7 to 1, 0, and 1/2. Example 4 (Problem 21 in the text). Consider a polygon P whose ver
tices are , in counterclockwise order, P1 (11:1, y1), P2(x2, yz),    ,Pn(xn, yn).
Clearly, the area A(73) of ’P is the sum, of areas of the triangles A1=P1P2Q1A2=P2P3Q7“')An—12Pn—1PnQiAn=PnP1Qa where Q is any point in the interior of 73. Since the area of the jth
triangle Aj, j = 1,2,   ,n, is given by //1dA=l]{ xdy—ydw
A 2 BA] .7 l
by (1.5) with 'y = 5, one has 11. 1
A0?) = ingA xdy—ydai. j=1 j 16.4: GREEN’S THEOREM 5 On the other hand, note that f xdy—ydm=/ xdy—ydx+/ xdy—ydx+/ azdy—ydm,
3A1 P1P2 P2Q QP1
{ xdy—yda:=/ :Itdy—yda:+/ xdy—ydm+/ xdy—ydac,
8A2 P2133 PaQ QP2 So where we have identiﬁed Pn+1 = P1. It is easy to show that / 33 d9 " 3/ d5” = xjyj+1 — $j+1yj
Pij+1 for each j, j = 1,2, ,n. Hence 1 n
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This note was uploaded on 11/11/2011 for the course MATH 254.01 taught by Professor Kwa during the Fall '10 term at Ohio State.
 Fall '10
 Kwa
 Calculus, Geometry

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