1、14.4 Greens Theorem in the PlaneCONTENTS* 1. Classification of domain* 2. Greens Theorem * 3. Examples and Applications* 4. Summary1. Classification of domain Suppose that D is a plane domain, if the part enclosed by anysimple closed curve in D belongs to D, D is called a simplyconnected domain, oth
2、erwise complex connected plex connected domainsimply connected domainDDwithout “holes” with “holes”When an observer walks along the Positive direction, D is always on his left.We suppose that L is a simple closed curve that forms the boundary of a region D in the xy-plane. DLlxyODLPositive direction
3、Counter-clockwise orientationBoundary L+ l L Counter-clockwise orientationl Clockwise directioncomplex connected domainTheorem Let L be a piecewise smooth, simple closed curve that forms theboundary of a region D in xy-plane. If Q(x,y) and P(x,y) arecontinuous and have continuous partial derivatives
4、 on D and itsboundary L, thenwhere L is the boundary curve in the positive direction of D. += LDyQxPyxyPxQ dddd)()1(2. Greens Theorem Greens formula(1) The first order partial derivative of P and Q are continuous in the closed region D;(2) The curve L is closed and takes positive direction .Notes +=
5、 LDyQxPyxyPxQ dddd)(),()(),( 21 bxaxyxyxD = ),()(),( 21 dycyxyyxD = D is both an x-simple and y-simple set which means that astraight line parallel to thecoordinate axis has at most twointersection points with L.(1) simple regionxyO a bdcD)(1 xy =)(2 xy =ABCE)(2 yx =)(1 yx =Proof of Greens Theorem T
6、he proof is divided into three cases, and the region gradually changes from special to general. () dd d dLDP x y PyQ Qyx x = + D)(2 yx =)(1 yx = dc yyyQ d),( 2= CBE yyxQ d),(Similarly, = LDxyxPyxyP d),(dddc yd dc yyyQ d),( 1= yyxQ d),(+ EAC yyxQ d),(= dc yy yyxQ d),( )( )(21xxQyy d)( )(21 CBE CAE yy
7、xQ d),(= L yyxQ d),(xyOdcABCEddDQx xy =Proof of Greens Theorem (2) simply connected regionAdditivity of integral regionD is simply connected. Divide D into three sets, , each of which is both x-simple and y-simple.1 2 3,D D D=DyxyPxQ dd)( yxyPxQ dd)(321 DDD +DL1L1D 2D3D2L3LProof of Greens Theorem MT
8、N= += L yQxP dd 1 , 2 3( , a r e po siti v e dir e c ti on s f or )L L L D DyxyPxQ dd)( + =321dd)(DDDyxyPxQ + yxyPxQ dd)( + yxyPxQ dd)( + yQxP dd + yQxP dd( ) d d d dLDQP x y P x Q yxy = + yxyPxQ dd)(1D 2D3D+ yQxP dd1L 2L 3LDL1L 2L3L1D 2D3DProof of Greens Theorem MTN(3) complex connected region ,.AB
9、 CE1L2L3LBy (2), we have DyxyPxQ dd)(= +3L( 0 , 0)A B B A CE E C+ = + = += L yQxP dd= 1 , 2 3( , a r e po siti v e dir e c ti on s f or )L L L D+ 2L BA+ AFC CE+)dd( yQxP + EC+ CGA )dd( yQxP +2L( +3L +1L)GFDCEABwe add two straight line segments,.The boundary curve of ismade up of ,2, ,3,and .Proof of
10、 Greens Theorem ABGreens theorem is a generalization of Fundamental Theorem of Calculus. Double integral is expressed as some quantity evaluated on the boundary of the region of integration.( ) d d d dLDQP x y P x Q yxy = + d d d dLDxy x y P x Q yPQ =+ Remark curve integraldouble integralanother sym
11、bol120113 4 201(68a) On , 2 and 2 , so4 2 8 2 4 CC y x dy dxx y dx y dy x dx xdx x x=+ = + = + =20302211884 2 833Cxx y dx y dy x dx+ = = = Example 1: Let C be the boundary of the triangle with vertices (0,0),(1,2) and (0,2). Calculate 242C x y dx y dy+(a) by the direct method, and (b) by Greens Theo
12、rem. Also,3. Examples and ApplicationsC1(0,2)C3y(1,2)x(0,0)C2134082=233xx + = ( )12 202 04x x dy dx( b ) By Gr e e n s T h e or e m 2 82Th u s, 4 2 6 4 .33Cx y dx y dy+ = = Example 1 242C x y dx y dy+= ( )211 2 2 300 2= 4 8 8xx y dx x x dx = +300222 24 2 2 4C x y dx y dy y dy y+ = = = ( ) d d d dLDQ
13、P x y P x Q yxy = + C1(0,2)C3y(1,2)x(0,0)C2Use Greens Theorem to calculate the line integral 3 +2 + 4 32 .Where is the ellipse 22 + 22 = 1.Example 2Let P , = 3 +2 and , = 4 32,So that = 2 and = 4.By Greens Theorem,3 +2 + 4 32 = 42 = 2 = 2( ) d d d dLDQP x y P x Q yxy = + 3. Examples and Applications
14、(1) Calculate the area of plane regionGreens formulaThe area of D: L xyyx dd( ) d d d dLDQP x y P x Q yxy = + 1 dd2 LA x y y x=y x=Dyx dd2Introduce three applications.Example 3Find the area enclosed by the ellipseByWe have tttabA d)s in( co s21 2202 += ab=c os , sin , 0 2x a t y b t t= = = L xyyxA d
15、d21O xyDApplication 2(2) Simplify the calculation of curve integral += LDyQxPyxyPxQ dddd)(I f QPxyis simple , line integral istransformed into Double integral.Find = + 3 + 2 , where is the positiveof circle 2 +2 = 2.Example 42.1,yeP = yxexyQy 23 +=,yeyP =3 yQ yex =+3yyPxQ =By Greens formula, we have
16、 =ISymmetryO xy= yxyDdd3 0Circle of from point to .Example 5Find22x y ax+=Analysis:( , 0)Aa(0, 0)Owhere is the upper halfThe integral path is not an closed curve!( sin ) d ( c os ) d ,xxAO e y m y x e y m y + AOAOO xy ( , 0)AaBut by myeQ x = c o s=xQ =yPObtain= yPxQVery simple.m,c o s ye x mye x c o
17、 s,s in myyeP x =Example 5In order to apply the Greens formula, we add a curve to form aclosed curve. To simplify the computation of integral, the addedcurve should be simple which usually be a straight line that parallelto coordinate axis. Here we add a straight lineBy Greens formula.OA=Dyxm dd ymy
18、exmyye xOAAOx d)co s(d)s in( +281 am = axy = 0,0OA:= a x0 d0So 0Then, =I .81 2am = 081 2am AO OA+ OA0QP mxy= dd( si c)n () osxxOA e y m ye y m y x + O xy( ,0)AaApplication 3(3) Simplify the double integralwhere D is the closed region enclosed by the triangle with 2F in d d d ,yD e x y( 0, 0 ) , ( 1
19、, 1 ) , ( 0, 1 ) .O A Bvertices O xy11 ABD0Then, = yPxQlet ,0=P 2yxeQ = Dy yxe dd2 +BOABOAy yxe d2 = OA y yxe d22 dyAB xe y+ 2yBO x ey+2ye)1(21 1= e = 10 d2 xxe x0+ 0+0Example 6( ) d d d dLDQP x y P x Q yxy = + 22If 0 w e havexy + ，Example 7Denote the region enclosed by L as D,where L is a piecewise
20、, closed curve that doesntpass through the origin. And the direction of L is counter-clockwise.22ddF in d ,Lx y y xxy+Let ,22 yx yP += 22 yx xQ +=xQ yP=22222)( yxxy+Example 7 =+L yx xyyx 22 ddIf , , which means that L isan arbitrary closed curve that doesnot encircled the origin .If , , which means
21、that L isan arbitrary closed curve thatencircled the origin. Then we add acircle in D and denotethe region enclosed by L and l as .By Greens Formula,( ) d d d dLDQP x y P x Q yxy = + 0QPxy=222: ryxl =+1DDLxyOLD1rlxyO=+L yx xyyx 22 dd += 20 22222ds inco s r rr += L yx xyyx 22 dd2= Pay attention to co
22、nditions of Greens formulayxyPxQ dd 0=0=22ddlx y y xxy+=s inc osryrx1D +l yx xyyx 22 dd222: ryxl =+l is along the counter-clockwiseL1DrlxyOExample 7QPxy=4. Summary += LDyQxPyxyPxQ dddd)(2. Three applications of Greens TheoremRelationship between closed curve integral and double integralConditions1. Greens TheoremGreen Formula and its application function
