Re: [張爸] 算物理遇到的
※ 引述《midarmyman (midarmyman)》之銘言：
: a/2 a/2
: ∫ ∫ (x^2+y^2+d^2)^(-3/2)dxdy
: -a/2 -a/2
: a.d都是常數 請問要怎麼積 我只想到三角代換
: 可是又做不出來@@
a/2 a/2
∫ ∫ (x^2+y^2+d^2)^(-3/2)dxdy
-a/2 -a/2
( Let x = √(y^2+d^2) tanθ = z tanθ )
( sinθ = x/√(x^2+z^2) => sinθ = x/√(x^2+y^2+d^2) )
a/2 x=a/2 2 2 2
= ∫ ∫ ( z tan θ + z )^(-3/2) d (z tanθ) dy
-a/2 x=-a/2
a/2 x=a/2 2 2 2
= ∫ ∫ ( z sec θ )^(-3/2) z sec θ dθ dy
-a/2 x=-a/2
a/2 x=a/2 -3 -3 2
= ∫ ∫ z sec θ z sec θ dθ dy
-a/2 x=-a/2
a/2 x=a/2 -2
= ∫ ∫ z cosθ dθ dy
-a/2 x=-a/2
a/2 -2 │x=a/2
= ∫ z sinθ│ dy
-a/2 │x=-a/2
a/2 1 x │x=a/2
= ∫ ───── ──────── │ dy
-a/2 y^2+d^2 √(x^2+y^2+d^2) │x=-a/2
a/2 1 a
= ∫ ───── ────────── dy
-a/2 y^2+d^2 √[y^2+(a^2/4+d^2)]
( Let b d^2 = a^2/4+d^2 => b = (a/2d)^2 + 1 => b - 1 = (a/2d)^2 )
a/2 1 1
= a ∫ ───── ─────── dy
-a/2 y^2+d^2 √(y^2+ b d^2)
( 偶函數，偶對稱範圍 )
a/2 1 1
= 2a ∫ ───── ─────── dy
0 y^2+d^2 √(y^2+ b d^2)
a/2 1 1 dy
= 2a ∫ ───── ──────── ──
0 1+ (d/y)^2 √[1+ b (d/y)^2] y^3
( Let t = (d/y)^2 => dt = -2 (d^2)/(y^3) dy => dy/(y^3) = (-1/2)(1/d^2) dt )
( y = 0 => t → ∞；y = a/2 => t = (2d/a)^2 )
(2d/a)^2 1 1
= (2a/d^2) (-1/2) ∫ ─── ────── dt
∞ 1 + t √ (1 + b t)
(2d/a)^2 1 1
= (a/d^2) (-1) ∫ ─── ────── dt
∞ 1 + t √ (1 + b t)
∞ 1 1
= (a/d^2) ∫ ─── ────── dt
(2d/a)^2 1 + t √ (1 + b t)
( Let τ = √ (1 + b t) => τ^2 = 1 + b t => 2τdτ = b dt => dt = (2τ/b)dτ)
( b t = τ^2 - 1 => t = (τ^2 - 1)/b => 1 + t = [ (b-1) + τ^2 ]/b
=> 1 + t = [ (a/2d)^2 + τ^2 ]/b )
( t = (2d/a)^2 => τ = √ [ 1 + b (2d/a)^2 ] ; t → ∞ => τ→ ∞ )
∞ 1 1 2τ
= (a/d^2) ∫ ────────── ─ ─ dτ
√ [ 1 + b (2d/a)^2 ] [ (a/2d)^2 + τ^2]/b τ b
∞ 1
= (2a/d^2) ∫ ──────── dτ
√ [ 1 + b (2d/a)^2 ] (a/2d)^2 + τ^2
2a/d^2 ∞ 1
= ──── ∫ ───────── d( τ/(a/2d) )
a/2d √ [1 + b (2d/a)^2 ] 1 + [τ/(a/2d)]^2
4 -1 │τ→ ∞
= - tan ( 2dτ/a ) │
d │τ= √ [ 1 + b (2d/a)^2 ]
4 π -1
= - [ ─ - tan ( 2d √ [ 1 + b (2d/a)^2 ] /a ]
d ２
( where b = (a/2d)^2 + 1 => 1 + b (2d/a)^2 => 2 + (2d/a)^2 )
4 π -1
= - [ ─ - tan ( 2d √ [ 2 + (2d/a)^2 ] /a ) ]
d ２
4 -1
= - cot ( 2d √ [ 2 + (2d/a)^2 ] /a )
d
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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 118.161.245.60
※ 編輯: Frobenius 來自: 118.161.245.60 (08/19 23:33)
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08/19 23:37, , 1F
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