Terry
Lv 5
Terry 發問時間: 科學數學 · 1 0 年前

有關 Green's Theorem 格林定理一題證明

Let R be the region in the plane bounded by a continuously differentiable simple closed curve C.

Show that the area of R is given by

(1/2)∮c (xdy - ydx)

please show the work in detail, thanks a lot!

已更新項目:

∮ 是亂碼, 更正為封閉曲線積分符號

1 個解答

評分
  • prime
    Lv 4
    1 0 年前
    最佳解答

    令 P(x,y)= -y, Q(x,y)= x

    則 P(x,y), Q(x,y)在整個R^2皆為C^(1), 故在R上為C^(1).

    根據Green's Theorem

    ∮(on C) P(x,y)dx + Q(x,y)dy

    =∫∫(on R) ( ∂ Q / ∂x - ∂ P/ ∂ y) dx dy

    =∫∫(on R) 2 dx dy

    = 2 ∫∫(on R) 2 dx dy

    = 2 * The Area of R

    故 (1/2) *∮(on C) P(x,y)dx + Q(x,y)dy = The Area of R

    2006-12-13 04:52:06 補充:

    令 P(x,y)= -y, Q(x,y)= x

    則 P(x,y), Q(x,y)在整個R^2皆為C^(1), 故在R上為C^(1).

    根據Green's Theorem

    ∮(on C) P(x,y)dx Q(x,y)dy

    =∫∫(on R) ( ∂ Q / ∂x - ∂ P/ ∂ y) dx dy

    =∫∫(on R) 2 dx dy

    = 2 ∫∫(on R) 2 dx dy

    = 2 * The Area of R

    故 (1/2) *∮(on C) P(x,y)dx Q(x,y)dy = The Area of R

    2006-12-13 04:52:34 補充:

    令 P(x,y)= -y, Q(x,y)= x

    則 P(x,y), Q(x,y)在整個R^2皆為C^(1), 故在R上為C^(1).

    根據Green's Theorem

    ∫(on C) P(x,y)dx Q(x,y)dy

    =∫∫(on R) ( ∂ Q / ∂x - ∂ P/ ∂ y) dx dy

    =∫∫(on R) 2 dx dy

    = 2 ∫∫(on R) 2 dx dy

    = 2 * The Area of R

    故 (1/2) *∫(on C) P(x,y)dx Q(x,y)dy = The Area of R

    2006-12-13 04:53:44 補充:

    令 P(x,y)= -y, Q(x,y)= x

    則 P(x,y), Q(x,y)在整個R^2皆為C^(1), 故在R上為C^(1).

    根據Green's Theorem

    ∫(on C) P(x,y)dx +Q(x,y)dy

    =∫∫(on R) ( ∂ Q / ∂x - ∂ P/ ∂ y) dx dy

    =∫∫(on R) 2 dx dy

    = 2 ∫∫(on R) 2 dx dy

    = 2 * The Area of R

    故 (1/2) *∫(on C) P(x,y)dx+ Q(x,y)dy = The Area of R

    2006-12-13 04:54:12 補充:

    對不起,經常有符號會消失....

    2006-12-13 05:00:28 補充:

    令 P(x,y)= -y, Q(x,y)= x

    則 P(x,y), Q(x,y)在整個R^2皆為C^(1), 故在R上為C^(1).

    根據Green's Theorem

    ∫(on C) P(x,y)dx +Q(x,y)dy

    =∫∫(on R) ( ∂ Q / ∂x - ∂ P/ ∂ y) dx dy

    =∫∫(on R) 2 dx dy

    = 2 ∫∫(on R) dx dy

    = 2 * The Area of R

    故 (1/2) *∫(on C) P(x,y)dx+ Q(x,y)dy = The Area of R

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