Questions on Green's Theorem
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From: aegis on 10 Jun 2010 21:53 With the given problem: F = (-yi + xj)/(x^2 + y^2), show that int_C(F.dr) = 2*Pi for every positively oriented simple closed path that encloses the origin. We can't compute it directly because C is arbitrary. But what motivates us to consider a smaller circle: C_1 where it is a positively oriented circle with center the origin and some radius R where C_1 lies in C? Is this to create a region D between C and C_1 so that P, Q are continuous partial derivatives on D so that we may apply Green's Theorem? -- aegis
From: fernando revilla on 10 Jun 2010 23:48 aegis wrote: > Is this to create a region D between C and C_1 so > that > P, Q are continuous partial derivatives on D so that > we may apply Green's Theorem? Yes, as a consequence of the Green's Theorem for not simply connected regions we obtain: Int_{C} ( P dx +Q dy ) = Int_{C_1} ( P dx +Q dy ) --- http://ficus.pntic.mec.es/~frej0002/
From: Rob Johnson on 11 Jun 2010 11:40 In article <34a2a2ab-ba86-405f-b1df-ec2f6c227ace(a)u7g2000yqm.googlegroups.com>, aegis <aegis(a)mad.scientist.com> wrote: >With the given problem: F = (-yi + xj)/(x^2 + y^2), show that >int_C(F.dr) = 2*Pi for every positively oriented simple closed path >that encloses the origin. > >We can't compute it directly because C is arbitrary. But >what motivates us to consider a smaller circle: C_1 where >it is a positively oriented circle with center >the origin and some radius R where C_1 lies in C? In article <270061298.316473.1276242546537.JavaMail.root(a)gallium.mathforum.org>, fernando revilla <frej0002(a)ficus.pntic.mec.es> wrote: >aegis wrote: > >> Is this to create a region D between C and C_1 so >> that >> P, Q are continuous partial derivatives on D so that >> we may apply Green's Theorem? > >Yes, as a consequence of the Green's Theorem for not >simply connected regions we obtain: > >Int_{C} ( P dx +Q dy ) = Int_{C_1} ( P dx +Q dy ) Not for all P and Q. One sufficient condition is that dP dQ -- = -- (d/dx = partial derivative) [1] dy dx holds for all points through which the path is deformed. Condition [1] is necessary in a region if the integral equation is to hold for all paths deformed through that region. In aegis' integral, we have |\ x dy - y dx |\ x dy - y dx O ----------- = O ----------- \|C x^2 + y^2 \|C_1 x^2 + y^2 So we need to have d x d y -- --------- = - -- --------- dx x^2 + y^2 dy x^2 + y^2 y^2 - x^2 x^2 - y^2 ------------- = - ------------- (x^2 + y^2)^2 (x^2 + y^2)^2 Each side is equal and continuous except at (0,0), so the path of integration can be deformed over any region except one that contains the origin. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: fernando revilla on 11 Jun 2010 14:32
Rob Johnson wrote: > Not for all P and Q. Of course. I meant for P and Q in this problem and as a consequence of the theorem I mentioned. -- http://ficus.pntic.mec.es/~frej0002/ |