Solução_Calculo_Stewart_6e

F.
Then
C xy dx +(x − y) dy = C 1
xy dx +(x − y) dy + C 2
xy dx +(x − y) dy
272 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
5. If we choose x as the parameter, parametric equations for C are x = x, y = √ x for 1 ≤ x ≤ 4 and
x 2 y 3 − √
x dy =
4
x 2 · ( √ x ) 3 − √
1
x
C
1
2 √ x dx = 1 4
2 1 x 3 − 1 dx
= 1 2
1
4 x4 − x 4
= 1
1 2 64 − 4 −
1
+1 = 243
4 8
7. C = C 1 + C 2
On C 1: x = x, y =0 ⇒ dy =0dx, 0 ≤ x ≤ 2.
On C 2: x = x, y =2x − 4 ⇒ dy =2dx, 2 ≤ x ≤ 3.
9. x =2sint, y = t, z = −2cost, 0 ≤ t ≤ π.ThenbyFormula9,
C xyz ds =
π
(2 sin t)(t)(−2cost) dx
2
0 dt + dy
2
dt + dz
2
dt dt
= 2
0 (0 + 0) dx + 3
2 [(2x2 − 4x)+(−x +4)(2)]dx
= 3
2 (2x2 − 6x +8)dx = 17 3
= π
−4t sin t cos t (2 cos t)
0 2 +(1) 2 +(2sint) 2 dt = π
−2t sin 2t 4(cos
0 2 t +sin 2 t)+1dt
= −2 √ 5 π
0 t sin 2tdt= −2 √ 5 − 1 2 t cos 2t + 1 4 sin 2t π
0
= −2 √ 5 − π 2 − 0 = √ 5 π
11. Parametric equations for C are x = t, y =2t, z =3t, 0 ≤ t ≤ 1. Then
C xeyz ds = 1
te(2t)(3t)√ 1
0 2 +2 2 +3 2 dt = √ 14 1
0 te6t2 dt = √
14
13.
C x2 y √ zdz= 1
0 (t3 ) 2 (t) √ t 2 · 2tdt= 1
0 2t9 dt = 1 5 t10 1
0 = 1 5
1
e6t2 1
12
0
integrate by parts with
u = t, dv =sin2tdt
= √ 14
12 (e6 − 1).
15. On C 1 : x =1+t ⇒ dx = dt, y =3t ⇒
dy =3dt, z =1 ⇒ dz =0dt, 0 ≤ t ≤ 1.
On C 2 : x =2 ⇒ dx =0dt, y =3+2t ⇒
dy =2dt, z =1+t ⇒ dz = dt, 0 ≤ t ≤ 1.
Then
(x + yz) dx +2xdy+ xyz dz
C
= C 1
(x + yz) dx +2xdy+ xyz dz + C 2
(x + yz) dx +2xdy+ xyz dz
= 1
(1 + t +(3t)(1)) dt +2(1+t) · 3 dt +(1+t)(3t)(1) · 0 dt
0
+ 1
(2 + (3 + 2t)(1 + t)) · 0 dt +2(2)· 2 dt +(2)(3+2t)(1 + t) dt
0
= 1
0 (10t +7)dt + 1
0 (4t2 +10t + 14) dt = 5t 2 +7t 1
0 + 4
3 t3 +5t 2 +14t 1
0 =12+61 3 = 97 3