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Problems: Extended Green’s Theorem y dx − x dy exact? If so, ﬁnd a potential function. y2 1 x Answer: M = and N = − 2 are continuously diﬀerentiable whenever y y y = 0, i.e. in the two half-planes R1 and R2 – both simply connected. Since My = −1/y 2 = Nx in each half-plane the ﬁeld is exact where it is deﬁned. To ﬁnd a potential function f for which F = df we use method 2. fx = 1/y ⇒ f = x/y + g(y). fy = −x/y 2 + g / (y) = −x/y 2 ⇒ g / (y) = 0 ⇒ g(y) = c. ⇒ f (x, y) = x/y + c. 1. Is F = (continued) 1 y R1 x 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 = r n x, N = r n y ⇒ My = nr n−2 xy = Nx ⇔ curlF = 0. We show F is conservative by showing C F · dr = 0 for all simple closed curves C. If C1 is a simple closed curve not around 0 then Green’s Theorem implies C1 F · dr = 0. F · dr = If C3 is a circle centered on (0,0) then, since F is radial C3 C3 F · T ds = 0. y C1 If C3 completely surrounds C2 then extended Green’s Theorem F · dr = implies Thus C2 R C3 F · dr = 0. C3 x F · dr = 0 for all closed loops ⇒ F is conservative. C C2 To ﬁnd the potential function we use method 1 over the curve C shown. The calculation works for n = 2. For n = 2 everything is the same except we’d 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.) 1 f (x1 , y1 ) = r n x dx + rn y dy 1Cy1 1 x1 2 n/2 = (1 + y ) y dy + (x2 + y12 )n/2 x dx 1 = = = (1 + y 2 )(n+2)/2 n+2 y12 )(n+2)/2 y1 1 + 1 (x2 2(n+2)/2 − + n+2 (x12 + y12 )(n+2)/2 − 2(n+1)/2 n+2 (1 + y x1 + y12 )(n+2)/2 n + 2 (x21 + 1 y12 )(n+2)/2 − (1 + (n + 2)/2 C2 y12 ) (x1 , y1 ) C1 (1, 1) x rn+2 + C. n+2 If n = −2 we get f (x, y) = ln r + C. ⇒ f (x, y) = 2 MIT OpenCourseWare http://ocw.mit.edu 18.02SC Multivariable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.