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21. INTEGRALI CURVILINEI, TEOREMA DI STOKES, FORMULE DI GAUSS-GREEN - CONTINUAZIONE <strong>10</strong>7<br />

Pertanto sarà necessario invertire il segno, si ha allora che il flusso richiesto vale:<br />

Φ(S, ⎛<br />

2π F1 ◦ ϕ −<br />

1 ⎜<br />

F ) = − det ⎜<br />

⎝<br />

0 −1<br />

y2 y cos θ<br />

+ 1 sin θ √<br />

y2 +1<br />

F2 ◦ ϕ 0 1<br />

F3 ◦ ϕ y2 ⎞<br />

⎟<br />

⎠ dy dθ<br />

y sin θ<br />

+ 1 cos θ √<br />

y2 +1<br />

⎛<br />

2π (y<br />

1 ⎜<br />

= − det ⎜<br />

⎝<br />

0 −1<br />

2 + 1) cos2 θ − y2 ⎞<br />

y cos θ<br />

+ 1 sin θ √<br />

y2 +1 ⎟<br />

<br />

y/2 0 1 ⎟<br />

⎠ dy dθ<br />

y2 + 1 cos θ y2 y sin θ<br />

+ 1 cos θ √<br />

y2 +1<br />

⎛<br />

2π 1<br />

= − (−y/2)det ⎝ −y2 ⎞<br />

y cos θ<br />

+ 1 sin θ √<br />

y2 +1 ⎠ dy dθ+<br />

y2 + 1 cos θ<br />

−<br />

= −<br />

0 −1<br />

2π 1<br />

0 −1<br />

2π 1<br />

0<br />

= − 2<br />

3 π.<br />

−1<br />

(−1)det<br />

√y sin θ<br />

y2 +1<br />

(y 2 + 1) cos 2 θ − y 2 + 1 sin θ<br />

y 2 + 1 cos θ y 2 + 1 cos θ<br />

y 2 1<br />

/2 dy dθ +<br />

−1<br />

Nell’ultimo passaggio è sfruttato il fatto che:<br />

2π<br />

0<br />

cos θ sin θ dθ = 1<br />

2<br />

2π<br />

0<br />

cos 3 θ dθ =<br />

=<br />

=<br />

2π<br />

0<br />

3π/2<br />

−π/2<br />

π/2<br />

−π/2<br />

1<br />

−1<br />

2π<br />

0<br />

sin(2θ) dθ = 1<br />

4<br />

cos 3 θ dθ =<br />

<br />

dy dθ<br />

<br />

(y 2 + 1) 3/2 cos 3 θ + (y 2 <br />

+ 1) sin θ cos θ dθ dy<br />

4π<br />

0<br />

3π/2<br />

−π/2<br />

(1 − sin 2 θ) cos θ dθ +<br />

(1 − w 2 ) dw +<br />

−1<br />

1<br />

sin w dw = 0.<br />

(1 − sin 2 θ) cos θ dθ<br />

3/2π<br />

π/2<br />

(1 − w 2 ) dw = 0.<br />

(1 − sin 2 θ) cos θ dθ

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