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Jac ϕ(θ, y) =
⎛
⎜
⎝
− y 2 + 1 sin θ
y cos θ
√ y 2 +1
0 1
y2 + 1 cos θ
√y sin θ
y2 +1
ω2 = det 2 B1 + det 2 B2 + det 2 B3
B1 =
− y 2 + 1 sin θ
⎛
y cos θ
√ y 2 +1
0 1
B2 = ⎝ −y2 + 1 sin θ
y2 + 1 cos θ
B3 =
0 1
y2 + 1 cos θ
√y cos θ
y2 +1
√y sin θ
y2 +1
√y sin θ
y2 +1
⎞
⎞
⎟
⎠ .
, det 2 B1 = (y 2 + 1) sin 2 θ.
⎠ , det 2 B2 = y 2 ,
, det 2 B3 = (y 2 + 1) cos 2 θ.
ω2 = 2y2 + 1
ϕ
(1, 0, 0) = ϕ(0, 0) (0, 1, 0) (0, 0, 1)
(±1, 0, 0)
⎛
det ⎝
±1 0 0
0 0 1
0 1 0
⎞
= ⎠ = ∓1.
(1, 0, 0) (−1, 0, 0)
Φ(S, ⎛
2π F1 ◦ ϕ −
1 ⎜
F ) = det ⎜
⎝
0 −1
y2 y cos θ
+ 1 sin θ √
y2 +1
F2 ◦ ϕ 0 1
F3 ◦ ϕ y2 ⎞
⎟
⎠ dy dθ
y sin θ
+ 1 cos θ √
y2 +1
⎛
2π (y
1 ⎜
= det ⎜
⎝
0 −1
2 + 1) cos2 θ − y2 ⎞
y cos θ
+ 1 sin θ √
y2 +1 ⎟
y/2 0 1 ⎟
⎠ dy dθ
y2 + 1 cos θ y2 y sin θ
+ 1 cos θ
=
=
2π 1
0
2π
+
−1
1
0 −1
2π 1
0
= 2
3 π.
−1
⎛
(−y/2)det ⎝ −y2 + 1 sin θ
y2 + 1 cos θ
√y cos θ
y2 +1
√y sin θ
y2 +1
⎞
√ y 2 +1
⎠ dy dθ+
(y2 + 1) cos2 θ − y2 + 1 sin θ
(−1)det
y 2 /2 dy dθ +
y 2 + 1 cos θ y 2 + 1 cos θ
1 2π
−1
0
dy dθ
(y 2 + 1) 3/2 cos 3 θ + (y 2
+ 1) sin θ cos θ dθ dy