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

1
Jac ϕ(θ) 1
ds = ˙x 2 + ˙y 2 = (A sin(θ) + Aθ cos(θ)) 2 + (A cos(θ) − Aθ sin(θ)) 2 = A 2 (1 + θ 2 ) = |A| 1 + θ 2 .
t = θ 2 dt = 2θ dθ z = 1 + t dz = dt
I =
4π
0
= |A|
2
= |A|
2
θ 3 |A| 1 + θ2 dθ = |A|
2
16π 2 +1
1
z 5/2
5/2
z=16π 2 +1
z=1
16π 2
0
16π2 +1
z 3/2 dz − |A|
2 1
− |A|
z
2
3/2
3/2
t √ 1 + t dt = |A|
2
z 1/2 dz
z=16π 2 +1
z=1
= |A|
5 ((16π2 + 1) 5/2 − 1) − |A|
3 ((16π2 + 1) 3/2 − 1).
16π 2 +1
F · n dσ, F := (xz, xy, yz)
S
1
(z − 1) √ z dz
S := ∂{(x, y, z) ∈ R3 : x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1}
f dσ, f(x, y, z) := (z + 1) 1 + x2 + y2 + x 3 + y 2
C
C := {(x, y, z) ∈ R 3 : x 2 + y 2 = z 2 , 0 < z < 1}
α > 0
⎧
| sin(xy) − xy|
⎪⎨
f(x, y) =
⎪⎩
α
(x2 + y2 ) 3 (x, y) = (0, 0),
0 (x, y) = (0, 0).
α f (0, 0)
sin(xy)−xy
β > 0
sin(s) − s
lim
s→0 sβ
sin(s) − s
lim
s→0 sβ cos(s) − 1
= lim
s→0 βsβ−1 = lim
s→0
− sin(s)
,
β(β − 1)sβ−2 β − 2 = 1 β = 3
sin(s) − s
lim
s→0 s3 = − 1
6 .
α ≤ 0 α ≤ 0 (x, y) → 0
| sin(xy) − xy| α
(x 2 + y 2 ) 3
≥
1
(x 2 + y 2 ) 3