18 02A Topic 41 Extensions and applications of Green s theorem Author Jeremy Orloff Read SN V5 V6 pp 1 2 Review of Green s Theorem I I ZZ Tangential work form F dr F T ds curlF k dA C C R ZZ I or M dx N dy Nx My dA C R ZZ I F n ds divF dA Normal flux form C R I ZZ or M dy N dx Mx Ny dA C y T C R n x R Both forms require for now C is a positively oriented interior on left simple closed curve and R is its interior T unit tangent vector n unit normal CW rotation of T it points out from R Application Definition A region D in the plane is simply connected if it has no holes Examples y y x D1 D2 D4 D3 x D5 whole plane D1 D5 are simply connected For any simple closed curve inside them its interior is entirely inside the region Sometimes we say any curve can be shrunk to a point without leaving the region The regions at right are not simply connected That is the interior of C the curve C is not entirely in the region Annulus y C Puntured plane Theorem If D is a simply connected region and My Nx in D the F hM N i is conservative in D i e F f for some f defined in D Proof Since D is simply connected if C is a simple closed curve in D its interior R I is also inZD Z Therefore using Green s Theorem Nx My curlF 0 y F dr curlF k dA 0 C R D R C 1 x x 18 02A topic 41 2 Summary F M i N j hM N i on simply connected region D The following statements are equivalent Z Q Z Q 1 F dr is path independent 1 M dx N dy is path independent P P I I 2 F dr 0 for any closed path C 2 M dx N dy 0 C C 3 F f for some f in D 3 M dx N dy df If F is continuously differentiable then 1 2 3 4 4 My Nx in D 4 curlF 0 in D C D Why we need D simply connected in the theorem If there is a hole then F might not be defined on the interior of C See example 1 below Extended Green s Theorem If R has multiple boundary curves we can extend Green s Theorem Suppose R region between C1 and C2 note R is always to the left as you traverse either curve I I ZZ C1 R F dr F dr curlF k dA C2 C1 C2 C1 R Likewise for Imore than two I I curves I F dr F dr F dr C1 R C2 C3 C2 C3 C4 ZZ F dr C4 curlF k dA R Proof Make a cut I so the new curve is simple ZZ Green s Thm F dr curlF k dA C1 C3 C2 C3 R I ZZ F dr curlF k dA QED C1 C2 C3 C3 C1 C2 R yi xj tangential field y r2 F is defined on D plane 0 0 punctured plane It s easy to compute we ve done it before that curlF 0 in D I Question For the tangential field F what values can F dr take Example 1 Let F C for C a simple closed curve positively oriented We have two cases i C1 not around 0 ii C2 around 0 I ZZ i Green s Theorem F dr curlF k dA 0 C1 R x y R C2 x C1 3 y 18 02A topic 41 I F dr 2 ii We show that R C3 C2 C2 x Let C3 be a small circle of radius a entirely inside C2 By I extendedIGreen s Theorem ZZ F dr F dr C3 I I F dr F dr C2 C2 curlF k dA 0 R C3 On the circle C3 Iwe can easily compute the line integral Z 1 2 a F T 1 a F T ds ds 2 QED a C3 C3 a y Answer to question 0 or 2 I If C is not simple F dr 2 n where n is the number of times x C C C goes CCW around 0 0 Not for class n is called the winding number of C around 0 n also equals the number of times C crosses the positive x axis counting 1 from below and 1 from above Example 2 Is y dx x dy exact If so find a potential function y2 y 1 x M N 2 are continuosly differentiable whenever y 6 0 y y i e in the two half planes R1 and R2 both simply connected My 1 y 2 Nx in each half plane M dx N dy df R1 x We use method 2 fx 1 y f x y g y fy x y 2 g 0 y x y 2 g 0 y 0 g y c f x y x y c R2 Example 3 Let F rn xi yj Use extended Green s Theorem to show that F is conservative for all integers n Find a potential function First note M rn x N rn y My nrn 2 xy Nx curlF 0 R We show F is conservative by showing C F dr 0 for all simple closed curves C R If C1 is a simple closed curve not around 0 then Green s Theorem implies C1 F dr 0 I I If C3 is a circle centered on 0 0 then since F is radial F dr F T ds 0 C3 C3 18 02A topic 41 4 If C3 completely surrounds C2 then extended Green s Theorem implies sinceI curlF 0 between the curves I y F dr F dr 0 C2 C3 I Thus F dr 0 for all closed loops F is conservative R C1 C3 C2 C x To find the potential function we use method 1 over the curve C shown The calculation works for n 6 2 For n 2 everything is the same except we get natural logs instead of powers We also ignore the fact that if x1 y1 is on the negative x axis we shoud use a different path that doesn t go through the origin This isn t really an issue since we already know a potential function exists so continuity would handle these points without using an integral Z f x1 y1 rn x dx rn y dy ZCy1 Z x1 2 n 2 1 y y dy x2 y12 n 2 x dx 1 2 n 2 2 y1 1 2 x1 1 y x y12 n 2 2 n 2 n 2 1 1 2 n 2 2 n 2 2 1 y1 2 x21 y12 n 2 2 1 y12 n 2 n 2 2 2 2 n 2 2 n 1 2 x1 y1 2 n …
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