Esercizi svolti di analisi reale e complessa - Dipartimento di ...

Area(E) =
=
= 1
2
= 1
4
E
π
4
− π
4
= 1
2 .
dx dy =
dθ
π
4
− π
4
E (pol)
√ cos 2θ
0
[sin 2θ] π
4
− π
4
cos 2θ dθ =
=
ρ dρ =
ρ dρ dθ =
Ek
E (pol)
k
=
(ρ, θ) : − π π
≤ θ ≤
4 4 , ρ2 ≤ k 2 cos 2θ
Ek = {(x, y) ∈ R 2 : x ≥ 0, (x 2 + y 2 ) 2 ≤ k 2 (x 2 − y 2 )} .
E
Area(Ek) =
=
= 1
2
Ek
π
4
− π
4
dx dy =
dθ
π
4
− π
4
E (pol)
k
√ k cos 2θ
0
= 1
4 [k2 sin 2θ] π
4
k 2 cos 2θ dθ =
− π
4
ρ dρ =
ρ dρ dθ =
= k2
2 .
G G
G = {(x, y, z) ∈ R 3 : x ≥ 0, (x 2 + y 2 ) 2 ≤ (1 − z 2 )(x 2 − y 2 ), |z| ≤ 1}
= {(x, y, z) ∈ R 3 : |z| ≤ 1, (x, y) ∈ E√ 1−z2} .
=
,