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

SECTION 16.4 GREEN'S THEOREM D 313
Since these aren't equal, the line integral ofF isn't independent of path. (Or notice that J~
3
F · dr = f 0
2 1r dt = 21r where.
Cs is the circle x 2 + y 2 = 1, and apply the contrapositive of Theorem 3.) This doesn't contradict Theorem 6, since the
domain ofF , which is IR 2 except the origin, isn't simply-connected.
16.4 Green's Theorem
1. (a) Parametric equations for C are x = 2 cos t, y = 2 sin t, 0 :::; t :::; 27r. Then
fc(x - y) dx + (x + y) dy = J;" [(2 cost- 2sin t )(-2 sin t) + (2 cos t+ 2 sin t)(2 cost)] dt
= J;"(4sin 2 t+ 4 cos 2 t)dt = J~" 4dt = 4 t]~"' = 81r
(b) Note that Cas given in part (a) is a positively oriented, smooth, simple closed curve. Then by Green's Theorem,
fc(x - y)dx + (x + y) dy= ffv [: ., (x+y) - : 11
(x - y)] dA = ffv [1- (-1)] dA = 2ffvdA
= 2A(D) = 27r(2) 2 = 81r
3. (a) y
(1,2)
Ct: x = t => dx = dt, y = D => dy = Ddt, D :::; t :::; 1.
C2: x = 1 => dx =Ddt, y = t => dy = dt, D:::; t $ 2.
C3: x= 1-t => dx = -dt, y=2-2t => dy = - 2dt, D$ t$ 1.
o c 1
(I,OJ
X
Thus fcxydx + x 2 y 3 dy = f xydx+x 2 y 3 dy
C1 + C2 + C3
= f 0
1
Ddt + f 0
2
t 3 dt + J; [-(1- t)(2 - 2t)- 2(1- t?(2 - 2t) 3 ] dt
= D + [~t 4 ]~ + [t(1 - t? + ~( 1- t) 6 ]~ = 4- ¥ = ~
(b) f c xydx +x 2 y 3 dy = ffv [:x (x 2 y 3 ) - /J~ (xy)] dA = f 0
1
J0 2 ~(2xy 3 - x)dydx
rt [ 1 4 ] y=2z
=
d rt ( s
Jo 2xy - xy 2 2) d y=O X= Jo 8x - X X = 4 f - 3 2 = 3'
2
5.
y
4 (2,4)
The region D enclosed by Cis given by {(x, y) I D :::; x:::; 2, x :::; y $ 2x }, so
f c xy 2 dx + 2x 2 ydy = ffv [ :, (2xiy) - : 11
(xy 2 )] dA
~ I: I:"' ( 4xy - 2xy) dy dx
= r2 [xl 2] y=2x d~
J o Y y= x ""
= f 2 3x 3 dx = !!.x 4 ] 2 = 12
J o 4 o
7. fc (v + e-fi) dx+ (2x +cosy 2 ) dy = Jfv [:x (2x + cosy 2 ) - :v· (v + e-fi)] dA
= fol Jyfi (2 - 1) dx dy = fol (yl/2 - y2) dy = ~
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