ch10

Advanced Engineering Mathematics

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Unformatted text preview: CHAPTER 10 Vector Integral Calculus. Integral Theorems SECTION 10.1. Line Integrals, page 420 Purpose. To explain line integrals in space and in the plane conceptually and technically with regard to their evaluation by using the representation of the path of integration. Main Content, Important Concepts Line integral (3), (3 9 ), its evaluation Its motivation by work done by a force (“work integral”) General properties (5) Dependence on path (Theorem 2) Background Material. Parametric representation of curves (Sec. 9.5); a couple of review problems may be useful. Comments on Content The integral (3) is more practical than (8) (more direct in view of subsequent material), and work done by a force motivates it sufficiently well. Independence of path is settled in the next section. SOLUTIONS TO PROBLEM SET 10.1, page 425 2. r ( t ) 5 [ t , 10 t , 0], 0 % t % 2, r 9 5 [1, 10, 0], F ( r ( t )) 5 [1000 t 3 , t 3 , 0]. Hence the integral is E 2 1010 t 3 dt 5 4040. The path of integration is shorter than that in Prob. 1, but the value of the integral is greater. This is not unusual. 4. r 5 [2 cos t , 2 sin t , 0], 0 % t % p , r 9 5 [ 2 2 sin t , 2 cos t ]. Hence F ( r ( t )) 5 [4 cos 2 t , 4 sin 2 t , 0]. This gives the integral E p ( 2 8 cos 2 t sin t 1 8 sin 2 t cos t ) dt 5 2 _ 16 3 1 0. 6. Clockwise integration is requested, so that we take, say, r 5 [cos t , 2 sin t ], 0 % t % 1 _ 2 p . By differentiation, r 9 5 [ 2 sin t , 2 cos t ]. Hence F ( r ( t )) 5 [ e cos t , e 2 sin t ], and the integral is E p / 2 ( 2 e cos t sin t 2 e 2 sin t cos t ) dt 5 [ e cos t 1 e 2 sin t ] p /2 . This gives the answer 2 2 sinh 1. 8. r 5 [ t , t 2 , t 3 ], 0 % t % 1 _ 2 , r 9 5 [1, 2 t , 3 t 2 ]. Hence F ( r ( t )) 5 [cosh t , sinh t 2 , e t 3 ]. 190 im10.qxd 9/21/05 12:49 PM Page 190 This gives the integral E 1/2 (cosh t 1 2 t sinh t 2 1 3 t 2 e t 3 ) dt 5 sinh 1 _ 2 1 cosh 1 _ 4 1 e 1/8 2 2 5 0.6857. 10. Here we integrate around a triangle in space. For the three sides and corresponding integrals we obtain r 1 5 [ t , t , 0], r 9 1 5 [1, 1, 0], F ( r 1 ( t )) 5 [ t , 0, 2 t ], E 1 t dt 5 1 _ 2 r 2 5 [1, 1, t ], r 9 2 5 [0, 0, 1], F ( r 2 ( t )) 5 [1, 2 t , 2], E 1 2 dt 5 2 r 3 5 [1 2 t , 1 2 t , 1 2 t ], r 9 3 5 [ 2 1, 2 1, 2 1], F ( r 3 ( t )) 5 [1 2 t , 2 1 1 t , 2 2 2 t ], E 1 ( 2 2 1 2 t ) dt 5 2 2 1 1. Hence the answer is 3/2. 12. r 5 [cos t , sin t , t ], r 9 5 [ 2 sin t , cos t , 1]. From F 5 [ y 2 , x 2 , cos 2 z ] we obtain F ( r ( t )) 5 [sin 2 t , cos 2 t , cos 2 t ]. Hence the integral is E 4 p ( 2 sin 3 t 1 cos 3 t 1 cos 2 t ) dt 5 0 1 0 1 2 p 5 2 p . 14. Project. (a) r 5 [cos t , sin t ], r 9 5 [ 2 sin t , cos t ]. From F 5 [ xy , 2 y 2 ] we obtain F ( r ( t )) 5 [cos t sin t , 2 sin 2 t ]. Hence the integral is 2 2 E p / 2 cos t sin 2 t dt 5 2 ....
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ch10 - CHAPTER 10 Vector Integral Calculus. Integral...

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