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[email protected] CURSOS LIVRES DE 3º GRAU CÁLCULO III INTEGRAIS DE LINHA – EXERCÍCIOS RESOLVIDOS 1. Calcule a integral de linha
∫ ( x + 2y ) ds, onde C é uma semicircunferência centrada na origem de raio igual a 3
C
e orientada no sentido positivo. Solução: A parametrização dessa semicircunferência será dada por:
r r r r(t) = 3cos ti + 3sent j, 0 ≤ t ≤ π ⇒ ds = π
∫ ( 3 cos t + 6sent ) 3dt = 3 ( 3sent − 6cos t )
( −3sent ) π 0
2
+ ( 3cos t ) dt ⇔ ds = 9 dt = 3dt . Substituindo: 2
= 3 × ( 12 ) = 36
0
2. Calcular a integral
∫ ( x² + y² − z ) ds,
C
onde C é a hélice circular dada por :
r r r r r(t) = cos ti + sent j + tk de P(1,0,0) a Q(1,0,2 π) Solução:
ds =
( −sent )
2π
2
+ ( cos t ) ² + 1dt = 2 dt. Assim, podemos escrever:
∫ ( cos ²t + sen²t − t ) 0
2π
2π
t² 2 dt = 2 ∫ ( 1 − t )dt = 2 t − 2 0 0
2π
4 π² 2 ∫ ( 1 − t ) dt = 2 2π − = 2π 2 ( 1 − π ) 2 0 3. Calcule
∫ ( 2x − y + z ) ds , onde C é o segmento de reta que liga A(1, 2, 3) a B(2, 0, 1).
C
Solução: Parametrização do segmento de reta AB:
x(t) = 2 + t uuur r r r suur 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 1 AFONSO CELSO – FONE: (62) 3092-2268 / CEL: (62) 9216-9668
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[email protected] r r r(t) = x ( t ) ˆi + y ( t ) ˆj + z ( t ) kˆ ⇔ r(t) = ( 2 + t ) ˆi − 2tjˆ + ( 1 − 2t ) kˆ Assim : ur r r r r 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
−1
−1
0 ∫ ( 2x − y + z ) ds = ∫ ( 5 + 2t ) 3dt = 3∫ (5 + 2t) dt = 3(5t + t²) | −1
C
∫ ( 2x − y + z ) ds = 0 − 3( −5 + 1) = (−3)(−4) = 12
C
Resp.: 12 4. Calcule
∫ xz ds , onde C é a interseção da esfera x² + y² + z² = 4 com o plano x = y.
C
Solução: Vamos parametrizar a curva dada:
x = y = t ⇒ t² + t² + z² = 4 ⇒ z² = 4 − 2t² ∴ z = 4 − 2t² 4 − 2t² ≥ 0 ⇔ 2t² − 4 ≤ 0 ⇒ − 2 ≤ t ≤ 2 r r r r r(t) = x ( t ) ˆi + y ( t ) ˆj + z ( t) kˆ ⇔ r(t) = ti + t j + r r ' ( t ) = ˆi + ˆj −
2t 4 − 2t 2
r 4 − 2t² k
(
)
kˆ 2
ur 2t 4t2 8 − 4t2 + 4t 2 r '(t) = 12 + 12 + − = 2+ = = 4 − 2t 2 4 − 2t 2 4 − 2t 2 e f ( x, y,z ) = xz ⇔ f ( t ) = t 4 − 2t²
8 8 = 2 4 − 2t 4 − 2t 2
( 1)
(2)
Substituindo (1) e (2) na integral dada:
∫ xz ds =
C
2
∫
t 4 − 2t² ⋅
− 2
∫ xz ds = 8 ×
C
t2 2
2
= − 3
8 4 − 2t 2 8 × 2
2
dt = 8
∫
t dt
− 2
( 2) − ( − 2) 2
2
= 8 × 2−2 =0 ( ) 2
Resp.: 0
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[email protected] Outra Solução:
C : x 2 + y 2 + z2 = 4 Assim :
x=y
y 2 + y 2 + z 2 = 4 ⇔ 2y 2 + z2 = 4 ∴
y 2 z2 + =1 2 4
Parametrizando: x ( t ) = 2 cos t
y ( t ) = 2 cos t
z( t) = 2sent
Assim : r r r ( t ) = 2 cos t, 2 cos t, 2sent ⇒ r '( t) = − 2sent, − 2sent, 2cos t
(
)
e r r '( t) =
(−
2sent
) +( − 2
(
2sent
)
2
)
r + ( 2cos t) ⇔ r '( t) = 2sen2 t + 2sen2 t + 4 cos2 t
r r r r r ' ( t ) = 4sen2 t + 4cos2 t ⇔ r '( t) = 4( sen2 t + cos2 t) ⇔ r '( t) = 4 ∴ r '( t) = 2 Substituindo : 2π
∫ xzds =
∫
C
2π
b
0
a
2 cos t × 2sent × 2dt = 4 2 ∫ sent cos tdt = 4 2∫ udu
0
Onde : u = sent ⇒ du = cos tdt Assim: b
∫ xzds = 4 2 ∫ udu = 4 2 ×
C
a
u2 2
b
= 2 2 ( sent)
a
2 2π 0
= 2 2 sen2 ( 2π) − sen2 ( 0) = 0
Resp: 0 5. Calcule
∫ xyds , onde C é a elipse
C
x² y² + = 1. a² b²
Solução: A parametrização da elipse é dada por:
x(t) = a cos t e y(t) = bsen t t ∈ [ 0, 2π ] r r r r(t) = acos ti + bsen t j, 0 ≤ t ≤ 2 π e r r ' ( t ) = −asentiˆ + bcos tjˆ ur r '(t) = a²sen²t + b² cos ²t, mas sen²t = 1 − cos ²t ur ur ur r '(t) = a² ( 1 − cos2 t ) + b² cos2 t ⇔ r '(t) = a² − a 2 cos2 t + b² cos 2 t ∴ r '(t ) = (b² − a²)cos ²t + a²
r ds = r '(t) dt ∴ ds = (b² − a²)cos ²t + a² dt
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[email protected] Substituindo na integral dada: 2π
∫ xyds =
∫ a cos t ⋅ bsent ⋅
C
(b² − a²)cos ²t + a² dt
0
2π
∫ xyds = ab ∫ cos t ⋅ sent ⋅
C
(b² − a²)cos ²t + a² dt
0
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 du ∫C xyds = ab∫ cos t ⋅ sent ⋅ u ⋅ 2(a² − b²) ⋅ cos t ⋅ sent 1 [ (b² − a²)cos ²t + a²] |2π ab ab 0 ∫C xyds = 2(a² − b²) ∫ u2 du = 2(a² − b²) 3 2 3 2
3 ab 2 ⋅ ( b² − a² ) cos2 t + a2 2 2 (a² − b²) 3
∫ xyds =
C
2π
= 0
{
ab ( b2 − a2 + a2 ) − ( b2 − a2 + a2 ) = 0 ∴ xyds = 0 ∫ 2 − b2 ) C
∫ xyds = 3 ( a
C
Resp.:0 6.
∫ ( 3y −
C
)
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:
x ( t ) = 1 C = y ( t ) = t 2 z ( t ) = t
0≤t≤2
Assim:
r r r r ( t ) = 1,t,t 2 ⇒ r ' ( t ) = ( 0,1,2t ) ∴ r ' ( t ) = 1 + 4t2
(
}
ab ( b2 − a2 ) cos2 ( 2π) + a2 − ( b2 − a2 ) cos2 ( 0) + a2 3 ( a2 − b2 )
)
Assim:
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∫ ( 3y −
2
)
(
z ds = ∫ 3t − t
C
0
2
)
2
2
1 + 4t dt = ∫ ( 3t − t ) 1 + 4t dt = ∫ 2t 1 + 4t2 dt 2
2
0
0
Fazendo : u = 1 + 4t 2 ⇒
du du = 8t ∴ dt = dt 8t
e 0 ≤ t ≤ 2 ⇔ 1 ≤ u ≤ 17 Substituindo :
∫(
C
∫(
C
∫(
C
)
0
3 1 u2 3y − z ds = 4 3 2 1 3y − z ds = 17 6
) )
(
(
17
17
1 du 2t 2 2t u = u 8t 8t ∫ du ∫1 1
17
1 2 32 32 1 = × × 17 − 1 = 4 3 6 1
(
)
173 − 1
)
17 − 1
)
1 17 17 − 1 6
Resp:
7.
2
3y − z ds = ∫ 2t 1 + 4t 2 dt =
∫ y ds , onde C é a curva dada por y = x³ de (-1,-1) a (1, 1).
C
Solução: Sabemos que:
y, y = − y,
se y ≥ 0 ⇔ −1 ≤ y ≤ 0 se y < 0 ⇔ 0 < y < 1
Parmetrizando C:
C:
x ( t ) = t; y ( t ) = t 3
Assim:
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[email protected] r r r ( t ) = x ( t ) ˆi + y ( t ) ˆj ∴ r ( t ) = t,t 3
(
)
Assim : r r r ' ( t ) = 1,3t 2 ⇒ r ' ( t ) = 1 + 3t2
(
)
( )
2
r ∴ r ' ( t ) = 1 + 9t4
Assim :
∫
∫ -yds +
y ds =
C
C1
∫
C2
0
1
−1
0
yds = ∫ −t 3 1 + 9t 4 dt + ∫ t3 1 + 9t4 dt
Fazendo : du du = 36t3 ∴ dt = dt 36t 3 Se − 1 ≤ t ≤ 0 ⇔ 10 ≤ u ≤ 1e 0 ≤ t ≤ 1 ⇔ 1 ≤ u ≤ 10 Substituindo :
u = 1 + 9t 4 ⇒
0
∫ y ds = ∫ −t
C
−1
∫
y ds = −
C
∫
C
1 + 9t dt + ∫ t 4
1 3
1 + 9t dt = 4
0
1
C
∫
1
3
10
∫ −t
3
10 10
10
du du u + ∫ t3 u 3 3 36t 1 36t 10
3 2
1 1 u 1 2 32 32 1 y ds = u du = × = × × 10 − 1 = 18 ∫1 18 3 18 3 27 2 10 10 1 10 10 − 1 y ds = − = 27 27 27
Resp:
10
1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 u du + u du = u du + u du = 2 × u du 36 10∫ 36 1∫ 36 1∫ 36 1∫ 36 1∫ 10
1 2
(
)
103 − 1 =
(
)
1 10 10 − 1 27
10 10 − 1 27
8. Calcule
∫ y(x − z)ds , onde C é a interseção das superfícies x² + y² + z² = 9 e x + z = 3.
C
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[email protected] Solução: Parametrizando C:
x2 + y 2 + z2 = 9 x 2 + y 2 + z 2 = 9 C: ⇔C: x + z = 3 z = 3 − x Assim : x2 + y 2 + z2 = 9 ⇔ x 2 + y 2 + ( 3 − x ) = 9 2
x 2 + y 2 + 9 − 6x + x 2 = 9 ⇔ 2x 2 − 6x + y 2 = 0 Comple tando o quadrado : 2
2
9 9 3 9 3 2 x 2 − 3x + − + y 2 = 0 ⇔ 2 x − + y 2 = ⇔ 4 x − + 2y 2 = 9 4 2 2 2 2 2
3 4 x − 2y 2 2 + = 1⇔ 9 9
2
3 x − 2 2 + y =1 9 9 4 2
Assim: 3 3 + cos t 2 2 Mas :
y=
3
x=
e
z = x −3 ⇔ z =
3 3 3 3 + cos t − 3 = − + cos t 2 2 2 2
2
sent
Assim : r 3 3 3 3 3 r ( t ) = + cos t, sent, − + cos t 2 2 2 2 2 e
0 ≤ t ≤ 2π
r 3 3 3 r ' ( t ) = − sent, cos t, − sent 2 2 2 Então : 2
2
2
r r 3 3 3 r ' ( t ) = − sent + cos t + − sent ⇔ r ' ( t ) = 2 2 2 r 9 9 9 r '( t) = sen 2t + cos 2 t = sen 2t + cos 2 t 2 2 2
(
)
1
=
9 9 9 sen 2t + cos 2 t + sen 2t 4 2 4
r 9 3 3 = ∴ r '( t) = 2 2 2
Assim:
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[email protected] 2π
∫ y(x − z)ds = ∫
C
3 3 3 3 3 sent + cos t − + cos t − 3 dt 2 2 2 2 2 2
3
0
3 3 × 2 2
∫ y(x − z)ds =
C
∫ y(x − z)ds =
C
2π
2π
3
3
3
3
∫ sent 2 + 2 cost - 2 − 2 cos t + 3 dt 0
2π
9 27 27 3sentdt = sentdt = ( − cos t ) ∫ ∫ 20 2 0 2
2π
=− 0
27 27 ( cos 2π − cos0 ) = − ( 1 − 1) = 0 2 2
Assim :
∫ y(x − z)ds = 0
C
Resp: 0 9. Calcule
∫ (x + y)ds , onde C é a interseção das superfícies z = x² + y² e z = 4.
C
Solução: A curva C é a circunferência x² + y² = 4, cuja parametrização é dada por:
x = 2cos t C: 0 ≤ t ≤ 2π y = 2sent Assim : r r r ( t ) = ( 2cos t, 2sent ) ⇒ r ' ( t ) = ( −2sent, 2cos t ) e r r ' ( t ) = 4sen2 t + 4 cos2 t = 4 sen2 t + cos2 t
(
)
1
r = 4 = 2∴ r '( t) = 2
Substituindo :
∫ (x + y)ds =
C
2π
2π
0
0
2π
∫ ( 2cos t + 2sent ) 2dt = 4 ∫ ( cos t + sent ) dt = 4 ( sent − cos t ) 0
∫ (x + y)ds = 4 sen ( 2π ) − sen ( 0 ) − ( cos ( 2π ) − cos ( 0 ) ) = 4 ( 0 − 0 − 1 + 1) = 4 × 0 = 0
C
Logo :
∫ (x + y)ds = 0
C
10. Calcule
∫ (x + y + z)ds , onde C é o quadrado de vértices (1,0,1), (1,1,1),(0,1,1) e (0,0,1).
C
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[email protected] Solução: Parametrizando os segmentos de reta que formam os lados do quadrado, temos:
A(1,0,1), B(1,1,1), C(0,1,1) e D(0,0,1) suur Reta AB : r uAB = B − A = ( 0,1,0 ) Assim : x = 1 C1 : y = t z = 1 r r r r ( t ) = ( 1,t,1) ⇒ r ' ( t ) = ( 0,1,0 ) ∴ r ' ( t ) = 1 0 ≤ t ≤ 1 Assim : 1
1
t2 x + y + z ds = 1 + t + 1 dt = 2 + t dt = 2t + ( ) ( ) ( ) ∫ ∫ ∫ 2 C1 0 0
1
= 2+ 0
1 5 = 2 2
A(1,0,1), B(1,1,1), C(0,1,1) e D(0,0,1) suur Reta BC : r uBC = C − B = ( −1,0,0 ) Assim : x = −t C2 : y = 1 z = 1 r r r r ( t ) = ( -t,1,1) ⇒ r ' ( t ) = ( −1,0,0 ) ∴ r ' ( t ) = 1 − 1 ≤ t ≤ 0 Assim : 0
0
t2 x + y + z ds = − t + 1 + 1 dt = 2 − t dt = 2t − ( ) ( ) ( ) ∫ ∫ ∫ 2 C2 −1 −1
0
1 5 = 0 − −2 − = 2 2 −1
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[email protected] A(1,0,1), B(1,1,1), C(0,1,1) e D(0,0,1) suur Reta CD : r uCD = D − C = ( 0, −1,0 ) Assim : x = 0 C3 : y = − t z = 1 r r r r ( t ) = ( 0,-t,1) ⇒ r ' ( t ) = ( 0, −1,0 ) ∴ r ' ( t ) = 1 − 1 ≤ t ≤ 0 Assim : 0
0
t2 x + y + z ds = 0 − t + 1 dt = 1 − t dt = t − ( ) ( ) ( ) ∫ ∫ ∫ 2 C3 −1 −1
0
1 3 = 0 − −1 − = 2 2 −1
A(1,0,1), B(1,1,1), C(0,1,1) e D(0,0,1) suur Reta DA : r uDA = A − D = ( 1,0,0 ) Assim : x = 1 + t C4 : y = 0 z = 1 r r r r ( t ) = ( 1+t,0,1) ⇒ r ' ( t ) = ( 1,0,0 ) ∴ r ' ( t ) = 1 − 1 ≤ t ≤ 0 Assim : 0
0
t2 ∫ ( x + y + z ) ds = −∫1 ( 1 + t + 0 + 1) dt = −∫1 ( 2 + t ) dt = 2t + 2 C4
0
1 3 = 0 − −2 + = 2 2 −1
Assim:
∫ (x + y + z)ds = ∫ (x + y + z)ds + ∫ (x + y + z)ds + ∫ (x + y + z)ds + ∫ (x + y + z)ds
C
C1
5
C2
5
3
3
∫ (x + y + z)ds = 2 + 2 + 2 + 2 =
C
C3
C4
5 + 5 + 3 + 3 16 = = 8 ∴ ∫ (x + y + z)ds = 8 2 2 C
Resp: 8
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[email protected] 11. Calcular a integral
∫ xyds, onde C é a interseção das superfícies x² + y² = 4 e y + z = 8.
C
∫ 3xyds , sendo C o triângulo de vértices A(0,0), B(1,0) e C(1,2), no sentido anti-horário.
12. Calcular
C
13. Calcule
∫ y(x − z)ds , onde C é a interseção das superfícies x² + y² + z² = 9 e x + z = 3.
C
14. Calcule
∫ (x + y)ds , onde C é a interseção das superfícies z = x² + y² e z = 4.
C
15. Calcule
∫ ( x² + y² − z ) ds , onde C é a interseção das superfícies x² + y² + z² = 8z e z = 4. c
16. Calcule
∫ xy²(1 − 2x²)ds , onde C é a parte da curva de Gauss y = e
C
17.
∫ ds , onde C : rr ( t ) = ( t cos t, tsent )
−C
−x²
de A(0,1) a
1 1 B − . e 2
t ∈ 0,1 .
Solução: t
r
∫ ds = ∫ ds = ∫ r ' ( t ) dt ( 1)
−C
C
t0
Assim : r r ' ( t ) = ( cos t − tsent,sent + t cos t ) r 2 2 r ' ( t ) = ( cos t − tsent ) + ( sent + t cos t ) r r ' ( t ) = cos2 t − 2t cos tsent + t2sen2 t + sen2 t + 2tsent cos t + t2 cos2 t r r ' ( t ) = 1 + t2 sen2 t + cos2 t r r ' ( t ) = 1 + t2
(
)
Substituindo em ( 1) : t
r
∫ ds = ∫ ds = ∫ r ' ( t ) dt
−C
C 1
∫ ds = ∫
−C
t0
1 + t2 dt
0
Resolvendo
∫
−C
1
ds =
∫
1 + t2 dt :
0
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∫
1
ds =
−C
∫
1 + t 2 dt
0
Mas : t = tgθ ⇒ de = sec 2 θdθ Se t = 0 ⇒ θ = 0
Se t = 1 ⇒ θ =
π 4
Assim : 1
∫ ds = ∫
−C
1 + t 2 dt =
0
π 4
∫
1 + tg 2θ sec 2 θdθ
Mas :1 + tg 2θ = sec 2θ
0
Substituindo: 1
∫ ds = ∫
−C
1 + t2 dt =
0
−C
∫ ds = ∫
−C
∫
2
1 + t dt =
0
1
0
1
ds =
−C
∫ 0
∫
1 + tg2θ sec 2 θdθ
0
1
∫ ds = ∫
π 4
π 4
∫
sec 2 θ sec 2 θdθ
0
π 4
1 + t2 dt = ∫ sec θ ⋅ sec 2 θdθ 0
π 4
1 + t2 dt = ∫ sec 3 θ dθ 0
Utilizando :
∫ sec
n
udu =
1 n−2 sec n− 2 u ⋅ tgu + sec n− 2udu n −1 n−1 ∫
Assim : 1
∫ ds = ∫
−C
0
1
∫ ds = ∫
−C
0
π 4
1 + t dt = ∫ sec3 θ dθ 2
0
π
1 14 1 + t dt = sec θ ⋅ tgθ + ∫ sec θdθ 2 2 0 2
Mas : ∫ sec ud = ln sec u + tgu + c
Substituindo : 1
∫ ds = ∫
−C
0
1
∫ ds = ∫
−C
∫
0
1
ds =
−C
∫
1 1 1 + t dt = sec θ ⋅ tgθ + ln sec θ + tgθ 2 2 2
π 4 0
1 1 π π 1 π π 1 1 + t2 dt = sec ⋅ tg + ln sec + tg − sec ( 0 ) ⋅ tg ( 0 ) + ln sec ( 0 ) + tg ( 0 ) 2 4 4 2 4 4 2 2 1 + t2 dt =
0
1 1 1 × 2 × 1 + ln 2 + 1 − sec ( 0 ) ⋅ tg ( 0 ) 2 2 2
0
+
1 ln 1 + 0 2
0
Logo :
∫
−C
1
ds =
∫ 0
1 + t2 dt =
2 1 + ln 2 + 1 2 2
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[email protected] 18.
∫ x ds , onde C : x 2
C
2 3
2
2
+ y 3 = a3
1º quadrante .
a>0
Solução: Uma equação vetorial para a hipociclóide
2
2
2
x 3 + y 3 = a3
é:
r ˆ r ( t ) = a cos3 tˆi + asen3 tj
r ˆ r ( t ) = a cos3 tˆi + asen3 tj Mas : r r ' ( t ) = −3acos2 t ⋅ sent,3asen2 t ⋅ cos t
(
Assim : r r ' ( t) =
( −3acos t ⋅ sent ) 2
2
)
(
+ 3asen2 t ⋅ cos t
r r ' ( t ) = 9a2 cos2 t ⋅ sen2 t cos2 t + sen2 t r r ' ( t ) = 3a cos t ⋅ sent
(
)
1
)
2
= 9a2 cos4 t ⋅ sen2 t + 9a2 sen4 t ⋅ cos2 t
= 9a2 cos2 t ⋅ sen2 t = 3acos t ⋅ sent
Assim:
r r ' ( t ) = 3a cos t ⋅ sent t
r 2 x ds = f t r ' ( t ) dt = ( ) ∫ ∫
C
t0 π 2
∫ x ds = ∫ a 2
C
2
0
π 2
∫ ( a cos t ) ( 3a cos t ⋅ sent ) dt 2
3
0
π 2
cos6 t ( 3a cos t ⋅ sent ) dt = 3a3 ∫ cos7 t ⋅ sentdt 0
Fazendo : du du = −sent ∴ dt = − dt sent π Se t = 0 ⇒ u = 1 Se t = ⇒ u = 0 2 Substituindo :
u = cos t ⇒
π 2
0 du 3 7 ∫C x ds = 3a 0∫ cos t ⋅ sentdt = 3a ∫1 u sent − sent = −3a 1∫ u du 2
3
7
3
0
7
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0
8 u8 18 1 3a3 3 0 3 ∫ x ds = −3a 1∫ u du = −3a 8 = −3a 8 − 8 = −3a × − 8 = 8 C 1 Logo : 2
3
2 ∫ x ds =
C
19.
7
(
3
)
3a3 8
∫ x ds , onde C : rr ( t ) = ( 2 cos 2
3
C
t,2sen3 t
)
π t ∈ 0, . 2
Solução:
r ˆ r ( t ) = 2 cos3 tˆi + 2sen3 tj Mas : r r ' ( t ) = −6cos2 t ⋅ sent, 6sen2 t ⋅ cos t
(
Assim : r r ' ( t) =
( −6cos t ⋅ sent) 2
2
)
(
+ 6sen2 t ⋅ cos t
r r ' ( t ) = 36 cos2 t ⋅ sen2 t cos2 t + sen2 t r r ' ( t ) = 6 cos t ⋅ sent
(
)
1
)
2
= 36 cos4 t ⋅ sen2 t + 36sen4 t ⋅ cos2 t
= 36 cos2 t ⋅ sen2 t = 6 cos t ⋅ sent
Assim:
r r ' ( t ) = 6 cos t ⋅ sent t
r ∫ x ds = ∫ f ( t ) r ' ( t ) dt = 2
C
t0 π 2
π 2
∫ ( 2 cos t ) ( 6 cos t ⋅ sent ) dt 3
2
0
π 2
∫ x ds = ∫ 4 cos t ( 6 cos t ⋅ sent ) dt = 24 ∫ cos 2
C
6
0
7
t ⋅ sentdt
0
Fazendo : du du = −sent ∴ dt = − dt sent π Se t = 0 ⇒ u = 1 Se t = ⇒ u = 0 2 Substituindo :
u = cos t ⇒
π 2
0 0 du 2 7 7 x ds = 24 cos t ⋅ sentdt = 24 u sent − = − 24 u7du ∫C ∫0 ∫1 ∫ sent 1
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0
u8 08 18 1 24 ∫C x ds = −241∫ u du = −24 8 = −24 8 − 8 = ( −24) × − 8 = 8 = 3 1 Logo : 2
7
∫ x ds = 3 2
C
20.
∫ ( x − y ) ds , onde C é o triângulo da figura abaixo:
C
Solução: Parametrizando os segmentos de reta
AB, BC e CA .
3 A 1, ; B ( 2,2 ) e C ( 2,1) 2 x = 2 + t AB ⇔ C1 : −1 ≤ t ≤ 0 1 y = 2 + 2 t Assim : r 1 1 r ( t ) = 2 + t, 2 + t ⇒ r ' ( t ) = 1, 2 2 e r ' ( t) = 1 +
1 = 4
5 5 ∴ r ' ( t) = 4 2
Assim : 0
1 5 5 dt = 2 2
∫ ( x − y ) ds = ∫ 2 + t − 2 − 2 t −1
C1
∫ ( x − y ) ds =
C1
5 4
0
∫ tdt =
−1
2
5 t × 4 2
0
= −1
0
1
∫ 2 t dt
−1
5 0 1 5 − =− ∴ 8 2 2 8
∫ ( x − y ) ds = −
C1
5 8
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[email protected] 3 A 1, ; B ( 2,2 ) e C ( 2,1) 2 x = 2 BC ⇔ C2 : 0 ≤ t ≤1 y = 2 − t Assim : r r ( t ) = ( 2, 2 − t ) ⇒ r ' ( t ) = ( 0, − 1) e r ' ( t) = 0 + 1 = 1 ∴ r ' ( t) = 1 Assim :
∫
( x − y ) ds =
C2
1
∫( 0
1
)
2 − 2 + t ( 1) dt = ∫ t dt = 0
t2 2
1
= 0
1 ∴ 2
1
∫ ( x − y ) ds = 2
C2
3 A 1, ; B ( 2,2) e C ( 2,1) 2 x = 2 − t CA ⇔ C3 : 0 ≤ t ≤1 1 y = 1 + 2 t Assim : r 1 1 r ( t ) = 2 − t, 1 + t ⇒ r ' ( t ) = −1, 2 2 e r ' ( t) = 1 +
1 = 4
5 5 ∴ r ' ( t) = 4 2
Assim : 1
1 5 5 3 dt = ∫0 1 − 2 t dt 2 2
1
∫ ( x − y ) ds = ∫ 2 − t − 1 − 2 t
C3
0
1
∫ ( x − y ) ds =
C3
5 3 t2 5 3 5 1 5 5 1− = × = ∴ ∫ ( x − y ) ds = t − × = 2 2 2 0 2 4 2 4 8 8 C3
Assim:
∫ ( x − y ) ds = ∫ ( x − y ) ds + ∫ ( x − y ) ds + ∫ ( x − y ) ds
C
C1
∫ ( x − y ) ds = −
C
C2
C3
5 1 5 1 1 + + = ∴ ∫ ( x − y ) ds = 8 2 8 2 C 2
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∫ y ds , onde C é a semicircunferência da figura abaixo: 2
21.
C
Solução: Parametrizando a semicircunferência, temos:
x = 2cos t C: 0 ≤ t ≤ 2π y = 2sent Assim : r r r ( t ) = ( 2cos t, 2sent ) ⇒ r ' ( t ) = ( −2sent, 2cos t ) e r r ' ( t ) = 4sen2 t + 4 cos2 t = 4 sen2 t + cos2 t
(
)
1
r = 4 = 2∴ r '( t) = 2
Substituindo : π
π
π
π
2 1 1 2 2 2 ∫C y ds = ∫0 ( 2sent ) 2dt = 20∫ 4sen tdt = 80∫ sen tdt = 80∫ 2 − 2 cos ( 2t ) dt
π
π
0
0
2 ∫ y ds = 4∫ dt − 4∫ cos ( 2t ) dt
C
Mas : ∫ cos ( mx ) dx =
1 ⋅ sen ( mx ) + C m
Assim : π
1 1 y ds = 4t − 4 × sen 2t = 4 π − 2 sen 2 π − sen 0 = 4π∴ ∫ y 2ds = 4π ( ) ( ) ( ) ∫C 2 0 C 2
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∫ y ds , onde C é o 1º arco da ciclóide: 2
C
r r ( t ) = 2 ( t − sent ) ˆi + 2 ( 1 − cos t ) ˆj . Solução:
r r ( t ) = 2 ( t − sent ) ˆi + 2 ( 1 − cos t ) ˆj r r ( t ) = ( 2t − 2sent,2 − 2 cos t ) 0 ≤ t ≤ 2π Derivando : r r' ( t ) = ( 2-2cost,2sent ) Mas :
r ds = r ' ( t ) dt Assim : r r ' ( t) =
( 2 − 2 cos t )
r r ' ( t) =
4 − 8 cos t + 4 cos2 t+sen2 t
2
+ ( 2sent )
2
4 − 8 cos t + 4 cos2 t + 4sen2 t
=
(
)
1
=
4 − 8 cos t + 4 = 8 − 8 cos t
Assim : r r r ' ( t ) = 8 ( 1 − cos t ) = 8 1 − cos t ∴ r ' ( t ) = 2 2 1 − cos t Substituindo na integral: 2π
∫ y ds = ∫ ( 2 − 2 cos t) 2
C
2
× 2 2 1 − cos t dt
0
2π
(
)
2 2 ∫ y ds = 2 2 ∫ 4 − 8 cos t + 4 cos t 1-cost dt
C
0
2π
2π
2π
0
0
2π
2π
0
0
2π
2π
2π
0
0
0
∫ y ds = 8 2 ∫ 2
C
1 − cos t dt − 16 2 ∫ cos t 1 − cos t dt + 8 2 ∫ cos2 t 1 − cos t dt
0
Mas : cos2 t = 1 − sen2 t Assim : 2π
∫ y ds = 8 2 ∫ 2
C
(
)
1 − cos t dt − 16 2 ∫ cos t 1 − cos t dt + 8 2 ∫ 1 − sen2 t
0
1 − cos t dt 2π
2 2 ∫ 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
C
∫ y ds = 16 2
C
0
2π
2π
2π
0
0
0
2 ∫ 1 − cos t dt − 16 2 ∫ cos t 1 − cos t dt − 8 2 ∫ sen t 1 − cos t dt 2
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2π
2π
0
0
0
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
C
Fazendo : t = 2θ ⇒ dt = 2dθ e se t = 0 ⇒ θ = 0 e se t=2π ⇒ θ = π e mais : cos ( 2θ ) = cos2 θ − sen2 θ
1 − cos t = 1 − cos 2θ Assim :
1 − cos t = 1 − cos2 θ + sen2 θ =
sen2 θ + sen2 θ = 2sen2 θ
Logo : 1 − cos t = 2 senθ Substituindo : 2π
2π
2π
0
0
0
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
C
π
π
0
0
2 ∫ y ds = 16 2 ∫ 2 senθ ( 2dθ) − 16 2∫ cos ( 2θ) ×
C
π
π
π
C
0
0
0
Resolvendo
64∫ cos ( 2θ ) senθ dθ :
(
)
π
2 senθ × ( 2dθ ) − 8 2∫ sen2 ( 2θ ) × 0
(
2 2 ∫ y ds = 64∫ senθdθ − 64∫ cos ( 2θ) senθ dθ − 32∫ sen ( 2θ) senθ dθ
π
0
π
π
0
0
π
π
π
0
0
0
π
π
π
0
0
0
π
π
π
0
0
(
)
64∫ cos ( 2θ ) senθ dθ = 64∫ cos2 θ − sen2 θ senθ dθ 64∫ cos ( 2θ ) senθ dθ = 64∫ cos2 θsenθdθ − 64∫ sen2 θsenθ dθ
(
)
64∫ cos ( 2θ ) senθ dθ = 64∫ cos2 θsenθdθ − 64∫ 1 − cos2 θ senθ dθ π
64∫ cos ( 2θ ) senθ dθ = 64∫ cos2 θsenθdθ − 64∫ senθdθ+64∫ cos2 θsenθ dθ 0
0
π
π
π
0
0
0
)
2 senθ × ( 2dθ)
64∫ cos ( 2θ ) senθ dθ = 128∫ cos2 θsenθdθ − 64∫ senθdθ
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π
π
64∫ cos ( 2θ ) senθ dθ = 128∫ cos θsenθdθ − 64∫ senθdθ 0
2
0
0
π
Re solvendo 128∫ cos2 θsenθdθ : 0
π
−1 du 2 128∫ cos θsenθdθ = 128 ∫ u senθ − = −128 ∫ u du senθ 0 1 1 Onde : du du u = cos θ → = −senθ ∴ dθ = − dθ senθ e
se θ = 0 ⇒ u = 1 Logo : π
−1
2
2
se θ = π ⇒ u = −1
e
−1
−1
u3 128∫ cos θsenθdθ = −128 ∫ u du = −128 × 31 0 1 2
2
( −1) 3 ( 1) 3 128 128 256 = 128∫ cos θsenθdθ = −128 × − + = 3 3 3 3 3 0 Assim : π
2
π
128∫ cos2 θsenθdθ = 0
256 3
Substituindo: π
π
π
0
0
0
64∫ cos ( 2θ ) senθ dθ = 128∫ cos2 θsenθdθ − 64∫ senθdθ π
64∫ cos ( 2θ ) senθ dθ = 0 π
64∫ cos ( 2θ ) senθ dθ = 0 π
64∫ cos ( 2θ ) senθ dθ = 0 π
π 256 − 64 × ( − cos θ ) 0 3 π 256 256 + 64 × ( cos θ ) 0 = + 64 × ( cos π − cos 0 ) 3 3
256 256 256 + 64 × ( −1 − 1) = + 64 × ( −2 ) = − 128 3 3 3
64∫ cos ( 2θ ) senθ dθ = − 0
128 3
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Resolvendo
32∫ sen2 ( 2θ ) senθ dθ : 0
π
π
0
0
32∫ sen2 ( 2θ ) senθ dθ = 32∫ ( 2senθ cos θ ) senθ dθ π
2
π
32∫ sen ( 2θ ) senθ dθ = 128∫ sen2 θ cos2 θ senθ dθ 2
0
0
π
π
0
0
(
)
32∫ sen2 ( 2θ ) senθ dθ = 128∫ 1 − cos2 θ cos2 θ senθ dθ π
π
π
32∫ sen ( 2θ ) senθ dθ = 128∫ cos θsenθdθ − 128∫ cos4 θsenθdθ 2
0
2
0
0
Fazendo : u = cos θ →
du du = −senθ ∴ dθ = − dθ senθ
e se θ = 0 ⇒ u = 1 Assim : π
e
se θ = π ⇒ u = −1 π
π
32∫ sen ( 2θ ) senθ dθ = 128∫ cos θsenθdθ − 128∫ cos4 θsenθdθ 2
2
0
0
π
−1
0
−1 du 32∫ sen2 ( 2θ ) senθ dθ = 128 ∫ u2 senθ − − 128 u4 senθ ∫ senθ 0 1 1 π
−1
du − senθ
−1
32∫ sen ( 2θ ) senθ dθ = −128 ∫ u du + 128 ∫ u4 du 2
0 π
2
1
1
-1
−1
u3 u5 32∫ sen ( 2θ ) senθ dθ = −128 × + 128 × 3 1 5 1 0 2
( −1) 3 ( 1) 3 ( −1) 5 ( 1) 5 + 128 × 32∫ sen ( 2θ ) senθ dθ = −128 × − − 3 5 5 3 0 π
2
π
1 1 1 1 256 256 512 32∫ sen2 ( 2θ ) senθ dθ = −128 × − − + 128 × − − = − = 3 5 15 3 3 5 5 0 Assim : π
32∫ sen2 ( 2θ ) senθ dθ = 0
512 15
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π
π
0
0
0
2 2 ∫ y ds = 64∫ senθdθ − 64∫ cos ( 2θ ) senθ dθ − 32∫ sen ( 2θ) senθ dθ
C
Onde : π
64∫ senθdθ = 64 ( −cosθ ) 0 = −64 ( cos π − cos 0 ) = ( −64 ) × ( −1 − 1) = 128 π
0 π
64∫ cos ( 2θ ) senθ dθ = − 0 π
32∫ sen2 ( 2θ ) senθ dθ = 0
128 3
512 15
Substituindo : π
π
π
0
0
0
2 2 ∫ y ds = 64∫ senθdθ − 64∫ cos ( 2θ ) senθ dθ − 32∫ sen ( 2θ ) senθ dθ
C
128 512 2048 128 512 − = 128 + − = 3 15 3 15 15
∫ y ds = 128 − − 2
C
Logo :
∫ y ds = 2
C
2048 15
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