※ 引述《aacvbn ( 姿勢+)》之銘言： : ^ : -> aR -> -> : 1. V = ----- , then ▽˙V = ? , for | R |≠0 : -> 2 : |R | : ∞ ∞ ∞ 1 -> : and ∫ ∫ ∫ ---- ▽˙V dxdydz =? : -∞ -∞ -∞ 2π : ^ -> : 2. Find ∫[aR 3sinΘ]˙dS , among S is origin(0,0,0) used as center, : spherical surface radius of 5 . : (please use divergence theorem to solve and prove) : 3. Check the divergence theorem for the function : 2 ^ 2 ^ 2 ^ : v=r sinΘaR + 4r cosΘaΘ + r tanΘaφ : Using the volume of the “ice-cream cone” shown in Fig.1 : (the top surface is spherical , with radius R and centered at the orgin) : Figure: http://tinyurl.com/y8sz88d : 這三題是算電磁學當中的向量分析部份最難的 ...不知道如何解題 有請高手賜教 : 謝謝~:) assume R=xi+yj+zk e_R 1. ▽˙----- =4πδ^3(R)=4πδ(x)δ(y)δ(z) |R|^2 if R=/=0 ▽˙V=0 ∞ ∞ ∞ 1 -> ∫ ∫ ∫ ---- ▽˙V dxdydz -∞ -∞ -∞ 2π ∞ ∞ ∞ 1 =∫ ∫ ∫ ---- 4πδ(x)δ(y)δ(z) dxdydz -∞ -∞ -∞ 2π =2 (all space include R=zero) 3. 2 ^ 2 ^ 2 ^ v=r sinΘaR + 4r cosΘaΘ + r tanΘaφ ∫v‧ds=∫v‧ds∫v‧ds s s1 s2 2ππ/6 ∫v‧ds=∫ ∫R^2sinθR^2sinθdθdφ s1 0 0 π 3^1/2 =2πR^4(---- - ------) 12 8 2πR ∫v‧ds=∫∫4r^2cosθrsinθdrdφ s2 0 0 3^1/2 =2πR^4cosθsinθ=2πR^4(------) 4 1 @r^2V_r 1 @sinθV_θ 1 @v_φ ▽˙V=--- ------- + ------ ---------- +------ ------- r^2 @r rsinθ @θ rsinθ @φ 1 =2rsinθ + 2rsinθ + ------{4r^2cos^2θ-4r^2sin^2θ} rsinθ 4rcos^2θ =4rsinθ + -------- - 4rsinθ=4rcos^2θ/sinθ sinθ 2π R π/6 ∫▽˙Vdv=∫ ∫ ∫ (4rcos^2θ)r^2dθdrdφ v 0 0 0 π 3^1/2 =2πR^4(---- + -------) 12 8 =∫v‧ds s 觀念cheng都有,不要被數學卡住的話... -- ───────────────╮ 操千曲而後曉聲 ∣ ╭───────────────╯ │ 觀千劍而後識器 ╰------------------------------- --

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