pdf (it, 2.898,97 KB, 25/01/13)

1
A =
√ 1−y2 √
2/2 − √ 1−y2 π/2
π/2
√ √
1 √
2dxdy = 2 2 √ 1 − y2 dy = 2 2
2/2
π/2
= 2 √ 2 cos
π/4
2 θ dθ = 2 √ cos(2θ) + 1
2
dθ =
π/4 2
√ 2
√ √ √
π
2 2
2
= π + cos z dz = (π − 2).
4 2
4
π/2
π/4
π/2
π/4
1 − sin 2 θ cos θ dθ
(cos(2θ) + 1) dθ
S = {(x, y, z) : x = s cos t, y = t, z = s sin t, t ∈ (0, π), s ∈
(1, 2)} f : R3 → R f(x, y, z) = yz
f dσ.
S
ϕ(t, s) = (s cos t, t, s sin t)
⎛
−s sin t
Jac ϕ = ⎝ 1
⎞
cos t
0 ⎠ .
s cos t sin t
2
2
ω2(∂tϕ, ∂sϕ) = det 2
−s sin t cos t
+ det
1 0
2
−s sin t cos t
+ det
s cos t sin t
2
1
s cos t
0
sin t
= 1 + s 2
π 2
fdσ = f ◦ ϕ(t, s)ω2(∂tϕ, ∂sϕ) ds dt
S
0 1
π 2
= ts sin t
0 1
1 + s2
ds dt = 1
π
t sin t dt ·
2 0
= π
3 (5√5 − 2 √ 2).
4
1
√ 1 + w dw
D = {(u, v) ∈ R 2 : u 2 + v 2 < 1} S ⊂ R 3
ϕ : D → R 3 ϕ(u, v) = (u + v, u − v, u 2 + v 2 )
(x
S
2 + y 2 ) √ 1 + 2z dσ.
⎛ ⎞
1 1
Jac ϕ = ⎝ 1 −1 ⎠ .
2u 2v
2
2
ω2(∂uϕ, ∂vϕ) = det 2
1 1
+ det
1 −1
2
1 1
+ det
2u 2v
2
1 −1
2u 2v
= 4 + (2v − 2u) 2 + (2v + 2u) 2 = 2 1 + u 2 + v 2