16 - Teorema de Green - Alum
16 - Teorema de Green - Alum
16 - Teorema de Green - Alum
ENUNCIADO
Si
𝑦
- 𝐹 : 𝐴 ⊂ ℝ2 ⟶ ℝ2 , 𝐷 ⊂ 𝐴 (𝐴 abierto)
Entonces
𝜕𝑄 𝜕𝑃
𝑃𝑑𝑥 + 𝑄𝑑𝑦 = − 𝑑𝑥𝑑𝑦
𝐶 𝐷 𝜕𝑥 𝜕𝑦
𝑥-simple a la vez)
𝑦
d
A
D
c C
𝑥
a b
El teorema de Green quedará demostrado si se prueba que:
𝜕𝑃
𝟏 𝑃𝑑𝑥 = − 𝑑𝑥𝑑𝑦
𝐶 𝐷 𝜕𝑦
𝜕𝑄
𝟐 𝑄𝑑𝑦 = 𝑑𝑥𝑑𝑦
𝐶 𝐷 𝜕𝑥
Demostración de (1)
𝐷= 𝑥, 𝑦 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑔1 (𝑥) ≤ 𝑦 ≤ 𝑔2 (𝑥)
𝑦
𝒚 = 𝒈𝟐 (𝒙)
𝑪𝟐
𝑏 𝑔2 𝑥
𝜕𝑃 𝜕𝑃
− 𝑑𝑥𝑑𝑦 = − 𝑥, 𝑦 𝑑𝑦 𝑑𝑥 𝑪−
𝟐
𝐷 𝜕𝑦 𝑎 𝑔1 𝑥 𝜕𝑦
D
𝐏𝐨𝐫 𝐞𝐥 𝐭𝐞𝐨𝐫𝐞𝐦𝐚 𝐟𝐮𝐧𝐝𝐚𝐦𝐞𝐧𝐭𝐚𝐥
𝐝𝐞𝐥 𝐜á𝐥𝐜𝐮𝐥𝐨
𝑪𝟏
𝒚 = 𝒈𝟏 (𝒙) 𝑥
a b
𝑏
𝑦=𝑔2 𝑥
=− 𝑃(𝑥, 𝑦) 𝑦=𝑔1 𝑥 𝑑𝑥
𝑎
𝑏
=− 𝑃(𝑥, 𝑔2 𝑥 ) − 𝑃(𝑥, 𝑔1 𝑥 ) 𝑑𝑥
𝑎
𝑏 𝑏
𝜕𝑃
− 𝑑𝑥𝑑𝑦 = 𝑃 𝑥, 𝑔1 𝑥 𝑑𝑥 − 𝑃 𝑥, 𝑔2 𝑥 𝑑𝑥
𝐷 𝜕𝑦 𝑎 𝑎
= 𝑃 𝑥, 𝑦 𝑑𝑥 − 𝑃 𝑥, 𝑦 𝑑𝑥
𝐶1 𝐶2−
𝐩𝐨𝐫 𝐩𝐫𝐨𝐩𝐢𝐞𝐝𝐚𝐝 𝐝𝐞
𝐢𝐧𝐯𝐞𝐫𝐬𝐢ó𝐧 𝐝𝐞𝐥 𝐜𝐚𝐦𝐢𝐧𝐨
= 𝑃 𝑥, 𝑦 𝑑𝑥 + 𝑃 𝑥, 𝑦 𝑑𝑥
𝐶1 𝐶2
= 𝑃(𝑥, 𝑦)𝑑𝑥
𝐶
Y queda demostrado 𝟏 .
𝑥-simple.
Ejemplo 1
𝑪𝟑− C1 𝜕𝑄 𝜕𝑃
𝑃𝑑𝑥 + 𝑄𝑑𝑦 = − 𝑑𝑥𝑑𝑦
𝜕𝑥 𝜕𝑦
C2 𝑪 = 𝑪𝟏 ∪ 𝑪𝟐 ∪ 𝑪𝟑 𝐶 𝐷
0,0 1 𝑥 𝑃 = 3𝑦 + 𝑥 2 ; 𝑄 = 4𝑥
𝐷 = (𝑥, 𝑦) 0 ≤ 𝑥 ≤ 1, 𝑥 ≤ 𝑦 ≤ 1
1 1 1 1
𝜕𝑄 𝜕𝑃
− 𝑑𝑥𝑑𝑦 = 4 − 3 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑥
𝐷 𝜕𝑥 𝜕𝑦 0 𝑥 0 𝑥
1
𝑦=1
= 𝑦 𝑦 =𝑥 𝑑𝑥
0
1
= 1 − 𝑥 𝑑𝑥
0
1
𝑥2 1 1
= 𝑥− =1− =
2 0
2 2
Parametrización para C1
𝑥=𝑡 ; 𝑑𝑥 = 𝑑𝑡
𝑦 = 𝑥; 0≤𝑡≤1
𝑦=𝑡 ; 𝑑𝑦 = 𝑑𝑡
1 1 1
2 2
7 1
3𝑦 + 𝑥 𝑑𝑥 + 4𝑥𝑑𝑦 = 3𝑡 + 𝑡 𝑑𝑡 + 4𝑡𝑑𝑡 = 7𝑡 + 𝑡 𝑑𝑡 = 𝑡 2 + 𝑡 3
2
𝐶1 0 0 2 3 0
23
=
6
Parametrización para 𝑪−
𝟐
𝑥=𝑡 ; 𝑑𝑥 = 𝑑𝑡
𝑦 = 1; 0≤𝑡≤1
𝑦=1 ; 𝑑𝑦 = 0𝑑𝑡
1 0 0
2 2 2
𝑡3
− 3𝑦 + 𝑥 𝑑𝑥 + 4𝑥𝑑𝑦 = − 3 + 𝑡 𝑑𝑡 + 0 = 3 + 𝑡 𝑑𝑡 = 3𝑡 +
𝐶2− 0 1 3 1
10
=−
3
𝑥 = 0 ; 𝑑𝑥 = 0𝑑𝑡
𝑥 = 0; 0≤𝑡≤1
𝑦 = 𝑡 ; 𝑑𝑦 = 𝑑𝑡
1
2
− 3𝑦 + 𝑥 𝑑𝑥 + 4𝑥𝑑𝑦 = − 0+ 0 =0
𝐶3− 0
Luego
3𝑦 + 𝑥 2 𝑑𝑥 + 4𝑥𝑑𝑦
𝐶
1 2
𝑥𝑦 𝑑𝑥 + 𝑥 2 𝑦𝑑𝑦
𝐶 2
Solución
𝑦
5,4
𝑦2 C2
𝑥= −3
2
D 𝑪 = 𝑪𝟏 ∪ 𝑪𝟐
C1
𝑥 =𝑦+1
−1, −2
1
𝑃 = 𝑥𝑦 2 ; 𝑄 = 𝑥2𝑦
2
𝑦2
𝐷= 𝑥, 𝑦 − 2 ≤ 𝑦 ≤ 4, − 3 ≤ 𝑥 ≤ 𝑦 + 1
2
1 2 𝜕 2 𝜕 1 2
𝑥𝑦 𝑑𝑥 + 𝑥 2 𝑦𝑑𝑦 = 𝑥 𝑦 − 𝑥𝑦 𝑑𝑥𝑑𝑦
𝐶 2 𝐷 𝜕𝑥 𝜕𝑦 2
4 𝑦 +1
= 2𝑥𝑦 − 𝑥𝑦 𝑑𝑥 𝑑𝑦
𝑦2
−2 −3
2
4 𝑦 +1
= 𝑥𝑦 𝑑𝑥 𝑑𝑦
𝑦2
−2 −3
2
4 𝑥=𝑦 +1
𝑥2
= 𝑦 𝑑𝑦
−2 2 𝑦2
𝑥= −3
2
2
1 4 2 𝑦2
= 𝑦+1 𝑦− −3 𝑦 𝑑𝑦
2 −2 2
4
1 1
= 𝑦 3 + 2𝑦 2 + 𝑦 − 𝑦 5 + 3𝑦 3 − 9𝑦 𝑑𝑦
2 −2 4
4
1 1
= 4𝑦 3 + 2𝑦 2 − 8𝑦 − 𝑦 5 𝑑𝑦
2 −2 4
4
1 2 1
= 𝑦 4 + 𝑦 3 − 4𝑦 2 − 𝑦 6
2 3 24 −2
= 36
Ejemplo 3
1 1
𝑒 𝑎𝑟𝑐 𝑡𝑔 𝑥
− 𝑦 3 𝑑𝑥 + 𝑥 3 𝑑𝑦
𝑐 3 3
𝑥
𝑥2 + 𝑦2 = 1
Solución
1 1 𝜕 1 3 𝜕 1
𝑒 𝑎𝑟𝑐 𝑡𝑔 𝑥
− 𝑦 3 𝑑𝑥 + 𝑥 3 𝑑𝑦 = 𝑥 − 𝑒 𝑎𝑟𝑐 𝑡𝑔 𝑥
− 𝑦3 𝑑𝐴
𝑐 3 3 𝐷 𝜕𝑥 3 𝜕𝑦 3
Cambiando a coordenadas polares
𝜋 2
2 2
𝑥 + 𝑦 𝑑𝐴 = 𝑟 2 𝑟𝑑𝑟 𝑑𝜃
𝐷 0 1
𝜋 2
= 𝑟 3 𝑑𝑟 𝑑𝜃
0 1
𝜋 2
𝑟4
= 𝑑𝜃
0 4 1
𝜋
1
= 4− 𝑑𝜃
0 4
𝜋
15 15
= 𝜃 = 𝜋
4 0 4
EJERCICIOS RESUELTOS
Ejercicio 1
Solución
𝑦
1,1
𝑪𝟐 𝜕𝑄 𝜕𝑃
𝑃𝑑𝑥 + 𝑄𝑑𝑦 = − 𝑑𝑥𝑑𝑦
𝑦=𝑥 𝐶 𝐷 𝜕𝑥 𝜕𝑦
𝑪𝟏
𝑫
𝑦 = 𝑥2 𝑃=𝑦 ; 𝑄 = 2𝑥 + 𝑦
𝑥
𝑪 = 𝑪𝟏 ∪ 𝑪𝟐
1 𝑥 1 𝑥 1 1 1
𝑥 2
𝑥2 𝑥3 1 1 1
2 − 1 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑥 = 𝑦 𝑥2 𝑑𝑥 = 𝑥−𝑥 𝑑𝑥 = − = − =
0 𝑥2 0 𝑥2 0 0 2 3 0
2 3 6
𝜕 𝜕 1
𝐷
2𝑥 + 𝑦 − 𝑦 𝑑𝐴 =
𝜕𝑥 𝜕𝑦 6
= 𝑦𝑑𝑥 + 2𝑥 + 𝑦 𝑑𝑦 + 𝑦𝑑𝑥 + 2𝑥 + 𝑦 𝑑𝑦
𝐶1 𝐶2
= 𝑦𝑑𝑥 + 2𝑥 + 𝑦 𝑑𝑦 − 𝑦𝑑𝑥 + 2𝑥 + 𝑦 𝑑𝑦
𝐶1 𝐶2−
Parametrización para C1
𝑥=𝑡 ; 𝑑𝑥 = 𝑑𝑡
𝑦 = 𝑥2; 0≤𝑡≤1
𝑦 = 𝑡2 ; 𝑑𝑦 = 2𝑡𝑑𝑡
1 1
2 2
𝑦𝑑𝑥 + 2𝑥 + 𝑦 𝑑𝑦 = 𝑡 𝑑𝑡 + 2𝑡 + 𝑡 2𝑡𝑑𝑡 = 5𝑡 2 + 2𝑡 3 𝑑𝑡
𝐶1 0 0
1
5𝑡 3 𝑡4 5 1 13
= + = + =
3 2 0 3 2 6
1
1
=− 4𝑡𝑑𝑡 = − 2𝑡 2 0 = −2
0
𝑄 𝑄
𝜕 𝜕 𝑃 1 𝑃
2𝑥 + 𝑦 − 𝑦 𝑑𝐴 = = 𝑦 𝑑𝑥 + 2𝑥 + 𝑦 𝑑𝑦
𝐷 𝜕𝑥 𝜕𝑦 6 𝐶
Ejercicio 2
Solución
𝑦
0,1
𝑪 = 𝑪𝟏 ∪ 𝑪𝟐 ∪ 𝑪𝟑 𝜕𝑄 𝜕𝑃
𝑪𝟐
𝑃𝑑𝑥 + 𝑄𝑑𝑦 = − 𝑑𝑥𝑑𝑦
𝑪𝟑 𝑦 =1−𝑥 𝐶 𝐷 𝜕𝑥 𝜕𝑦
D
𝑥
𝑃 = 𝑦2 ; 𝑄 = 𝑥𝑦
0,0 𝑪𝟏 1,0
Integral doble:
1 1−𝑥
𝜕𝑄 𝜕𝑃
− 𝑑𝑥𝑑𝑦 = 𝑦 − 2𝑦 𝑑𝑦 𝑑𝑥
𝐷 𝜕𝑥 𝜕𝑦 0 0
1 1−𝑥
𝑦2
=− 𝑑𝑥
0 2 0
1
1 2
=− 1−𝑥 𝑑𝑥
2 0
1
1
=− 1 − 2𝑥 + 𝑥 2 𝑑𝑥
2 0
1
1 2
𝑥3 1 1 1
=− 𝑥−𝑥 + =− 1−1+ =−
2 3 0
2 3 6
𝑃
𝜕 𝑄 𝜕 1
𝑥𝑦 − 𝑦2 𝑑𝐴 = −
𝐷 𝜕𝑥 𝜕𝑦 6
Integral de línea:
Parametrización para C1
𝑥=𝑡 ; 𝑑𝑥 = 𝑑𝑡
𝑦 = 0; 0≤𝑡≤1
𝑦=0 ; 𝑑𝑦 = 0𝑑𝑡
1
2
𝑦 𝑑𝑥 + 𝑥𝑦𝑑𝑦 = 0+0=0
𝐶1 0
𝑥=𝑡 ; 𝑑𝑥 = 𝑑𝑡
𝑦 = 1 − 𝑥; 0≤𝑡≤1
𝑦 =1−𝑡 ; 𝑑𝑦 = −𝑑𝑡
1
2 2
𝑦 𝑑𝑥 + 𝑥𝑦𝑑𝑦 = − 𝑦 𝑑𝑥 + 𝑥𝑦𝑑𝑦 = − (1 − 𝑡)2 𝑑𝑡 + 𝑡(1 − 𝑡)(−𝑑𝑡)
𝐶2 𝐶2− 0
1
=− 1 − 2𝑡 + 𝑡 2 − 𝑡 + 𝑡 2 𝑑𝑡
0
1
=− 1 − 3𝑡 + 2𝑡 2 𝑑𝑡
0
1
3 2
= − 𝑡 − 𝑡2 + 𝑡3
2 3 0
3 2 1
= −1 + − = −
2 3 6
𝑥=0 ; 𝑑𝑥 = 0
𝑥 = 0; 0≤𝑡≤1
𝑦=𝑡 ; 𝑑𝑦 = 𝑑𝑡
1
2 2
𝑦 𝑑𝑥 + 𝑥𝑦𝑑𝑦 = − 𝑦 𝑑𝑥 + 𝑥𝑦𝑑𝑦 = − 𝑡 2 . 0 + 0. 𝑑𝑡 = 0
𝐶3 𝐶3− 0
𝑃 𝑄
1 1
𝑦2 𝑑𝑥 + 𝑥𝑦 𝑑𝑦 = 𝑦 2 𝑑𝑥 + 𝑥𝑦𝑑𝑦 − 𝑦 2 𝑑𝑥 + 𝑥𝑦𝑑𝑦 − 𝑦 2 𝑑𝑥 + 𝑥𝑦𝑑𝑦 = 0 − +0=−
𝐶 𝐶1 𝐶2− 𝐶3− 6 6
EJERCICIOS PROPUESTOS
Verifique el teorema de Green para cada uno de los siguientes campos vectoriales 𝐹
en la región 𝐷 de ℝ2 indicada.
−1 ≤ 𝑦 ≤ 𝑥
I) 𝐹 𝑥, 𝑦 = 3𝑥𝑦, 2𝑥 2 ; 𝐷:
0≤𝑥≤1
5
Resp: 𝐷
= 𝐶=𝜕𝐷
=
6
𝑥 ≤ 𝑦 ≤ 2 − 𝑥2
II) 𝐹 𝑥, 𝑦 = 𝑥𝑦, 𝑥 2 ; 𝐷:
0≤𝑥≤1
5
Resp: 𝐷
= 𝐶=𝜕𝐷
=
12
1 1
− 𝑥 ≤ 𝑦 ≤ 1 − 𝑥2
III) 𝐹 𝑥, 𝑦 = 𝑥, 𝑥 ; 2
𝐷: 2 2
0≤𝑥≤1
1
Resp: 𝐷
= 𝐶=𝜕𝐷
=
3
0 ≤ 𝑦 ≤ 𝑥2
IV) 𝐹 𝑥, 𝑦 = 𝑥𝑦, 𝑥 2 ; 𝐷:
0≤𝑥≤2
Resp: 𝐷
= 𝐶=𝜕𝐷
=4
𝑥 2 − 2𝑥 ≤ 𝑦 ≤ 2𝑥
V) 𝐹 𝑥, 𝑦 = 3𝑦 + 1, 4𝑥 ; 𝐷:
0≤𝑥≤3
Resp: 𝐷
= 𝐶=𝜕𝐷
=−9
1 − 𝑥 ≤ 𝑦 ≤ 𝑥 2 + 2𝑥 + 1
VI) 𝐹 𝑥, 𝑦 = 𝑦, 2𝑥 ; 𝐷:
−1 ≤ 𝑥 ≤ 1
14
Resp: 𝐷
= 𝐶=𝜕𝐷
=
3
𝑥 ≤ 𝑦 ≤ 𝑥3 + 2
VII) 𝐹 𝑥, 𝑦 = 3𝑥 2 𝑦 − 𝑦, 𝑥 3 ; 𝐷:
−1 ≤ 𝑥 ≤ 1
Resp: 𝐷
= 𝐶=𝜕𝐷
=4
0 ≤ 𝑦 ≤ 𝑥2
VIII) 𝐹 𝑥, 𝑦 = 2𝑦, −𝑥 ; 𝐷:
0≤𝑥≤2
Resp: 𝐷
= 𝐶=𝜕𝐷
= −8
IX) 𝐹 𝑥, 𝑦 = 𝑥, 2𝑥𝑦 ; 𝐷: 𝑦 2 ≤ 𝑥 ≤ 2 − 𝑦2
Resp: 𝐷
= 𝐶=𝜕𝐷
=0
−𝑥 − 1 ≤ 𝑦 ≤ 𝑥 + 1
X) 𝐹 𝑥, 𝑦 = 𝑥𝑦, 𝑥 2 ; 𝐷:
−1 ≤ 𝑥 ≤ 1
4
Resp: 𝐷
= 𝐶=𝜕𝐷
=
3