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※ 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...
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