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UBA - CienciaS :: Ver Tema - Pedido Final 28/02/2014
UBA - CienciaS
http://ubacs.com.ar/ubacs/

Pedido Final 28/02/2014
http://ubacs.com.ar/ubacs/viewtopic.php?f=62&t=3031
Página 1 de 1

Autor:  inkosoft [ 28 Feb 2014, 01:53 ]
Asunto:  Pedido Final 28/02/2014

Dejo pedido el final. Suerte a los que rinda y gracias.

Autor:  aflukes [ 28 Feb 2014, 14:32 ]
Asunto:  Re: Pedido Final 28/02/2014

1) Sea la curva resultante de intersecar con el plano . Hallar dos parametrizaciones diferentes de C que tengan la misma orientación.

2)Consideremos el campo .
a) Sea Mostrar que si la frontera de está orientada con la normal exterior, entonces:
acá es la integral sobre la frontera de B
b) Calcular la divergencia de en
c)Sea una región elemental de para la cual vale el teorema de gauss tal que está en el interior de Omega. Mostrar que si la frontera de Omega está orientada con la normal exterior, entonces:

3)
a) Probar que lla solución general de la ecuación:
con a y b constante tiene la forma:
donde una es la solución del sistema homogéneo y la otra una solución particular.

b) Determinar la solución general de la ecuación:

Autor:  inkosoft [ 28 Feb 2014, 15:27 ]
Asunto:  Re: Pedido Final 28/02/2014

Como hiciste el 2)a y c?. En el 3)a con reemplazar alcanza?

Autor:  Quimey [ 28 Feb 2014, 15:41 ]
Asunto:  Re: Pedido Final 28/02/2014

Para el 3 no alcanza con reemplazar:
Reemplazando ves que las soluciones de esa forma son, efectivamente, soluciones. Pero te falta demostrar que todas las soluciones son de esa forma.

Autor:  inkosoft [ 28 Feb 2014, 15:49 ]
Asunto:  Re: Pedido Final 28/02/2014

Y como veo eso?

Autor:  aflukes [ 28 Feb 2014, 19:28 ]
Asunto:  Re: Pedido Final 28/02/2014

Para el 3.A) tenés que parametrizar la superficie de la bola de radio e, luego, al reemplazar en la integral, se te va a cancelar el radio y muchos senos y cosenos quedando una integral sencilla que da 4pi.
Luego, para el c, tenés que definir una bola centrada en el (0,0,0) que esté contenida en la región omega, ponele que llamo a esta región W. La frontera de W es la frontera de la bola mas la frontera de omega. Como la divergencia de f es 0, entonces podés hacer Gauss con W, y por lo tanto la integral sobre la frontera de W y la frontera de omega son iguales, y como en el punto A chequeaste que la frontera de todas las bolas es 4pi, entonces la integral sobre la frontera de omega es 4pi

Autor:  Yossarian [ 28 Feb 2014, 19:29 ]
Asunto:  Re: Pedido Final 28/02/2014

Así:



Podrían haberse puesto las pilas con el 1, francamente...

Autor:  inkosoft [ 28 Feb 2014, 19:33 ]
Asunto:  Re: Pedido Final 28/02/2014

Gracias! Deja que el 1 esta bien ;)

Autor:  quino [ 01 Mar 2014, 17:22 ]
Asunto:  Re: Pedido Final 28/02/2014


Autor:  aflukes [ 01 Mar 2014, 20:52 ]
Asunto:  Re: Pedido Final 28/02/2014

Perdón! En 2a es la integral sobre la frontera de b

Autor:  Ylde [ 15 Jul 2014, 11:06 ]
Asunto:  Re: Pedido Final 28/02/2014

Una boludes seguro, pero en el uno la unica que se me ocurre es x=2 Cos (\theta), y=3 Sin(\theta), z=0.
Cual seria otra?

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