標題:
line integral
發問:
let phi(t) = acost +ibsint 0<=t<=2pi evaluate [ integal of (1/z) dz ] in two ways show this = 2pi / ab thx 更新: let phi(t) = acost +ibsint 0 < = t < =2pi evaluate [ integal of (1/z) dz ] in two ways show this = 2pi / ab 更新 2: the integral is calculate at the region of phi
最佳解答:
Method1: integrate directly (1/z)dz=[(xdx+ydy)+i(xdy-ydx)]/(x^2+y^2) =[ (-a^2 cost sint+b^2 sint cost)+i(ab)]dt/[(a cost)^2+(b sint)^2] line integral= ∫[0~2π] (b^2-a^2)sint cost/[(acost)^2+(bsint)^2] dt +iab∫[0~2π] 1/[(acost)^2+(bsint)^2] dt = ln| (acost)^2+(bsint)^2 | sub. t=0~2π +2iab∫[-π/2~π/2] (sect)^2/[a^2+(btant)^2] dt = 0 + 2iab/(ab)*arctan(btant /a) for t=-π/2 ~ π/2 = 0 + 2i*(π/2 + π/2)= 2πi Method2: by Residue thm. line integral∫ dz/z =2πi*(esidue of 1/z at the pole z=0) = 2πi * 1= 2πi
其他解答:
法3 自創路徑 令C：re^iθ r→0+ ∮phi( 1／z )dz = ∮c( 1／z )dz|||||this is closed line integral. the answer is 2πi
line integral
發問:
let phi(t) = acost +ibsint 0<=t<=2pi evaluate [ integal of (1/z) dz ] in two ways show this = 2pi / ab thx 更新: let phi(t) = acost +ibsint 0 < = t < =2pi evaluate [ integal of (1/z) dz ] in two ways show this = 2pi / ab 更新 2: the integral is calculate at the region of phi
最佳解答:
Method1: integrate directly (1/z)dz=[(xdx+ydy)+i(xdy-ydx)]/(x^2+y^2) =[ (-a^2 cost sint+b^2 sint cost)+i(ab)]dt/[(a cost)^2+(b sint)^2] line integral= ∫[0~2π] (b^2-a^2)sint cost/[(acost)^2+(bsint)^2] dt +iab∫[0~2π] 1/[(acost)^2+(bsint)^2] dt = ln| (acost)^2+(bsint)^2 | sub. t=0~2π +2iab∫[-π/2~π/2] (sect)^2/[a^2+(btant)^2] dt = 0 + 2iab/(ab)*arctan(btant /a) for t=-π/2 ~ π/2 = 0 + 2i*(π/2 + π/2)= 2πi Method2: by Residue thm. line integral∫ dz/z =2πi*(esidue of 1/z at the pole z=0) = 2πi * 1= 2πi
其他解答:
法3 自創路徑 令C：re^iθ r→0+ ∮phi( 1／z )dz = ∮c( 1／z )dz|||||this is closed line integral. the answer is 2πi
此文章來自奇摩知識+如有不便請留言告知
文章標籤
全站熱搜
留言列表
