Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 16.4 GREEN'S THEOREM 0 315
21. (a) Using Equation 16.2.8, we write parametric equations of the lin_e segment as x = (1 - t)x1 + tx2, y = (1- t)y1 + ty 2 ,
0 ~ t ~ 1. Then dx = (x2- xl) dt and dy = (y2- yl) dt, so
1
fc x dy- y dx = f 0 [(1 - t)x1 + tx2](y2- Yl) dt + [(1 - t)y1 + ty2](x2- x1) dt
= J; (x1(Y2 - Yl) - y1(x2- xl) + t((y2- Y1)(x2- x1)-:- (x2 - x1)(y2- Y1)]) dt
1
= f 0 (x1y2- x2yl) dt = x~y2- X2Y1
(b) We apply Green's Theorem to the path C = C1 U C2 U · · · U Cn, where Ci is the line segment that joins (xi , Yi) to
(xi+l,Yi+l) fori= 1, 2, ..., n -1, and Gn is the linesegmentthatjoins (x,.,yn) to (xl,Yl). From (5),
~ J~ x dy - y dx = JJ 0
dA, where D is tite polygon bounded by C. Therefore
area of polygon= A( D)= ffv dA = ~ J~ xdy- ydx
= ~ (J~ 1 x dy- ydx + fc 2
xdy - ydx + .... + J~n-l xdy- ydx +fen xdy- ydx)
To evaluate these integrals we use the formula from (a) to get
A(D) = ~[(x1y2- X2Yl) + (X2Y3 - XaY2) + ' · · + (Xn-;-lYn- XnYn- 1) + (XnYl - XlYn)].
(c) A = ~ [(0 · 1 - 2 · 0) + (2 · 3 - 1 · 1) + (1 · 2 - 0 · 3) + (0 · 1 - (-1) · 2) + ( ~ 1 · 0- 0 · 1)]
= Ho+5+2+2) = ~
23. We orient the quarter-circular region as shown in the figure.
A = {7ra 2 sox = 1l'a! / 2
fc x 2 dy and y = - 1ra! 12
fc y 2 dx.
y
a
Here c = c l + c2 + Cs where cl: X = t, y = 0, 0 ~ ·t ~ a;
C2: x = a cost, y =a sin t, 0 ~ t. ~ ~;and
c3: X = 0, y = a - t, 0 ~ t ~ a. Then
fcx 2 dy = fc, x 2 dy + fc 2
x 2 dy +.fc 3
x 2 dy = foa Odt + J 12 0 " (acost) 2 (acost) dt + J; Odt
= J~"/ 2 a 3 cos 3 tdt = a 3 J 0
"/\1- sin 2 t) costdt = a 3 [sint -.i sin 3 t]; 12 = ~a~
_ 1 J 2
4a
so x = 7ra2 /2 Jc x dy = 311'.
c.
a
X
fcy 2 dx = fc, y 2 dx + )~ 2 Y2 dx + fc 3
Y 2 dx = foa 0 dt + f 12 " 0
(a sin t) 2 (- a sin t) dt + Ja 0
0 dt
= J 0 " 12 (- a 3 sin 3 t) dt = - a 3 J 0 " 1 \ 1 - cos 2 t) sin tdt = -a 3 [ ~ cos 3 t- cos t]; 12 = - ~a 3 ,
so 'ii = - 1l'a!/ 2
£y2 dx = ~;.Thus (x,y) = ( ::. ::)·
25. By Green's Theorem, - ~pfcy 3 dx = - kp J.f'v( - 3y 2 ) dA = ffv y 2 pdA = I ., and
tPfcx 3 dy = ~p ff 0 (3x 2 ) dA = ffv x 2 pdA = Iu .
27. As in Example 5, let C' be a counterclockwise-oriented circle with center the origin and radius a, where a is chosen to
be small enough so that C' lies inside C, and D the region bounded by C and C'. Here
=>
EJP = 2x(x 2 + y 2 ) 2 . - 2xy · 2(x 2 + y 2 ) · 2y = 2x 3 - 6x~ 2 and
oy (x2 + y2)4 (x2 + y2)3
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