CALCULO-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 2r(t) 3cos ti 3sent j, 0 t ds 3sent 3cos t dt ds 9 dt 3dt= + ≤ ≤ pi ⇒ = − + ⇔ = =r r r . Substituindo:
( ) ( ) ( )0
0
3cos t 6sent 3dt 3 3sent 6cos t 3 12 36
pi
pi
+ = − = × =∫
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 )= + + pi
r r r r
Solução:
( ) ( )2ds sent cost ² 1dt 2 dt.= − + + = Assim, podemos escrever:
( ) ( )
( ) ( )
22 2
00 0
2
0
t²cos ²t sen²t t 2 dt 2 1 t dt 2 t
2
4 ²2 1 t dt 2 2 2 2 1
2
pipi pi
pi
 
+ − = − = −  
pi 
− = pi − = pi − pi  
∫ ∫
∫
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
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( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
= + + ⇔ = + − + −
= − − ⇒ = + + = = ∴ =
= − + ⇔ = + − − + − = + + + − = + ∴ = +
r r
ur r r r r
ˆ ˆ ˆ ˆ ˆ ˆr(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
22
x y t t² t² z² 4 z² 4 2t² z 4 2t²
4 2t² 0 2t² 4 0 2 t 2
ˆ ˆ ˆr(t) x t i y t j z t k r(t) ti t j 4 2t² k
2tˆ ˆ ˆr ' t i j k
4 2t
2t 4t 8 4tr '(t) 1 1 2
4 2t4 2t
+ 24t ( )
( ) ( )
= =
− −
−
= ⇔ = −
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
22 2 2
C 3
dt 8 t dt
t 8 8xz ds 8 2 2 2 2 0
2 2 2
Resp.: 0
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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 zy 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 ( ) ( ) ( ) ( )
( ) ( ) ( )
pi pi
pi
+ ⇔ = + ⇔ = ∴ =
= × × = =
= ⇒ =
 = = × = = pi − = 
∫ ∫ ∫ ∫
∫ ∫
r r r2 2 2 2
2 2 b
C 0 0 a
bb 2 22 2 2
0C 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:
uxzds 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:
[ ]
( )
( )
= = ∈ pi
= + ≤ ≤ pi
= − +
= + = −
= − + ⇔ = − + ∴ = − +
r r r
r
ur
ur ur ur
2 2 2 2 2
x(t) acos t e y(t) bsen t t 0, 2
r(t) acosti bsen t j, 0 t 2
e
ˆ ˆr ' t asent i bcos tj
r '(t) a²sen²t b²cos ²t, mas sen²t 1 cos²t
r '(t) a² 1 cos t b²cos t r '(t) a² a cos t b²cos t r '(t ) (b² a²)cos²t a²
= ∴ = − +
r
ds r '(t) dt ds (b² a²)cos ²t a² dt
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Substituindo na integral dada:
pi
pi
= ⋅ ⋅ − +
= ⋅ ⋅ − +
= − + ⇒ = − ⋅ − = − − ⋅ ⋅
= − ⋅ ⋅ ⋅ ∴ =
− ⋅ ⋅
=
∫ ∫
∫ ∫
∫
2
C 0
2
C 0
C
xyds acos t bsent (b² a²)cos ²t a² dt
xyds ab cos t sent (b² a²)cos²t a² dt
u (b² a²)cos ²t a² du 2(b² a²)cos t ( sent) 2(b² a²) cos t sent
dudu 2(a² b²) cos t sent dt dt
2(a² b²) cos t sent
xyds ab co ⋅s t sent ⋅ ⋅
− ⋅ ⋅
∫ duu 2(a² b²) cos t sent
[ ] pi− +
= =
− −
=
∫ ∫
∫
3
2
1
2 2
0
C
C
(b² a²)cos ²t a²ab abxyds u du |32(a² b²) 2(a² b²)
2
abxyds
2
⋅
−
2
(a² b²)
( ) ( ) ( ) ( ) ( ) ( ){ }
( ) ( ) ( )
pi
    
− + = − pi + − − +     
−
 = − + − − + = ∴ = 
−
∫ ∫
23
2 2 2 2 2 2 2 2 2 22
2 2
0
2 2 2 2 2 2
2 2
C C
abb² a² cos t a b a cos 2 a b a cos 0 a
3 3 a b
abxyds 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 r r2 2r 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 duu 1 4t 8t dt
dt 8t
e
0 t 2 1 u 17
Substituindo :
du 2t3y z ds 2t 1 4t dt 2t u u du
8t 8t
1 u 1 23y z ds 1734 4 3
2
( )
( ) ( )
 
− = −  
− = −∫
3 3
32 2
C
11 17 1
6
13y z ds 17 17 1
6
Resp: ( )−1 17 17 16
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:
( ) ( )= = 3C : x t t; y t t
Assim:
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( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
−
= + ∴ =
= ⇒ = + ∴ = +
= + = − + + +
= + ⇒ = ∴ =
− ≤ ≤ ⇔ ≤ ≤ ≤ ≤ ⇔ ≤ ≤
= −
∫ ∫ ∫ ∫ ∫
∫
r r
r r r
1 2
3
22 2 4
0 1
3 4 3 4
C C C 1 0
4 3
3
3
C
ˆ ˆr 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 duu 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 10 10 10 101 1 1 1 1
2 2 2 2 2
C 10 1 1 1 1
3
10 1 3 32
32 2 2
C 1
du du1 9t dt t 1 9t dt t u t u
36t 36t
1 1 1 1 1yds u du u du u du u du 2 u du
36 36 36 36 36
1 1 u 1 2 1 1yds u du 10 1 10 1318 18 18 3 27 2
2
( )−
−
= − =∫
C
10 10 1
7
10 10 1 10 10 1yds
27 27 27
Resp: 
−10 10 1
27
8. Calcule 
C
y(x z)ds−∫ , onde C é a interseção das superfícies x² + y² + z² = 9 e x + z = 3.
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Solução:
Parametrizando C:
( )
 + + = + + =
⇔ 
+ = = − 
+ + = ⇔ + + − =
+ +
2 2 2 2 2 2
22 2 2 2 2
2 2
x y z 9 x y z 9C : C :
x z 3 z 3 x
Assim :
x y z 9 x y 3 x 9
x y 9 − + =26x x 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 32 x 3x y 0 2 x y 4 x 2y 9
4 2 2 2 2
3 34 x x
2y y2 21 19 99 9
4 2
Assim:
3 3 3x cost e y sent
2 2 2
Mas :
3 3 3 3z x 3 z cost 3 cos t
2 2 2 2
Assim
( )
( )
( ) ( )
( ) ( )
 
= + − + ≤ ≤ pi  
 
= − −  
    
= − + + − ⇔ = + +        
= + = +
r
r
r r
r
22 2
2 2 2
2 2 2 2
:
3 3 3 3 3r t cos t, sent, cos t 0 t 2
2 2 2 22
e
3 3 3r ' t sent, cos t, sent
2 22
Então :
3 3 3 9 9 9r ' t sent cos t sent r ' t sen t cos t sen t
2 2 4 2 42
9 9 9r ' t sen t cos t sen t cos t
2 2 2
( )= = ∴ =r1 9 3 3r ' t
2 2 2
Assim:
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pi   
− = + − + −    
− = ×
∫ ∫
∫
2
C 0
C
3 3 3 3 3 3y(x z)ds sent cos t cos t 3 dt
2 2 2 22 2
3 3 3y(x z)ds sent
22 2
+
3 cost
2
3-
2
−
3 cos t
2
( ) ( ) ( )
pi
pipi pi
 
+  
− = = = − = − pi − = − − =
− =
∫
∫ ∫ ∫
∫
2
0
22 2
0C 0 0
C
3 dt
9 27 27 27 27y(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:
( ) ( ) ( ) ( )
( ) ( )
=
≤ ≤ pi
=
= ⇒ = −
= + = +
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 ( )
( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( )
pi pi
pi
= = ∴ =
+ = + = + = −
 + = pi − − pi − = − − + = × = 
+ =
∫ ∫ ∫
∫
∫
r1
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
10. Calcule 
C
(x y z)ds+ +∫ , onde C é o quadrado de vértices (1,0,1), (1,1,1),(0,1,1) e (0,0,1).
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Solução:
Parametrizando os segmentos de reta que formam os lados do quadrado, temos:
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
= − =
=
=
=
= ⇒ = ∴ = ≤ ≤
+ + = + + = + = + = + =∫ ∫ ∫
suur
r
r r r
1
AB
1
11 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 5x y z ds 1 t 1 dt 2 t dt 2t 2
2 2 2
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
− −
−
= − = −
= −
=
=
= ⇒ = − ∴ = − ≤ ≤
 
+ + = − + + = − = − = − − − =  ∫ ∫ ∫
suur
r
r r r
2
BC
2
00 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 5x y z ds t 1 1 dt 2 t dt 2t 0 2
2 2 2
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( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
− −
−
= − = −
=
= −
=
= ⇒ = − ∴ = − ≤ ≤
 
+ + = − + = − = − = − − − =  ∫ ∫ ∫
suur
r
r r r
3
CD
3
00 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 3x y z ds 0 t 1 dt 1 t dt t 0 1
2 2 2
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
− −
−
= − =
= +
=
=
= ⇒ = ∴ = − ≤ ≤
 
+ + = + + + = + = + = − − + =  ∫ ∫ ∫
suur
r
r r r
4
DA
4
00 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 3x y z ds 1 t 0 1 dt 2 t dt 2t 0 2
2 2 2
Assim:
+ + = + + + + + + + + + + +
+ + +
+ + = + + + = = = ∴ + + =
∫ ∫ ∫ ∫ ∫
∫ ∫
1 2 3 4C 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 x²y e−= de A(0,1) a 1 1B
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 2 2 2t sen t sen t 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
−
= +∫ ∫ :
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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
−
pi
−
= +
= θ ⇒ = θ θ
pi
= ⇒ 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
pi
−
pi
−
pi
−
pi
−
− −
−
= + = + θ θ θ
= + = θ θ θ
= + = θ ⋅ θ θ
= + = θ θ
−
= ⋅ +
− −
= + =
∫ ∫ ∫
∫ ∫ ∫
∫ ∫ ∫
∫ ∫ ∫
∫ ∫
∫ ∫
( )
4
3
0
1 4
2
C 0 0
1 4
2
0C 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
pi
pi
−
pi
−
−
θ θ
= + = θ ⋅ θ + θ θ = + +
= + = θ ⋅ θ + θ + θ
pi pi pi pi       
= + = ⋅ + + −              
∫
∫ ∫ ∫ ∫
∫ ∫
∫ ∫ ( ) ( ) ( )
( ) ( )
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
−
= + = + +∫ ∫
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18. 
2
C
x ds∫ , onde 2 2 23 3 3C : x y a a 0 1º quadrante+ = > .
Solução:
Uma equação vetorial para a hipociclóide 
2 2 2
3 3 3x y a+ = é: ( ) 3 3ˆ ˆr t acos ti asen tj= +r
( )
( ) ( )
( ) ( ) ( )
( ) ( )
3 3
2 2
2 22 2 2 4 2 2 4 2
2 2 2 2 2
ˆ ˆr t acos t i 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 29a cos t sen t 3acos t sent
r ' t 3acos t sent
= ⋅ = ⋅
= ⋅
r
Assim:
( )
( ) ( ) ( ) ( )
( )
0
t 2 22 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
pi
pi pi
= ⋅
= = ⋅
= ⋅ = ⋅
= ⇒ = − ∴ = −
pi
= ⇒ = = ⇒ =
= ⋅
∫ ∫ ∫
∫ ∫ ∫
∫
r
r
2
3 7dt 3a u sent
pi
=∫ dusent−
0 0
3 7
1 1
3a u du
 
= −  ∫ ∫
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( )
00 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 ( ) ( )3 3C : r t 2cos t,2sen t t 0,
2
pi 
= ∈   
r
.
Solução:
( )
( ) ( )
( ) ( ) ( )
( ) ( )
3 3
2 2
2 22 2 4 2 4 2
2 2 2 2
ˆ ˆr t 2cos ti 2sen tj
Mas :
r ' t 6cos t sent,6sen t cos t
Assim:
r ' t 6cos t sent 6sen t cos t 36cos t sen t 36sen t cos t
r ' t 36 cos t sen t cos t sen t
= +
= − ⋅ ⋅
= − ⋅ + ⋅ = ⋅ + ⋅
= ⋅ +
r
r
r
r
( )
1
2 236cos t sen t 6 cos t sent
r ' t 6 cos t sent
= ⋅ = ⋅
= ⋅
r
Assim:
( )
( ) ( ) ( ) ( )
( )
0
t 2 22 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 2cos t 6cos t sent dt
x ds 4cos t 6cos 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
pi
pi pi
pi
= ⋅
= = ⋅
= ⋅ = ⋅
= ⇒ = − ∴ = −
pi
= ⇒ = = ⇒ =
= ⋅ =
∫ ∫ ∫
∫ ∫ ∫
∫ ∫
r
r
724 u sent
du
sent
−
0 0
7
1 1
24 u du
 
= −  ∫ ∫
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( )
00 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 t 01
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
00 2
C 1 C1
1 5 5 1
t dt 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
− −
−
−
    
− =         
 
− = = × = − = − ∴ − = −  
∫ ∫
∫ ∫ ∫
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( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( )
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
11 1 2
0 0 C0
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 : 0 t 11
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
( )
3C
5 3 5 1 5 5
1 x y ds
2 4 2 4 8 8
 
= − = × = ∴ − =   ∫
Assim:
( ) ( ) ( ) ( )
( ) ( )
1 2 3C 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
− = − + − + −
− = − + + = ∴ − =
∫ ∫ ∫ ∫
∫ ∫
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21. 
2
C
y ds∫ , onde C é a semicircunferência da figura abaixo:
Solução:
Parametrizando a semicircunferência, temos:
( ) ( ) ( ) ( )
( ) ( )
=
≤ ≤ pi
=
= ⇒ = −
= + = +
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 ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
pi pi pi pi
pi pi
pi
= = ∴ =
 
= = = = −  
= − = ⋅ +
 = − × = pi − pi − 
∫ ∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
∫
r1
22 2 2
C 0 0 0 0
2
C 0 0
2
0C
4 2 r ' t 2
Substituindo :
1 1y ds 2sent 2dt 2 4sen tdt 8 sen tdt 8 cos 2t dt
2 2
1y ds 4 dt 4 cos 2t dt Mas : cos mx dx sen mx C
m
Assim :
1y ds 4t 4 sen 2t 4 2 sen 2 sen 0
2
= pi∴ = pi∫1 2
C
4 y ds 4
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22. 
2
C
y ds∫ , onde C é o 1º arco da ciclóide:
( ) ( ) ( )ˆ ˆr t 2 t sent i 2 1 cos t j= − + −r .
Solução:
( ) ( ) ( )
( ) ( )
( ) ( )
( )
( ) ( ) ( )
( ) ( )
2 2 2 2
2 2
ˆ ˆr t 2 t sent i 2 1 cos t j
r t 2t 2sent,2 2cos t 0 t 2
Derivando :
r' t 2-2cost,2sent
Mas :
ds r ' t dt
Assim:
r ' t 2 2cos t 2sent 4 8cos t 4cos t 4sen t
r ' t 4 8cos t 4 cos t+sen t
= − + −
= − − ≤ ≤ pi
=
=
= − + = − + +
= − +
r
r
r
r
r
r
( ) ( ) ( )
1
4 8cos t 4 8 8cos 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
22
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 2cos t 2 2 1 cos t dt
y ds 2 2 4 8cos t 4cos 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
pi
pi
pi pi pi
pi pi
= − × −
= − +
= − − − + −
= −
= − − − + −
∫ ∫
∫ ∫
∫ ∫ ∫ ∫
∫ ∫ ∫ 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
pi
pi pi pi pi
pi pi pi
−
= − − − + − − −
= − − − − −
∫
∫ ∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
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( )
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 cos2 cos 2 cos sen
Assim:
1 cos t 1 cos sen sen sen 2sen
Logo :
1 cos t 2 sen
pi pi pi
= − − − − −
= θ ⇒ = θ
= ⇒ θ = pi ⇒ θ = pi
− = − 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
pi pi pi
pi pi pi
pi pi pi
θ
= − − − − −
= θ θ − θ × θ × θ − θ × θ × θ
= θ θ − θ θ θ − θ θ θ
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
Resolvendo ( )
0
64 cos 2 sen d
pi
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
pi pi
pi pi pi
pi pi pi
pi pi pi
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
pi
pi pi pi
θ θ θ = θ θ θ − θ θ
∫
∫ ∫ ∫
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( ) 2
0 0 0
2
0
2 2
0
64 cos 2 sen d 128 cos sen d 64 sen d
Resolvendo 128 cos sen d :
128 cos sen d 128 u sen
pi pi pi
pi
pi
θ θ θ = θ θ θ − θ θ
θ θ θ
θ θ θ = θ
∫ ∫ ∫
∫
∫ dusen− θ
( ) ( )
1 1
2
1 1
11 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
− −
−pi −
pi
pi
 
= −  
= θ → = − θ ∴ θ = −
θ θ
θ = ⇒ = θ = pi ⇒ = −
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 cos0
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
pi pi pi
pi
pi
pi
pi
pi
pi
θ θ θ = θ θ θ − θ θ
θ θ θ = − × − θ
θ θ θ = + × θ = + × pi −
θ θ θ = + × − − = + × − = −
θ θ θ = −
∫ ∫ ∫
∫
∫
∫
∫
20 AFONSO CELSO – FONE: (62) 3092-2268 / CEL: (62) 9216-9668
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 Resolvendo ( )2
0
32 sen 2 sen d
pi
θ θ θ∫ :
( ) ( )
( )
( ) ( )
( )
22
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
pi pi
pi pi
pi pi
pi pi pi
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
pi pi pi
pi
θ = −
θ
θ = ⇒ = θ = pi ⇒ = −
θ θ θ = θ θ θ − θ θ θ
θ θ θ = θ
∫ ∫ ∫
∫ dusen− θ 4128 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
− −
pi − −
−
pi
pi
pi
   
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
pi
  
+ × − − = − =    
θ θ θ =∫
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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 cos0 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
pi pi pi
pi
pi
pi
pi
pi pi
= θ θ − θ θ θ − θ θ θ
θ θ = − θ = − pi − = − × − − =
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
pi
θ θ θ
   
= − − − = + − =      
=
∫
∫
∫
22 AFONSO CELSO – FONE: (62) 3092-2268 / CEL: (62) 9216-9668