CÁLCULO III - INTEGRAIS DE LINHA RESOLVIDAS EM 04 MAI 2011

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CURSOS LIVRES DE 3º GRAU CÁLCULO III

INTEGRAIS DE LINHA – EXERCÍCIOS RESOLVIDOS

1. Calcule a integral de linha

(

)

C

x 2y ds,+

∫

onde C é uma semicircunferência centrada na origem de raio igual a 3 e orientada no sentido positivo.

Solução:

A parametrização dessa semicircunferência será dada por:

(

) (

2

)

2

r(t) 3cos ti 3sent j, 0 t= + ≤ ≤ π ⇒ds= −3sent + 3cos t dt⇔ds= 9 dt 3dt=

r r r

. Substituindo:

(

)

(

)

0

( )

0

3cos t 6sent 3dt 3 3sent 6cos t 3 12 36

π π + = − = × =

∫

2. Calcular a integral

(

)

C x² y² z ds,+ −

∫

onde C é a hélice circular dada por :

r(t) cos ti sent j tk de P(1,0,0) a Q(1,0,2 )= + + π

r r r r

Solução:

(

) (

2

)

ds= −sent + cost ² 1dt+ = 2 dt. Assim, podemos escrever:

(

)

(

)

(

)

(

)

2 2 2 0 0 0 2 0 t² cos ²t sen²t t 2 dt 2 1 t dt 2 t 2 4 ² 2 1 t dt 2 2 2 2 1 2 π π π π   + − = − = _ − _   π   − = _ π − _= π − π  

∫

3. Calcule

(

)

C 2x y z ds− +

∫

, onde C é o segmento de reta que liga A(1, 2, 3) a B(2, 0, 1). Solução:

Parametrização do segmento de reta AB:

= +   = − − = − − ⇔ _ = −  _{= −}  = ⇒ = − = ⇒ = ∴− ≤ ≤ uuur r r r suur x(t) 2 t AB (1, 2, 2) i 2j 2k; B(2,0,1) AB : y(t) 2t z(t) 1 2t y 2 t 1; y 0 t 0 1 t 0

(2)

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = + + ⇔ = + − + − = − − ⇒ = + + = = ∴ = = − + ⇔ = + − − + − = + + + − = + ∴ = + r r ur r r r r ˆ ˆ ˆ ˆ ˆ ˆ r(t) x t i y t j z t k r(t) 2 t i 2tj 1 2t k Assim : r '(t) i 2j 2k r(t) 1 4 4 9 3 ds 3dt (1) f x,y,z 2x y z f t 2(2 t) ( 2t) 1 2t 4 2t 2t 1 2t 5 2t f t 5 2t (2)

Substituindo (1) e (2) na integral dada:

(

)

(

)

(

)

0 0 0 1 C 1 1 C 2x y z ds 5 2t 3dt 3 (5 2t) dt 3(5t t²) | 2x y z ds 0 3( 5 1) ( 3)( 4) 12 − − − − + = + = + = + − + = − − + = − − =

∫

Resp.: 12 4. Calcule C xz ds

∫

, onde C é a interseção da esfera x² + y² + z² = 4 com o plano x = y. Solução:

Vamos parametrizar a curva dada:

( )

(

)

( )

= = ⇒ + + = ⇒ = − ∴ = − − ≥ ⇔ − ≤ ⇒ − ≤ ≤ = + + ⇔ = + + − = + − −   − = + + −_ _ = + = − −   r r r r r r ur 2 2 ₂ ₂ 2 2 2 2 x y t t² t² z² 4 z² 4 2t² z 4 2t² 4 2t² 0 2t² 4 0 2 t 2 ˆ ˆ ˆ r(t) x t i y t j z t k r(t) ti t j 4 2t² k 2t ˆ ˆ ˆ r ' t i j k 4 2t 2t 4t 8 4t r '(t) 1 1 2 4 2t 4 2t +_4t2

( )

(

)

( )

= = − − ₋ = ⇔ = − 2 2 ₂ 8 8 1 4 2t 4 2t _{4 2t} e f x,y,z xz f t t 4 2t² (2)

Substituindo (1) e (2) na integral dada:

= −

∫

C xz ds t 4 2t² − ⋅ −

∫

2 2 2 8 4 2t

( ) ( )

(

)

− − =   = × = ×_ − − _= × − =  

∫

2 2 2 2 ₂ ₂ C 3 dt 8 t dt t 8 8 xz ds 8 2 2 2 2 0 2 2 2 Resp.: 0

(3)

Outra Solução:

( )

(

)

( )

(

)

( )

(

) (

)

(

)

( )

+ + = = + + = ⇔ + = ∴ + = = = = = ⇒ = − − = − + − + ⇔ = + + = r r r r r 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 C : x y z 4 x y Assim : y z y y z 4 2y z 4 1 2 4 Parametrizando:

x t 2 cos t y t 2 cos t z t 2sent Assim :

r t 2 cos t, 2 cos t, 2sent r ' t 2sent, 2sent, 2cos t e

r ' t 2sent 2sent 2cos t r ' t 2sen t 2sen t 4cos t

r ' t 4

( )

(

)

( )

(

)

( )

π π π + ⇔ = + ⇔ = ∴ = = × × = = = ⇒ =   = = × = = _ π − _ =

∫

r r r 2 2 2 2 2 2 b C 0 0 a b b 2 ₂2 2 2 0 C a a

sen t 4cos t r ' t 4 sen t cos t r ' t 4 r ' t 2 Substituindo :

xzds 2 cos t 2sent 2dt 4 2 sent cos tdt 4 2 udu Onde :

u sent du cos tdt Assim:

u

xzds 4 2 udu 4 2 2 2 sent 2 2 sen 2 sen 0 0 2

Resp: 0 5. Calcule

C xyds

∫

, onde C é a elipse x² y² 1 a² b²+ = . Solução:

A parametrização da elipse é dada por:

[

]

( )

(

)

= = ∈ π = + ≤ ≤ π = − + = + = − = − + ⇔ = − + ∴ = − + r r r r ur ur ₂ ₂ ur ₂ ₂ ₂ ur

x(t) acos t e y(t) bsen t t 0, 2 r(t) acos ti bsen tj, 0 t 2 e

ˆ ˆ

r ' t asenti bcos tj

r '(t) a²sen²t b² cos ²t, mas sen²t 1 cos ²t

r '(t) a² 1 cos t b² cos t r '(t) a² a cos t b² cos t r '(t ) (b² a²)cos ²t a²

(4)

Substituindo na integral dada: π π = ⋅ ⋅ − + = ⋅ ⋅ − + = − + ⇒ = − ⋅ − = − − ⋅ ⋅ = − ⋅ ⋅ ⋅ ∴ = − ⋅ ⋅ =

∫

2 C 0 2 C 0 C

xyds acos t bsent (b² a²)cos ²t a² dt xyds ab cos t sent (b² a²)cos ²t a² dt

u (b² a²)cos ²t a² du 2(b² a²)cos t ( sent) 2(b² a²) cos t sent du

du 2(a² b²) cos t sent dt dt

2(a² b²) cos t sent xyds ab cos t sent⋅ ⋅ ⋅

− ⋅ ⋅

∫

u du

2(a² b²) cos t sent

[ − + ] π = = − − =

∫

3 2 1 2 2 0 C C (b² a²)cos ²t a² ab ab xyds u du | 3 2(a² b²) 2(a² b²) 2 ab xyds 2 − ⋅ 2 (a² b²) ( )

(

) (

{

)

( )

(

)

( )

}

(

) (

)

π      − +  = − π + − − +   ₋       = _ − + − − + _= ∴ = −

∫

2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 C C ab

b² a² cos t a b a cos 2 a b a cos 0 a

3 3 a b ab xyds b a a b a a 0 xyds 0 3 a b Resp.:0 6.

∫

(

−

)

C

3y z ds_{, onde C é o arco da parábola z = y² e x = 1 de A(1,0,0) a B(1,2,4).}

Solução: Parametrizando C:

( )

 =  =_ = ≤ ≤  ₌  2 x t 1 C y t t 0 t 2 z t t Assim:

( )

=

(

)

⇒

( ) (

=

)

∴

( )

= + r ₂ r r ₂ r t 1,t,t r ' t 0,1,2t r ' t 1 4t Assim:

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(

)

(

)

(

)

(

)

(

)

− = − + = − + = + = + ⇒ = ∴ = ≤ ≤ ⇔ ≤ ≤   − = + = _ _=       − = _ _ = × ×    

∫

2 2 2 2 2 2 2 C 0 0 0 2 2 17 17 1 2 2 C 0 1 1 17 3 2 C 1 3y z ds 3t t 1 4t dt 3t t 1 4t dt 2t 1 4t dt Fazendo : du du u 1 4t 8t dt dt 8t e 0 t 2 1 u 17 Substituindo : du 2t 3y z ds 2t 1 4t dt 2t u u du 8t 8t 1 u 1 2 3y z ds 17 3 4 4 3 2

(

)

(

)

(

)

  − = −     − = −

∫

3 3 3 2 2 C 1 1 17 1 6 1 3y z ds 17 17 1 6 Resp: 1

(

17 17 1−

)

6 7.

∫

C

y ds_{, onde C é a curva dada por y = x³ de (-1,-1) a (1, 1).}

Solução: Sabemos que: ≥ ⇔ − ≤ ≤  = ₋ _{< ⇔ < <}  y, se y 0 1 y 0 y y, se y 0 0 y 1 Parmetrizando C:

( )

=

( )

= 3 C : x t t; y t t Assim:

(6)

( )

(

)

( )

− = + ∴ = = ⇒ = + ∴ = + = + = − + + + = + ⇒ = ∴ = − ≤ ≤ ⇔ ≤ ≤ ≤ ≤ ⇔ ≤ ≤ = −

∫

r r r r r 1 2 3 2 2 2 4 0 1 3 4 3 4 C C C 1 0 4 3 3 3 C ˆ ˆ r t x t i y t j r t t,t Assim : r ' t 1,3t r ' t 1 3t r ' t 1 9t Assim :

yds -yds yds t 1 9t dt t 1 9t dt Fazendo : du du u 1 9t 36t dt dt 36t Se 1 t 0 10 u 1e 0 t 1 1 u 10 Substituindo : yds t

(

)

−     + + + = − _ _+ _ _     = − + = + = ×   = = × = × ×_ − _= − =  

∫

0 1 1 10 4 3 4 3 3 3 3 1 0 10 1 1 1 10 1 10 1 10 1 10 1 2 2 2 2 2 C 10 1 1 1 1 3 10 1 ₂ 3 3 3 2 2 2 C 1 du du 1 9t dt t 1 9t dt t u t u 36t 36t 1 1 1 1 1 yds u du u du u du u du 2 u du 36 36 36 36 36 1 1 u 1 2 1 1 yds u du 10 1 10 1 3 18 18 18 3 27 2 2

(

−

)

− = − =

∫

C 10 10 1 7 10 10 1 10 10 1 yds 27 27 27 Resp: 10 10 1− 27 8. Calcule C y(x z)ds−

(7)

Solução: Parametrizando C:

(

)

 + + = _⇔  + + =  _{+ =}  _{= −}   + + = ⇔ + + − = + + 2 2 2 2 2 2 2 2 2 2 2 2 2 2 x y z 9 x y z 9 C : C : x z 3 z 3 x Assim : x y z 9 x y 3 x 9 x y 9 −_{6x x}+ 2 = ₉ _⇔ ₋ ₊ ₌  ₋ ₊ _{− +} _{= ⇔}  ₋  ₊ _{= ⇔}  ₋  ₊ ₌              ₋   ₋        ₊ _{= ⇔}  ₊ ₌ = + = = − ⇔ = + − = − + 2 2 2 2 2 2 2 2 2 2 2 2 2x 6x y 0

Comple tando o quadrado :

9 9 3 9 3 2 x 3x y 0 2 x y 4 x 2y 9 4 2 2 2 2 3 3 4 x x 2y y 2 ₁ 2 ₁ 9 9 9 9 4 2 Assim: 3 3 3 x cos t e y sent 2 2 2 Mas : 3 3 3 3 z x 3 z cost 3 cos t 2 2 2 2 Assim

( )

(

)

  =_ + − + _ ≤ ≤ π     = −_ − _         = _− _ +_ _ + −_ _ ⇔ = + +       = + = + r r r r r 2 2 2 2 2 2 2 2 2 2 : 3 3 3 3 3

r t cos t, sent, cos t 0 t 2

2 2 2 2 2

e

3 3 3

r ' t sent, cos t, sent

2 2 2

Então :

3 3 3 9 9 9

r ' t sent cos t sent r ' t sen t cos t sen t

2 2 2 4 2 4

9 9 9

r ' t sen t cos t sen t cos t

2 2 2 = = ∴

( )

= r 1 ₉ ₃ ₃ r ' t 2 2 2 Assim:

(8)

π _ _   − = _ + −_ + − __     − = ×

∫

2 C 0 C 3 3 3 3 3 3

y(x z)ds sent cos t cos t 3 dt

2 2 2 2 2 2 3 3 3 y(x z)ds sent 2 2 2 + 3 cost 2 3 -2 − 3 cos t 2

(

)

(

)

(

)

π π π π   +     − = = = − = − π − = − − = − =

∫

2 0 2 2 2 0 C 0 0 C 3 dt 9 27 27 27 27

y(x z)ds 3sentdt sentdt cos t cos2 cos0 1 1 0

2 2 2 2 2 Assim : y(x z)ds 0 Resp: 0 9. Calcule C (x y)ds+

∫

, onde C é a interseção das superfícies z = x² + y² e z = 4. Solução:

A curva C é a circunferência x² + y² = 4, cuja parametrização é dada por:

( ) (

)

( ) (

)

( )

(

)

=  _{≤ ≤ π}  =  = ⇒ = − = + = + r r r ₂ ₂ ₂ ₂ x 2cos t C : 0 t 2 y 2sent Assim :

r t 2cos t, 2sent r ' t 2sent, 2cos t e

r ' t 4sen t 4cos t 4 sen t cos t

( )

(

)

(

)

(

)

( )

₍

( )

₎

(

)

π π π = = ∴ = + = + = + = −   + = _ π − − π − _= − − + = × = + =

∫

r 1 2 2 2 0 C 0 0 C C 4 2 r ' t 2 Substituindo :

(x y)ds 2cos t 2sent 2dt 4 cos t sent dt 4 sent cos t

(x y)ds 4 sen 2 sen 0 cos 2 cos 0 4 0 0 1 1 4 0 0 Logo :

(x y)ds 0

(9)

Solução:

Parametrizando os segmentos de reta que formam os lados do quadrado, temos:

(

)

( ) (

)

( ) (

)

( )

(

)

(

)

(

)

= − = =   =   =  = ⇒ = ∴ = ≤ ≤ + + = + + = + = + = + =

∫

suur r r r r 1 AB 1 1 1 1 2 C 0 0 0 A(1,0,1), B(1,1,1), C(0,1,1) e D(0,0,1) Reta AB : u B A 0,1,0 Assim : x 1 C : y t z 1 r t 1,t,1 r ' t 0,1,0 r ' t 1 0 t 1 Assim : t 1 5 x y z ds 1 t 1 dt 2 t dt 2t 2 2 2 2

(

)

( ) (

)

( ) (

)

( )

(

)

(

)

(

)

− − − = − = − = −   =   =  = ⇒ = − ∴ = − ≤ ≤   + + = − + + = − = − = − − −_ _=  

∫

suur r r r r 2 BC 2 0 0 0 2 C 1 1 1 A(1,0,1), B(1,1,1), C(0,1,1) e D(0,0,1) Reta BC : u C B 1,0,0 Assim : x t C : y 1 z 1 r t -t,1,1 r ' t 1,0,0 r ' t 1 1 t 0 Assim : t 1 5 x y z ds t 1 1 dt 2 t dt 2t 0 2 2 2 2

(10)

(

)

( ) (

)

( ) (

)

( )

(

)

(

)

(

)

− − − = − = − =   = −   =  = ⇒ = − ∴ = − ≤ ≤   + + = − + = − = − = − − −_ _=  

∫

suur r r r r 3 CD 3 0 0 0 2 C 1 1 1 A(1,0,1), B(1,1,1), C(0,1,1) e D(0,0,1) Reta CD : u D C 0, 1,0 Assim : x 0 C : y t z 1 r t 0,-t,1 r ' t 0, 1,0 r ' t 1 1 t 0 Assim : t 1 3 x y z ds 0 t 1 dt 1 t dt t 0 1 2 2 2

(

)

( ) (

)

( ) (

)

( )

(

)

(

)

(

)

− − − = − = = +   =   =  = ⇒ = ∴ = − ≤ ≤   + + = + + + = + = + = − − +_ _=  

∫

suur r r r r 4 DA 4 0 0 0 2 C 1 1 1 A(1,0,1), B(1,1,1), C(0,1,1) e D(0,0,1) Reta DA : u A D 1,0,0 Assim : x 1 t C : y 0 z 1 r t 1+t,0,1 r ' t 1,0,0 r ' t 1 1 t 0 Assim : t 1 3 x y z ds 1 t 0 1 dt 2 t dt 2t 0 2 2 2 2 Assim: + + = + + + + + + + + + + + + + + + + = + + + = = = ∴ + + =

∫

1 2 3 4 C C C C C C C (x y z)ds (x y z)ds (x y z)ds (x y z)ds (x y z)ds 5 5 3 3 5 5 3 3 16 (x y z)ds 8 (x y z)ds 8 2 2 2 2 2 2 Resp: 8

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11. Calcular a integral C

xyds,

∫

onde C é a interseção das superfícies x² + y² = 4 e y + z = 8. 12. Calcular

C 3xyds

∫

, sendo C o triângulo de vértices A(0,0), B(1,0) e C(1,2), no sentido anti-horário. 13. Calcule

C

y(x z)ds−

∫

, onde C é a interseção das superfícies x² + y² + z² = 9 e x + z = 3. 14. Calcule

C

(x y)ds+

∫

, onde C é a interseção das superfícies z = x² + y² e z = 4. 15. Calcule

(

)

c

x² y² z ds+ −

∫

, onde C é a interseção das superfícies x² + y² + z² = 8z e z = 4. 16. Calcule

C

xy²(1 2x²)ds−

∫

, onde C é a parte da curva de Gauss y e₌ −x²_{de A(0,1) a}_B 1 1

2 e  ₋     . 17. C ds

−

∫

, onde C : r t

( ) (

= t cos t, tsent

)

t∈ 0,1

r . Solução: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 t C C t 2 2 2 ds ds r ' t dt 1 Assim :

r ' t cos t tsent,sent t cos t r ' t cos t tsent sent t cos t r ' t cos t 2t cos tsent

− = = = − + = − + + = − ∫ ∫ ∫ r r r

r ₊_{t sen t sen t}2 2 ₊ 2 ₊ _{2tsent cos t}

( )

(

)

( ) ( ) ( ) 0 2 2 2 2 2 2 t C C t 1 2 C 0 t cos t r ' t 1 t sen t cos t r ' t 1 t Substituindo em 1 : ds ds r ' t dt ds 1 t dt − − + = + + = + = = = + ∫ ∫ ∫ ∫ ∫ r r r Resolvendo 1 2 C 0 ds 1 t dt − = +

∫

:

(12)

1 2 C 0 2 1 4 2 2 2 2 2 C 0 0 ds 1 t dt Mas : t tg de sec d Se t 0 0 Se t 1 4 Assim :

ds 1 t dt 1 tg sec d Mas :1 tg sec

− π − = + = θ ⇒ = θ θ π = ⇒ θ = = ⇒ θ = = + = + θ θ θ + θ = θ

∫

Substituindo: 1 4 2 2 2 C 0 0 1 4 2 2 2 C 0 0 1 4 2 2 C 0 0 1 4 2 3 C 0 0 n n 2 n 2 1 2 C 0 ds 1 t dt 1 tg sec d ds 1 t dt sec sec d ds 1 t dt sec sec d ds 1 t dt sec d Utilizando : 1 n 2

sec udu sec u tgu sec udu

n 1 n 1 Assim : ds 1 t dt se π − π − π − π − − − − = + = + θ θ θ = + = θ θ θ = + = θ ⋅ θ θ = + = θ θ − = ⋅ + − − = + =

∫

( ) 4 3 0 1 4 2 C 0 0 1 ₄ 2 0 C 0 1 2 C 0 c d 1 1

ds 1 t dt sec tg sec d Mas : secud ln secu tgu c

2 2 Substituindo : 1 1 ds 1 t dt sec tg ln sec tg 2 2 1 1 1

ds 1 t dt sec tg ln sec tg sec 0

2 4 4 2 4 4 2 π π − π − − θ θ = + = θ ⋅ θ + θ θ = + + = + = θ ⋅ θ + θ + θ π π π π         = + = _{ }⋅ _{ }+ _{ }+ _{ } −        

∫

( ) ( ) ( ) ( ) ( ) 1 2 C 0 1 tg 0 ln sec 0 tg 0 2 1 1 1 ds 1 t dt 2 1 ln 2 1 sec 0 tg 0 2 2 2 − ⋅ + + = + = × × + + − ⋅

∫

0 1 ln 1 0 2 + + 0 1 2 C 0 Logo : 2 1 ds 1 t dt ln 2 1 2 2 − = + = + +

∫

(13)

18. 2 C

x ds

∫

, onde _{C : x}32 ₊_y23 ₌ _a23 _{a 0}_> _{1º quadrante}. Solução:

Uma equação vetorial para a hipociclóide _x23 ₊ _y23 ₌_a23 é: r tr

( )

=acos ti asen tj3 ˆ+ 3ˆ

( )

(

)

( )

(

) (

)

( )

(

)

3 3 2 2 2 2 2 2 2 4 2 2 4 2 2 2 2 2 2 ˆ ˆ r t acos ti asen tj Mas :

r ' t 3acos t sent,3asen t cos t Assim :

r ' t 3acos t sent 3asen t cos t 9a cos t sen t 9a sen t cos t r ' t 9a cos t sen t cos t sen t

= + = − ⋅ ⋅ = − ⋅ + ⋅ = ⋅ + ⋅ = ⋅ + r r r r

( )

1 2 2 2

9a cos t sen t 3acos t sent r ' t 3acos t sent = ⋅ = ⋅ = ⋅ r Assim:

( )

( ) ( )

(

)

(

)

(

)

0 t 2 ₂ 2 3 C t 0 2 2 2 2 6 3 7 C 0 0 2 3 7 C 0 r ' t 3acos t sent

x ds f t r ' t dt acos t 3acos t sent dt

x ds a cos t 3acos t sent dt 3a cos t sentdt Fazendo : du du u cos t sent dt dt sent Se t 0 u 1 Se t u 0 2 Substituindo : x ds 3a cos t sent π π π = ⋅ = = ⋅ = ⋅ = ⋅ = ⇒ = − ∴ = − π = ⇒ = = ⇒ = = ⋅

∫

r r 2 3 7 dt 3a u sent π =

∫

0 − _sentdu 30 7 1 1 3a u du   _{= −}    

∫

(14)

(

)

0 0 8 8 8 3 2 3 7 3 3 3 C 1 1 3 2 C u 0 1 1 3a x ds 3a u du 3a 3a 3a 8 8 8 8 8 Logo : 3a x ds 8       = − = − _ _ = − _ − _ = − × −_ _ =       =

∫

19. 2 C x ds

∫

, onde C : r t

( )

(

2 cos t,2sen t3 3

)

t 0, 2 π   = _{∈ } _   r . Solução:

( )

(

)

( )

(

) (

)

( )

(

)

3 3 2 2 2 2 2 2 4 2 4 2 2 2 2 2 ˆ ˆ r t 2 cos ti 2sen tj Mas :

r ' t 6cos t sent,6sen t cos t

Assim :

r ' t 6cos t sent 6sen t cos t 36 cos t sen t 36sen t cos t

r ' t 36 cos t sen t cos t sen t

= + = − ⋅ ⋅ = − ⋅ + ⋅ = ⋅ + ⋅ = ⋅ + r r r r

( )

1 2 2

36 cos t sen t 6 cos t sent

r ' t 6 cos t sent = ⋅ = ⋅ = ⋅ r Assim:

( )

( ) ( )

(

)

(

)

(

)

0 t 2 ₂ 2 3 C t 0 2 2 2 6 7 C 0 0 2 2 7 C 0 r ' t 6 cos t sent

x ds f t r ' t dt 2 cos t 6 cos t sent dt

x ds 4 cos t 6 cos t sent dt 24 cos t sentdt Fazendo : du du u cos t sent dt dt sent Se t 0 u 1 Se t u 0 2 Substituindo : x ds 24 cos t sentdt π π π π = ⋅ = = ⋅ = ⋅ = ⋅ = ⇒ = − ∴ = − π = ⇒ = = ⇒ = = ⋅ =

∫

r r 7 24 u sent du sent − 0 0 7 1 1 24 u du   = −    

∫

(15)

(

)

0 0 8 8 8 2 7 C 1 1 2 C u 0 1 1 24 x ds 24 u du 24 24 24 3 8 8 8 8 8 Logo : x ds 3       = − = − _ _ = − _ − _ = − × −_ _ = =       =

∫

20.

(

)

C x y ds−

∫

, onde C é o triângulo da figura abaixo:

Solução:

Parametrizando os segmentos de reta AB, BC e CA.

( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 C 3 A 1, ; B 2,2 e C 2,1 2 x 2 t AB C : ₁ 1 t 0 y 2 t 2 Assim : 1 1 r t 2 t, 2 t r ' t 1, 2 2 e 1 5 5 r ' t 1 r ' t 4 4 2 Assim : x y ds 2       = +   ⇔ _{ = +} − ≤ ≤      =_ + + _⇒ =_ _     = + = ∴ = − = ∫ r t 2 + − ( ) ( ) 1 1 0 0 1 1 0 0 2 C 1 ₁ C 1_t 5 _dt 5 1_{t dt} 2 2 2 2 5 5 t 5 0 1 5 5 x y ds tdt x y ds 4 4 2 8 2 2 8 8 − − − ₋    ₋ _ _ ₌    _ _    _ _     − = = × = _ − _= − ∴ − = −   ∫ ∫ ∫ ∫ ∫

(16)

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 C 3 A 1, ; B 2,2 e C 2,1 2 x 2 BC C : 0 t 1 y 2 t Assim : r t 2, 2 t r ' t 0, 1 e r ' t 0 1 1 r ' t 1 Assim : x y ds 2       =  ⇔ _{ = −} ≤ ≤  = − ⇒ = − = + = ∴ = − = ∫ r 2 −

(

)

( ) ( ) 2 1 1 1 2 0 0 0 C t 1 1 t 1 dt t dt x y ds 2 2 2 + = = = ∴ − = ∫ ∫ ∫ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 1 1 C 0 0 1 2 C ₀ 3 A 1, ; B 2,2 e C 2,1 2 x 2 t CA C : ₁ 0 t 1 y 1 t 2 Assim : 1 1 r t 2 t, 1 t r ' t 1, 2 2 e 1 5 5 r ' t 1 r ' t 4 4 2 Assim : 1 5 5 3 x y ds 2 t 1 t dt 1 t dt 2 2 2 2 5 3 t x y ds t 2 2 2       = −   ⇔ _{ = +} ≤ ≤      =_ − + _⇒ = −_ _     = + = ∴ =       − = _ − − − __ _ = _ − _     − = _ − × _   ∫ ∫ ∫ ∫ r ( ) 3 C 5 ₁ 3 5 1 5 _{x y ds} 5 2 4 2 4 8 8   = _ − _= × = ∴ − =   ∫ Assim:

(

)

(

)

(

)

(

)

(

)

(

)

1 2 3 C C C C C C x y ds x y ds x y ds x y ds 5 1 5 1 1 x y ds x y ds 8 2 8 2 2 − = − + − + − − = − + + = ∴ − =

∫

(17)

21. 2 C

y ds

∫

, onde C é a semicircunferência da figura abaixo:

Solução:

Parametrizando a semicircunferência, temos:

( ) (

)

( ) (

)

( )

(

)

=  _{≤ ≤ π}  =  = ⇒ = − = + = + r r r ₂ ₂ ₂ ₂ x 2cos t C : 0 t 2 y 2sent Assim :

r t 2cos t, 2sent r ' t 2sent, 2cos t e

r ' t 4sen t 4cos t 4 sen t cos t

( )

(

)

( )

π π π π π π π = = ∴ =   = = = = _ − _   = − = ⋅ +   = − × = π − _ π − _

∫

r 1 2 2 2 2 C 0 0 0 0 2 C 0 0 2 0 C 4 2 r ' t 2 Substituindo : 1 1

y ds 2sent 2dt 2 4sen tdt 8 sen tdt 8 cos 2t dt 2 2

1

y ds 4 dt 4 cos 2t dt Mas : cos mx dx sen mx C m

Assim :

1

y ds 4t 4 sen 2t 4 2 sen 2 sen 0

2 = π∴

∫

= π

1

2 C

(18)

22. 2 C

y ds

∫

, onde C é o 1º arco da ciclóide:

( )

(

)

ˆ

(

)

ˆ r tr =2 t sent i 2 1 cos t j− + − . Solução:

( )

(

)

(

)

( ) (

)

( ) (

)

( )

(

) (

)

( )

(

)

2 2 2 2 2 2 ˆ ˆ r t 2 t sent i 2 1 cos t j r t 2t 2sent,2 2 cos t 0 t 2 Derivando : r' t 2-2cost,2sent Mas : ds r ' t dt Assim :

r ' t 2 2 cos t 2sent 4 8 cos t 4 cos t 4sen t

r ' t 4 8 cos t 4 cos t+sen t

= − + − = − − ≤ ≤ π = = = − + = − + + = − + r r r r r r

( )

(

)

( )

1 4 8 cos t 4 8 8 cos t Assim :

r ' t 8 1 cos t 8 1 cos t r ' t 2 2 1 cos t

= − + = − = − = − ∴ = − r r Substituindo na integral: ( )

(

)

(

)

2 2 2 C 0 2 2 2 C 0 2 2 2 2 2 C 0 0 0 2 2 2 2 2 2 C 0 0 y ds 2 2 cos t 2 2 1 cos t dt y ds 2 2 4 8 cos t 4 cos t 1-cost dt

y ds 8 2 1 cos t dt 16 2 cos t 1 cos t dt 8 2 cos t 1 cos t dt Mas :

cos t 1 sen t Assim :

y ds 8 2 1 cos t dt 16 2 cos t 1 cos t dt 8 2 1 sen t

π π π π π π π = − × − = − + = − − − + − = − = − − − + −

∫

2 0 2 2 2 2 2 2 C 0 0 0 0 2 2 2 2 2 C 0 0 0 1 cos t dt

y ds 8 2 1 cos t dt 16 2 cos t 1 cos t dt 8 2 1 cos t dt 8 2 sen t 1 cos t dt y ds 16 2 1 cos t dt 16 2 cos t 1 cos t dt 8 2 sen t 1 cos t dt

π π π π π π π π − = − − − + − − − = − − − − −

∫

(19)

( )

2 2 2 2 2 C 0 0 0 2 2 2 2 2 2 2

y ds 16 2 1 cos t dt 16 2 cos t 1 cos t dt 8 2 sen t 1 cos t dt Fazendo : t 2 dt 2d e se t 0 0 e se t=2 e mais :

1 cos t 1 cos 2 cos 2 cos sen Assim :

1 cos t 1 cos sen sen sen 2sen

Logo : 1 cos t 2 sen π π π = − − − − − = θ ⇒ = θ = ⇒ θ = π ⇒ θ = π − = − θ θ = θ − θ − = − θ + θ = θ + θ = θ − =

∫

(

)

( )

(

)

(

)

( )

(

)

(

)

( )

2 2 2 2 2 C 0 0 0 2 2 C 0 0 0 2 2 C 0 0 0 Substituindo :

y ds 16 2 1 cos t dt 16 2 cos t 1 cos t dt 8 2 sen t 1 cos t dt

y ds 16 2 2 sen 2d 16 2 cos 2 2 sen 2d 8 2 sen 2 2 sen 2d

y ds 64 sen d 64 cos 2 sen d 32 sen 2 sen d

π π π π π π π π π θ = − − − − − = θ θ − θ × θ × θ − θ × θ × θ = θ θ − θ θ θ − θ θ θ

∫

Resolvendo

( )

0 64 cos 2 sen dπ

∫

θ θ θ: ( )

(

)

( ) ( )

(

)

( ) 2 2 0 0 2 2 0 0 0 2 2 0 0 0 2 2 0 0 0

64 cos 2 sen d 64 cos sen sen d

64 cos 2 sen d 64 cos sen d 64 sen sen d 64 cos 2 sen d 64 cos sen d 64 1 cos sen d

64 cos 2 sen d 64 cos sen d 64 sen d +64 cos sen d

π π π π π π π π π π π θ θ θ = θ − θ θ θ θ θ θ = θ θ θ − θ θ θ θ θ θ = θ θ θ − − θ θ θ θ θ θ = θ θ θ − θ θ θ θ θ

∫

( ) 0 2 0 0 0

64 cos 2 sen d 128 cos sen d 64 sen d

π

π π π

θ θ θ = θ θ θ − θ θ

∫

(20)

( )

2 0 0 0 2 0 2 2 0

64 cos 2 sen d 128 cos sen d 64 sen d Re solvendo 128 cos sen d :

128 cos sen d 128 u sen

π π π π π θ θ θ = θ θ θ − θ θ θ θ θ θ θ θ = θ

∫

− _sendu θ

( )

1 1 2 1 1 1 1 3 2 2 0 1 1 3 3 2 0 2 0 128 u du Onde : du du u cos sen d d sen e se 0 u 1 e se u 1 Logo : u 128 cos sen d 128 u du 128 3 1 1 128 128 256 128 cos sen d 128 3 3 3 3 3 Assim : 128 cos sen d − − − π − π π   _{= −}     = θ → = − θ ∴ θ = − θ θ θ = ⇒ = θ = π ⇒ = − θ θ θ = − = − ×  ₋    θ θ θ = − × − = + =     θ θ θ

∫

256 3 =

∫

Substituindo:

( )

(

)

( )

(

)

(

)

( )

(

)

( )

2 0 0 0 0 0 0 0 0 0

64 cos 2 sen d 128 cos sen d 64 sen d 256

64 cos 2 sen d 64 cos 3

256 256

64 cos 2 sen d 64 cos 64 cos cos 0

3 3 256 256 256 64 cos 2 sen d 64 1 1 64 2 128 3 3 3 128 64 cos 2 sen d 3 π π π π π π π π π θ θ θ = θ θ θ − θ θ θ θ θ = − × − θ θ θ θ = + × θ = + × π − θ θ θ = + × − − = + × − = − θ θ θ = −

∫

(21)

Resolvendo 2

( )

0 32 sen 2 sen d

∫

π θ θ θ:

( )

(

)

( )

(

)

( )

2 2 0 0 2 2 2 0 0 2 2 2 0 0 2 2 4 0 0 0

32 sen 2 sen d 32 2sen cos sen d 32 sen 2 sen d 128 sen cos sen d 32 sen 2 sen d 128 1 cos cos sen d

32 sen 2 sen d 128 cos sen d 128 cos sen d Fazendo : du u cos sen d π π π π π π π π π θ θ θ = θ θ θ θ θ θ θ = θ θ θ θ θ θ θ = − θ θ θ θ θ θ θ = θ θ θ − θ θ θ = θ → = − θ ∴ θ

∫

( )

2 2 4 0 0 0 2 2 0 du d sen e se 0 u 1 e se u 1 Assim :

32 sen 2 sen d 128 cos sen d 128 cos sen d 32 sen 2 sen d 128 u sen

π π π π θ = − θ θ = ⇒ = θ = π ⇒ = − θ θ θ = θ θ θ − θ θ θ θ θ θ = θ

∫

− _sendu θ 4 128 u sen  ₋ _θ     du sen − θ

( )

1 1 1 1 1 1 2 2 4 0 1 1 -1 1 3 5 2 0 ₁ ₁ 3 3 5 5 2 0 2 0 32 sen 2 sen d 128 u du 128 u du u u 32 sen 2 sen d 128 128 3 5 1 1 1 1 32 sen 2 sen d 128 128 3 3 5 5 1 1 32 sen 2 sen d 128 3 3 − − π − − − π π π       θ θ θ = − +     θ θ θ = − ×_ _ + ×_ _      ₋   ₋      θ θ θ = − × − + × −          θ θ θ = − × − −_

∫

( )

2 0 1 1 256 256 512 128 5 5 3 5 15 Assim : 512 32 sen 2 sen d 15 π ₊ _{× − −}  ₌ ₋ ₌       θ θ θ =

∫

(22)

Substituindo na integral:

( )

(

)

(

) (

)

( )

2 2 C 0 0 0 0 0 0 2 0 2 C 0 0

y ds 64 sen d 64 cos 2 sen d 32 sen 2 sen d Onde :

64 sen d 64 cos 64 cos cos 0 64 1 1 128 128 64 cos 2 sen d 3 512 32 sen 2 sen d 15 Substituindo :

y ds 64 sen d 64 cos 2 sen d 3

π π π π π π π π π = θ θ − θ θ θ − θ θ θ θ θ = − θ = − π − = − × − − = θ θ θ = − θ θ θ = = θ θ − θ θ θ −

∫

2

( )

0 2 C 2 C 2 sen 2 sen d 128 512 128 512 2048 y ds 128 128 3 15 3 15 15 Logo : 2048 y ds 15 π θ θ θ     = − −_ _{ }− _ = + − =     =

∫