弧长 定义 性质 例子 第11章曲线积分和曲面积分 §11.1第一型曲线积分 11.1.1曲线的弧长 设Ⅰ是空间中一条光滑曲线，其参数方程为 r=(t)=(x(t),y(t),z(t)):[a,B]CRR3. 此曲线切向量为 r'(t)=(x'(t),y'(t),z'(t). 这段曲线的弧长为 al)=1 t eonat=1 vearile)+we 2ntict. (11.1) 弧长微元是 ds=|r'(t)|dt=√xn(t)+y2(t)+zn(t)dt. ‖返回全屏关闭退出 1/14
l ½Â 5 ~f 1 11 Ù È©Ú¡È© §11.1 1.È© 11.1.1 l L ´m¥^1w, Ùëê§ ~r = ~r(t) = (x(t), y(t), z(t)) : [α, β] ⊂ R −→ R 3 . dþ ~r 0 (t) = (x 0 (t), y0 (t), z0 (t)). ùãl s(L) = Z β α |~r 0 (t)|dt = Z β α p x02(t) + y02(t) + z 02(t)dt. (11.1) l´ ds = |~r 0 (t)|dt = p x02(t) + y02(t) + z 02(t)dt. 1/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f 11.1.2 1.È©½Â áþ L ´3 R3 ¥þ!á, Ùþ©Ù (Ý) ëY¼ê ρ(x, y, z). ¦dáþ. Ün{´kòá L ©¤k¿©áã: L1, L2, · · · , Ln, zãþþÝCq~ê ρ(ξi, ηi, ζi), ùp (ξi, ηi, ζi) ã Li þ:, ¤±ãþCq ρ(ξi, ηi, ζi)∆si , Ù¥ ∆si ´ã Li Ý. ò¤kùCq\Ò´áoþCq: X n i=1 ρ(ξi, ηi, ζi)∆si. ©©[, ùÚª4ATÒ´áþ. 2/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f ½Â 1 L ´ R3 ¥^¦, å:Úª:©O A Ú B. f ´½Â3 L þ¼ê. l A B 3 L þg: A = A0, A1, · · · , An = B, §ò L ©¤ n ã, /¤ L © T , P1 i ã Li = ) Ai−1Ai l ∆si (i = 1, 2, · · · , n), ¿P λ = max 16i6n ∆si, ¡© T °Ý. 3 Li þ?: Pi = (ξi, ηi, ζi), A A A A A A A 1 2 3 n-1 i i-1 B Pi 3/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f Úª S(f, T ) = X n i=1 f(Pi)∆si. (11.2) XJ3ê I, ¦é?¿ ε > 0, 3 δ > 0, λ < δ, ØØ Pi ∈ Li XÛÀJÑk |S(r, T ) − I| < ε, K¡ f 3 L þÈ, ê I ¡ f 3 L þ1.È©, P Z L f(x, y, z) ds ½ Z L f ds 5 ~¼ê c 3 L þÈ, Z L c ds = cs(L), Ù¥ s(L) ´ L l. 4/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f 11.1.3 È©5 ½n 1 L ´ R3 ¥^1w, ÙëêL« ϕ(t) = (x(t), y(t), z(t)), (α 6 t 6 β). XJ f ´ L þëY¼ê, @ok Z L fds = Z β α f ◦ ϕ(t)|ϕ 0 (t)| dt. (11.3) y² T1 : α = t0 < t1 < · · · < tn = β ´ [α, β] ©. 3N ϕ epÑ L © T : A0, A1, · · · , An, Ù¥ Ai = ϕ(ti), (i = 1, 2 · · · , n). ÷^½Â 1 PÒ, âlúªÚÈ© 5/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f ¥½n, k ∆si = Z ti ti−1 |ϕ 0 (t)| dt = |ϕ 0 (t˜i)|∆ti, Ù¥ t˜i ∈ (ti−1, ti). (ξi, ηi, ζi) = ϕ(t˜i). Ïd S(f, T ) = X n i=1 f ◦ ϕ(t˜i)|ϕ 0 (t˜i)|∆ti. ù´¼ê f ◦ ϕ(t)|ϕ0 (t)| 3«m [α, β] þ Riemann Ú. - kT1k → 0, Ò1.È©Oúª (11.3). íØ 1 L ´ R3 ¥^1w, ÙëêL« ϕ(t) = (x(t), y(t), z(t)), (α 6 t 6 β). K L l´ s(L) = Z L ds = Z β α |ϕ 0 (t)| dt. 6/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f ½n 2 XJ f 3 L þÈ, @o f 3 L þk. ½n 3 e f Ú g Ñ3 L þÈ, c1, c2 ´~ê, K c1f + c2g 3 L þ È, Z L (c1f + c2g) ds = c1 Z L f ds + c2 Z L g ds. ½n 4 f Ú g Ñ3 L þÈ. (1) e f > 0, K Z L f ds > 0; (2) e f > g, K Z L f ds > Z L g ds. 7/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f ½n 5 L1 Ú L2 ´ü^¦ã, L1 ª:´ L2 å:. XJ f 3 L1 Ú L2 þÑÈ, @o f 3 L = L1 ∪ L2 þÈ, ¿ Z L f ds = Z L1 f ds + Z L2 f ds. ½Â 2 8Ü E ⊂ R3 ¡"ÿ8, e31w ϕ : [0, 1] → R3 ¦ ϕ CX E, E 3 ϕ e´ [0, 1] ¥"ÿ8. ½n 6 L ´^¦ã, f ´ L þk.¼ê, XJ f 3 L þ mä: N´"ÿ8, K f 3 L þÈ. ½n 7 L ´^¦ã. e f ´ L þëY¼ê, K3 P0 ∈ L ¦ Z L f ds = f(P0)s(L). 8/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f ~ 1 ¦È© Z L p x2 + y2 ds, Ù¥ L ´ ± x 2 + y 2 = ax (a > 0). ) L ëê§ x = a 2 + a 2 cos t, y = a 2 sin t, (0 6 t 6 2π). Kþ ~r 0 (t) = − a 2 sin t, a 2 cos t . l´ ds = |~r 0 (t)| dt = a 2 dt. Ïd, Z L p x2 + y2 ds = Z 2π 0 a 2 2 r a(1 + cos t) 2 dt = a 2 2 Z 2π 0 cos t 2 dt = a 2 Z π 0 | cos t| dt = 2a 2 . 9/14 kJ Ik J I £ ¶ '4 òÑ
l ½Â 5 ~f ~ 2 L ý x 2 a2 + y 2 b 2 = 1 (a > 0, b > 0) 31lã, O È© Z L xyds. ) L §¤ x = a cos θ, y = b sin θ, 0 6 θ 6 π 2 , ¤± Z L xyds = ab Z π/2 0 cos θ sin θ p a2 sin2 θ + b 2 cos2 θ dθ = ab 2 Z π/2 0 q b 2 + (a2 − b 2) sin2 θ d sin2 θ = ab 3(a2 − b 2) b 2 + (a 2 − b 2 ) sin2 θ 3/2 π/2 0 = ab(a 2 + ab + b 2 ) 3(a + b) . 10/14 kJ Ik J I £ ¶ '4 òÑ