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F(x, y) = 2xyi + (y² – x²)j (x2 + y2)2 and C is any positively oriented simple closed curve that encloses the origin. F. dr =

F(x, y) = 2xyi + (y² – x²)j (x2 + y2)2 and C is any positively oriented simple closed curve that encloses the origin. F. dr =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Use the method of this example to calculate F. dr, where = 2xyi + (y2 – x²)j (x2 + y2)² F(x, y) = and C is any positively oriented simple closed curve that encloses the origin. F. dr = /c
Transcribed Image Text:Use the method of this example to calculate F. dr, where = 2xyi + (y2 – x²)j (x2 + y2)² F(x, y) = and C is any positively oriented simple closed curve that encloses the origin. F. dr = /c
EXAMPLE If F(x, y) = (-yi+x j)/(x² + y³), show that f. F • dr = 27 for every positively oriented simple closed path that encloses the origin. SOLUTION Since C is an arbitrary closed path that encloses the origin, it's difficult to compute the given integral directly. So let's consider a counterclockwise-oriented circle C' with center the origin and radius a, where a is chosen to be small enough that C' lies inside C. (See Figure 1.) Let D be the region bounded by C and C'. Then its positively oriented boundary is CU (-C') and so the general version of Green's Theorem gives C õe ( P dx + Q dy + L Pds + edy - ( -) da D ду D y² – x² (x² + y²)² y? - x? (x² + y?)? dA = 0 FIGURE 1 S. P dx + Q dy = P dx + Q dy Therefore F • dr = {_ F that is, We now easily compute this last integral using the parametrization given by r(t) = a cos ti + a sin t j, 0 < t < 27. Thus F. dr = F· dr = " F(r(?)) · r'(1) dt (2= (-a sin t)(-a sin t) + (a cos t)(a cos t) a² cos?t + a² sin²t *27 dt = " dt = 27
Transcribed Image Text:EXAMPLE If F(x, y) = (-yi+x j)/(x² + y³), show that f. F • dr = 27 for every positively oriented simple closed path that encloses the origin. SOLUTION Since C is an arbitrary closed path that encloses the origin, it's difficult to compute the given integral directly. So let's consider a counterclockwise-oriented circle C' with center the origin and radius a, where a is chosen to be small enough that C' lies inside C. (See Figure 1.) Let D be the region bounded by C and C'. Then its positively oriented boundary is CU (-C') and so the general version of Green's Theorem gives C õe ( P dx + Q dy + L Pds + edy - ( -) da D ду D y² – x² (x² + y²)² y? - x? (x² + y?)? dA = 0 FIGURE 1 S. P dx + Q dy = P dx + Q dy Therefore F • dr = {_ F that is, We now easily compute this last integral using the parametrization given by r(t) = a cos ti + a sin t j, 0 < t < 27. Thus F. dr = F· dr = " F(r(?)) · r'(1) dt (2= (-a sin t)(-a sin t) + (a cos t)(a cos t) a² cos?t + a² sin²t *27 dt = " dt = 27
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