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

|y|
x 2 + y 2
D √ 2
(±1, 0) 1
D f
D ∩ {x ≥ 0, y ≥ 0} 4 x 2 + y 2 = 2
(x − 1) 2 + y 2 = 1 x 2 − (x − 1) 2 = 1 x 2 − x 2 + 2x − 1 − 1 = 0 x = 1 y = ±1
(1, 1) x
≤ 1,
D ∩ {x ≥ 0, y ≥ 0} = {(x, y) : 0 ≤ x ≤ 1, 1 − (x − 1) 2 ≤ y ≤ 2 − x 2 }
I = 4
= 4
= 4
= 4
1 √ 2−x2 √ 1−(x−1) 2
0
1 √ 2−x2 0
1
0
1
0
y
dy dx = 4
x2 + y2 1 √ 2−x2 ∂
√ x2 + y2 dy dx
1−(x−1) 2 ∂y
x2 + 2 − x2 − x2 + 1 − (x − 1) 2 dx
√ √
2 − 2x dx = 4√2 3 .
I = dx dy dz,
T
0
2y
√
1−(x−1) 2 2 x2 dy dx
+ y2 T z = a 2 x 2 + b 2 y 2 z = k 2 k = 0
T := {(x, y, z) : (ax) 2 + (by) 2 ≤ z ≤ k 2 }
2 2 =
(x, y, z) :
x
√z/a
+
y
√z/b
≤ 1, 0 < z ≤ k 2
x
√z/a = s cos θ
√ √
zs cos θ/a y(s, θ, z) = zs sin θ/b z(s, θ, z) = z
⎛
a
⎜
Jac(ψ)(s, θ, z) = ⎝
√ z cos θ − √ zs sin θ/a s cos θ
2a √ z
⎞
⎟
⎠ ,
zs/(ab)
y
√ z/b = s sin θ z = z 0 ≤ s ≤ 1 0 < z ≤ k 2 x(s, θ, z) =
I =
b √ z sin θ √ zs cos θ/b s sin θ
2b √ z
0 0 1
2π 1 k2 0
0
0
zs
πk4
dz ds dθ =
ab 2ab .
a > 0
x(x 2 + y 2 ) = a(x 2 − y 2 )