Solução_Calculo_Stewart_6e

F.
276 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
17.3 The Fundamental Theorem for Line Integrals ET 16.3
1. C appears to be a smooth curve, and since ∇f is continuous, we know f is differentiable. Then Theorem 2 says that the value
of ∇f · dr is simply the difference of the values of f at the terminal and initial points of C. From the graph, this is
C
50 − 10 = 40.
3. ∂(2x − 3y)/∂y = −3 =∂(−3x +4y − 8)/∂x and the domain of F is R 2 which is open and simply-connected, so by
Theorem 6 F is conservative. Thus, there exists a function f such that ∇f = F, thatis,f x(x, y) =2x − 3y and
f y (x, y) =−3x +4y − 8. Butf x (x, y) =2x − 3y implies f(x, y) =x 2 − 3xy + g(y) and differentiating both sides of this
equation with respect to y gives f y (x, y) =−3x + g 0 (y). Thus−3x +4y − 8=−3x + g 0 (y) so g 0 (y) =4y − 8 and
g(y) =2y 2 − 8y + K where K is a constant. Hence f(x, y) =x 2 − 3xy +2y 2 − 8y + K is a potential function for F.
5. ∂(e x sin y)/∂y = e x cos y = ∂(e x cos y)/∂x and the domain of F is R 2 . Hence F is conservative so there exists a function f
such that ∇f = F. Thenf x(x, y) =e x sin y implies f(x, y) =e x sin y + g(y) and f y(x, y) =e x cos y + g 0 (y). But
f y (x, y) =e x cos y so g 0 (y) =0 ⇒ g(y) =K. Thenf(x, y) =e x sin y + K is a potential function for F.
7. ∂(ye x +siny)/∂y = e x +cosy = ∂(e x + x cos y)/∂x and the domain of F is R 2 . Hence F is conservative so there
exists a function f such that ∇f = F. Thenf x(x, y) =ye x +siny implies f(x, y) =ye x + x sin y + g(y) and
f y (x, y) =e x + x cos y + g 0 (y). Butf y (x, y) =e x + x cos y so g(y) =K and f(x, y) =ye x + x sin y + K is a potential
function for F.
9. ∂(ln y +2xy 3 )/∂y =1/y +6xy 2 = ∂(3x 2 y 2 + x/y)/∂x and the domain of F is {(x, y) | y>0} whichisopenandsimply
connected. Hence F is conservative so there exists a function f such that ∇f = F. Thenf x(x, y) =lny +2xy 3 implies
f(x, y) =x ln y + x 2 y 3 + g(y) and f y (x, y) =x/y +3x 2 y 2 + g 0 (y). Butf y (x, y) =3x 2 y 2 + x/y so g 0 (y) =0
⇒
g(y) =K and f(x, y) =x ln y + x 2 y 3 + K is a potential function for F.
11. (a) F has continuous first-order partial derivatives and ∂ ∂y
2xy =2x =
∂
∂x (x2 ) on R 2 , which is open and simply-connected.
Thus, F is conservative by Theorem 6. Then we know that the line integral of F is independent of path; in particular, the
value of F · dr depends only on the endpoints of C. Since all three curves have the same initial and terminal points,
C
F · dr willhavethesamevalueforeachcurve.
C
(b) We first find a potential function f,sothat∇f = F. Weknowf x (x, y) =2xy and f y (x, y) =x 2 .Integrating
f x(x, y) with respect to x, wehavef(x, y) =x 2 y + g(y). Differentiating both sides with respect to y gives
f y (x, y) =x 2 + g 0 (y), sowemusthavex 2 + g 0 (y) =x 2 ⇒ g 0 (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,
F · dr = f(3, 2) − f(1, 2) = 18 − 2=16for each curve.
C
13. (a) f x (x, y) =xy 2 implies f(x, y) = 1 2 x2 y 2 + g(y) and f y (x, y) =x 2 y + g 0 (y). Butf y (x, y) =x 2 y so g 0 (y) =0 ⇒
g(y) =K, a constant. We can take K =0,sof(x, y) = 1 2 x2 y 2 .