TEOREMA DE GREEN
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164 CAPÍTULO6. <strong>TEOREMA</strong><strong>DE</strong> <strong>GREEN</strong><br />
D1<br />
A<br />
D2<br />
C1<br />
B<br />
C31<br />
C2<br />
C 32<br />
i) Seja D 1 tal que ∂D + 1 = C+ 31 ∪ C− 1 , então ∫<br />
Figura6.23:<br />
∂D + 1<br />
∫<br />
F =<br />
C + 31<br />
∫<br />
F −<br />
C + 1<br />
F. Aplicando o teorema de<br />
Green:<br />
∫ ∫∫<br />
(∂F 2<br />
F =<br />
∂D + 1 D 1<br />
∂x − ∂F ∫ ∫<br />
1) dxdy = 0, logo F = F = 12.<br />
∂y<br />
C + 31 C + 1<br />
∫ ∫ ∫<br />
ii) Seja D 2 tal que ∂D 2 + = C+ 32 ∪ C− 2 , então F = F − F. Aplicando o teorema de<br />
Green:<br />
∫<br />
∂D + 2<br />
iii) Como C 3 + = C+ 31 ∪ C− 32 , temos:<br />
∫ ∫<br />
F =<br />
∂D + 2<br />
C + 32<br />
∫∫<br />
(∂F 2<br />
F =<br />
D 2<br />
∂x − ∂F ∫<br />
1) dxdy = 0, logo<br />
∂y<br />
C + 3<br />
C + 31<br />
C + 32<br />
C + 2<br />
∫<br />
F =<br />
∫<br />
F − F = 12 − 15 = −3.<br />
C + 32<br />
C + 2<br />
F = 15.