Marco Metodologico Proyecto 1 Apli 4.docx

Marco metodológico 1.Cuadratura gaussiana para 2 variables 𝑑

𝑏

∫ ∫ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦 𝑐

𝑥=

𝑎

𝑏−𝑎 𝑏+𝑎 𝑡+ 2 2 𝑑𝑥 =

𝑦=

𝑏−𝑎 𝑑𝑡 2

𝑑−𝑐 𝑑+𝑐 𝑢+ 2 2

𝑑𝑦 =

𝑑−𝑐 𝑑𝑢 2

(𝑏 − 𝑎)(𝑑 − 𝑐) 1 1 ∫ ∫ 𝑓(𝑥, 𝑦)𝑑𝑡𝑑𝑢 4 −1 −1 Ejercicio Utilizar la cuadratura gaussiana de dos variables para encontrar el área de la siguiente función 𝑧 = 𝑥 2 + 2𝑦 con n=5.

4

3

∫ ∫ (𝑥 2 + 2𝑦)𝑑𝑥𝑑𝑦 1

𝑥=

3−2 3+2 𝑡+ 2 2 𝑥=

1 5 𝑡+ 2 2

1 𝑑𝑥 = 𝑑𝑡 2

2

𝑦=

4−1 4+1 𝑢+ 2 2

3 5 𝑦= 𝑢+ 2 2 𝑑𝑦 =

3 𝑑𝑢 2

1 5 2 3 5 1 3 ∫ ∫ [( 𝑡 + ) + 2 ( 𝑢 + )] 𝑑𝑡 𝑑𝑢 2 2 2 2 2 −1 −1 2 1

1

Ecuación objetivo 3 1 1 5 ∫ ∫ [0.25𝑡 2 + 𝑡 + 3𝑢 + 11.25] 𝑑𝑡𝑑𝑢 4 −1 −1 2

Datos "Raíz" 𝑥 → −0.906179845938664 𝑥 → −0.5384693101056831 𝑥→0 𝑥 → 0.5384693101056831 𝑥 → 0.906179845938664

"Coeficientes" 0.236926885056 0.478628670499 0.568888888889 0.478628670499 0.236926885056

Resolviendo primera parte 3 1 5 ∫ [0.25𝑡 2 + 𝑡 + 3𝑢 + 11.25] 𝑑𝑡 4 −1 2 1

∫ 𝑓(𝑡)𝑑𝑡 = 0.236926885056 (0.25(−0.9061798459386642 ) −1

5 + (−0.906179845938664) + 3𝑢 + 11.25) 2 + 0.478628670499 (0.25(−0.53846931010568312 ) 5 + (−0.906179845938664) + 3𝑢 + 11.25) 2 5 + 0.568888888889 (0.25(02 ) + (0) + 3𝑢 + 11.25) 2 + 0.236926885056 (0.25(0.9061798459386642 ) 5 + (0.906179845938664) + 3𝑢 + 11.25) 2 + 0.478628670499 (0.25(0.53846931010568312 ) 5 + (0.906179845938664) + 3𝑢 + 11.25) 2

1

∫ 𝑓(𝑡)𝑑𝑡 = 22.666666666655285  + 5.999999999997𝑢 −1

Sustituyendo en ecuación objetivo 3 1 ∫ (22.666666666655285  + 5.999999999997𝑢)𝑑𝑢 4 −1

3 1 3 ∫ 𝑔(𝑢)𝑑𝑢 = (0.236926885056(22.666666666655285  4 −1 4 + 5.999999999997(−0.906179845938664))) + (0.478628670499(22.666666666655285  + 5.999999999997(−0.5384693101056831))) + (0.568888888889(22.666666666655285  + 5.999999999997(0))) + (0.236926885056(22.666666666655285  + 5.999999999997(0.906179845938664))) + (0.478628670499(22.666666666655285  + 5.999999999997(0.5384693101056831)))

3 1 ∫ 𝑔(𝑢)𝑑𝑢 = 33.99999999996593 4 −1 Resultado 4

3

∫1 ∫2 (𝑥 2 + 2𝑦)𝑑𝑥𝑑𝑦 = 33.99999999996593

Grafica

2.Cuadratura gaussiana para 3 variables 𝑓

𝑑

𝑏

∫ ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑥𝑑𝑦𝑑𝑧 𝑒

𝑥=

𝑐

𝑏−𝑎 𝑏+𝑎 𝑡+ 2 2 𝑑𝑥 =

𝑎

𝑦=

𝑏−𝑎 𝑑𝑡 2

𝑑−𝑐 𝑑+𝑐 𝑢+ 2 2

𝑑𝑦 =

𝑧=

𝑑−𝑐 𝑑𝑢 2

𝑒−𝑓 𝑒+𝑓 𝑣+ 2 2

𝑑𝜃 =

𝑒−𝑓 𝑑𝑣 2

(𝑏 − 𝑎)(𝑑 − 𝑐)(𝑒 − 𝑓) 𝑓 𝑑 𝑏 ∫ ∫ ∫ 𝑓(𝑥, 𝑦) 𝑑𝑥𝑑𝑦𝑑𝑧 6 𝑒 𝑐 𝑎 Ejercicio Utilizar cuadratura gaussiana para aproximar el volumen debajo de la curva acotada con la integral 4

(𝑥 2 +2𝑦)

3

(𝑥 2 + 2𝑦) 𝑑𝑧𝑑𝑥𝑑𝑦

∫ ∫ ∫ 1

1 5 𝑥= 𝑣+ 2 2

2

0

3 5 𝑦= 𝑢+ 2 2

1 𝑑𝑥 = 𝑑𝑣 2

3 𝑑𝑦 = 𝑑𝑢 2

𝑥2 + 𝑦2 𝑥2 + 𝑦2 𝑧= 𝑡+ 2 2 𝑥2 + 𝑦2 𝑑𝑧 = 𝑑𝑡 2

Función objetivo 1

1

1

∫ ∫ ∫ ((0.5𝑡 + (5⁄2))2 + 3𝑢 + 5) ∗ (0.5 ∗ ((0.5𝑡 + (5⁄2))2 ) + 3𝑢 + 5) 𝑑𝑡𝑑𝑢𝑑𝑣 −1 −1 −1

Datos "Raíz" 𝑥 → −0.906179845938664 𝑥 → −0.5384693101056831 𝑥→0 𝑥 → 0.5384693101056831 𝑥 → 0.906179845938664 Resolviendo

"Coeficientes" 0.236926885056 0.478628670499 0.568888888889 0.478628670499 0.236926885056

Primera parte 1

∫ ((0.5𝑡 + (5⁄2))2 + 3𝑢 + 5) ∗ (0.5 ∗ ((0.5𝑡 + (5⁄2))2 ) + 3𝑢 + 5) 𝑑𝑡 −1

= 0.236926885056 ∗ (((0.5 ∗ −0.906179845938664 + (5⁄2))2 + 3𝑢 + 5) ∗ (0.5 ∗ ((0.5 ∗ −0.906179845938664 + (5⁄2))2 ) + 3𝑢 + 5)) + 0.478628670499 ∗ (((0.5 ∗ −0.5384693101056831 + (5⁄2))2 + 3𝑢 + 5) ∗ (0.5 ∗ ((0.5 ∗ −0.5384693101056831 + (5⁄2))2 ) + 3𝑢 + 5)) + 0.568888888889 ∗ (((0.5 ∗ 0 + (5⁄2))2 + 3𝑢 + 5) ∗ (0.5 ∗ ((0.5 ∗ 0 + (5⁄2))2 ) + 3𝑢 + 5)) + 0.236926885056 ∗ (((0.5 ∗ 0.906179845938664 + (5⁄2))2 + 3𝑢 + 5) ∗ (0.5 ∗ ((0.5 ∗ 0.906179845938664 + (5⁄2))2 ) + 3𝑢 + 5)) + 0.478628670499 ∗ (((0.5 ∗ −0.5384693101056831 + (5⁄2))2 + 3𝑢 + 5) ∗ (0.5 ∗ ((0.5 ∗ 0.5384693101056831 + (5⁄2))2 ) + 3𝑢 + 5))

1

1

∫ ∫ (123.81577941575648  + 80.0674305832622𝑢 −1 −1

+ 168.88000000002046𝑢2 )𝑑𝑢𝑑𝑣

Segunda parte 1

∫ (123.81577941575648  + 80.0674305832622𝑢 + 168.88000000002046𝑢2 )𝑑𝑢 −1

= 0.236926885056 ∗ (123.81577941575648 + 80.0674305832622 ∗ −0.906179845938664 + 168.88000000002046 ∗ (−0.906179845938664)2 ) + 0.478628670499 ∗ (123.81577941575648 + 80.0674305832622 ∗ −0.5384693101056831 + 168.88000000002046 ∗ (−0.5384693101056831)2 ) + 0.568888888889 ∗ (123.81577941575648 + 80.0674305832622 ∗ 0. +168.88000000002046 ∗ (0)2 ) + 0.236926885056 ∗ (123.81577941575648 + 80.0674305832622 ∗ 0.906179845938664 + 168.88000000002046 ∗ (0.906179845938664)2 ) + 0.478628670499 ∗ (123.81577941575648 + 80.0674305832622 ∗ 0.5384693101056831 + 168.88000000002046 ∗ (0.5384693101056831)2 ) 1

∫ 360.2182254979811𝑑𝑣 −1

Tercera parte 1

∫ 360.2182254979811𝑑𝑣 −1

= 0.236926885056 ∗ 360.2182254979811 + 0.478628670499 ∗ 360.2182254979811 + 0.568888888889 ∗ 360.2182254979811 + 0.236926885056 ∗ 360.2182254979811 + 0.478628670499 ∗ 360.2182254979811 Respuesta

1

1

1

∫ ∫ ∫ ((0.5𝑡 + (5⁄2))2 + 3𝑢 + 5) ∗ (0.5 ∗ ((0.5𝑡 + (5⁄2))2 ) + 3𝑢 + 5) 𝑑𝑡𝑑𝑢𝑑𝑣 −1 −1 −1

= 720.436450995602