CÁLCULO III - INTEGRAIS DE LINHA RESOLVIDAS EM 04 MAI 2011

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Transcript of CÁLCULO III - INTEGRAIS DE LINHA RESOLVIDAS EM 04 MAI 2011

Rua 96 n 45 Setor Sul GoiniaEmail: [email protected] LIVRES DE 3 GRAUCLCULO IIIINTEGRAIS DE LINHA EXERCCIOS RESOLVIDOS1. Calcule a integral de linha ( )Cx 2y ds, + onde C uma semicircunferncia centrada na origem de raio igual a 3 e orientada no sentido positivo.Soluo:A parametrizao dessa semicircunferncia ser dada por: ( ) ( )2 2r(t) 3costi 3sent j, 0 t ds 3sent 3cost dt ds 9 dt 3dt + + r r r. Substituindo:( ) ( ) ( )003cost 6sent 3dt 3 3sent 6cost 3 12 36+ 2. Calcular a integral ( )Cx y z ds, + onde C a hlice circular dada por :r(t) costi sent j tk de P(1,0,0) a Q(1,0,2 ) + + r r r rSoluo:( ) ( )2ds sent cost 1dt 2 dt. + + Assim, podemos escrever:( ) ( )( ) ( )22 20 0 020tcost sent t 2 dt 2 1 t dt 2 t24 2 1 t dt 2 2 2 2 12 _+ , _ , 3. Calcule ( )C2x y z ds + , onde C o segmento de reta que liga A(1, 2, 3) a B(2, 0, 1).Soluo:Parametrizao do segmento de reta AB: + ' uuur r r r suur x(t) 2 tAB (1, 2, 2) i 2j 2k; B(2,0,1) AB: y(t) 2tz(t) 1 2ty 2 t 1; y 0 t 0 1 t 01 AFONSO CELSO FONE: (62) 3092-2268 / CEL: (62) 9216-9668Rua 96 n 45 Setor Sul GoiniaEmail: [email protected]tmail.com( ) ( ) ( ) ( ) ( )( ) ( ) ( ) + + + + + + + + + + + + + +r rur r r r r r(t) x t i y t j z t k r(t) 2 t i 2tj 1 2t kAssim:r '(t) i 2j 2k r(t) 1 4 4 9 3 ds 3dt (1)f x,y,z 2x y z f t 2(2 t) ( 2t) 1 2t 4 2t 2t 1 2t 5 2t f t 5 2t (2)Substituindo (1) e (2) na integral dada:( ) ( )( )0 001C 1 1C2x y z ds 5 2t 3dt 3 (5 2t) dt 3(5t t) |2x y z ds 0 3( 5 1) ( 3)( 4) 12 + + + + + + Resp.: 124. Calcule Cxz ds , onde C a interseo da esfera x + y + z = 4 com o plano x = y.Soluo:Vamos parametrizar a curva dada:( ) ( ) ( ) ( )( ) + + + + + + + _ + + + ,r r r r rrur22 2 22 222x y t t t z 4 z 4 2t z 4 2t4 2t 0 2t 4 0 2 t 2 r(t) x t i y t j z t k r(t) ti t j 4 2t k2t r ' t i j k4 2t2t 4t 8 4tr '(t) 1 1 24 2t4 2t + 24t( )( ) ( ) 2 2 28 814 2t 4 2t 4 2tef x, y,z xz f t t 4 2t (2)Substituindo (1) e (2) na integral dada: Cxz ds t 4 2t22284 2t( ) ( ) ( ) 1 1 ]2222 2 2C 3dt 8 t dtt 8 8xz ds 8 2 2 2 2 02 2 2Resp.: 02 AFONSO CELSO FONE: (62) 3092-2268 / CEL: (62) 9216-9668Rua 96 n 45 Setor Sul GoiniaEmail: [email protected] Soluo:( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )+ + + + + + + + + +r rr rr2 2 22 22 2 2 2 22 22 2 2C: x y z 4 x yAssim:y zy y z 4 2y z 4 12 4Parametrizando:x t 2 cost y t 2 cost z t 2sentAssim:r t 2 cost, 2 cost, 2sent r ' t 2sent, 2sent, 2coster ' t 2sent 2sent 2cost r ' t 2sen t 2sen t 4cos tr ' t 4 ( ) ( ) ( ) ( )( ) ( ) ( ) + + 1 ] r r r2 2 2 22 2 bC 0 0 abb 2 22 2 20C a asen t 4cos t r ' t 4 sen t cos t r ' t 4 r ' t 2Substituindo :xzds 2cost 2sent 2dt 4 2 sent costdt 4 2 uduOnde :u sent du costdtAssim:uxzds 4 2 udu 4 2 2 2 sent 2 2 sen 2 sen 0 02Resp: 05. Calcule Cxyds , onde C a elipse x y1a b+ .Soluo:A parametrizao da elipse dada por:[ ]( )( ) + + + + + +r r rrurur ur ur2 2 2 2 2x(t) acost e y(t) bsen t t 0, 2r(t) acosti bsen t j, 0 t 2e r ' t asent i bcostjr '(t) asent b cost, mas sent 1 costr '(t) a 1 cos t b cos t r '(t) a a cos t b cos t r '(t ) (b a)cost a +rds r '(t) dt ds (b a)cost a dt3 AFONSO CELSO FONE: (62) 3092-2268 / CEL: (62) 9216-9668Rua 96 n 45 Setor Sul GoiniaEmail: [email protected] na integral dada: + + + 2C 02C 0Cxyds acost bsent (b a)cost a dtxyds ab cost sent (b a)cost a dtu (b a)cost a du 2(b a)cost ( sent) 2(b a) cost sentdudu 2(a b) cost sent dt dt2(a b) cost sentxyds ab co st sent duu2(a b) cost sent[ ] + 3212 20CC(b a)cost aab abxyds u du |32(a b) 2(a b)2abxyds2 2(a b) ( )( ) ( ) ( ) ( ) ( ){ }( ) ( ) ( ) 1 1 1 + + + ] ] ] 1 + + ] 232 2 2 2 2 2 2 2 2 222 202 2 2 2 2 22 2C Cabb a cos t a b a cos 2 a b a cos 0 a3 3 a babxyds b a a b a a 0 xyds 03 a bResp.:06. ( )C3y z ds, onde C o arco da parbola z = y e x = 1 de A(1,0,0) a B(1,2,4).Soluo:Parametrizando C:( )( )( ) ' 2x t 1C y t t 0 t 2z t tAssim:( ) ( ) ( ) ( ) ( ) +r r r2 2r t 1,t,t r ' t 0,1,2t r ' t 1 4tAssim:4 AFONSO CELSO FONE: (62) 3092-2268 / CEL: (62) 9216-9668Rua 96 n 45 Setor Sul GoiniaEmail: [email protected]( ) ( ) ( )( )( ) + + + + _ + , _ , 2 2 22 2 2 2C 0 0 022 17 17 12 2C 0 1 11732C13y z ds 3t t 1 4t dt 3t t 1 4t dt 2t 1 4t dtFazendo :du duu 1 4t 8t dtdt 8te0 t 2 1 u 17Substituindo :du 2t3y z ds 2t 1 4t dt 2t u u du8t 8t1 u 1 23y z ds 1734 4 32 ( )( ) ( ) _ , 3 332 2C11 17 1613y z ds 17 17 16Resp: ( )117 17 167. Cy ds, onde C a curva dada por y = x de (-1,-1) a (1, 1).Soluo:Sabemos que: '