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Portafolio de ejercicios de Cálculo Vectorial Quiz integrales de línea - Corte II 𝐹(𝑥, 𝑦) = 𝑥5; 𝑦 = 𝑥4 , 𝑥 𝜖 [0,1] 𝑥 = 𝑡; 𝑥´(𝑡) = 1 𝑦 = 𝑡4; 𝑦´(𝑡) = 4𝑡3 𝑟(𝑡) = (𝑡, 𝑡4) ∫ 𝑡5√1 + 16𝑡6𝑑𝑡 1 0 𝑢 = 1 + 16𝑡6; 𝑑𝑢 = 96𝑡5𝑑𝑡 = 1 96 ∫ √𝑢 𝑑𝑢 1 0 = 1 96 [ 2𝑢 3 2 3 ] = 1 96 [ 2(1 + 16𝑡6) 3 2 3 ] 1 0 = (17) 3 2 144 − 1 144 = 17√17 − 1 144 ∫ [(2𝜋𝑡 + 𝑠𝑒𝑛 𝑡 + cos 𝑡 )√4𝜋2 + 𝑐𝑜𝑠2𝑡 + 𝑠𝑒𝑛2𝑡]𝑑𝑡 𝜋 0 = ∫ [(2𝜋𝑡 + 𝑠𝑒𝑛 𝑡 + cos 𝑡 )√4𝜋2 + 1]𝑑𝑡 𝜋 0 = √4𝜋2 + 1 ∫ 2𝜋𝑡 + 𝑠𝑒𝑛 𝑡 + cos 𝑡 𝑑𝑡 𝜋 0 = √4𝜋2 + 1⌊𝜋𝑡2 − cos 𝑡 + sen 𝑡⌋ 𝜋 0 = √4𝜋2 + 1⌊𝜋3 + 2⌋ 𝑟(𝑡) = (𝑃𝐹 − 𝑃𝐼)𝑡 + 𝑃𝐼 𝑟(𝑡) = ((1, 𝜋, −1) − (0,0,0))𝑡 + (0,0,0) 𝑟(𝑡) = (1, 𝜋, −1)𝑡 𝑟(𝑡) = (1𝑡, 𝜋𝑡, −1𝑡) Siempre que se usa esta parametrización t va de 0 a 1 𝑟(𝑡) = (𝑡, 𝜋𝑡, −𝑡) 𝑐𝑜𝑛 0 ≤ 𝑡 ≤ 1 ∫ −𝑡2 1 0 𝑠𝑒𝑛 𝜋𝑡 (√1 + 𝜋2 + 1) 𝑑𝑡 = ∫ −𝑡2 1 0 𝑠𝑒𝑛 𝜋𝑡 (√2+𝜋2) 𝑑𝑡 = (√2+𝜋2) ∫ (−𝑡2 1 0 𝑠𝑒𝑛 𝜋𝑡)𝑑𝑡 = (√2+𝜋2) ∫ (−𝑡2 1 0 𝑠𝑒𝑛 𝜋𝑡)𝑑𝑡 𝑢 = −𝑡2 ; 𝑑𝑢 = −2𝑡 𝑑𝑡 𝑑𝑣 = 𝑠𝑒𝑛 𝜋𝑡 𝑑𝑡; 𝑣 = −cos 𝜋𝑡 𝜋 𝑑𝑡 = −𝑡2 −cos 𝜋𝑡 𝜋 − 2 𝜋 ∫ t cos 𝜋𝑡 𝑑𝑡 Por partes nuevamente ∫ t cos 𝜋𝑡 𝑑𝑡 𝑢 = 𝑡; 𝑑𝑢 = 𝑑𝑡 𝑑𝑣 = cos 𝜋𝑡 𝑑𝑡 ; 𝑣 = sen 𝜋𝑡 𝜋 𝑑𝑡 = 𝑡 sen 𝜋𝑡 𝜋 + 1 𝜋 ∫ sen 𝜋𝑡 𝑑𝑡 = 𝑡 sen 𝜋𝑡 𝜋 + cos 𝜋𝑡 𝜋2 = [ 𝑡2cos 𝜋𝑡 𝜋 − 2𝑡 sen 𝜋𝑡 𝜋2 − 2cos 𝜋𝑡 𝜋3 ] 1 0 = − 1 𝜋 + 2 𝜋3 + 2 𝜋3 = − (√2+𝜋2) [ 1 𝜋 − 4 𝜋3 ] 𝑥2 + 𝑦2 = 4 𝑥 = 𝑟 cos 𝜃 = 2 cos 𝜃 𝑦 = 𝑟 sen 𝜃 = 2 sen 𝜃 [ 𝜋 2 ≤ 𝜃 ≤ 𝜋] 𝐹(𝑟(𝜃)) = 𝐹(2 cos 𝜃 , 2 sen 𝜃) ∫ [2( 𝜋 𝜋 2 2 cos 𝜃) + 3(2 sen 𝜃) + 4]√4 𝑐𝑜𝑠2 𝜃 + 4 𝑠𝑒𝑛2 𝜃 𝑑𝜃 = ∫ [ 𝜋 𝜋 2 8 cos 𝜃 + 12 𝑠𝑒𝑛 𝜃 + 8] = ∫ 8 cos 𝜃 𝑑𝜃 + ∫ 12 𝑠𝑒𝑛 𝜃 𝑑𝜃 + ∫ 8 𝜋 𝜋 2 𝜋 𝜋 2 𝜋 𝜋 2 𝑑𝜃 = [8𝑠𝑒𝑛 𝜃] 𝜋 𝜋 2 + [−12 𝑐𝑜𝑠 𝜃] 𝜋 𝜋 2 + [8𝜃] 𝜋 𝜋 2 𝑑𝜃 = −8 + 12 + 4𝜋 = 4[1 + 𝜋 ] 𝑦 = 𝑥3; 𝑥 𝜖 [0.1] 𝑥 = 𝑡; 𝑦 = 𝑡3; 𝑡 𝜖 [−1,1] 𝑟(𝑡) = (𝑡, 𝑡3); r´(t) = (1, 3𝑡2) 𝐹(𝑟(𝑡)) = (−𝑡3 , 𝑡) ∫ (−𝑡3 , 𝑡) ∙ (1, 3𝑡2) dt 1 −1 = ∫ (−𝑡3 + 3𝑡3) dt 1 −1 = [(− 𝑡4 4 + 3𝑡4 4 )] 1 −1 = [( 𝑡4 2 )] 1 −1 = 0 𝑥2 + 𝑦2 = 𝑟2 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 sen 𝜃 𝑟(𝜃) = (𝑟𝑐𝑜𝑠 𝜃, 𝑟𝑠𝑒𝑛 𝜃); 𝑐𝑜𝑛 𝜋 2 ≤ 𝜃 ≤ 𝜋 𝑟´(𝜃) = (−𝑟𝑠𝑒𝑛𝜃, 𝑟𝑐𝑜𝑠𝜃) 𝐹(𝑟(𝜃)) = ( 𝑟2𝑐𝑜𝑠 𝜃𝑠𝑒𝑛 𝜃, 𝑟 𝑐𝑜𝑠 𝜃) ∫ ( 𝑟2𝑐𝑜𝑠 𝜃𝑠𝑒𝑛 𝜃, 𝑟 𝑐𝑜𝑠 𝜃) ∙ (−𝑟𝑠𝑒𝑛𝜃, 𝑟𝑐𝑜𝑠𝜃) 𝜋 𝜋 2 𝑑𝜃 = ∫ (−𝑟3𝑐𝑜𝑠 𝜃𝑠𝑒𝑛2𝜃 + 𝑟2 𝑐𝑜𝑠2𝜃) 𝜋 𝜋 2 𝑑𝜃 = ∫ (−𝑟3𝑐𝑜𝑠 𝜃𝑠𝑒𝑛2𝜃 + 𝑟2 𝑐𝑜𝑠2𝜃) 𝜋 𝜋 2 𝑑𝜃 = ∫ (−𝑟3𝑐𝑜𝑠 𝜃𝑠𝑒𝑛2𝜃𝑑𝜃) 𝜋 𝜋 2 + ∫ 𝑟2 𝑐𝑜𝑠2𝜃 𝜋 𝜋 2 𝑑𝜃 𝑢 = sen 𝜃 ; 𝑑𝑢 𝑐𝑜𝑠 𝜃 = 𝑑𝜃 = −𝑟3 ∫ 𝑢2𝑑𝑢 𝜋 𝜋 2 = [ 𝑢3 3 ] 𝜋 𝜋 2 = [ 𝑠𝑒𝑛3𝜃 3 ] 𝜋 𝜋 2 = − −𝑟3 3 = 𝑟3 3 = ∫ 𝑟2 𝑐𝑜𝑠2𝜃 𝜋 𝜋 2 𝑑𝜃 Se usa esta identidad: = 𝑟2 ∫ 1 2 + 1 2 cos 2𝜃 𝜋 𝜋 2 𝑑𝜃 = 𝑟2 [ 𝜃 2 + 1 4 𝑠𝑒𝑛 2𝜃] 𝜋 𝜋 2 = 𝑟2 [ 𝜋 2 − ( 𝜋 4 )] = 𝜋𝑟2 4 = ∫ (−𝑟3𝑐𝑜𝑠 𝜃𝑠𝑒𝑛2𝜃 + 𝑟2 𝑐𝑜𝑠2𝜃) 𝜋 𝜋 2 𝑑𝜃 = 𝜋𝑟2 4 + 𝑟3 3 𝑥 = 𝑡; 𝑧 = 𝑡2; 𝑦 = 𝑡4 𝑟(𝑡) = (𝑡, 𝑡4 , 𝑡2); 𝑡 𝜖 [−1,1] 𝑟´(𝑡) = (1, 4𝑡3 , 2𝑡) 𝑓(𝑟(𝑡)) = (𝑡 − 𝑡4 + 𝑡2 , 2𝑡2 , 5 + 3𝑡6) ∫ (𝑡 − 𝑡4 + 𝑡2 , 2𝑡2, 5 + 3𝑡6) ∙ 1 −1 (1, 4𝑡3 , 2𝑡)𝑑𝑡 = ∫ (𝑡 − 𝑡4 + 𝑡2 + 8𝑡5 + 10𝑡 + 6𝑡7) 1 −1 𝑑𝑡 = ∫ (𝑡 − 𝑡4 + 𝑡2 + 8𝑡5 + 10𝑡 + 6𝑡7) 1 −1 𝑑𝑡 = [ 𝑡2 2 − 𝑡5 5 + 𝑡3 3 + 4𝑡6 3 + 5𝑡2 1 + 3𝑡8 4 ] 1 −1 = [ 1 2 − 1 5 + 1 3 + 4 3 + 5 1 + 3 4 ] − [ 1 2 + 1 5 − 1 3 + 4 3 + 5 1 + 3 4 ] = 463 60 − 149 20 = 4 15 𝑓𝑥 = 𝑒 𝑧𝑦 + 𝑧𝑦𝑒𝑥𝑦 + 𝑦𝑧𝑒𝑥𝑧 (1) 𝑓𝑦 = 𝑥𝑧𝑒 𝑧𝑦 + 𝑧𝑥𝑒𝑥𝑦 + 𝑒𝑥𝑧 (2) 𝑓𝑧 = 𝑒 𝑥𝑦 + 𝑥𝑦𝑒𝑧𝑦 + 𝑥𝑦𝑒𝑥𝑧 (3) ∫ 𝑒𝑧𝑦 + 𝑧𝑦𝑒𝑥𝑦 + 𝑦𝑧𝑒𝑥𝑧𝑑𝑥 = 𝑥𝑒𝑧𝑦 + 𝑧𝑒𝑥𝑦 + 𝑦𝑒𝑥𝑧 + ℎ(𝑦, 𝑧) (4) Derivamos 4 con respecto a y, z 𝑓𝑦 = 𝑥𝑧𝑒 𝑧𝑦 + 𝑧𝑥𝑒𝑥𝑦 + 𝑒𝑥𝑧 + ℎ(𝑧)(5) 𝑓𝑧 = 𝑥𝑦𝑒 𝑧𝑦 + 𝑒𝑥𝑦 + 𝑦𝑥𝑒𝑥𝑧 + ℎ(𝑥)(6) Igualamos (5) con (2) y (6) con (3) 𝑥𝑧𝑒𝑧𝑦 + 𝑧𝑥𝑒𝑥𝑦 + 𝑒𝑥𝑧 + ℎ(𝑧) = 𝑥𝑧𝑒𝑧𝑦 + 𝑧𝑥𝑒𝑥𝑦 + 𝑒𝑥𝑧 ℎ(𝑧) = 0 𝑥𝑦𝑒𝑧𝑦 + 𝑒𝑥𝑦 + 𝑦𝑥𝑒𝑥𝑧 + ℎ(𝑥) = 𝑒𝑥𝑦 + 𝑥𝑦𝑒𝑧𝑦 + 𝑥𝑦𝑒𝑥𝑧 ℎ(𝑥) = 0 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑒𝑧𝑦 + 𝑧𝑒𝑥𝑦 + 𝑦𝑒𝑥𝑧 ∫ 𝐹 ∙ 𝑑𝑟 = 𝑓(𝑝𝑓) − 𝑓(𝑝𝑖) ∫ 𝐹 ∙ 𝑑𝑟 = 𝑓(1,1,0) − 𝑓(0,0,0) 𝑓(1,1,0) − 𝑓(0,0,0) = 2 𝑓𝑥 = cos 𝑦 + 𝑧 cos 𝑥 − 𝑧 𝑠𝑒𝑛 𝑥 𝑠𝑒𝑛 𝑦 (1) 𝑓𝑦 = 𝑧 cos 𝑥 cos 𝑦 − 𝑥 sen 𝑦 (2) 𝑓𝑧 = 𝑠𝑒𝑛 𝑥 + cos 𝑥 𝑠𝑒𝑛 𝑦 (3) ∫ cos 𝑦 + 𝑧 cos 𝑥 − 𝑧 𝑠𝑒𝑛 𝑥 𝑠𝑒𝑛 𝑦 𝑑𝑥 = 𝑥 cos 𝑦 + 𝑧𝑠𝑒𝑛 𝑥 + 𝑧𝑐𝑜𝑠 𝑥 𝑠𝑒𝑛 𝑦 + 𝑔(𝑦, 𝑧) (4) Derivamos 4 con respecto a y, z: 𝑓𝑦 = −𝑥 sen 𝑦 + 𝑧 𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 𝑦 + 𝑔(𝑧) (5) 𝑓𝑧 = 𝑠𝑒𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 𝑠𝑒𝑛 𝑦 + 𝑔(𝑦) (6) Igualamos (5) con (2) y (6) con (3): 0 = 𝑔(𝑧) 𝑔(𝑦) = 0 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 cos 𝑦 + 𝑧𝑠𝑒𝑛 𝑥 + 𝑧𝑐𝑜𝑠 𝑥 𝑠𝑒𝑛 𝑦 𝑟(𝜃) = (𝜋 cos 𝜃 , 𝜋 𝑠𝑒𝑛 𝜃, 𝜃) 𝑟(0) = (𝜋, 0, 0) 𝑟(2𝜋) = (𝜋, 0, 2𝜋) ∫ 𝐹 ∙ 𝑑𝑟 = 𝑓(𝜋, 0, 2𝜋) − 𝑓(𝜋, 0, 0) = 0 𝑓𝑥 = 𝑦𝑧 (1) 𝑓𝑦 = 𝑥𝑧 (2) 𝑓𝑧 = 𝑥𝑦 (3) ∫ 𝑦𝑧 𝑑𝑥 = 𝑥𝑦𝑧 + ℎ(𝑦, 𝑧) (4) Derivamos 4 por y, z 𝑥𝑧 + ℎ(𝑧)(5) 𝑥𝑦 + ℎ(𝑦)(6) 𝑥𝑧 = 𝑥𝑧 + ℎ(𝑧); ℎ(𝑧) = 0 𝑥𝑦 = 𝑥𝑦 + ℎ(𝑦); ℎ(𝑦) = 0 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧 ∫ 𝐹 ∙ 𝑑𝑟 = 𝑓(𝑝𝑓) − 𝑓(𝑝𝑖) ∫ 𝐹 ∙ 𝑑𝑟 = 𝑓 (𝑒−1 , 0, 1 2 ) − 𝑓(1,0,1) = 0 𝐹(𝑥, 𝑦) = (𝑠𝑒𝑛 𝑦 + ln √𝑥2 + 9 , 𝑥 − 𝑒√𝑦 2+9) 𝑃 = 𝑠𝑒𝑛 𝑦 + ln √𝑥2 + 9 ; 𝑃𝑦 = cos 𝑦 𝑄 = 𝑥 − 𝑒√𝑦 2+9; 𝑄𝑥 = 1 𝜋 − 𝑥 ∫ ∫ 1 − cos 𝑦 𝑑𝑦𝑑𝑥 𝜋−𝑥 0 𝜋 0 = ∫ [𝑦 − sen 𝑦] 𝜋 − 𝑥 0 𝑑𝑥 𝜋 0 = ∫ [𝜋 − 𝑥 − sen(𝜋 − 𝑥)]𝑑𝑥 𝜋 0 = [𝜋𝑥 − 𝑥2 2 − cos (𝜋 − 𝑥)] 𝜋 0 = 𝜋2 − 𝜋2 2 − cos(𝜋 − 𝜋) − (− cos(𝜋)) = 𝜋2 − 𝜋2 2 − 2 𝑃 = (−3𝑥𝑦 + √4𝑥2 + 9) ; 𝑃𝑌 = −3𝑥 𝑄 = (𝑥 − 𝑦√𝑦2 + 9) ; 𝑄𝑥 = 1 ∬(1 + 3𝑥)𝑑𝐴 Región tipo 1 0 ≤ 𝑋 ≤ 2 𝑥3 ≤ 𝑦 ≤ 8 ∫ ∫ (1 + 3𝑥)𝑑𝑦𝑑𝑥 8 𝑥3 2 0 = ∫ [(𝑦 + 3𝑥𝑦)] 8 𝑥3 2 0 𝑑𝑥 = ∫ [(8 + 24𝑥) − (𝑥3 + 3𝑥4)] 8 𝑥3 2 0 𝑑𝑥 = [8𝑥 + 12𝑥2 − 𝑥4 4 − 3𝑥5 5 ] 2 0 = 16 + 48 − 4 − 96 5 = 204 5 𝑃 = (𝑒cos (𝑥 2)); 𝑃𝑌 = 0 𝑄 = (𝑠𝑒𝑛 (ln (𝑦2 + 20)) − 3𝑥); 𝑄𝑥 = −3 𝑥 = 𝑥2 0 = 𝑥2 − 𝑥 𝑥(𝑥 − 1) = 0 𝑥 = 0; 𝑥 = 1 0 ≤ 𝑥 ≤ 1 𝑥2 ≤ 𝑦 ≤ 𝑥 ∫ ∫ (−3)𝑑𝑦𝑑𝑥 𝑥 𝑥2 1 0 = ∫ [−3𝑦] 𝑥 𝑥2 1 0 = ∫ [−3𝑥 + 3𝑥2]𝑑𝑥 = 1 0 [ −3𝑥2 2 + 𝑥3] 1 0 = −3 2 + 1 = − 1 2 𝑃 = 𝑒𝑥 2 − cos 𝑥2 + 𝑦 ; 𝑃𝑌 = 1 𝑄 = sen 𝑦3 + ln(𝑦2 + 1) + 𝑥2 ; 𝑄𝑥 = 2𝑥 𝑥2 + 𝑦2 = 9; 𝑟 = 3 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 𝑠𝑒𝑛 𝜃 − 𝜋 2 ≤ 𝜃 ≤ 𝜋 2 ∫ ∫ 𝑟(2𝑟 cos 𝜃 − 1)𝑑𝜃𝑑𝑟 𝜋 2 − 𝜋 2 3 0 = ∫ ∫ (2𝑟2 cos 𝜃 − 𝑟)𝑑𝜃𝑑𝑟 𝜋 2 − 𝜋 2 3 0 = ∫ [2𝑟2 − 𝑟𝜃] 𝜋 2 − 𝜋 2 𝑑𝑟 3 0 = ∫ [2𝑟2 sen ( 𝜋 2 ) − 𝑟𝜋 2 ] − [2𝑟2 sen (− 𝜋 2 ) + 𝑟𝜋 2 ] 𝑑𝑟 3 0 = ∫ 4𝑟2 − 𝑟𝜋 𝑑𝑟 3 0 = [ 4 3 𝑟3 − 𝑟2𝜋 2 ] 3 0 = 36 − 9𝜋 2
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