Re: [張爸] 算物理遇到的

[Frobenius (▽．(▽×▽φ)=0)]

※ 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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◆ From: 118.161.245.60
※ 編輯: Frobenius       來自: 118.161.245.60       (08/19 23:33)
推 midarmyman:真有耐心 我還是考前背答案好了@@
推 gkaok2:謝原PO
推 andy2007:不推不行
推 ntust661:推強大原PO