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

SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS 0 311
(b) We first find a potential function f, so that \l f = F. We know ! :r:(x, y) = 2xy and jy(x, y) = x 2 • Integrating
fx(x, y) with respect to x, we have f(x, y) = x 2 y + g(y). Differentiating both sides with respect toy gives
· J 11
(x, y) = x 2 + g'(y), so we must have x 2 + g'(y) = x 2 =? g'(y) = 0 => g(y) = K, a constant.
Thus f(x, y) = x 2 y + K. All three curves start at (1, 2) and end at (3, 2), so by Theorem 2,
fe F · dr = !(3, 2) - !(1, 2) = 18 - 2 = 16 for each curve.
13. (a) f r(x, y) = xy 2 implies J(x, y) = ~x 2 y 2 + g(y) and fy (x, y) = x 2 y + g'(y). But /y(x, y) = x 2 y so g'(y) = 0 =>
g(y) = K, a constant. We can take K = 0, so f(x, y) = ~x 2 y 2 .
(b) The initial point of C is r (O) = (0, 1) and the terminal point is r(1) = (2, _1), so
fe F · dr = /(2, 1) - f(O, 1) = 2 - 0 = 2.
15. (a) f x(x, y, z ) = yz implies f(x, y, z) = xyz + g(y, z) and so jy(x, y , z) = xz + g 11 (y, z). But jy(x, y, z) = xz so
gy(y, z ) = 0 => g(y, z) = h(z). Thus f(x, y, z) = xyz + h(z) and f:(x, y, z) = X'!f + h'(z). But
f:(x, y, z) = xy + 2z, so h'(z) = 2z => h(z) = z 2 + K.. Hence f(x, y, z) = xyz + z 2 (taking J( = 0).
(b) fe F · dr = !(4,6,3)-!(1, 0, - 2) = 81 - 4 = 77.
17. (a) f:r:(x, y, z) = yze"'= implies f(x, y, z) = ye"'= + g(y, z) and so fv(x, y, z) = e"'= + gy(y, z). But jy(x, y, z ) = e"'= so.
gy{y,z) = 0 =? g(y,z) = h(z). Thus f(x,y,z)·= ye"'= + h(z) and fz(x, y,z) = xyex= + h'(z). But
f:.:(x,y,z) = xyex=, so h'(z) = 0 =? h(z) = K. Hence f(x,y,z) = ye"'= (taking!(= 0).
(b) r (O) = (1, -1, 0), r(2) = (5, 3, 0) so fe F · dr = /(5, 3, 0) - /(1, - 1, 0) = 3e 0 + e 0 = 4.
19. The functions 2xe-Y and 2y - x 2 e- v have continuous first-order derivatives on JR 2 and
~ (2xe-v) = -2xe- Y = ~ (2y- x 2 e:-v), so F(x, y) = 2xe-Y i + (2y- x 2 e-Y) j is a conservative vector field by
Ulj vx . .
Theor~m 6 and hence the line integral is independent of path. Thus a potential function f exists, and fx(x, y) = 2xe-v
implies f(x, y) = x 2 e- 11 + g(y) and fv(x, y) = - x 2 e-Y + g'(y). But fv(x, y) = 2y- x 2 e-v so
g'(y) = 2y =? g(y) = y 2 + K. We can take K = 0, so f(x,y) = x 2 e- 11 + y 2 • Then
.fe 2xe-v dx + (2y - x 2 e- v) dy = /(2, 1) - f(l, 0) = 4e- 1 + 1 - 1 = 4/e.
21. IfF is conservative, then fe F · dr is independent of path. This means that the work done along all piecewise-smooth curves
that have the described initial and terminal points is the same. Your reply: It doesn't matter which curve is chosen.
23. F (x, y) = 2y 3 1 2 i + 3x Jyj, W = .fe F · dr. Since 8(2y 3 1 2 )/8y = 3 y'Y = 8(3x Jy )/ ax, there exists a function f
suchthat\lf=F.Infact,fx(x, y)='2y 3 1 2 => f(x,y)=2xy 312 +g(y) => fv(x,y)=3xy 1 1 2 +g'(y). But
/y (x, y) = 3x y'Y so g'(y) = 0 or g(y) = K. We can take I< = 0 =? f(x, y) = 2xy 3 1 2 • Thus
W = fe F · dr = j(2, 4) :.... j(I,·l) = 2(2)(8)- 2(1) .= 30.
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