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∫∫∫ xyz dV , onde T é o paralelepípedo retângulo 0,1 × 0, 2 × 1, 3 2
1. Calcule
T
Solução: 1 2 3
1
2
3
dV = ∫ ∫ ∫ xyz dz dzdydx = ∫ xdx ∫ ydy ∫ z dz dz ∫∫∫ xyz dV 2
2
T
0 0 1
1
2
0
1
3
2
∫∫∫ ∫∫∫ T
0
z3 1 22 33 1 × = × × − 3 2 2 3 3 0 1 1 27 − 1 26 26 × 2 × = ∴ xyz2dV = xyz2dV = 3 2 3 3 T x2 y2 × xyz dV = 2 0 2
∫∫∫ ∫∫∫ T
2
∫∫∫
2. Calcule
∫∫∫ ∫∫∫ xdV , onde T é o tetraedro limitado pelos planos coordenados e pelo plano x + 2y + z = 4 . T
Solução:
Fig. 01
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Fig. 02
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] y 2
z
=4−x −
⇒0≤ z ≤4−x −
z
=0⇔4−x−
y 2
y 2
=0⇔
y 2
( Fig. 01)
= 4 − x ∴ y = −2x + 8
Assim:
0 ≤ x ≤ 4 0 y 2x 8 ≤ ≤− + 4 −2x + 8
Fig. 02) ( Fig.
4− x −
y 2
∫∫∫ xdV = ∫ ∫ ∫ xdzdydx T
∫∫∫
0
0
0
4−x −
4 −2x + 8
xdV
T
∫∫∫ ∫∫∫ xdV T
∫∫∫ xdV T
∫∫∫ xdV T
∫∫∫ xdV T
∫∫∫ ∫∫∫ xdV T
∫∫∫ xdV T
xz
∫∫∫ xdV = ∫ ( −4x T
0
∫∫∫ ∫∫∫ xdV = ∫∫∫ T
xdV
4 −2 x +8
− x − y dydx ∫0 0 ∫ 2 0 0 −2 x − + 4 −2x + 8 4 2 2x 8 xy x y dx = ∫ ∫ 4x − x2 − dydx = ∫ 4xy − x2y − ⋅ 2 2 2 0 0 0 0 −2x + 8 4 x 2 2 dx = ∫ 4xy − x y − ⋅ y 4 0 0 4 x 2 = ∫ 4x ( −2x + 8 ) − x2 ( −2x + 8 ) − ⋅ ( −2x + 8 ) dx 4 0 4 x dx = ∫ ( −2x + 8 ) 4x − x2 − ( −2x + 8 ) dx 4 0 4 x2 2 = ∫ ( −2x + 8 ) 4x − x + − 2x dx 2 0 4 x2 = ∫ ( −2x + 8 ) 2x − dx 2 0 =∫
4
T
y 2
x4 4
dydx
=∫
x 4
4
2
4x ) dx = ∫ ( x 3 + 16x − 8x 2 ) dx + x + 16x − 4x
16x2 + 2
= 64 + 128 −
3
2
0
−
3 4
8x 3
512 3
0
=
x4 4
+ 8x −
= 192 −
2
512 3
=
8x 3 3
4
=
44 4
∫∫∫
xdV
0
64 ∴ 3
T
+ 8×4 − 2
=
8 × 43 3
=
64 3
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dxdydz
3. (Exerc. 11) Calcule
∫∫∫ ∫∫∫ ( x + y + z + 1)
2
, onde T é o sólido delimitado pelos planos coordenados e pelo plano
T
x + y + z = 2. Solução:
Fig. 01 Fig. 02
z =2−x−y⇒0≤ z ≤2−x−y
( Fig. 01)
z = 0 ⇔ 2 − x − y = 0 ∴ y = −x + 2 Assim:
0 ≤ x ≤ 2 0 ≤ y ≤ −x + 2
Fig. 02) ( Fig.
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[email protected] 2
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
Re solvendo
=∫
− x +2 2 − x − y
0
dzdydx
∫ ∫ ( x + y + z + 1 ) 0
2
0
dz
∫ ( x + y + z + 1)
2
:
Fazend Fazendo o: u
= x + y + z +1 →
du dz
= 1 ∴ dz = du
Substituindo Substituindo : dz
∫ ( x + y + z + 1)
2
dz
∫ ( x + y + z + 1)
2
=∫ =−
du u2
= ∫ u du = −2
1 x+y
+z +1
u−1 ( −1)
1 u
+c=− +c
+c
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[email protected] 2
dxdydz
∫∫∫ ( x + y + z + 1)
2
=
T
dz
∫ ( x + y + z + 1)
2
dzdydx
∫ ∫ ∫ ( x + y + z + 1)
0
=−
− x +2 2 − x − y 0
2
0
1 x + y +z +1
+c
Assim: 2−x −y 1 ∫∫T∫ ( x + y + z + 1) 2 = 0∫ 0∫ − x + y + z + 1 0 dydx 2 − x +2 dxdydz 1 1 ∫∫∫ ∫∫T∫ ( x + y + z + 1) 2 = ∫0 0∫ − x + y + 2 − x − y + 1 − − x + y + 0 + 1 dydx 2 − x +2 1 dxdydz 1 = − − − ∫∫T∫ ( x + y + z + 1) 2 0∫ 0∫ 3 x + y + 1 dydx 2 − x +2 1 dxdydz 1 = − ∫∫T∫ ( x + y + z + 1) 2 0∫ 0∫ x + y + 1 3 dydx 2 − x +2 dxdydz dy 1 −x +2 = − d y ∫∫T∫ ( x + y + z + 1) 2 0∫ 0∫ x + y + 1 3 0∫ dx − x +2 2 dxdydz 1 ∫∫T∫ ( x + y + z + 1) 2 = 0∫ ln ( x + y + 1) − 3 y 0 dx 2
dxdydz
− x +2
2
1 ln x − x + 2 + 1 − 1 −x + 2 − ln = + + + ⋅0 l n x 0 1 ( ) ( ) ( ) ∫∫∫ ∫∫T∫ ( x + y + z + 1) 2 ∫ 0 3 3 2 dxdydz ln 3 + 1 x − 2 − ln x + 1 dx = ∫∫T∫ ( x + y + z + 1) 2 0∫ ( ) 3 3 ( ) dx dxdydz
∫∫∫ T
2
2
∫
∫
∫
∫
2
1 x2 2 = x ln ( 3) + ⋅ − x − ( x + 1) ln ( x + 1) − ( x + 1) 2 3 2 3 ( x + y + z + 1) 0 dxdydz
22 2 = 2 ln ( 3) + − × 2 − ( 2 + 1) ln ( 2 + 1) − ( 2 + 1) − 1 2 6 3 ( x + y + z + 1)
∫∫∫ ∫∫∫ T
2
1 2 = + − l n 3 d x x d x dx − ln ( x + 1) dx ( ) 2 3 3 + + + x y z 1 ( ) 0 0 0 0
∫∫∫ ∫∫∫ T
2
dxdydz
dxdydz dxdydz
∫∫∫ ∫∫∫ ( x + y + z + 1)
2
= 2 ln ( 3) + 4 − 4 − 3 ln ( 3) + 3 − 1 = 4 − ln ( 3)
2
= 4 − ln ( 3)
6
T
∫∫∫ ∫∫∫
dxdydz
3
3
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∫
OBS.: ln udu = u ⋅ ln u − u + c Exercícios 8.7
∫∫∫ ∫∫∫ ( x
2
1. (Exerc. 12) Calcular a integral
T
exterior ao cone
x2
+ y2 + z2 ) dV dV , sendo T a região interior à esfera x2 + y2 + z2 = 9
+ y2 = z2 .
Solução:
Em coordenadas esféricas, temos que:
x2
+ y2 + z 2 = 9 ⇔ r 2sen 2θ cos 2 φ + r 2sen 2θsen 2φ + r 2cos 2 θ = 9 ⇔ r 2sen 2θ (cos 2 φ + sen 2φ ) + r 2cos 2θ = 9
Assim: r2sen2θ + r 2 cos 2 θ
= 9 ⇔ r 2 ( cos 2 θ + sen 2θ ) = 9 ⇔ r 2 = 9 ∴ 0 ≤ r ≤ 3
e x2
+ y2 = z 2 ⇔ r 2sen 2θ cos 2 φ + r 2sen 2θsen 2φ = r 2cos 2 θ ⇔ r 2sen 2θ (cos 2 φ + sen 2φ ) = r 2cos 2θ
Assim : r2 sen2θ
= r 2 cos2 θ ⇔
sen2θ cos2 θ
= 1 ⇔ tan θ = ±1 ∴
3π 4
≤θ≤
π 4
Logo Logo :
0 ≤ r ≤ 3 3π π B: ≤θ≤ 4 4 0 ≤ φ ≤ 2π
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e
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Substituindo:
∫∫∫ ( x
2
+ y + z ) dV dV = 2
2
T
3π 2π 4 3
2π
3π 4
3
∫ ∫ ∫ r r senθdrdθdφ = ∫ dφ ∫ senθdθ∫ r d r 2 2
0
4
π 0
0
4
π
0
4
Integr Integrando ando : 3π 4
3
r5 3π π 35 2 2 2 + + = φ − θ = π − − + − x y z d V c o s 2 0 c o s c o s 0 ( ) ( ) ( ) π ∫∫∫ ∫∫T∫ 0 4 5 4 4 5 0 2 2 243 486 2π 2 2 2 + + = π × + x y z d V 2 ( ) ∫∫∫ ∫∫T∫ 2 × 5 = 2 5 486 2π ∫∫∫ ∫∫∫ ( x2 + y2 + z2 ) dV = 2π
5
T
2. (Exerc. 14) Calcular
∫∫∫ ∫∫∫ dV , sendo T a casca esférica delimitada por x
2
T
+ y 2 + z2 = 9
2 e x
Solução:
Em coordenadas esféricas, temos que:
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+ y2 + z2 = 16 .
Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] x2
+ y2 + z2 = 9 ⇔ r 2sen 2θ cos 2 φ + r 2sen 2θsen 2φ + r 2 cos 2 θ = 9 ⇔ r 2sen 2θ (cos 2φ + sen 2φ ) + r 2cos 2θ = 9
Assim: r2sen2θ + r 2 cos 2 θ
= 9 ⇔ r 2 ( cos 2 θ + sen 2θ ) = 9 ⇔ r 2 = 9 ∴ 0 ≤ r 1 ≤ 3
e x2
+ y2 + z2 = 16 ⇔ r 2sen 2θ co cos 2 φ + r 2sen 2θsen 2φ + r 2cos 2 θ = 16 ⇔ r 2sen 2θ (cos 2φ + sen 2φ ) + r 2cos 2 θ = 16
Assim: r2sen2θ + r 2 cos 2 θ
= 16 ⇔ r 2 ( cos 2 θ + sen 2θ ) = 16 ⇔ r 2 = 16 ∴ 0 ≤ r 2 ≤ 4
Logo Logo :
3 ≤ r ≤ 4 B : 0 ≤ θ ≤ π 0 ≤ φ ≤ 2π Substituindo: 2π
π 4
2π
π
4
∫∫∫ dV = ∫ ∫ ∫ r senθdrdθdφ = ∫ dφ∫ senθdθ∫ r dr 2
T
2
0 0 3
0
0
3
Integrando Integrando : 4
r3 43 33 = φ − θ = π − × − π + × d V c o s 2 0 c o s c o s 0 ( )0 ( ) ( ) ( ) − ∫∫∫ ∫∫T∫ 0 3 3 3 3 64 − 27 = 4π × 37 = 148π = π × + × d V 2 1 1 ( ) 3 ∫∫∫ ∫∫T∫ 3 3 3 148π dV = ∫∫∫ ∫∫∫
π
2π
3
T
3. (Exerc. 17) Calcular
∫∫∫ ∫∫∫ xdV , sendo T a região delimitada por x
2
T
2
Solução: 1ª Transformação:
2
+ ( y − 3) + ( z − 2 ) = 9 .
( x, y, z ) → ( u, v, w)
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Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] u = x ∴x = u dV =
v = y − 3 ∴y = v + 3
w = z − 2 ∴z = w + 2
∂ ( x,y,z ) dudvdw ∂ ( u,v,w )
Onde Onde :
∂x ∂u ∂ ( x,y,z ) ∂y = ∂ ( u, v, w) ∂u ∂z ∂u
∂x ∂v ∂y ∂v ∂z ∂v
∂x ∂w 1 0 0 ∂y = 0 1 0 =1 ∂w ∂z 0 0 1 ∂w
Assim:
∫∫∫
xdV =
T
∫∫∫
u
T'
∂ ( x,y,z) dudvdw = ∂ ( u,v,w )
∫∫∫ ud udvdw T'
Onde Onde : T ' = u2 + v2 + w2 = 9 Logo:
∫∫∫ xdV = ∫∫∫ u dudvdw T
T ' : u2 + v2 + w2 = 9
T'
→ ( r, φ, θ ) u = rsenθ cos φ v = rsenθsenφ w = r co cos θ u2 + v2 + w2 = 9 ⇔ r2sen2 θ cos2 φ + r2 sen2 θsen2φ + r2 cos2 θ = 9 ⇔ r2 sen2θ (cos2 φ + sen2φ ) + r2 cos2 θ = 9 2ª Tran ransfor sforma maçã ção o : ( u,v,w )
Assim: r2sen2 θ + r2 cos2 θ
= 9 ⇔ r2 ( cos2 θ + sen2θ ) = 9 ⇔ r2 = 9 ∴ 0 ≤ r ≤ 3
Logo Logo :
0 ≤ r ≤ 3 B : 0 ≤ φ ≤ π 0 ≤ θ ≤ 2π
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[email protected] Assim:
∫∫∫
xdV
∫∫∫
xdV
=
T
∫∫∫
=
ududvdw
T'
=
2π
∫
T
2π
π
3
∫ ∫∫
rsenθ cos φr2 senθdrdθ dφ
0 0 0
π
3
0
0
∫
2π
∫
∫
=
π
3
0
0
∫
∫
sen2 θ dθ cos φ r3 dr
0
sen2 θdθ cos φ r3 dr
0
=
2π
3 1 − 1 cos 2θ dθπ co φ φ c o s d ∫0 2 2 ( ) 0∫ ∫0 r3 d r
Integrand Integrando o: 3
2π
1 θ − 1 sen 2θ × senφ π × r 4 = x d V ( ) ( ) 0 ∫∫∫ 2 4 0 4 0 T 0 1 0 34 1 81 − 0 = π × 0 × =0 xdV = ( 2π − 0) − sen ( 4π ) − sen ( 0) × sen ( π ) − sen ( 0) × ∫∫∫ 4 4 4 T 2 ∫∫∫ xdV = 0 T
4. (Exerc. 18) Calcular
2 2 2 dV , onde T é elipsóide x + y + z = 1 . a2 b2 c2
∫∫∫ ∫∫∫ T
Solução: 1ª Trans ransfo form rma ação ção : ( x,y,z ) x
= au
y
= bv
z
→ ( u,v,w )
= cw
Assim Assim : x2 a2 e
+
y2 b2
+
z2 c2
=1⇔
∂x ∂u x,y,z ∂( ) ∂y = ∂ ( u, v, w ) ∂u ∂z ∂u
∂x ∂v ∂y ∂v ∂z ∂v
a2u2 a2
+
b2 v2 b2
+
c2 w2 c2
= 1 ∴ T ' : u2 + v2 + w2 = 1
∂x ∂w a 0 0 ∂y 0 b 0 abc = = ∂w ∂z 0 0 c ∂w
Assim Assim : dV
=
∂ ( x,y,z) dudvdw = abcdudvdw ∂ ( u,v,w)
Dessa forma:
∫∫∫ dV = ∫∫∫ a bcdudvdw T
Onde : T ' : u2 + v2 + w2 = 1
T'
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[email protected] → ( r, φ, θ ) u = rsenθ cos φ v = rsenθsenφ w = r co co s θ u2 + v2 + w2 = 1 ⇔ r2sen2 θ cos2 φ + r2 sen2θ sen2 φ + r2 cos2 θ = 9 ⇔ r2 sen2 θ (cos2 φ + sen2 φ )+ r2 cos2 θ = 1 2ª Tran ransfor sforma maç ção : ( u, v,w )
Assim: r2sen2 θ + r2 cos2 θ
= 1 ⇔ r2 ( cos2 θ + sen2θ ) = 1 ⇔ r2 = 1∴ 0 ≤ r ≤ 1
Logo Logo :
0 ≤ r ≤ 1 B : 0 ≤ φ ≤ π 0 ≤ θ ≤ 2π Assim: 2π
π 3
π
2π
3
∫∫∫ dV = abcdudvdw = ∫ ∫ ∫ abcr senθdrdθdφ = abc∫ senθdθ ∫ dφ∫ r d r 2
T
2
0 0 0
π
2π
0
0
0
3
∫∫∫ dV = abc∫ senθdθ ∫ dφ∫ r dr 2
T
0
0
0
Integrand Integrando o: 1
r3 = × − θ × φ × dV abc co cos ( ) cos ( π) + cos( 0) × ( 2π − 0) ( ) 0 = abc × − co ∫∫∫ 0 3 0 T 1 4π ∫∫∫ dV = abc × ( 1 + 1) × 2π × = abc π
2π
3
T
∫∫∫ dV = T
13 × − 0 3
3
4π abc 3
5. (Exerc. 20) Calcular
∫∫∫ ( x − 2y) dV , sendo T a região delimitada por: T
2
2
( x − 1 ) + ( y − 2 ) = 1, z = 0 e z = x + y . Solução:
→ ( u, v ) v = y − 2 ∴y = v + 2
1ª Transfo nsforrmação : ( x,y ) u
= x − 1∴x = u + 1
z
=0
e
z
=u + v +3
Assim: 2
2
( x − 1) + ( y − 2 ) = 1 ⇔ u2 + v2 = 1 ∴ R = u2 + v2 = 1 e
∂x ∂ ( x, y ) ∂u = ∂ ( u, v ) ∂x ∂v
∂y ∂u = 1 0 = 1 ∴ dS = ∂ ( x, y) dudv = dudv ∂ ( u, v) ∂y 0 1 ∂v
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Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] Assim:
u + v + 3 2y ) dV = ∫∫ ∫ ( u + 1 − 2v − 4) dz dudv ∫∫T∫ ( x − 2y R 0 u + v + 3 − = − − x 2y 2 y d V u 2v 2 v 3 ) ) ∫ d zdudv ∫∫T∫ ( ∫R∫ ( 0 u+v+ 3 ∫∫∫ ( x − 2y ) dV = ∫∫ ( u − 2v − 3) dudv ( z) 0 T
R
∫∫∫ ( x − 2y ) dV = ∫∫ ( u − 2v − 3) ( u + v + 3) dudv T
R
Mas: u=
ρ cos φ
v
= ρsenφ
0 R: 0
≤ ρ ≤1 ≤ φ ≤ 2π
Assim:
∂u ∂v ∂ ( u, v ) ∂ρ ∂ρ cos φ senφ = = = ρ cos2 φ + ρsen2 φ ∂ ( ρ, φ) ∂u ∂v −ρsenφ ρ cos φ ∂φ ∂φ 1 ∂ ( u, v ) = ρ ( cos2 φ + sen2φ) = ρ ∂ ( ρ, φ) Assim: dudv
=
∂ ( u, v ) dρdφ = ρdρd φ ∂ ( ρ, φ)
Substituin Substituindo do :
∫∫∫ ( x − 2y ) dV = ∫∫ ( u − 2v − 3) ( u + v + 3) dudv T
R
2π 1
∫∫∫ ( x − 2y ) dV = ∫ ∫ ( ρ cos φ − 2ρsenφ − 3) ( ρ cos φ + ρsenφ + 3) ρdρdφ I = ∫∫∫ ( x − 2y) dV T
0 0
T
2π 1
I
=
∫ ∫ ( ρ
2
cos2 φ + ρ2 cos φsenφ + 3ρ cos φ
senφ cos φ − 2ρ2se sen2φ − 6ρsen φ − 3ρ cos φ − 3ρsenφ − 9) ρdρd φ − 2ρ2se
0 0
I
=
2π 1
∫ ∫ ( ρ
2
cos2 φ − ρ2 cos φsenφ − 2ρ2sen2φ − 9ρsen φ − 9) ρd ρd φ
0 0
Resolvendo as seguintes integrais:
12 AFONSO AFONSO CELS CELSO O – FONE: FONE: (62) 3092-2 3092-2268 268 / CEL: CEL: (62) (62) 9216-9 9216-9668 668
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[email protected] 2π 1
I1 =
∫ ∫(ρ
2
cos φ ) ρd ρdφ = 2
0 0
2π 1
∫ ∫ ( co s
2
φ) ρ3 d ρd φ
0 0
2π
1
2π
ρ4 1 1 2 3 I1 = ∫ cos φdφ∫ ρ dρ = ∫ + cos ( 2φ) d φ × 2 2 4 0 0 0 0 1
2π
0 1 π 1 1 π 1 1 I1 = φ + sen ( 2φ ) × − 0 = × 2π × = ∴ I1 = 4 4 4 4 2 2 0 4
I2 =
2π 1
∫ ∫ ( −ρ
2
1
0
0
∫
∫
cos φsen φ) ρd ρd φ = − cos φsen φd φ ρ3 d ρ
0 0
2π
1
ρ4 sen2 φ I2 = − × 4 0 2 0 2π 1
I3 =
2π
1 = 4 × 0 = 0 ∴ I2 = 0 2π 1
∫ ∫ ( −2ρ sen φ) ρdρdφ = −2 ∫ ∫ sen φρ dρd φ 2
2
0 0
2
3
0 0
1
2π
ρ4 2 π 1 1 3 2 I3 = −2∫ ρ dρ ∫ sen φdφ = −2 × × ∫ − cos ( 2 φ) dφ 4 0 0 2 2 0 0 1
2π
0 π π 1 1 1 1 1 I3 = −2 × × φ − sen ( 2φ) = − × × 2π = − ∴ I3 = − 4 2 4 2 2 2 2 0 2π 1
I4 =
2π 1
∫ ∫ ( −9ρsenφ)ρdρdφ = −9 ∫ ∫ sen φρ d ρd φ 2
0 0
0 0
2π
1
ρ3 2π I4 = −9∫ ρ d ρ × ∫ senφd φ = −9 × × ( − cos φ) 0 3 0 0 0 1
2
I4 = −9 ×
1 × − cos ( 2π) + cos ( 0 ) = ( −3 ) × ( 0 ) = 0 ∴ I4 = 0 3
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Rua 96 nº 45 – Setor Sul – Goiânia Email:
[email protected] 2π 1
2π 1
2π
1
0 0
0
0
∫ ∫ ( −9)ρdρdφ = −9 ∫ ∫ ρdρdφ = −9 ∫ dφ∫ ρd ρ
I5 =
0 0
1
2π
I5 = −9 × ( φ ) 0
ρ2 × = −9 × 2π × 1 = −9 π ∴ I5 = −9 π 2 2 0
Assim: I=
2π 1
∫ ∫ ( ρ
2
cos2 φ − ρ2 cos φsen φ − 2 ρ2 sen2 φ − 9 ρsen φ − 9 ) ρd ρd φ
0 0
I = I1 + I2 + I3 + I4 + I5 I=
π
+0−
4 Logo Logo :
π
+ 0 − 9π = −
2
∫∫∫ ∫∫∫ ( x − 2y ) dV = − T
37π 4
37π 4
Exercícios 8.9 1. (Exerc.04) Calcular o volume do sólido limitado por
z = 8 − x2 − 2y2 , no primeiro octante.
Solução: Temos que:
z z
= 8 − x2 − 2y 2 = 0:
8−x
2
− 2y = 2
x2 0 ⇔ x + 2y = 8 ∴ R : 8 2
2
+
y2 4
=1
= 0, y = 0 : z = 8 − x 2 − 2y 2 ⇔ z = 8 − 0 − 0 = 8 ∴ z = 8 x
Assim: 8
∫∫∫ dV = ∫∫ dxdy ∫ dz = 8∫∫ d xdy T
R
0
R
Onde: R:
x2 8
+
y2 4
=1
∫∫
Re so solvendo 8 dxdy, onde R : R
x2 8
+
y2 4
=1
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[email protected] 1ª Transfo nsforrmação : ( x,y ) x = 2 2u Assim: x2 8 e
+
y2 4
y
→ ( u, v)
= 2v
=1 ⇔
∂x ∂ ( x, y ) ∂u = ∂ ( u, v ) ∂y ∂u
8u2 8
+
4v2 4
= 1 ∴ R ' : u2 + v2 = 1
∂x ∂v = 2 2 0 = 4 2 ∂y 0 2 ∂v
Assim: dxdy
=
∂ ( x, y ) dudv = 4 2dudv ∂ ( u, v )
Substituindo:
∫∫
8 dxdy R
= 8 ∫∫ 4 2dudv onde R ' : u2 + v 2 = 1 R'
2ª Transformação : ( u,v )
→ ( ρ, φ)
= ρ cos φ v = ρsenφ 0 ≤ ρ ≤ 1 R': π 0 ≤ φ ≤ 2
u
e
∂u ∂v ∂ ( u, v ) cos φ senφ ∂ρ ∂ρ dudv = dρdφ = d ρd φ = d ρd φ ∂ ( ρ, φ ) ∂u ∂v −ρsenφ ρ co cos φ ∂φ ∂φ dudv = ( ρ cos2 φ + ρsen2 φ ) d ρd φ = ρd ρd φ Assim: 1
π
∫∫
8 dxdy R
π ρ 1 π = 32 2 ∫ ∫ ρdρdφ = 32 2 × × ( φ ) 02 = 32 2 × × = 8 2 π 2 2 2 0 0 0 2 1
2
Logo:
∫∫∫ dV = 8∫∫ d xdy = 8 T
2π ∴ V = 8 2π
R
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2. (Exerc. 05) Calcular o volume do sólido acima do plano xy delimitado por
z = x2 + y2 e x2 + y2 = 16 .
Solução: Temos que:
= x2 + y 2 e x2 + y 2 = 16 = 0: x2 + y2 = 0 ∴ ( 0, 0, 0 ) x = 0, y = 0 : z = 16 ∴ ( 0, 0, 0,16 )
z z
Assim: 16
dz = 16 ∫∫ d xdy ∫∫∫ dV = ∫∫ dxdy ∫ dz T
R
0
R
Onde: R : x2
+ y2 = 16
∫∫
Re s ol olvendo 16 dxdy, onde R : x 2
+ y 2 = 16
R
Transformação : ( x, y )
→ ( ρ, φ )
= ρ cos φ y = ρsen φ 0 ≤ ρ ≤ 4 R': 0 ≤ φ ≤ π x
e
∂u ∂v ∂ ( u, v ) cos φ senφ ∂ρ ∂ρ dxdy = dρdφ = dρdφ = d ρdφ ∂ ( ρ, φ ) ∂u ∂v −ρsenφ ρ co cos φ ∂φ ∂φ dxdy = ( ρ cos2 φ + ρsen2φ ) d ρd φ = ρd ρd φ Assim: π 4
π
4
ρ2 16∫∫ dx dxdy = 16 ∫ ∫ ρdρd φ = 16 ∫ d φ∫ ρd ρ = 16 × ( φ ) 0 × 2 0 R 0 0 0 0
∫∫
16 dxdy
= 16 × π ×
R
42 2
4
π
= 128 π
Logo:
∫∫∫ dV = 16∫∫ d xdy = 128π ∴ V = 128π T
R
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3. (Exerc. 06) Calcular o volume do sólido acima do parabolóide
z = x2 + y2
e abaixo do cone
Solução: Temos que:
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z=
x2 + y2
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[email protected] z
= x2 + y2
z
= x2 + y2
Assim:
∫∫∫ dV = ∫∫∫ dxdydz T
T
= rsenθ cos φ y = rsen θsen φ z = r co cos θ z = x2 + y2 ⇔ r co cos θ = r 2sen2 θ co cos2 φ + r 2sen2 θsen2 φ x
r cos θ
= r 2sen2 θ ( cos2 φ + sen2 φ) ⇔ r cos θ = r 2sen2 θ ∴r =
cos θ sen2 θ
e z
= x2 + y2 ⇔ r cos θ = r 2sen2 θ cos2 φ + r 2sen2 θsen2 φ
r cos cos θ
= r2sen2θ ( cos2 φ + sen2 φ) = r 2sen2 θ = rsen θ
Assim: r cos θ
= rsenθ ⇔
senθ cos θ
= 1 ⇔ tan θ = 1 ∴ θ =
π 4
Assim:
0 ≤ θ ≤ π 4 cos θ T : 0 ≤ r ≤ sen2 θ 0 ≤ φ ≤ 2π Substit Substituindo uindo : π cos θ 2 π 4 sen2θ
∫∫∫ dV = ∫∫∫ dxdydz = ∫ ∫ ∫ r senθdrdθd φ T
T
0 0
cos θ
∫∫∫ ∫∫∫ dV = ( φ) T
2π 0
T
0
π
r3 sen θ 4 × × ∫ senθdθ 3 0 0 2
3
π
∫∫∫
2
π
1 cos θ sen θd θ dV = 2π × × 3 0 sen2 θ 4
∫
π
2π
4
an θ cos sec ∫∫∫ dV = 3 ∫ c ot an T
2
1
2π 3 − u du 3 0
∫
=
2π − 3
u4 × 4
1
0
∫
du θ − 2 cossec θ 2π 1 π π = − × = − ∴V =
θd θ =
0
∫∫∫ dV = T
3
2 π cos 3 θ dθ = 3 0 sen5θ 4
2π u3 cos sec 2 3
∫
3
4
6
6
4. (Exerc. 11) Calcular o volume do sólido delimitado pelas superfícies
x2 + y2 = 16, z = 2, x + z = 9
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[email protected] Solução: Temos que:
x2 + y2 = 16, z = 2, x + z = 9 ∴ z = 9 − x Assim: 9−x
∫∫∫ dV = ∫∫ ∫ dzdxdy = ∫∫ ( 9 − x − 2) dxdy T
R
2
R
2π 4
2π 4
2π 4
∫∫∫ dV = ∫ ∫ ( 7 − ρ cos φ) ρdρdφ = 7 ∫ ∫ ρdρdφ − ∫ ∫ ρ cos φd ρd φ 2
T
0 0
0 0
4 ρ3 4 2π ρ2 ∫∫∫ ∫∫T∫ dV = 7 × 2 0 × ( φ) 0 − 3 0
0 0
0 2π × ( senφ) 0
Assim: 42 × 2 π = 112 π ∴ V = 11 2 π dV = 7 × 2
∫∫∫ ∫∫∫ T
5. (Exerc.13) Calcular o volume do sólido delimitado pelas superfícies
z = 2x2 + y2 e z = 4 − 3x2 − y2 .
Solução: Temos que:
z = 2x2 + y2 e z Assim Assim :
3x 2 − y 2 = 4 − 3x
2x2 + y2 = 4 − 3x 2 Dividind Dividindo o por 4 : 5x2 4
+
2y2 4
=1⇔
− y 2 ⇔ 2x 2 + 3x 2 + y 2 + y 2 = 4 ⇔ 5x 2 + 2y 2 = 4
5x2 5 4 5
+
y2 2
4 − 3x2 − y2
∫∫∫ dV = ∫∫ ∫ T
R
2x2 + y2
dzdxdy
= 1 ∴R :
x2 4 5
+
y2 2
=1
= ∫∫ ( 4 2x 2 − y 2 )dxdy − 3x2 − y2 − 2x R
Assim Assim :
∫∫∫ ∫∫∫
dV
T
d xdy, on de R : = ∫∫ ( 4 − 5x2 − 2y2 ) dx R
x2 4 5
+
y2 2
=1
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[email protected] 1ª Tra Transfo nsform rma ação ( x, y ) x
=
2 5 u 5
dxdy
=
y
→ ( u, v )
= 2v 2v
∂ ( x, y ) dudv ∂ ( u, v )
Onde Onde :
∂x ∂ ( x, y ) ∂u = ∂ ( u, v ) ∂x ∂v
∂y 2 5 ∂u = 5 ∂y 0 ∂v
∫∫∫ dV = ∫∫ ( 4 − 5x T
0
=
2
− 2y 2 ) dxdy =
2
R
2 10 5
∫∫ ( 4 − 4u
2
R'
− 4v 2 )
2 10 dudv, onde R ' : u 2 + v 5
2
=1
Assim:
∫∫∫ dV = T
2 10 5
∫∫ ( 4 − 4u
2
− 4v 2 ) dudv, onde R ' : u 2 + v 2 = 1
R'
2ª Transformação : ( u, v ) → ( ρ, φ ) u = ρ cos φ
v = ρsen φ
0 ≤ ρ ≤ 1 R': 0 ≤ φ ≤ 2π Assim:
∫∫∫
dV =
T
∫∫∫
dV =
T
2 10 5 2 10 5
∫∫ ( 4 − 4u
− 4v2 ) dudv
2
R'
2π 1
∫ ∫ ( 4 − 4ρ
2
co cos2 φ − 4ρ2 sen2 φ) ρd ρd φ
0 0
Assim:
∫∫∫ T
2π 1 2π 1 2 10 2 π 1 3 2 dV = 4ρdρdφ − 4 ρ cos φdρd φ − 4 ρ3 sen2 φd ρd φ 5 0 0 0 0 0 0
∫∫
∫∫
∫∫
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Resolvendo as Integrais:
∫∫∫ T
π 2 10 π 4 π 3 2 ρ cos φd ρd φ − 4 ρ3 sen2 φd ρd φ dV = 4ρdρdφ − 5 0 0 500 0 0 2 1
2 1
∫∫
∫∫∫ ∫∫∫ dV = T
2 1
∫∫
∫∫
2 10 ( I1 + I2 + I3 ) 5
Onde Onde : 2π 1
2π
1
ρ2 12 = 4 π ∴ I1 = 4 π I1 = 4 ∫ ∫ ρdρdφ = 4 ∫ dφ ∫ ρd ρ = 4 × ( φ) 0 × = 4 × 2 π × 2 2 0 0 0 0 0 2π 1 1 2π 4 4 1 1 3 2 3 I2 = − ∫ ∫ ρ cos φd ρd φ = − × ∫ ρ d ρ × ∫ + cos ( 2φ ) dφ 5 00 5 0 2 2 0 1
2π
Integrand Integrando o: 2π
0 1 2 π 4 ρ 1 1 I2 = − × × φ + sen ( 2φ) = ( − 4 ) × × = −π ∴ I2 = −π 5 4 0 2 4 4 2 0 4
1
e 2π 1
I3 = −4
2π
1
1 1
∫ ∫ ρ sen φdρdφ = −4 × ∫ ρ dρ × ∫ 2 − 2 cos ( 2φ) d φ 3
2
0 0
3
0
0
Integrand Integrando o: 2π
0 ρ 1 1 2 π 1 I3 = −4 × × φ − sen ( 2φ) = ( − 4 ) × × = −π ∴ I3 = −π 4 2 4 4 2 0 0 4
1
Substituindo Substituindo :
∫∫∫ ∫∫∫
dV =
T
2 10 2 10 2 10 4 10 4 10 × 2π = ∴V = ( I1 + I2 + I3 ) = ( 4π − π − π ) = 5 5 5 5 5
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6. (Exerc. 10) Calcule o volume da parte da esfera,
x2 + y2 + z2 = 9 ,
entre entre os planos planos z= 1; z= 2. Esboc Esboce e a
região.
Solução:
z = 1:
x2 + y2 + z2 = 9 ⇔ x2 + y2 + 12 = 9 ⇔ x2 + y2 + 1 = 9 ∴ x2 + y2 = 8
z = 2:
x2 + y2 + z2 = 9 ⇔ x2 + y2 + 22 = 9 ⇔ x2 + y2 + 4 = 9 ∴ x2 + y2 = 5
(I) Cálculo de V 1:
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Utilizando coordenadas cilíndricas:
V1
V1
2 2π 5
2π 3
0 0 0
0
9 −r2
= ∫ ∫ ∫ rdrdθdz + ∫ ∫ ∫ rdzdrd θ =r
5
2
2
( 5) × 2π × 2 = 2
0
0
5
2 2π 3
× 4 π + ∫ ∫ r 9 − r2 drd θ 0
5
3 0 1 3 3 3 1 1 2 V1 = 10π + 2π − ( 9 − r ) = 11 π − 2 π ( 9 − 9 ) − ( 9 − 5 ) 3 3 3 5 = π + π × 8 = π + 16π = 46π ∴ = 46 π
V1
10
2
3
10
3
3
V1
3
(II) Cálculo de V 2:
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Utilizando coordenadas cilíndricas: 1 2π 8
V2
= ∫ ∫ ∫ rdrdθdz + ∫ ∫ ∫ rdzdrd θ 0 0 0
V2
9 −r2
2π 3
=r
8
2
2
0
× 2π × 1 = 0
(
0
8
8)
2
2
2π 3
× 2 π + ∫ ∫ r 9 − r2 drd θ 0
8
3 0 1 3 3 3 1 1 2 = 8 π − 2 π ( 9 − 9 ) − ( 9 − 8 ) V2 = 8π + 2 π − ( 9 − r ) 3 3 3 8 = π + π × 1 = π + 2π = 26π ∴ = 26π
V2
8
2
3
8
3
3
V2
3
Assim: V = V1 − V2 =
46π 26π 20π 20π − = u.v. ∴ V = u.v. 3 3 3 3
Resolução de
∫ r
9 − r2 dr dr
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∫ r
9 − r 2 dr = ?
Fazen Fazendo do : u = 9 − r2 ⇒ u2 = 9 − r2 → 2u
du = − 2r ∴ rdr = −udu dr
Substituindo Substituindo : u3 +c udu) = − ∫ u du = − ∫ r 9 − r dr = ∫ u ( −ud 3 Assim: 3 1 2 2 ∫ r 9 − r dr = − 3 9 − r + c 2
2
(
)
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