Re: [積分] 線積分

作者: Eliphalet (我大聲講嘢唔代表我冇禮)   2015-08-13 00:14:56
※ 引述《SamBetty (sam)》之銘言:
: 題目:
: → 2
: Let C be the oriented curve parameterized by r(t) = (cost, sint, 8-cos t-sint),
: → 2 2 2 √z
: 0≦t<2π, and let F be the vector field F(x,y,z) = (z - y ,-2xy , e cosz).
: → →
: Evaluate ∫ F‧dr.
: C
: 我有試著用Stokes' Theorem去解,但是卡住了。請問這題有什麼方法可以解嗎?
: 謝謝!
還是可以用 Stokes' Theorem 來處理啊

curl F = (0, 2z, 2y-2y^2)
令 S 為 x^2+y^2 ≦ 1 和 z = 8-x^2-y 之交集
∂_x (x,y,8-x^2-y) = (1,0,-2x)
∂_y (x,y,8-x^2-y) = (0,1,-1)
(1,0,-2x) x (0,1,-1) = (2x,1,1)
因此由 Stokes' 定理,
→ → → →
∫ F‧ dr = ∫ curl F‧ ds ,
C S
= ∫∫ (0, 2(8-x^2-y), 2y-2y^2 )‧(2x,1,1) dydx
x^2+y^2≦1
= ∫∫ 16-2(x^2+y^2) dydx
x^2+y^2≦1
= 16π - π
= 15π

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