# 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!

&conint; 是亂碼, 更正為封閉曲線積分符號

### 1 個解答

• 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

&conint;(on C) P(x,y)dx + Q(x,y)dy

=&int;&int;(on R) ( &part; Q / &part;x - &part; P/ &part; y) dx dy

=&int;&int;(on R) 2 dx dy

= 2 &int;&int;(on R) 2 dx dy

= 2 * The Area of R

故 (1/2) *&conint;(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