Solução_Calculo_Stewart_6e

F.
280 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
15. Here C = C 1 + C 2 where
C 1 can be parametrized as x = t, y =1, −1 ≤ t ≤ 1,and
C 2 is given by x = −t, y =2− t 2 , −1 ≤ t ≤ 1.
Thenthelineintegralis
C 1 +C 2
y 2 e x dx + x 2 e y dy = 1
−1 [1 · et + t 2 e · 0] dt
+ 1
−1 [(2 − t2 ) 2 e −t (−1) + (−t) 2 e 2−t2 (−2t)] dt
= 1
−1 [et − (2 − t 2 ) 2 e −t − 2t 3 e 2−t2 ] dt = −8e +48e −1
according to a CAS. The double integral is
∂Q
∂x − ∂P 1
2−x
2
dA = (2xe y − 2ye x ) dy dx = −8e +48e −1 , verifying Green’s Theorem in this case.
∂y
D
−1
1
17. By Green’s Theorem, W = C F · dr = C x(x + y) dx + xy2 dy = D (y2 − x) dy dx where C is the path described in the
question and D is the triangle bounded by C. So
W = 1 1−x
(y 2 − x) dy dx = 1
1
0 0 0 3 y3 − xy y =1−x
dx = 1
1 (1 − y =0 0 3 x)3 − x(1 − x) dx
= − 1 (1 − 12 x)4 − 1 2 x2 + 1 x3 1
=
3
− 1 + 1
0 2 3 − −
1
12 = −
1
12
19. Let C 1 be the arch of the cycloid from (0, 0) to (2π, 0), which corresponds to 0 ≤ t ≤ 2π,andletC 2 be the segment from
(2π, 0) to (0, 0),soC 2 is given by x =2π − t, y =0, 0 ≤ t ≤ 2π. ThenC = C 1 ∪ C 2 is traversed clockwise, so −C is
oriented positively. Thus −C encloses the area under one arch of the cycloid and from (5) we have
A = − −C ydx= C 1
ydx+ C 2
ydx= 2π
0 (1 − cos t)(1 − cos t) dt + 2π
0
0(−dt)
= 2π
(1 − 2cost 0 +cos2 t) dt +0= t − 2sint + 1 t + 1 sin 2t 2π
=3π
2 4 0
21. (a) Using Equation 17.2.8 [ ET 16.2.8], we write parametric equations of the line segment as x =(1− t)x 1 + tx 2 ,
y =(1− t)y 1 + ty 2 , 0 ≤ t ≤ 1. Thendx =(x 2 − x 1 ) dt and dy =(y 2 − y 1 ) dt,so
C xdy− ydx= 1
0 [(1 − t)x 1 + tx 2 ](y 2 − y 1 ) dt +[(1− t)y 1 + ty 2 ](x 2 − x 1 ) dt
= 1
(x1(y2 − y1) − y1(x2 − x1)+t[(y2 − y1)(x2 − x1) − (x2 − x1)(y2 − y1)]) dt
0
= 1
(x 0 1y 2 − x 2 y 1 ) dt = x 1 y 2 − x 2 y 1
(b) We apply Green’s Theorem to the path C = C 1 ∪ C 2 ∪ ···∪ C n ,whereC i is the line segment that joins (x i ,y i ) to
(x i+1 ,y i+1 ) for i =1, 2, ..., n − 1, andC n is the line segment that joins (x n ,y n ) to (x 1 ,y 1 ).From(5),
1
xdy− ydx= dA,whereD is the polygon bounded by C. Therefore
2 C D
area of polygon = A(D) = dA =
1 xdy− ydx
D 2 C
= 1 2 C 1
xdy− ydx+ C 2
xdy− ydx+ ···+ C n−1
xdy− ydx+
C n
xdy− ydx
To evaluate these integrals we use the formula from (a) to get
A(D) = 1 2 [(x 1y 2 − x 2 y 1 )+(x 2 y 3 − x 3 y 2 )+···+(x n−1 y n − x n y n−1 )+(x n y 1 − x 1 y n )].