CÁLCULO III – INTEGRAIS TRIPLAS
Exercícios 8.5.
1. Calcule
∫∫∫ xyz dV , onde T é o paralelepípedo retângulo 0,1 × 0,2 × 1,3
2
T
Solução:
2
∫∫∫ xyz dV =
T
1 2 3
1
2
3
0
0
1
2
2
∫ ∫ ∫ xyz dzdydx = ∫ xdx ∫ ydy ∫ z dz
0 0 1
1
2
3
z3
1 22  33 1 
= ×
×
− 
3
2
2
3
3

T
0
0
1
26
1
 27 − 1  26
=
∴ ∫∫∫ xyz2dV =
× 2 ×
xyz2dV =

∫∫∫
3
3
2
 3 
T
T
2
∫∫∫ xyz dV =
2. Calcule
x2
2
×
y2
2
×
y
∫∫∫ xdV , onde T é o tetraedro limitado pelos planos coordenados e pelo plano x + 2 + z = 4 .
T
Solução:
Fig. 01
Fig. 02
y
y
⇒0≤z≤4−x−
(Fig. 01)
2
2
y
y
z = 0 ⇔ 4 − x − = 0 ⇔ = 4 − x ∴ y = −2x + 8
2
2
Assim :
z=4−x−
0 ≤ x ≤ 4

0 ≤ y ≤ −2x + 8
4 −2x + 8
4−x−
∫∫∫ xdV = ∫ ∫
T
∫∫∫ xdV =
T
∫∫∫ xdV =
T
0
(Fig. 02)
∫
0
0
4−x −
∫ ∫
0
0
y
2
xz
0
4 −2x + 8
xdzdydx
0
4 −2x + 8
∫ ∫
y
2
4 −2x + 8
dydx =
∫ ∫
−2x
0
0
y

x  4 − x −  dydx
2

4
x y2
xy 

2
2

4x
−
x
−
dydx
=
4xy
−
x
y
−
⋅

∫0 
2 
2 2


−2x + 8
0

 dx


−2x + 8
4

x 2
2
xdV
=
4xy
−
x
y
−
⋅
y

 dx
∫∫∫
∫


4
0
T
0

4
∫∫∫ xdV =
T
x

∫ 4x ( −2x + 8 ) − x ( −2x + 8) − 4 ⋅ ( −2x + 8)
2
2
0
4



x
( −2x + 8)   dx
4

∫∫∫ xdV =
∫ ( −2x + 8)  4x − x
∫∫∫ xdV =



x2
2
2x
8
4x
x
−
+
−
+
− 2x   dx
(
)


∫0
2



∫∫∫ xdV =


x2  
−
+
−
2x
8
2x
)
  dx
∫0 (
2




T
T
T
2
0
−
4
4
4
∫∫∫ xdV = ∫ ( −4x
2
T
4
+ x + 16x − 4x
3
2
0
∫∫∫ xdV = 64 + 128 −
T
3. (Exerc. 11) Calcule
3
)
+ 16x − 8x2 dx
4
0
x4
8x3
=
+ 8x2 −
4
3
4
=
0
44
8 × 43
+ 8 × 42 −
=
4
3
512
512 64
64
= 192 −
=
∴ ∫∫∫ xdV =
3
3
3
3
T
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
x + y + z = 2.
) dx = ∫ ( x
0
x4 16x2 8x3
xdV
=
+
−
∫∫∫
4
2
3
T
Solução:

 dx

, onde T é o sólido delimitado pelos planos coordenados e pelo plano
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)
2 −x +2 2 − x − y
dxdydz
∫∫∫ ( x + y + z + 1)
2
=
T
Re solvendo
dzdydx
∫ ∫ ∫ ( x + y + z + 1)
2
0
0
0
dz
∫ ( x + y + z + 1)
2
:
Fazendo :
u = x + y + z +1 →
du
= 1 ∴ dz = du
dz
Substituindo :
dz
∫ ( x + y + z + 1)
2
dz
∫ ( x + y + z + 1)
2
=
du
u−1
1
−2
=
u
du
=
+c=− +c
∫ u2 ∫
u
( −1)
=−
1
+c
x + y + z +1
2 − x +2 2 − x − y
dxdydz
∫∫∫ ( x + y + z + 1)
2
=
T
dz
∫ ( x + y + z + 1)
2
0
=−
2
dzdydx
∫ ∫ ∫ ( x + y + z + 1)
0
0
1
+c
x + y + z +1
Assim :
dxdydz
∫∫∫ ( x + y + z + 1)
2
2 − x +2
=
T
dxdydz
∫∫∫ ( x + y + z + 1)
2
0
dxdydz
2
∫∫∫ ( x + y + z + 1)
2
∫ ∫
0
0
2 − x +2
=
T
dxdydz
0
2 − x +2
=
T
∫∫∫ ( x + y + z + 1)
∫ ∫
∫ ∫
0
0
2 − x +2
=
T
∫ ∫
0
0
2−x−y


1
−
 dydx
 x + y + z +10






1
1
− −
−
  dydx
 x + y + 2 − x − y + 1  x + y + 0 + 1  
 1 

1
− −  −
  dydx
 3  x + y + 1 

1
1
−  dydx

 x + y + 1 3
− x +2
 − x +2

dy
1
= ∫ ∫
−
dy
 dx
x + y + 1 3 ∫0
0 0

2
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
dxdydz
∫∫∫ ( x + y + z + 1)
2
−x +2


1
ln
x
+
y
+
1
−
y
 dx
)
∫0  (
3
0


2
=
2
=
2
∫∫∫ ( x + y + z + 1)
2
=
T
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
)
− x +2 +1 −
0
T
dxdydz

∫ ln ( x

1

2
∫ ln (3) + 3 x − 3 − ln ( x + 1) dx
0
2
= ln (3) ∫ dx +
0
2
2
2
1
2
xdx − ∫ dx − ∫ ln ( x + 1)dx
∫
30
30
0
2
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
dxdydz
∫∫∫ ( x + y + z + 1)
2
1 x2 2
= x ln (3) + ⋅
− x − ( x + 1) ln ( x + 1) − ( x + 1) 
3 2 3
0
= 2 ln (3) +
22 2
− × 2 − (2 + 1) ln (2 + 1) − (2 + 1)  − 1
6 3
= 2 ln (3) +
4 4
4
− − 3ln (3) + 3 − 1 = − ln (3)
6 3
3
T
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
dxdydz
∫∫∫ ( x + y + z + 1)
2
T
OBS.:
1
1
( −x + 2 ) − ln ( x + 0 + 1) + ⋅ 0
3
3 
=
4
− ln (3)
3
∫ lnudu = u ⋅ lnu − u + c
Exercícios 8.7
1. (Exerc. 12) Calcular a integral
∫∫∫ ( x
T
e exterior ao cone
x +y =z
2
2
2
.
2
)
+ y2 + z2 dV , sendo
T a região interior à esfera
x2 + y2 + z2 = 9
Solução:
Em coordenadas esféricas, temos que:
(
)
x2 + y2 + z2 = 9 ⇔ r 2sen2θ cos2 φ + r2sen2θsen2φ + r2 cos2 θ = 9 ⇔ r 2sen2 θ cos2 φ + sen2φ + r2 cos2 θ = 9
Assim :
(
)
r 2sen2θ + r2 cos2 θ = 9 ⇔ r 2 cos2 θ + sen2θ = 9 ⇔ r2 = 9 ∴ 0 ≤ r ≤ 3
e
(
)
x2 + y2 = z2 ⇔ r2sen2θ cos2 φ + r 2sen2 θsen2φ = r2 cos2 θ ⇔ r2sen2θ cos2 φ + sen2φ = r2 cos2 θ
Assim :
r2 sen2θ = r2 cos2 θ ⇔
Logo :
0 ≤ r ≤ 3
 3π
π

≤θ≤
B:
4
4

0 ≤ φ ≤ 2π
Substituindo:
sen2θ
3π
π
= 1 ⇔ tan θ = ±1 ∴
≤θ≤
cos2 θ
4
4
∫∫∫ ( x
3π
2π 4 3
2
+y +z
2
2
T
2π
3π
4
3
) dV = ∫ ∫ ∫ r r senθdrdθdφ = ∫ dφ ∫ senθdθ∫ r dr
2 2
4
π 0
4
0
0
π
4
0
Integrando :
∫∫∫ ( x
2
+y +z
2
2
) dV = φ ( − cos θ)
2π
0
T
∫∫∫ ( x
2
T
∫∫∫ ( x
2
3
3π
4
π
4
 r5 


π   35
 3π 
=
π
−
−
+
− 0
2
0
cos
cos
(
)
 





4  5
 4 

 5 0

 2
2  243 486 2π
+ y2 + z2 dV = 2π × 
+
=
×
 2
2 
5
5

)
486 2π
5
)
+ y2 + z2 dV =
T
2. (Exerc. 14) Calcular
∫∫∫ dV , sendo T a casca esférica delimitada por x
2
+ y2 + z2 = 9 e x2 + y2 + z2 = 16 .
T
Solução:
Em coordenadas esféricas, temos que:
(
)
x2 + y2 + z2 = 9 ⇔ r 2sen2 θ cos2 φ + r 2sen2 θsen2φ + r2 cos2 θ = 9 ⇔ r2sen2 θ cos2 φ + sen2φ + r 2 cos2 θ = 9
Assim :
(
)
r2sen2 θ + r2 cos2 θ = 9 ⇔ r2 cos2 θ + sen2 θ = 9 ⇔ r2 = 9 ∴ 0 ≤ r1 ≤ 3
e
(
)
x2 + y2 + z2 = 16 ⇔ r2sen2 θ cos2 φ + r 2sen2θsen2 φ + r2 cos2 θ = 16 ⇔ r2sen2 θ cos2 φ + sen2φ + r2 cos2 θ = 16
Assim :
(
)
r2sen2 θ + r2 cos2 θ = 16 ⇔ r 2 cos2 θ + sen2 θ = 16 ⇔ r 2 = 16 ∴ 0 ≤ r2 ≤ 4
Logo :
3 ≤ r ≤ 4

B : 0 ≤ θ ≤ π
0 ≤ φ ≤ 2π

Substituindo:
∫∫∫ dV =
2π π 4
2π
0 0 3
0
∫
T
2
∫ ∫ r senθdrdθdφ =
∫
π
4
0
3
dφ ∫ senθdθ∫ r2dr
Integrando :
4
∫∫∫ dV = φ
2π
0
T
 r3 
 43 33 
( − cos θ ) 0   = (2π − 0 ) ×  − cos ( π) + cos (0 ) ×  − 
3
 3 3
 3
π
37 148π
 64 27 
−
= 4π ×
=

3
3 
3
3
∫∫∫ dV = 2π × (1 + 1) × 
T
∫∫∫ dV =
T
148π
3
3. (Exerc. 17) Calcular
∫∫∫ xdV , sendo T a região delimitada por x
2
+ ( y − 3) + ( z − 2 ) = 9 .
2
2
T
Solução:
1ª Transformação:
( x, y, z ) → (u, v, w )
u = x∴x = u
v = y − 3∴y = v + 3
∂ ( x, y, z )
dV =
∂ (u, v, w )
w = z − 2∴z = w + 2
dudvdw
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
0 0 1
∂z
∂w
Assim :
∂ ( x, y, z )
∫∫∫ xdV = ∫∫∫ u ∂ (u, v, w ) dudvdw = ∫∫∫ ududvdw
T
T'
T'
Onde :
T ' = u2 + v2 + w2 = 9
Logo:
∫∫∫ xdV = ∫∫∫ ududvdw
T
T ' : u2 + v2 + w2 = 9
T'
2ª Transformação : (u, v, w ) → (r, φ, θ )
u = rsenθ cos φ
v = rsenθsenφ
w = r cos θ
(
)
u + v + w = 9 ⇔ r sen θ cos φ + r 2sen2 θsen2 φ + r 2 cos2 θ = 9 ⇔ r2sen2θ cos2 φ + sen2 φ + r 2 cos2 θ = 9
2
2
2
2
2
2
Assim :
(
)
r 2sen2 θ + r 2 cos2 θ = 9 ⇔ r 2 cos2 θ + sen2θ = 9 ⇔ r 2 = 9 ∴ 0 ≤ r ≤ 3
Logo :
0 ≤ r ≤ 3

B : 0 ≤ φ ≤ π
0 ≤ θ ≤ 2π

Assim :
∫∫∫ xdV =
T
2π π 3
∫∫∫ ududvdw =
∫
T'
2π
2
∫ ∫ rsenθ cos φr senθdrdθdφ =
0 0 0
π
2π
3
1

1
2π
∫
0
π
3
0
0
sen2θdθ∫ cos φ∫ r3dr
π
3
∫∫∫ xdV = ∫ sen θdθ∫ cos φ∫ r dr = ∫  2 − 2 cos (2θ ) dθ∫ cos φdφ∫ r dr
2
T
3
0
0
0
0
3
0
0
Integrando :
3
2π
π
 r4 
1
1

xdV
=
θ
−
sen
2
θ
×
sen
φ
×
(
)
(
)
 
∫∫∫
2

0
4

0
 4 0
T
0
 1

0
 34

1
81




xdV
=
2
π
−
0
−
sen
4
π
−
sen
0
×
sen
π
−
sen
0
×
− 0 = π × 0 ×
=0
(
)
(
)
(
)
(
)
(
)



∫∫∫




2
4
4
 4

T


∫∫∫ xdV = 0
T
4. (Exerc. 18) Calcular
∫∫∫ dV , onde T é elipsóide
T
x2 y2 z2
+
+
= 1.
a2 b2 c2
Solução:
1ª Transformação : ( x, y, z ) → (u, v, w )
x = au y = bv
Assim :
z = cw
x2 y2 z2
a2u2 b2v2 c2w2
+ 2 + 2 = 1 ⇔ 2 + 2 + 2 = 1 ∴ T ' : u2 + v2 + w2 = 1
2
a
b
c
a
b
c
e
∂x
∂u
∂ ( x, y, z )
∂y
=
∂ (u, v, w )
∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
a 0 0
∂y
= 0 b 0 = abc
∂w
0 0 c
∂z
∂w
Assim :
dV =
∂ ( x, y, z )
∂ (u, v, w )
dudvdw = abcdudvdw
Dessa forma:
∫∫∫ dV = ∫∫∫ abcdudvdw
T
Onde : T ' : u2 + v2 + w2 = 1
T'
2ª Transformação : (u, v, w ) → (r, φ, θ )
u = rsenθ cos φ
v = rsenθsenφ
w = r cos θ
(
)
u + v + w = 1 ⇔ r sen θ cos φ + r 2sen2 θsen2φ + r2 cos2 θ = 9 ⇔ r 2sen2 θ cos2 φ + sen2 φ + r2 cos2 θ = 1
2
2
2
2
2
2
Assim :
(
)
r 2sen2 θ + r 2 cos2 θ = 1 ⇔ r 2 cos2 θ + sen2 θ = 1 ⇔ r 2 = 1 ∴ 0 ≤ r ≤ 1
Logo :
0 ≤ r ≤ 1

B : 0 ≤ φ ≤ π
0 ≤ θ ≤ 2π

Assim :
∫∫∫ dV = abcdudvdw =
T
2π π 3
π
2π
3
0 0 0
0
0
0
∫
2
2
∫ ∫ abcr senθdrdθdφ = abc∫ senθdθ ∫ dφ∫ r dr
π
2π
3
0
0
0
∫∫∫ dV = abc∫ senθdθ ∫ dφ∫ r dr
2
T
Integrando :
1
π
2π
 r3 
 13

dV = abc ×  − cos ( θ ) × ( φ ) 0 ×   = abc ×  − cos ( π ) + cos ( 0 )  × (2π − 0 ) × 
− 0
∫∫∫
0
3
3
 0


T
1
∫∫∫ dV = abc × (1 + 1) × 2π × 3 =
T
∫∫∫ dV =
T
4π
abc
3
4π
abc
3
5. (Exerc. 20) Calcular
∫∫∫ ( x − 2y ) dV , sendo T a região delimitada por:
T
( x − 1)
+ ( y − 2 ) = 1, z = 0 e z = x + y .
2
2
Solução:
1ª Transformação : ( x, y ) → (u, v )
u = x − 1∴ x = u + 1
Assim :
( x − 1)
2
v = y −2∴y = v + 2
z=0
e
+ ( y − 2 ) = 1 ⇔ u2 + v2 = 1 ∴ R = u2 + v2 = 1
2
e
∂x
∂ ( x, y )
∂u
=
∂ (u, v )
∂x
∂v
Assim:
∂y
∂ ( x, y )
1 0
∂u
=
= 1 ∴ dS =
dudv = dudv
∂ (u, v )
∂y
0 1
∂v
z =u+v+3
u + v + 3

∫∫∫ ( x − 2y ) dV = ∫∫  ∫ (u + 1 − 2v − 4) dzdudv

u + v +3 
( x − 2y ) dV = ∫∫ (u − 2v − 3)  ∫ dz dudv
∫∫∫
T
R
 0

T

R
0
∫∫∫ ( x − 2y ) dV = ∫∫ (u − 2v − 3)dudv ( z )
T
R
u+ v + 3
0
∫∫∫ ( x − 2y ) dV = ∫∫ (u − 2v − 3) (u + v + 3) dudv
T
R
Mas :
u = ρ cos φ
v = ρsenφ
0 ≤ ρ ≤ 1
R:
0 ≤ φ ≤ 2π
Assim :
∂u
∂ρ
=
∂ ( ρ, φ )
∂u
∂φ
∂ (u, v )
∂ (u, v )
∂ ( ρ, φ )
∂v
cos φ
senφ
∂ρ
=
= ρ cos2 φ + ρsen2 φ
∂v
−ρsenφ ρ cos φ
∂φ
(
= ρ cos2 φ + sen2φ
)
1
=ρ
Assim :
dudv =
∂ (u, v )
∂ ( ρ, φ )
dρdφ = ρdρdφ
Substituindo :
∫∫∫ ( 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
T
I=
0 0
∫∫∫ ( x − 2y ) dV
T
2π 1
I=
∫ ∫ (ρ
2
)
cos2 φ + ρ2 cos φsenφ + 3ρ cos φ − 2ρ2senφ cos φ − 2ρ2sen2φ − 6ρsenφ − 3ρ cos φ − 3ρsenφ − 9 ρdρdφ
0 0
2π 1
I=
∫ ∫ (ρ
2
)
cos2 φ − ρ2 cos φsenφ − 2ρ2sen2φ − 9ρsenφ − 9 ρdρdφ
0 0
Resolvendo as seguintes integrais:
2π 1
I1 =
∫ ∫ (ρ
2
2π 1
)
cos φ ρdρdφ =
2
0 0
∫ cos
2
0
2
3
0 0
2π
I1 =
∫ ∫ ( cos φ )ρ dρdφ
1
φdφ∫ ρ dρ =
3
0
2π
∫
0
1
 ρ4 
1 1

+
cos
2
φ
d
φ
×
(
)
 
2 2



 4 0
2π
0
1

1
1 π
π
1
 1
I1 =  φ + sen (2φ )  ×  − 0  = × 2π × = ∴ I1 =
4 4
4
4
4
 2
 2

0
2π 1
I2 =
∫ ∫ ( −ρ
2
0 0
)
2π
1
0
0
cos φsenφ ρdρdφ = − ∫ cos φsenφdφ∫ ρ3dρ
2π
1
 ρ4 
 sen2φ 
1
I2 =  −  × 
 = × 0 = 0 ∴ I2 = 0
4
 4 0  2 0
2π 1
I3 =
∫ ∫(
0 0
2π 1
)
−2ρ2sen2 φ ρdρdφ = −2 ∫ ∫ sen2 φρ3dρdφ
0 0
1
2π
2π
 ρ4 
1 1

I3 = −2∫ ρ dρ ∫ sen φdφ = −2 ×   × ∫  − cos (2φ )  dφ

 4  0 0 2 2
0
0
1
3
2
2π
0

1 1
1
1 1
π
π
I3 = −2 × ×  φ − sen (2φ )  = − × × 2π = − ∴ I3 = −
4 2
4
2 2
2
2


0
2π 1
I4 =
∫
2π 1
2
∫ ( −9ρsenφ )ρdρdφ = −9 ∫ ∫ senφρ dρdφ
0 0
0 0
1
I4 = −9∫ ρ dρ ×
2
0
2π
∫
0
1
2π
 ρ3 
senφdφ = −9 ×   × ( − cos φ ) 0
 3 0
1
×  − cos (2π ) + cos ( 0 )  = ( −3) × ( 0 ) = 0 ∴ I4 = 0
3 
I4 = −9 ×
2π 1
I5 =
2π 1
2π
1
0 0
0
0
∫ ∫ ( −9)ρdρdφ = −9 ∫ ∫ ρdρdφ = −9 ∫ dφ∫ ρdρ
0 0
1
2π
 ρ2 
1
I5 = −9 × ( φ ) 0 ×   = −9 × 2π × = −9π ∴ I5 = −9π
2
 2 0
Assim :
2π 1
I=
∫ ∫ (ρ
2
)
cos2 φ − ρ2 cos φsenφ − 2ρ2sen2φ − 9ρsenφ − 9 ρdρdφ
0 0
I = I1 + I2 + I3 + I4 + I5
π
π
37π
+ 0 − + 0 − 9π = −
4
2
4
Logo :
37π
( x − 2y ) dV = −
∫∫∫
4
T
I=
Exercícios 8.9
1. (Exerc.04) Calcular o volume do sólido limitado por
Solução:
Temos que:
z = 8 − x2 − 2y2 , no primeiro octante.
z = 8 − x2 − 2y2
z = 0:
8 − x2 − 2y2 = 0 ⇔ x2 + 2y2 = 8 ∴ R :
x2 y2
+
=1
8
4
x = 0, y = 0 :
z = 8 − x2 − 2y2 ⇔ z = 8 − 0 − 0 = 8 ∴ z = 8
Assim :
8
∫∫∫ dV = ∫∫ dxdy ∫ dz = 8∫∫ dxdy
T
R
0
R
Onde :
R:
x2 y2
+
=1
8
4
Re solvendo 8∫∫ dxdy, onde R :
R
x2 y2
+
=1
8
4
1ª Transformação : ( x, y ) → (u, v )
x = 2 2u
Assim :
y = 2v
x2 y2
8u2 4v2
+
=1⇔
+
= 1 ∴ R ' : u2 + v2 = 1
8
4
8
4
e
∂x
∂u
=
∂ (u, v )
∂y
∂u
Assim :
∂x
2 2 0
∂v
=
=4 2
∂y
0
2
∂v
∂ ( x, y )
dxdy =
∂ ( x, y )
∂ (u, v )
dudv = 4 2dudv
Substituindo:
8∫∫ dxdy = 8∫∫ 4 2dudv onde R ' : u2 + v2 = 1
R
R'
2ª Transformação : (u, v ) → ( ρ, φ )
u = ρ cos φ
v = ρsenφ
0 ≤ ρ ≤ 1

R': 
π
0 ≤ φ ≤ 2
e
dudv =
∂ (u, v )
∂ ( ρ, φ )
dρdφ =
(
∂u
∂ρ
∂v
∂ρ
∂u
∂φ
∂v
∂φ
d ρd φ =
cos φ
senφ
−ρsenφ ρ cos φ
d ρd φ
)
dudv = ρ cos2 φ + ρsen2φ dρdφ = ρdρdφ
Assim :
π
2 1
1
π
 ρ2 
1 π
8∫∫ dxdy = 32 2 ∫ ∫ ρdρdφ = 32 2 ×   × ( φ ) 02 = 32 2 × × = 8 2π
2
2 2
 0
0 0
R
Logo:
∫∫∫ dV = 8∫∫ dxdy = 8
T
2π ∴ V = 8 2π
R
2. (Exerc. 05) Calcular o volume do sólido acima do plano xy delimitado por
Solução:
Temos que:
z = x2 + y2 e x2 + y2 = 16
z = 0:
x2 + y2 = 0 ∴ ( 0, 0, 0 )
x = 0, y = 0 :
z = 16 ∴ ( 0, 0,16 )
Assim :
16
∫∫∫ dV =
∫∫ dxdy ∫ dz = 16∫∫ dxdy
T
R
0
R
Onde :
R : x2 + y2 = 16
Re solvendo 16 ∫∫ dxdy, onde R : x2 + y2 = 16
R
Transformação : ( x, y ) → ( ρ, φ )
x = ρ cos φ
y = ρsenφ
0 ≤ ρ ≤ 4
R': 
0 ≤ φ ≤ π
e
dxdy =
∂ (u, v )
∂ ( ρ, φ )
dρdφ =
(
∂u
∂ρ
∂v
∂ρ
∂u
∂φ
∂v
∂φ
dρdφ =
cos φ
senφ
−ρsenφ ρ cos φ
dρdφ
)
dxdy = ρ cos2 φ + ρsen2φ dρdφ = ρdρdφ
Assim :
4
π 4
π
4
π
 ρ2 
16 ∫∫ dxdy = 16 ∫ ∫ ρdρdφ = 16 ∫ dφ∫ ρdρ = 16 × ( φ ) 0 ×  
 2 0
R
0 0
0
0
16 ∫∫ dxdy = 16 × π ×
R
42
= 128π
2
Logo :
∫∫∫ dV = 16∫∫ dxdy = 128π ∴ V = 128π
T
R
z = x2 + y2 e x2 + y2 = 16 .
3. (Exerc. 06) Calcular o volume do sólido acima do parabolóide
z = x2 + y2
e abaixo do cone
z=
x2 + y2
Solução:
Temos que:
z = x2 + y2
z=
x2 + y2
Assim :
∫∫∫ dV = ∫∫∫ dxdydz
T
T
x = rsenθ cos φ
y = rsenθsenφ
z = r cos θ
z = x + y ⇔ r cos θ = r sen θ cos φ + r 2sen2 θsen2 φ
2
2
2
2
(
2
)
r cos θ = r 2sen2 θ cos2 φ + sen2φ ⇔ r cos θ = r 2sen2θ ∴ r =
cos θ
sen2θ
e
z=
x2 + y2 ⇔ r cos θ = r 2sen2 θ cos2 φ + r2sen2θsen2 φ
(
)
r cos θ = r 2sen2θ cos2 φ + sen2 φ = r 2sen2θ = rsenθ
Assim :
r cos θ = rsenθ ⇔
senθ
π
= 1 ⇔ tan θ = 1 ∴ θ =
cos θ
4
Assim :
π

0 ≤ θ ≤ 4

cos θ

T : 0 ≤ r ≤
sen2 θ

0 ≤ φ ≤ 2π


Substituindo :
π cos θ
2 π 4 sen2 θ
∫∫∫ dV = ∫∫∫ dxdydz = ∫ ∫ ∫
T
T
0 0
cos θ
∫∫∫ dV = ( φ )
T
2π
0
r2senθdrdθdφ
0
π
4
 r3  sen2θ
× 
× ∫ senθdθ
 3 0
0
π
4
3
π
1
2π 4 cos3 θ
 cos θ 
dV = 2π × × ∫ 
senθdθ =
dθ
2 
∫∫∫
3 0  sen θ 
3 ∫0 sen5θ
T
π
2π 4
2π
dV =
cot an3θ cos sec2 θdθ =
u3 cos sec2 θ
∫∫∫
∫
∫
3
3
T
0
∫∫∫ dV = −
T
1
2π 3
2π u4
u du = −
×
∫
3 0
3
4
1
=−
0

du
 −
2
 cos sec θ
2π 1
π
π
× = − ∴V =
3 4
6
6
4. (Exerc. 11) Calcular o volume do sólido delimitado pelas superfícies
Solução:
Temos que:



x2 + y2 = 16, z = 2, x + z = 9
x2 + y2 = 16, z = 2, x + z = 9 ∴ z = 9 − x
Assim :
∫∫∫ dV =
T
∫∫∫ dV =
T
9−x
∫∫ ∫
R
dzdxdy =
2
∫∫ (9 − x − 2) dxdy
R
2π 4
2π 4
2π 4
0 0
0 0
0 0
∫
2
∫ (7 − ρ cos φ )ρdρdφ = 7 ∫ ∫ ρdρdφ − ∫ ∫ ρ cos φdρdφ
4
0
 ρ3  4
2π
2π
 ρ2 

dV
=
7
×
×
φ
−
×
sen
φ
( )0   (
)0 
 
∫∫∫
2
3


 0
T
  0

Assim :
42
dV = 7 ×
× 2π = 112π ∴ V = 112π
∫∫∫
2
T
5. (Exerc.13) Calcular o volume do sólido delimitado pelas superfícies
Solução:
Temos que:
z = 2x2 + y2 e z = 4 − 3x2 − y2
Assim :
2x2 + y2 = 4 − 3x2 − y2 ⇔ 2x2 + 3x2 + y2 + y2 = 4 ⇔ 5x2 + 2y2 = 4
Dividindo por 4 :
5x2
5x
2y
y2
x2 y2
+
=1⇔ 5 +
= 1∴R :
+
=1
4
4
4
4
2
2
5
5
2
2
4 − 3x2 − y2
∫∫∫ dV =
T
∫∫ ∫
R
2
dzdxdy =
2
∫∫ ( 4 − 3x
2
)
− y2 − 2x2 − y2 dxdy
R
2x + y
Assim :
∫∫∫ dV =
T
∫∫ ( 4 − 5x
2
R
)
− 2y2 dxdy, onde R :
x2 y2
+
=1
4
2
5
z = 2x2 + y2 e z = 4 − 3x2 − y2 .
1ª Transformação ( x, y ) → (u, v )
2 5
u y = 2v
5
∂ ( x, y )
dxdy =
dudv
∂ (u, v )
x=
Onde :
∂x
∂u
=
∂ (u, v )
∂x
∂v
∂ ( x, y )
∂y
2 5
∂u
= 5
∂y
0
∂v
∫∫∫ dV = ∫∫ ( 4 − 5x
T
2
0
=
2
)
− 2y2 dxdy =
R
2 10
5
∫∫ ( 4 − 4u
2
R'
− 4v2
) 2 510 dudv, onde R ' : u
2
+ v2 = 1
Assim :
∫∫∫ dV =
T
2 10
5
∫∫ ( 4 − 4u
2
)
− 4v2 dudv, onde R ' : u2 + v2 = 1
R'
2ª Transformação : (u, v ) → ( ρ, φ )
u = ρ cos φ
v = ρsenφ
0 ≤ ρ ≤ 1
R': 
0 ≤ φ ≤ 2π
Assim :
∫∫∫ dV =
T
2 10
5
2 10
dV =
∫∫∫
5
T
∫∫ ( 4 − 4u
)
− 4v2 dudv
2
R'
2π 1
∫ ∫ ( 4 − 4ρ
2
)
cos2 φ − 4ρ2sen2φ ρdρdφ
0 0
Assim :
∫∫∫ dV =
T
2π 1
2π 1
2π 1

2 10 
3
2
3
2
 ∫ ∫ 4ρdρdφ − 4 ∫ ∫ ρ cos φdρdφ − 4 ∫ ∫ ρ sen φdρdφ 
5 0 0
0 0
0 0

Resolvendo as Integrais:
∫∫∫ dV =
2 10 
4
 ∫ ∫ 4ρdρdφ −
5 0 0
5
∫∫∫ dV =
2 10
( I1 + I2 + I3 )
5
2π 1
T
T
2π 1

3
2
3
2
ρ
cos
φ
d
ρ
d
φ
−
4
∫0
∫0 ∫0 ρ sen φdρdφ 
2π 1
∫
0
Onde :
2π 1
2π
1
0 0
0
0
1
2π
I1 = 4 ∫ ∫ ρdρdφ = 4 ∫ dφ∫ ρdρ = 4 × ( φ ) 0
I2 = −
4
5
2π 1
∫
3
2
∫ ρ cos φdρdφ = −
0 0
1
 ρ2 
12
×   = 4 × 2π ×
= 4π ∴ I1 = 4π
2
 2 0
4
× ∫ ρ3dρ ×
5 0
2π
1

1
∫  2 + 2 cos (2φ )dφ
0
Integrando :
2π
1
0
1

1
4  ρ4 
1
I2 = − ×   ×  φ + sen (2φ )  = − 4 × 
5  4  0 2
4

4

0
( )
  2π 
 = −π ∴ I2 = −π
×
  2 
e
2π 1
1
2π
0 0
0
0
I3 = −4 ∫ ∫ ρ3sen2 φdρdφ = −4 × ∫ ρ3dρ ×
1
1

∫  2 − 2 cos (2φ )dφ
Integrando :
2π
1
0
1

 ρ4 
1
1
I3 = −4 ×   ×  φ − sen (2φ )  = − 4 × 
4

4
 4  0  2
0
( )
  2π 
 = −π ∴ I3 = −π
×
  2 
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
6. (Exerc. 10) Calcule o volume da parte da esfera,
região.
Solução:
x2 + y2 + z2 = 9 ,
entre os planos z= 1; z= 2. Esboce a
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 V1:
Utilizando coordenadas cilíndricas:
2 2π 5
9 − r2
2π 3
V1 = ∫ ∫ ∫ rdrdθdz + ∫ ∫
0 0 0
r2
V1 =
2
5
0
∫ rdzdrdθ
0
5
( 5)
× 2π × 2 =
2
2π 3
× 4π + ∫ ∫ r 9 − r2 drdθ
2
0
0
5
1
9−r
= 11π − 2π 
9−9
3
5

8
16π 46π
46π
V1 = 10π + 2π × = 10π +
=
∴ V1 =
3
3
3
3
 1
V1 = 10π + 2π  −
 3
(
2
)
3



3
(
)
3
0
−
1
3
(
3
9−5 


)
(II) Cálculo de V2:
Utilizando coordenadas cilíndricas:
1 2π 8
9 − r2
2π 3
V2 = ∫ ∫ ∫ rdrdθdz + ∫ ∫
0 0 0
r2
V2 =
2
8
0
∫ rdzdrdθ
0
8
( 8)
× 2π × 1 =
2
2π 3
× 2π + ∫ ∫ r 9 − r 2 drdθ
2
0
0
8
1
9−r
= 8π − 2π 
9−9
3
8

1
2π 26π
26π
V2 = 8π + 2π × = 8π +
=
∴ V2 =
3
3
3
3
 1
V2 = 8π + 2π  −
 3
(
2
)
3



3
(
)
3
0
−
1
3
(
Assim :
V = V1 − V2 =
46π 26π 20π
20π
−
=
u.v. ∴ V =
u.v.
3
3
3
3
Resolução de
2
∫ r 9 − r dr
2
∫ r 9 − r dr = ?
Fazendo :
u = 9 − r2 ⇒ u2 = 9 − r2 → 2u
du
= − 2r ∴ rdr = −udu
dr
Substituindo :
u3
∫ r 9 − r dr = ∫ u ( −udu) = − ∫ u du = − 3 + c
Assim :
2
2
∫ r 9 − r dr = −
2
1
3
(
9 − r2
)
3
+c
9−8
)
3



