Portafolio de cálculo vectorial

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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