A2-CINTEGRAL-RAÚL ANDRÉS GUILLÉN RANGEL-IMKT-MARZO-JULIO 2021

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Raúl Andrés Guillén Rangel 20030941 
Page | 1 
 
 
 
 
 
 
 
 
INSTITUTO TECNOLÓGICO DE CELAYA 
INGENIERÍA MECATRÓNICA 
GRUPO A 
CÁLCULO INTEGRAL 
SARA MARCELA ARELLANO DÍAZ 
RAÚL ANDRÉS GUILLÉN RANGEL 
No. De Control 20030941 
VOLUMEN UTILIZANDO CAPAS CIILÍNDRICAS 
Raúl Andrés Guillén Rangel 20030941 
Page | 2 
 
1. 𝑦 =
1
√2𝜋
𝑒−
𝑥2
2 , 𝑦 = 0, 𝑥 = 0, 𝑥 = 1 
 
𝑟 = 𝑥 − 0 
ℎ =
1
√2𝜋
𝑒−
𝑥2
2 − 0 
𝑉 = 2𝜋 ∫ [(𝑥)(
1
√2𝜋
𝑒−
𝑥2
2 )] 𝑑𝑥
1
0
 
𝑉 =
2𝜋
√2𝜋
∫ (𝑥𝑒−
𝑥2
2 )
1
0
𝑑𝑥 
𝑉 =
2𝜋
√2𝜋
[𝑒−
𝑥2
2 ]|0
1 
𝑉 = 0.98𝑢3 
 
 
 
 
 
 
Raúl Andrés Guillén Rangel 20030941 
Page | 3 
 
2. 𝑦 =
1
2
𝑥2+ 1 
 
𝑟 = 𝑥 − 0 
ℎ = 3 − (
1
2
𝑥2+ 1) 
𝑉 = 2𝜋 ∫ [(𝑥) (3 − (
1
2
𝑥2 + 1))] 𝑑𝑥
2
0
 
𝑉 = 2𝜋 ∫ [−
1
2
𝑥3 + 2𝑥] 𝑑𝑥
2
0
 
𝑉 = 2𝜋[−
1
8
𝑥4+ 𝑥2]|0
2 
𝑉 = 12.56𝑢3 
 
 
 
 
 
 
Raúl Andrés Guillén Rangel 20030941 
Page | 4 
 
3. 𝑦 = 1 − 𝑥 
 
𝑟 = 0 − 𝑦 
ℎ = 3 − (1 − 𝑦) 
𝑉 = 2𝜋 ∫ [(−𝑦)(3 − (1 − 𝑦))]𝑑𝑦
0
−2
 
𝑉 = 2𝜋 ∫ [(−𝑦)(2 + 𝑦)]𝑑𝑦
0
−2
 
𝑉 = 2𝜋∫ [−2𝑦 − 𝑦2)]𝑑𝑦
0
−2
 
𝑉 = [−𝑦2 −
1
3
𝑦3]|−2
0 
𝑉 = 8.377𝑢3 
 
 
 
 
 
Raúl Andrés Guillén Rangel 20030941 
Page | 5 
 
4. (𝑦 − 2)2 = 4 − 𝑥 
 
𝑟 = 𝑦 − 0 
ℎ = 4 − (𝑦 − 2)2 
𝑉 = 2𝜋 ∫ [(𝑦)(4 − (𝑦 − 2)2)]𝑑𝑦
4
0
 
𝑉 = 2𝜋 ∫ [−𝑦3 + 4𝑦2]𝑑𝑦
4
0
 
𝑉 = [−
1
4
𝑦2 +
4
3
𝑦3]|0
4 
𝑉 = 134.04𝑢3 
 
 
 
 
 
 
Raúl Andrés Guillén Rangel 20030941 
Page | 6 
 
5. 
I. 
 
a. 
𝑉 = 𝜋 ∫ [(𝑥3)2]𝑑𝑥
2
0
 
𝑉 = [
1
7
𝑥7]|0
2 
𝑉 = 57.44𝑢3 
 
b. 
𝑉 = 2𝜋∫ [(𝑥)(𝑥3)]𝑑𝑥
2
0
 
𝑉 = [
1
5
𝑥5]|0
2 
𝑉 = 40.21𝑢3 
 
Raúl Andrés Guillén Rangel 20030941 
Page | 7 
 
c. 
𝑉 = 2𝜋 ∫ [(4 − 𝑥)(𝑥3)]𝑑𝑥
2
0
 
𝑉 = [𝑥4−
1
5
𝑥5] |0
2 
𝑉 = 60.31𝑢3 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Raúl Andrés Guillén Rangel 20030941 
Page | 8 
 
II. 
 
a. 
𝑉 = 𝜋 ∫ [(
10
𝑥2
)2] 𝑑𝑥
5
1
 
𝑉 = 𝜋 ∫ [
100
𝑥4
] 𝑑𝑥
5
1
 
𝑉 = [−
100
3𝑥3
] |1
5 
𝑉 = 103.88𝑢3 
 
 
 
 
b. 
𝑉 = 2𝜋 ∫ [(𝑥)(
10
𝑥2
)] 𝑑𝑥
5
1
 
Raúl Andrés Guillén Rangel 20030941 
Page | 9 
 
𝑉 = 2𝜋 ∫ [
10
𝑥
] 𝑑𝑥
5
1
 
𝑉 = 20𝜋[ln⁡(𝑥)]|1
5 
𝑉 = 101.12𝑢3 
 
c. 
𝑉 = 𝜋∫ [(10)2 − (10 −
10
𝑥2
)2] 𝑑𝑥
5
1
 
𝑉 = 𝜋 ∫ [
200
𝑥2
+
100
𝑥4
] 𝑑𝑥
5
1
 
𝑉 = 𝜋[−
200
𝑥
+
100
3𝑥3
]|1
5 
𝑉 = 398.98𝑢3 
 
 
 
 
 
 
 
 
 
 
 
III. 
Raúl Andrés Guillén Rangel 20030941 
Page | 10 
 
 
a. 
𝑉 = 2𝜋 ∫ [(𝑦)(√𝑎 − √𝑦)2]𝑑𝑥
9
0
 
𝑉 = 2𝜋 ∫ [𝑎𝑦 − 2𝑦√𝑎√𝑦 + 𝑦2]𝑑𝑥
9
0
 
𝑉 = 2𝜋[
1
2
𝑎𝑦2 −
4
5
√𝑎𝑦
5
2 +
1
3
𝑦3]|0
9 
𝑉 =
𝜋𝑎3
15
𝑢3 
 
b. 
𝑉 = 2𝜋 ∫ [(𝑥)(√𝑎 − √𝑥)2]𝑑𝑥
9
0
 
𝑉 = 2𝜋∫ [𝑎𝑥 − 2𝑥√𝑎√𝑥 + 𝑥2]𝑑𝑥
9
0
 
𝑉 = 2𝜋[
1
2
𝑎𝑥2−
4
5
√𝑎𝑥
5
2+
1
3
𝑥3]|0
9 
𝑉 =
𝜋𝑎3
15
𝑢3 
 
 
Raúl Andrés Guillén Rangel 20030941 
Page | 11 
 
c. 
𝑉 = 2𝜋 ∫ [(𝑎 − 𝑥)(√𝑎 − √𝑥)2]𝑑𝑥
9
0
 
𝑉 = 2𝜋 ∫ [𝑎2 − 2𝑎√𝑎√𝑥 + 2𝑥√𝑎√𝑥 − 𝑥2]𝑑𝑥
9
0
 
𝑉 = 2𝜋 [𝑥𝑎2 −
4
3
𝑎√𝑎𝑥
3
2+
4
5
√𝑎𝑥
5
2−
1
3
𝑥3] |0
9 
𝑉 =
4𝜋𝑎3
15
𝑢3 
 
6. 
a. 
𝑉 = 𝜋 ∫ [√𝑥(4 − 𝑥)2
2
] 𝑑𝑥
4
0
 
𝑉 = 𝜋∫ [16𝑥 − 8𝑥2+ 𝑥3]𝑑𝑥
4
0
 
𝑉 = 𝜋[8𝑥2−
8
3
𝑥3+
1
4
𝑥4]|0
4 
𝑉 = 67.02𝑢3 
 
b. 
𝑉 = 2𝜋∫ [(𝑥)(√𝑥(4 − 𝑥)2)] 𝑑𝑥
4
0
 
𝑉 = 2𝜋 ∫ [4𝑥
3
2− 𝑥
5
2] 𝑑𝑥
4
0
 
𝑉 = 2𝜋[
8
5
𝑥
5
2−
2
7
𝑥
7
2]|0
4 
𝑉 = 183.82𝑢3 
Raúl Andrés Guillén Rangel 20030941 
Page | 12 
 
c. 
𝑉 = 2𝜋 ∫ [(4 − 𝑥)(√𝑥(4 − 𝑥)2)] 𝑑𝑥
4
0
 
𝑉 = 2𝜋 ∫ [16𝑥
1
2− 8𝑥
3
2+ 𝑥
5
2]𝑑𝑥
4
0
 
𝑉 = 2𝜋[
32
3
𝑥
3
2−
16
5
𝑥
5
2 +
2
7
𝑥
7
2]|0
4 
𝑉 = 245𝑢3