Řešené úlohy na trojný integrál ve válcových a sférických souřadnicích

Trojné integrály - substituce do válcových a sférických souřadnic
Válcové souřadnice (ρ, ϕ, z) bodu (x, y, z) jsou definovány vztahy
(1) x = ρ cos ϕ, y = ρ, sin ϕ, z = z,
kde ρ > 0, α ≤ ϕ ≤ α + 2π a −∞ < z < ∞. Transformace objemu provedeme dosazením
dxdydz = ρdρdϕdz. Poznamenejme, že je x 2 + y 2 = ρ 2 .
Sférické souřadnice (r, θ, ϕ) bodu (x, y, z) jsou definovány vztahy
(2) x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ,
kde r > 0, 0 ≤ θ ≤ π, α ≤ ϕ ≤ α + 2π. Transformaci objemu provedeme dosazením
dxdydz = r2 sin θdrdθdϕ. Poznamenejme, že je x2 + y2 + z2 = r2 .
58. yz dxdydz; A = {(x, y, z); x
A
2 + y 2 + z 2 ≤ 1, z ≥ 0, y ≥ 0}; [ 2
15 ]
(2) r2 ≤ 1, r sin θ sin ϕ ≥ 0, r cos θ ≥ 0 ⇒ 0 < r ≤ 1, 0 ≤ θ ≤ π,
0 ≤ ϕ ≤ π;
2
60.
A
π
yz dxdydz =
=
r 5
5
0
1
.
0
π
2
0
1
sin 3 θ
3
0
π
2
0
r 4 sin 2
θ cos θ sin ϕ dr dθ dϕ =
. [− cos ϕ] π
0
= 2
15 .
(x
A
2 + y 2 ) dxdydz; A = {(x, y, z); x 2 + y 2 − z ≤ 1, z ≤ 0}; [ π
(1) ρ 2 ≤ 1 + z, z ≤ 0 ⇒ ρ 2 − 1 ≤ z ≤ 0, 0 < ρ ≤ 1, 0 ≤ ϕ ≤ 2π;
(x
A
2 + y 2 2π 1 0
) dxdydz =
0 0 ρ2−1
61.
A
= 2π
1
0
ρ 3
dz dρ dϕ = 2π
ρ 3 − ρ 5
dρ = π
6 .
1
0
ρ 3 z 0
ρ 2 −1
x dxdydz; A = {(x, y, z); x 2 + y 2 ≤ 2x, 0 ≤ z ≤ xy} [ 4
5 ]
(1) ρ 2 ≤ 2ρ cos ϕ, 0 ≤ z ≤ ρ 2 cos ϕ sin ϕ ⇒
0 ≤ z ≤ ρ 2 cos ϕ sin ϕ, 0 < ρ ≤ 2 cos ϕ, 0 ≤ ϕ ≤ π/2;
=
A
π
2
0
x dxdydz =
π
2
2 cos ϕ ρ2 cos ϕ sin ϕ
0 0
0
2 cos ϕ
ρ
0
4 cos 2
ϕ sin ϕ dρ
20
dϕ = 32
5
π
2
0
ρ 2 cos ϕ dz
dρ
dϕ =
cos 7 ϕ sin ϕ dϕ = 4
5 .
6 ]
dρ =

62.
yz dxdydz;
A
A = {(x, y, z); x 2 + y 2 ≤ 4, 0 ≤ z ≤ y}; [ 64
15 ]
(1) ρ2 ≤ 4, 0 ≤ z ≤ ρ sin ϕ ⇒ 0 ≤ z ≤ ρ sin ϕ, 0 < ρ ≤ 2, 0 ≤ ϕ ≤ π;
A
63.
π
yz dxdydz =
0
2 ρ sin ϕ
0
0
= 16
5
π
0
ρ 2
z sin ϕ dz dρ dϕ = 1
π 2
ρ
2 0 0
4 sin 3
ϕ dρ dϕ =
sin ϕ(1 − cos 2 ϕ) dϕ = 64
15 .
(x
A
2 + y 2 + z 2 ) 2 dxdydz; A = {(x, y, z); x 2 + y 2 + z 2 ≤ 1, z 2 ≤ x 2 + y 2 , z > 0};
[ π√2 7 ]
(2) r2 ≤ 1, r2 cos2 θ ≤ r2 sin2 θ, r cos θ ≥ 0 ⇒ 0 < r ≤ 1, π/4 ≤ θ ≤ π/2, 0 ≤ ϕ ≤ 2π;
(x
A
2 +y 2 +z 2 ) 2 2π
dxdydz =
0
64.
A
π
2
π
4
1
0
r 6
sin θ dr dθ dϕ = 2π
r 7
7
1
0
. [− cos θ] π
2
π
4
x 2 y dxdydz; A = {(x, y, z); 1 ≤ x 2 + y 2 ≤ 4, y > 0, |z| < 2}; [ 248
15 ]
(1) 1 ≤ ρ 2 ≤ 4, |z| ≤ 2, ρ sin ϕ ≥ 0 ⇒ 1 ≤ ρ ≤ 2, −2 ≤ z ≤ 2, 0 ≤ ϕ ≤ π;
A
x 2 π
y dxdydz =
65.
A
0
2 2
1
−2
ρ 4 cos 2
ϕ sin ϕ dz dρ dϕ =
ρ 5
5
2 . [z]
1
2
−2 .
− cos3 ϕ
3
xy dxdydz; A = {(x, y, z); x 2 + y 2 ≤ 1, 0 < z < 2, x > 0, y > 0}; [ 1
4 ]
π
= π√ 2
7 .
0
= 248
15 .
(1) 0 ≤ ρ 2 ≤ 1, 0 ≤ z ≤ 2, ρ cos ϕ ≥ 0, ρ sin ϕ ≥ 0 ⇒ 0 < ρ ≤ 1, 0 ≤ z ≤ 2, 0 ≤ ϕ ≤ π
2 ;
A
68.
xy dxdydz =
π
2
0
1 2
0
0
ρ 3
cos ϕ sin ϕ dz dρ dϕ =
ρ 4
4
1 . [z]
0
2
0 .
− cos2 ϕ
2
dxdydz; A = {(x, y, z); z ≥ 0, x + y + z ≤ 2, x
A
2 + y 2 ≤ 1}; [2π]
(1) ρ(cos ϕ + sin ϕ) + z ≤ 2, ρ 2 ≤ 1, z ≥ 0 ⇒
0 ≤ z ≤ 2 − ρ(cos ϕ + sin ϕ), 0 < ρ ≤ 1, 0 ≤ ϕ ≤ 2π;
69.
A
=
dxdydz =
2π 1
0
0
2π
1 2−ρ(cos ϕ+sin ϕ)
0
0
0
ρ dz
dρ
(2ρ − ρ 2
(cos ϕ + sin ϕ)) dρ dϕ = 2π.
dϕ =
π
2
0
= 1
4 .
(x
A
2 + y 2 ) dxdydz; A = {(x, y, z); x 2 + y 2 − 2z + 2 ≤ 0, x 2 + y 2 + z ≤ 4}; [2π]
21

(1) ρ 2 − 2z + 2 ≤ 0, ρ 2 + z ≤ 4 ⇒ 1 + ρ2
2 ≤ z ≤ 4 − ρ2 , 0 < ρ ≤ √ 2, 0 ≤ ϕ ≤ 2π;
70.
A
(x
A
2 + y 2
2π √
2 4−ρ2 ) dxdydz =
0 0 1+ ρ2
2
ρ 3 dz
dρ
dϕ = 2π.
dxdydz; A = {(x, y, z); x 2 + y 2 + z 2 ≤ 9, 1 ≤ z ≤ 2}; [ 20π
3 ]
(1) ρ 2 + z 2 ≤ 9, 1 ≤ z ≤ 2 ⇒ 0 < ρ ≤ √ 9 − z 2 , 1 ≤ z ≤ 2, 0 ≤ ϕ ≤ 2π;
72.
A
A
dxdydz =
2π
2 √
9−z2 0
1
0
ρ dρ
dz
2
dϕ = π (9 − z
1
2 ) dz = 20π
3 .
dxdydz; A = {(x, y, z); x 2 + y 2 + z 2 ≤ 2, x 2 + y 2 ≤ 1}; [ 4
3 π(2√ 2 − 1)]
(1) ρ 2 + z 2 ≤ 2, ρ 2 ≤ 1 ⇒ − √ 2 − ρ 2 ≤ z ≤ √ 2 − ρ 2 , 0 < ρ ≤ 1, 0 ≤ ϕ ≤ 2π;
73.
A
dxdydz =
⎛ ⎛
2π 1
⎝ ⎝
0 0 −
√ 2−ρ2 ⎞ ⎞
√ ρ dz⎠
dρ⎠
dϕ = 2π
2−ρ2
= 2π − 2
3 (2 − ρ2 ) 3
1
2 =
0
2 √ 2 − 1 4π
3 .
A
(1) ρ2 ≤ z2 , 0 ≤ z ≤ 1 ⇒ ρ ≤ z ≤ 1, 0 ≤ z ≤ 1, 0 ≤ ϕ ≤ 2π;
A
80.
1
0
2ρ 2 − ρ2 dρ =
x 2 + y 2 dxdydz; A = {(x, y, z); x 2 + y 2 ≤ z 2 , 0 ≤ z ≤ 1}; [ π
6 ]
x2 + y2 2π
dxdydz =
0
1 1
0
ρ
ρ 2 1
dz dρ dϕ = 2π (ρ
0
2 − ρ 3 ) dρ = π
6 .
(x
A
2 + y 2 ) dxdydz; A = {(x, y, z); x 2 + y 2 + z 2 ≤ 3} : [ 24π√3 (1) ρ 2 + z 2 ≤ 3 ⇒ − √ 3 − ρ 2 ≤ z ≤ √ 3 − ρ 2 , 0 < ρ ≤ √ 3, 0 ≤ ϕ ≤ 2π;
(x
A
2 + y 2 ⎛
2π
) dxdydz = ⎝
0
√ ⎛
3
⎝
0 −
81.
A
=
3 − ρ 2 = t 2 , ρ 2 = 3 − t 2
−ρdρ = tdt,
√ 3−ρ2 √ ρ
3−ρ2 3 ⎞ ⎞
dz⎠
dρ⎠
dϕ = 2π
= 4π
√ 3
0
√ 3
(3t 2 − t 4 ) dt = 24π√3 .
5
0
5 ]
2ρρ 2
3 − ρ 2 dρ =
dxdydz; A = {(x, y, z); x 2 + y 2 + z 2 ≤ 1, x 2 + y 2 ≥ z 2 }; [ 2√ 2
3 π]
(2) r 2 ≤ 1, r 2 sin 2 θ ≥ r 2 cos 2 θ ⇒ 0 < r ≤ 1, π/4 ≤ θ ≤ 3π/4, 0 ≤ ϕ ≤ 2π;
22

A
82.
dxdydz =
A
2π
0
3π
4
π
4
1
0
r 2
sin θ dr dθ dϕ = 2π
r 3
3
1
0
. [− cos θ] 3π
4
π
4
= 2π√2 .
3
xyz dxdydz; A = {(x, y, z); x 2 + y 2 + z 2 ≤ 2, x ≥ 0, y ≥ 0, z ≥ 0}; [ 1
48 ]
(2) r 2 ≤ 2, r sin θ cos ϕ ≥ 0, r sin θ sin ϕ ≥ 0, r cos θ ≥ 0 ⇒
0 < r ≤ √ 2, 0 ≤ θ ≤ π/2, 0 ≤ ϕ ≤ π/2;
83.
A
xyz dxdydz =
=
π
2
0
r 6
6
π
2
√ 2
0
0
.
√
2
0
r 5 sin 3
θ cos θ cos ϕ sin ϕ dr
π
2
π
2
sin 4 θ
4
0
.
sin 2 ϕ
2
0
= 1
48 .
dθ
dϕ =
(x
A
2 + y 2 ) dxdydz; A = {(x, y, z); x 2 + y 2 + z 2 ≤ 4, x 2 + y 2 ≤ 2x} : [ 128 26 (π − 15 15 )]
(1) ρ2 + z2 ≤ 4, ρ2 ≤ 2ρ cos ϕ ⇒
− √ 4 − ρ2 ≤ z ≤ √ 4 − ρ2 , 0 < ρ ≤ 2 cos ϕ, −π/2 ≤ ϕ ≤ π/2;
=
= −2
85.
π
2
(x
A
2 + y 2 π
2
) dxdydz =
− π
⎛ ⎛
2 cos ϕ
⎝ ⎝
0 2
√ 4−ρ2 √ ρ
− 4−ρ2 3 ⎞ ⎞
dz⎠
dρ⎠
dϕ =
2 cos ϕ
2ρ
0
3
4 − ρ2
4 − ρ
dρ dϕ =
2 = t2
, 0 → 2
−ρdρ = tdt, 2 cos ϕ → 2| sin ϕ| =
− π
2| sin ϕ|
(4t
2 2
2 − t 4 π
2
)dt dϕ = 2
− π
64 32
−
15 3 2
| sin3 ϕ| + 32
5 | sin5
ϕ| dϕ =
π
2 64 32
= 4 −
0 15 3 sin3 ϕ + 32
5 sin5
ϕ dϕ = 128
π −
15
26
.
15
− π
2
π
2
dxdydz; A = {(x, y, z); x
A
2 + y 2 ≤ 4, z ≥ 0, z + y ≤ 2}; [8π]
(1) ρ 2 ≤ 4, z ≥ 0, ρ sin ϕ + z ≤ 2 ⇒ 0 ≤ z ≤ 2 − ρ sin ϕ, 0 < ρ ≤ 2, 0 ≤ ϕ ≤ 2π;
A
86.
dxdydz =
2π 2 2−ρ sin ϕ
0
0
0
=
2π
0
2π
ρ dz dρ dϕ =
4 − 8
sin ϕ dϕ = 8π.
3
0
2
0
(2ρ − ρ 2
sin ϕ)dρ dϕ =
z dxdydz; A = {(x, y, z); x
A
2 + y 2 + z 2 ≤ 9, 1 ≤ z ≤ 2}; [ 39
4 π]
(1) ρ2 + z2 ≤ 9, 1 ≤ z ≤ 2 ⇒ 0 < ρ ≤ √ 9 − z2 , 1 ≤ z ≤ 2, 0 ≤ ϕ ≤ 2π;
A
z dxdydz =
2π
2 √
9−z2 0
1
0
ρz dρ
23
dz
2
dϕ = π (9z − z
1
3 ) dz = 39π
4 .