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

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SECTION 16.2 LINE INTEGRALS 0 307<br />

(b) All of the (nonzero) field vectors along the circle with radius 3 are point<strong>ed</strong> in the clockwise direction, that is, opposite the<br />

direction to the path. So F · Tis negative, and therefore fc 2<br />

F · dr = fc 2<br />

F · T ds is negative.<br />

19. r(t) = llt 4 i + t 3 j ,, so F (r (t)) = (llt 4 )(t 3 ) i + 3(t 3 ) 2 j = llt 7 i + 3t 6 j and r'(t) =· 44t 3 i + 3t 2 j . Then<br />

fc F · dr = J; F(r(t)) · r'(t)dt = J;(ne · 44e + 3t 6 · 3t 2 )dt = J;(484t 10 + 9t 8 )dt = [44t 11 +t 9 )~ = 45.<br />

21 . fc F · dr = J; (sin t\ cos( -t 2 ), t 4 ) · (3t 2 , - 2t, 1) dt<br />

= J;(3t2 sint 3 - 2tcose + t 4 ) dt = [- cost 3 - sint 2 + tt 5 )~ =~-cos 1-sinl<br />

fc ~ ·<br />

d~ = i<br />

2<br />

= i<br />

2<br />

2<br />

F (r (t )) · r'(t)dt = i [et-t 2 et :rsm(e-t 2 )<br />

[ e 2 t- t 2 2 2<br />

- 2te - t sin ( e - t ) J dt ::::J 1.9633<br />

. ( - '2te- t 2 ) ] dt<br />

25. x = t 2 , y = t 3 , z = t 4 so by Formula 9,<br />

fc xsin(y + z) ds = J;(e) sin(t 3 + t 4 )vf(2t)2 + (3t2)2 + {4t 3 )2 dt<br />

= f 0<br />

5 t<br />

2<br />

sin(t 3 + t 4 ) J4t 2 + 9t 4 + 16t6 dt ::::: 15.0074<br />

27. We graph F ( x, y) = ( x - y) i + x y j and the curve C. We see that most of\he vectors starting on C poi~t in roughly the same<br />

direction as C, so for these portions of C the tangential compon~nt F · T is positive. Although some vectors in the third<br />

quadrant which start on C point in roughly the opposite direction, and hence give negative tangential components, it seems<br />

reasonable that the effect of these portions of C is outweigh<strong>ed</strong> by the positive tangential components. Thus, we would expect<br />

fc F · dr = fc F · T ds to be positive.<br />

To verifY, we evaluate fc F · dr: The curve C can be represent<strong>ed</strong> by r (t) = 2 cost i + 2 sin t j , 0 $ t $ 3 ;,<br />

so F (r(t)) = (2 cost- 2 sin t) i + 4 cos tsin t j and r ' (t) = - 2 sin t i + 2 cos t j . Then<br />

fc F · dr = ./;" 12 F (r (t)) · r ' (t) dt<br />

= J;" 12 [-2 sin t(2 cost - 2 sin t) + 2 cos t(4cos t sin t )) dt<br />

3<br />

= 4J ,.<br />

12<br />

0<br />

(sin<br />

2<br />

t- sintcost + 2sintcos 2 t) dt<br />

[using a CAS)<br />

@) 2012 Ccngogc Lcruning. All Rights Reserv<strong>ed</strong>. May nol be 5Cl!Mcd. copi<strong>ed</strong>, or duplicnl<strong>ed</strong>, or poslcd lo • publicly accessible wcbsilc, in whole or in part.

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