CÁLCULO III - INTEGRAIS TRIPLAS EXERC. RESOLVIDOS - PARTE I EM 26 SET 2012

Rua 96 nº 45 – Setor Sul – Goiânia Email: [email protected] CÁLCULO III – INTEGRAIS TRIPLAS Exercícios 8.5.

∫∫∫ 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

1 AFONSO AFONSO CELS CELSO O – FONE: FONE: (62) 3092-2 3092-2268 268 / CEL: CEL: (62) (62) 9216-9 9216-9668 668

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


Rua 96 nº 45 – Setor Sul – Goiânia Email: [email protected]

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.


Rua 96 nº 45 – Setor Sul – Goiânia Email: [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


Rua 96 nº 45 – Setor Sul – Goiânia Email: [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



∫

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


e


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:


+ 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


= 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


= 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


∫∫∫ ∫∫∫ 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)


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 


Rua 96 nº 45 – Setor Sul – Goiânia Email: [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


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'


Rua 96 nº 45 – Setor Sul – Goiânia Email: [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


∫∫∫ ( 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


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 zdudv ∫∫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:


Rua 96 nº 45 – Setor Sul – Goiânia Email: [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 


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


Rua 96 nº 45 – Setor Sul – Goiânia Email: [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



2. (Exerc. 05) Calcular o volume do sólido acima do plano xy delimitado por

z = x2 + y2 e x2 + y2 = 16 .


= 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



3. (Exerc. 06) Calcular o volume do sólido acima do parabolóide

z = x2 + y2

e abaixo do cone



z=

x2 + y2

Rua 96 nº 45 – Setor Sul – Goiânia Email: [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


Rua 96 nº 45 – Setor Sul – Goiânia Email: [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 .


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


Rua 96 nº 45 – Setor Sul – Goiânia Email: [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 

∫∫

∫∫

∫∫



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



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:



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:



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



∫ 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

(

)


CÁLCULO III - INTEGRAIS TRIPLAS EXERC. RESOLVIDOS - PARTE I EM 26 SET 2012

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