Ch 18 Review Exercises

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

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Chapter Review Exercises 1271 =− Z 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 = Z y 0 x 2 2 t 2 + z 2 ( x 2 + t 2 + z 2 ) 5 / 2 = y ( x 2 + y 2 + z 2 ) 3 / 2 = Q ( x , y , z ) The last integral can be verifed by showing that y à 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 Z y 0 z ( x 2 + t 2 + z 2 ) 3 / 2 z ( x 2 + y 2 + z 2 ) 3 / 2 R ( x , y , z ) We conclude that the vector A = h f , 0 , g i is a vector potential oF F in D ,since curl ( A ) = ¿ g y , f z g x , f y À = h P , Q , R i = F . (b) Suppose we remove the x -axis. In this case, we let A = h 0 , f , g i g ( x , y , z ) Z x x 0 Q ( t , y , z ) + Z y y 0 P ( x 0 , t , z ) f ( x , y , z ) = Z x x 0 R ( t , y , z ) 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 modifcations 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 ±ux 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 I 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 0 ( t ) = h− sin t , cos t i and F ( c ( t )) = ( cos t + sin 2 t , cos 2 t sin t ) . We compute the dot product: F ( c ( t )) · c 0 ( t ) = D cos t + sin 2 t , cos 2 t sin t E · h− sin t , cos t i = ( sin t )( cos t + sin 2 t ) + cos t ( cos 2 t sin t ) = cos 3 t sin 3 t 2sin t cos t The line integral is thus Z C F ( c ( t )) · c 0 ( t ) = Z 2 0 ³ cos 3 t sin 3 t t cos t ² = Z 2 0 cos 3 tdt Z 2 0 sin 3 Z 2 0 sin 2 = cos 2 t sin t 3 + t 3 ¯ ¯ ¯ ¯ 2 0 + à sin 2 t cos t 3 + 2cos t 3 ! ¯ ¯ ¯ ¯ 2 0 + cos 2 t 2 ¯ ¯ ¯ ¯ 2 0 = 0 We now compute the integral using Green’s Theorem. We compute the curl oF F .S ince P = x + y 2 and Q = x 2 y , we have Q x P y = 2 x 2 y
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1272 CHAPTER 18 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (ET CHAPTER 17) Thus, Z C F · d s = ZZ D ( 2 x 2 y ) dx dy 1 x y C D We compute the double integral by converting to polar coordinates. We get Z C F · d s = Z 2 π 0 Z 1 0 ( 2 r cos θ 2 r sin ) rdrd = Z 2 0 Z 1 0 2 r 2 ( cos sin ) dr d = Ã Z 1 0 2 r 2 dr Z 2 0 ( cos sin ) d ! = Ã 2 3 r 3 ¯ ¯ ¯ ¯ 1 0 sin + cos ¯ ¯ ¯ ¯ 2 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 I R 1 F · d s if I R F · d s = 4and I R 2 F · d s =− 2. (b) What is the value of I 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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This note was uploaded on 04/13/2008 for the course MATH 32B taught by Professor Rogawski during the Winter '08 term at UCLA.

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

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