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Ch 18 Review Exercises

Ch 18 Review Exercises - Chapter Review Exercises 1271 = =...

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Chapter Review Exercises 1271 = − y 0 x 2 + t 2 2 z 2 + t 2 + z 2 2 x 2 ( x 2 + t 2 + z 2 ) 5 / 2 dt = y 0 x 2 2 t 2 + z 2 ( x 2 + t 2 + z 2 ) 5 / 2 dt = y ( x 2 + y 2 + z 2 ) 3 / 2 = Q ( x , y , z ) The last integral can be verified by showing that y y ( x 2 + y 2 + z 2 ) 3 / 2 = x 2 2 y 2 + z 2 ( x 2 + y 2 + z 2 ) 5 / 2 and f y = − y y 0 z ( x 2 + t 2 + z 2 ) 3 / 2 dt = − z ( x 2 + y 2 + z 2 ) 3 / 2 = − R ( x , y , z ) We conclude that the vector A = f , 0 , g is a vector potential of F in D , since curl ( A ) = g y , f z g x , f y = P , Q , R = F . (b) Suppose we remove the x -axis. In this case, we let A = 0 , f , g g ( x , y , z ) = − x x 0 Q ( t , y , z ) dt + y y 0 P ( x 0 , t , z ) dt f ( x , y , z ) = x x 0 R ( t , y , z ) dt Using similar procedure to that in Exercise 34, one can show that F = curl ( A ). In removing the z -axis the proof is similar, with corresponding modifications of the functions in Exercise 34. (c) The ball inside any sphere containing the origin must intersect the x , y , and z axes; therefore, F does not have a vector potential in the ball, and the flux of F through the sphere may differ from zero, as in our example. CHAPTER REVIEW EXERCISES 1. Let F ( x , y ) = x + y 2 , x 2 y and let C be the unit circle, oriented counterclockwise. Evaluate C F · d s directly as a line integral and using Green’s Theorem. SOLUTION We parametrize the unit circle by c ( t ) = ( cos t , sin t ) , 0 t 2 π . Then, c ( t ) = sin t , cos t and F ( c ( t )) = ( cos t + sin 2 t , cos 2 t sin t ) . We compute the dot product: F ( c ( t )) · c ( t ) = cos t + sin 2 t , cos 2 t sin t · − sin t , cos t = ( sin t )( cos t + sin 2 t ) + cos t ( cos 2 t sin t ) = cos 3 t sin 3 t 2 sin t cos t The line integral is thus C F ( c ( t )) · c ( t ) dt = 2 π 0 cos 3 t sin 3 t 2 sin t cos t dt = 2 π 0 cos 3 t dt 2 π 0 sin 3 t dt 2 π 0 sin 2 t dt = cos 2 t sin t 3 + 2 sin t 3 2 π 0 + sin 2 t cos t 3 + 2 cos t 3 2 π 0 + cos 2 t 2 2 π 0 = 0 We now compute the integral using Green’s Theorem. We compute the curl of F . Since P = x + y 2 and Q = x 2 y , we have Q x P y = 2 x 2 y
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1272 C H A P T E R 18 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (ET CHAPTER 17) Thus, C F · d s = D ( 2 x 2 y ) dx dy 1 x y C D We compute the double integral by converting to polar coordinates. We get C F · d s = 2 π 0 1 0 ( 2 r cos θ 2 r sin θ ) r dr d θ = 2 π 0 1 0 2 r 2 ( cos θ sin θ ) dr d θ = 1 0 2 r 2 dr 2 π 0 ( cos θ sin θ ) d θ = 2 3 r 3 1 0 sin θ + cos θ 2 π 0 = 2 3 ( 1 1 ) = 0 2. Let R be the boundary of the rectangle in Figure 1 and let R 1 and R 2 be the boundaries of the two triangles, all oriented counterclockwise. (a) Determine R 1 F · d s if R F · d s = 4 and R 2 F · d s = − 2. (b) What is the value of R F d s if R is oriented clockwise? x Rectangle R y R 1 R 2 FIGURE 1 SOLUTION (a) Since all boundaries are oriented counterclockwise, the segment DB is oriented in opposite directions as part of the boundaries R 1 and R 2 .
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