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Cálculo de volumen de un solido
mediante doble integrales
Ramírez León Hugo Fernando
Hallar el volumen de la región sólida dada por:
𝑧 = 4 − 𝑥 − 𝑦
Δ𝑦 𝑉 =
0
2
0
𝑦
(4 − 𝑥 − 𝑦) ⅆ𝑥 ⅆ𝑦
𝑉 =
0
2
[4𝑥 − 𝑥2
2
−𝑥𝑦 ]0
𝑦
ⅆ𝑦
𝑉 =
0
2
4𝑦 −
𝑦2
2
− 𝑦2
ⅆ𝑦
𝑉 = [4(y) −
(𝑦)2
2
− (y)y]
= [2𝑦2 −
𝑦3
6
−
𝑦3
3
]0
2
= [2(2)2
−
2 3
6
−
(2)3
2
]- [2(0)2
−
0 3
6
−
(0)3
3
]
= 8 -
8
6
−
8
3
=
4 𝑢𝑛𝑖ⅆ𝑎ⅆ𝑒𝑠 𝑐𝑢𝑏𝑖𝑐𝑎𝑠
▪ Hallar el volumen de
la región solida R
acotada por:
▪ 𝑧 = 4 − 𝑥2
− 𝑦2
e inferior por el
plano:
▪ 𝑧 = 1 − 𝑦
𝑥2
+ 𝑦2
= 4
▪ Igualando los valores de z ,
sabremos la intercesión de las dos
superficies
4 − 2𝑥 = 4 − 𝑥2 − 𝑦2
Despejando “y”
∴ 𝑦 = 1 − (𝑥 − 1)2
𝑦 = 2𝑥 − 𝑥2 →
𝑉 =
0
2
0
1−(𝑥−1)2
4 − 𝑥2 − 𝑦2 − 4 − 2𝑥 ⅆ𝑦ⅆ𝑥
𝑉 =
0
2
0
1−(𝑥−1)2
(2𝑥 − 𝑥2 − 𝑦2)ⅆ𝑦ⅆ𝑥
𝑉= 0
2
[2xy−𝑥2 −
𝑦3
3
]0
1−(𝑥−1)2
𝑉 =
0
2
2𝑥 1 − (𝑥 − 1)2− 𝑥2 1 − 𝑥 − 1 2 −
( 1 − 𝑥 − 1 2)3
3
ⅆ𝑥
𝑉
=
1
2
𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 +
1
2
𝑠𝑒𝑛 2𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 +
1
2
𝑠𝑒𝑛 2𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1
−
1
8
𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 −
1
4
sen 4arcsen x − 1
−
1
3
((𝑥 − 1)(−𝑥2 + 2𝑥)
3
2 +
3
8
(𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 −
1
4
(4arcsen(x-1))) ]0
2
Escriba aquí la ecuación.
𝑉 =
𝜋
8
− −
𝜋
8
=
𝜋
4

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Área de solido integrales multiples, ron larson

  • 1. Cálculo de volumen de un solido mediante doble integrales Ramírez León Hugo Fernando
  • 2. Hallar el volumen de la región sólida dada por: 𝑧 = 4 − 𝑥 − 𝑦
  • 3. Δ𝑦 𝑉 = 0 2 0 𝑦 (4 − 𝑥 − 𝑦) ⅆ𝑥 ⅆ𝑦
  • 4. 𝑉 = 0 2 [4𝑥 − 𝑥2 2 −𝑥𝑦 ]0 𝑦 ⅆ𝑦 𝑉 = 0 2 4𝑦 − 𝑦2 2 − 𝑦2 ⅆ𝑦 𝑉 = [4(y) − (𝑦)2 2 − (y)y]
  • 5. = [2𝑦2 − 𝑦3 6 − 𝑦3 3 ]0 2 = [2(2)2 − 2 3 6 − (2)3 2 ]- [2(0)2 − 0 3 6 − (0)3 3 ] = 8 - 8 6 − 8 3 = 4 𝑢𝑛𝑖ⅆ𝑎ⅆ𝑒𝑠 𝑐𝑢𝑏𝑖𝑐𝑎𝑠
  • 6. ▪ Hallar el volumen de la región solida R acotada por: ▪ 𝑧 = 4 − 𝑥2 − 𝑦2 e inferior por el plano: ▪ 𝑧 = 1 − 𝑦 𝑥2 + 𝑦2 = 4
  • 7. ▪ Igualando los valores de z , sabremos la intercesión de las dos superficies 4 − 2𝑥 = 4 − 𝑥2 − 𝑦2 Despejando “y” ∴ 𝑦 = 1 − (𝑥 − 1)2 𝑦 = 2𝑥 − 𝑥2 →
  • 8. 𝑉 = 0 2 0 1−(𝑥−1)2 4 − 𝑥2 − 𝑦2 − 4 − 2𝑥 ⅆ𝑦ⅆ𝑥 𝑉 = 0 2 0 1−(𝑥−1)2 (2𝑥 − 𝑥2 − 𝑦2)ⅆ𝑦ⅆ𝑥 𝑉= 0 2 [2xy−𝑥2 − 𝑦3 3 ]0 1−(𝑥−1)2
  • 9. 𝑉 = 0 2 2𝑥 1 − (𝑥 − 1)2− 𝑥2 1 − 𝑥 − 1 2 − ( 1 − 𝑥 − 1 2)3 3 ⅆ𝑥 𝑉 = 1 2 𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 + 1 2 𝑠𝑒𝑛 2𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 + 1 2 𝑠𝑒𝑛 2𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 − 1 8 𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 − 1 4 sen 4arcsen x − 1 − 1 3 ((𝑥 − 1)(−𝑥2 + 2𝑥) 3 2 + 3 8 (𝑎𝑟𝑐𝑠𝑒𝑛 𝑥 − 1 − 1 4 (4arcsen(x-1))) ]0 2 Escriba aquí la ecuación. 𝑉 = 𝜋 8 − − 𝜋 8 = 𝜋 4