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Having Hard Time using Greens Formulas

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0
1. Homework Statement
Evaluate:

[tex]
\int\int \phi \nabla(\frac{1}{R}) - \frac{1}{R}\nabla\phi * \vec{n} dS
[/tex]

a) over the surface of the ellipsoid
[tex]1/9\,{x}^{2}+1/16\,{y}^{2}+1/25\,{z}^{2}=1
>
[/tex]
where
[tex]
\phi={x}^{2}+{y}^{2}-2\,{z}^{2}+4
[/tex]


2. Homework Equations



3. The Attempt at a Solution
We are supposed to use one greens formulas for this, but I have no clue how to even approach this one. The formulas aren't stated anywhere clearly because the section is just one long proof. If anyone could please help me get started on this I would appreciate it greatly.
 

Answers and Replies

olgranpappy
Homework Helper
1,271
3
[tex]
\int_{\partial M}\left(\phi\nabla(\frac{1}{R})-\frac{1}{R}\nabla\phi\right)\cdot\hat n dS
=\int_{M}\left(\phi\nabla^2(\frac{1}{R})-\frac{1}{R}\nabla^2\phi\right) d^3x
[/tex]
is green's identity.

you also might like to use [itex]\nabla^2(1/R)=-4\pi\delta^3(\vec R)[/itex]
 
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thank you for the response! That looks like the proofs, and it is similar to what I was going to do, but the back of the book says to do this:

[tex] -4\pi\phi(0,1,0) = -20\pi [/tex].

Do you know how they might have come to this?
 
olgranpappy
Homework Helper
1,271
3
well... maybe. it would be easier to say for sure if you could tell me exactly what "R" is in your equations--a difference between two vectors, one of which I supposed is integrated over, and one of which is... perhaps the point (0,1,0)?... I don't know, because you didn't say...
 
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Sorry about that; R is the magnitude of the position vector:
[tex] \vec{R} = x\vec{i} + y\vec{j} + z\vec{k}[/tex]
so
[tex] R = \sqrt{x^{2} + y^{2}+z^{2}} [/tex]

Edit: I'm also not sure if this will be helpful at all (incase someone has the same book) but it is from the book 'Introduction to Vector Analysis, Seventh Edition, Henrry Davis, Arthur David Snider'
 
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I've been trying to figure this out, but i still cannot understand it at all. The Greens formula which I am trying to use is:

[tex]\phi(\vec{R}) = \frac{-1}{4\pi}\int\int\int \frac{ \nabla^{2}\phi(\vec{R'})}{|\vec{R}-\vec{R'}|}dV'
+
\frac{1}{4\pi}\int\int \frac{ \nabla\phi(\vec{R'})}{|\vec{R}-\vec{R'}|} - \phi(\vec{R'})\nabla'(\frac{1}{|\vec{R}-\vec{R'}|)}dS'

[/tex]

This way I can throw out the term with the laplacian:
[tex]

\nabla^{2}(\phi)=\nabla^{2}({x}^{2}+{y}^{2}-2\,{z}^{2}+4)=0

[/tex]
. Then I have the integral that I'm looking for equal to:

[tex]4\pi\phi(\vec{R}))[/tex]
But, i don't know what R is in this case. I have tried reading the section but am having a really hard time following them. If anyone could please help me out I would appreciate it greatly. The answer to the quesiton is give a couple lines above also. Thank you for any help!
 
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I know I'm using the right relation, but how they why they used phi (0,1,0) is completely beyond me. If anyone can see why they did this and could please explain it to me I would appreciate it greatly.
 
olgranpappy
Homework Helper
1,271
3
maybe you should figure our what "R" is. Is it (0,1,0)? Obviously, you can not do the problem and wind up with a numerical answer if you dont know the numerical value of "R".
 
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I don't know what R is because the book is very vague about it. It says its just the position vector. How they got (0,1,0) I cannot figure out -- I've tried looking at the given scalar fields but I cannot find any relationship. I was hoping that someone on these forums had already seen this and could help me out in using the formula.
 
olgranpappy
Homework Helper
1,271
3
first of all, you changed the problem you were asking halfway through the thread... so, it's hard for people to help when you dont correctly state the problem. in fact, I think you have still not fully stated the problem. there is missing information.
 
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The whole problem is definitely here, but it is broken up so I apolagize. I will restate it:
Evaluate:
[tex]

\int\int \phi \nabla(\frac{1}{R}) - \frac{1}{R}\nabla\phi * \vec{n} dS

[/tex]
Over the surface of the ellipsoid:
[tex]
1/9\,{x}^{2}+1/16\,{y}^{2}+1/25\,{z}^{2}=1
[/tex]
where
[tex]

\phi={x}^{2}+{y}^{2}-2\,{z}^{2}+4

[/tex]

To do this, I am supposed to use one of greens formulas. I believe it is the third. Here it is:

[tex]
\phi(\vec{R}) = \frac{-1}{4\pi}\int\int\int \frac{ \nabla^{2}\phi(\vec{R'})}{|\vec{R}-\vec{R'}|}dV'
+
\frac{1}{4\pi}\int\int \frac{ \nabla\phi(\vec{R'})}{|\vec{R}-\vec{R'}|} - \phi(\vec{R'})\nabla'(\frac{1}{|\vec{R}-\vec{R'}|)}dS'


[/tex]

First, I multiply everything by 4*pi to get the following expression:

[tex]
4*\pi \phi(\vec{R}) = \int\int\int \frac{ \nabla^{2}\phi(\vec{R'})}{|\vec{R}-\vec{R'}|}dV'
+
\int\int \frac{ \nabla\phi(\vec{R'})}{|\vec{R}-\vec{R'}|} - \phi(\vec{R'})\nabla'(\frac{1}{|\vec{R}-\vec{R'}|)}dS'


[/tex]

Now, I will evaluate the laplacian of phi in the first integral on the right:
[tex]


\nabla^{2}(\phi)=\nabla^{2}({x}^{2}+{y}^{2}-2\,{z}^{2}+4)=0


[/tex]
This takes care of the first term, but taking it to zero. I then multiply everythig by a minus 1 to make it look like the expression in the question.I now have the following expression:



[tex]
-4*\pi \phi(\vec{R}) =
- \int\int \frac{ \nabla\phi(\vec{R'})}{|\vec{R}-\vec{R'}|} + \phi(\vec{R'})\nabla'(\frac{1}{|\vec{R}-\vec{R'}|)}dS'


[/tex]

I now have the same expression that the question is asking me to evaluate where:

[tex] R = |\vec{R}-\vec{R'}| = sqrt{x^{2}+y^{2}+z^{2}}[/tex]
[tex] R' is the variable of integration

The problem now is what R is. The book claims it is (0,1,0). But I have not a clue how they got this. From reading the section, they made it sound like R is the distance from the origin to any point on the surface, which seems incorrect because (0,1,0) is clearly not on the ellipsoid. Can anyone please help me figure this out? Any help at all is greatly appreciated!
 

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