Re: [微積] Stokes' Theorem

[air11 (拿出一張白紙...)]

※ 引述《alex800813 (L)》之銘言：
: use Stokes' Theorem to calculate ∫c [(x-z)dx+(x+y)dy+(y+z)dz]
: where C is the ellipse where plane z=y intersects the cylinder
: x^2+y^2=1 , oriented counterclockwise as viewed from above.
: 知道題目大概的意思 應該是要用curl F 下去解的吧..
: 但是我解出來有點複雜 希望有高手願意幫忙


∫  Fdr = ∫∫ curlFdS
  C          S

S:h(r,θ) = (rcosθ,rsinθ,rsinθ) 0≦r≦1 and 0≦θ≦2pi

  h_r X h_θ = (0,-r,r)  (Note that "X" is cross )
By calculating, we get curlF = (1,-1,1)

by Stokes' thm, we have

 2pi  1
∫  ∫  2rdrdθ = 2pi*1=2pi
  0   0


BTY, we check ans by line integral as below :

C:(cosθ,sinθ,sinθ)  0≦θ≦2pi

so ∫(x-z)dx+(x+y)dy+(y+z)dz
    C
    2pi
 = ∫  (cosθ-sinθ)*(-sinθ)+(cosθ+sinθ)*(cosθ)+(2sinθ)*(cosθ)dθ
    0

 =2pi
     #

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推 alex800813  :非常謝謝你的幫忙! 我會了!
推 peicachu    :剛剛眼花，現在手殘，curlF算成(1,1,1)XD
推 alex800813  :突然有個地方想不通 他的h_r X h_θ是怎麼算出來的@@
推 peicachu    :cylinder被z=y截面的法向量，取逆時鐘方向
推 peicachu    :h_r的意思是對r偏微分，直接外積就好