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17.2 Ex 40-49 - 1084 C H A P T E R 17 L I N E A N D S U R...

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1084 C H A P T E R 17 LINE AND SURFACE INTEGRALS (ET CHAPTER 16) 40. Let C be the path from P to Q in Figure 17 that traces C 1 , C 2 , and C 3 in the orientation indicated. Suppose that C F · d s = 5 , C 1 F · d s = 8 , C 3 F · d s = 8 Determine: (a) C 3 F · d s (b) C 2 F · d s (c) C 1 C 3 F · d s x y P Q C 1 C 3 C 2 FIGURE 17 (d) What is the value of the line integral of F over the path that traverses the loop C 2 four times in the clockwise direction? SOLUTION x y P Q C 1 C 3 C 2 (a) If the orientation of the path is reversed, the line integral changes sign, thus: C 3 F · d s = − C 3 F · d s = − 8 (b) By additivity of line integrals, we have C F · d s = C 1 F · d s + C 2 F · d s + C 3 F · d s Substituting the given values we obtain 5 = 8 + C 2 F · d s + 8 or C 2 F · d s = 5 16 = − 11 (c) Using properties of line integrals gives C 1 C 3 F · d s = C 1 F · d s + C 3 F · d s = − C 1 F · d s C 3 F · d s = − 8 8 = − 16 (d) Using additivity and the integral over the curve with the reversed orientation, the line integral of F over the path that traverses the loop C 2 four times in the clockwise direction is: 4 C 2 F · d s = 4 · C 2 F · d s = − 4 C 2 F · d s = − 4 · ( 11 ) = 44 In Exercises 41–44, let F be the vortex vector field (so-called because it swirls around the origin as shown in Figure 18) F = y x 2 + y 2 , x x 2 + y 2
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S E C T I O N 17.2 Line Integrals (ET Section 16.2) 1085 FIGURE 18 Vector field F = y x 2 + y 2 , x x 2 + y 2 . 41. Let I = C F · d s , where C is the circle of radius 2 centered at the origin oriented counterclockwise (Figure 18). (a) Do you expect I to be positive, negative, or zero? (b) Evaluate I . (c) Verify that I changes sign when C is oriented in the clockwise direction. SOLUTION (a) When the circle is oriented counterclockwise, the dot product of F with the unit tangent vector at each point along the circle is positive. Therefore, we expect the vector line integral I to be positive. (b) 2 x y P The circle of radius 2 oriented counterclockwise has the parametrization: c ( t ) = ( 2 cos t , 2 sin t ), 0 t 2 π Hence, F ( c ( t )) = 2 sin t 4 cos 2 t + 4 sin 2 t , 2 cos t 4 cos 2 t + 4 sin 2 t = 1 2 sin t , cos t c ( t ) = 2 sin t , 2 cos t Therefore, the integrand is the dot product, F ( c ( t )) · c ( t ) = 1 2 sin t , cos t · − 2 sin t , 2 cos t = sin 2 t + cos 2 t = 1
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