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17 Feb 2010 15:29:23 IST

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what is the general proof of gauss divergence theorm

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what is the general proof of gauss divergence theorm

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Black Gene

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Joined: 25 Dec 2009

Posts: 65

17 Feb 2010 15:31:16 IST

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Just see if u can understand it .............

Given a closed surface S, with an outward pointing unit normal, n, the Divergence Theorem says:

We take

F = Mi + Nj + Pk

We will perform this proof only for Convex regions. That is for regions like the one pictured below:

Using the geometric definition of the dot product, and because n is a unit normal, we can dot it with the basis vectors to get the angles between n and the axes. That is:

n (dot) i = cos A - The angle from n to the x-axis
n (dot) j = cos B - The angle from n to the y-axis
n (dot) k = cos C - The angle from n to the z-axis

n = cos A i + cos B j + cos C k since the dot product of n with each basis vector is that component of n.

Now we restate the Divergence Theorem using the components of the vector field F. Next we begin the proof.

The last step involves taking the integral of the derivitive of P. This is like multiplying by one. It doesn't change anything, you can do it anytime.

Looking at the last term, what we really have is a volume integral. Evaluating the inner integral, we have [ f1(x,y) - f2(x,y) ] * dA. This is the volume of a differential element of the region. Adding all of them up gives us the volume of the region:

This is the result we required. Thus we have proved the z terms are equal.
The other terms work out in a similar mannar. If the region we have is not a convex region, we can costruct it by placing convex regions next to each other and summing up the results. Thus, we have proved the Divergence Theorem for a large number of regions.

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