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Prove. The Divergence theorem. If V is the volume bounded by a closed surface S and A is a vector function of position with continuous derivatives, then

where n is the positive (outward drawn) normal to S.

Proof. The Divergence theorem in the full generality in which it is stated is not easy to prove. However given a sufficiently simple region it is quite easily proved.

Let S be a closed surface so shaped that any line parallel to any coordinate axis cuts the surface in at most two points. We will now proceed to prove the following assertion:

Denote the projection of the surface on the xy plane by R. A line erected from within R
perpendicular to the xy plane intersects the surface at two points, at a lower surface and an upper
surface. Denote the lower surface by S_{1} and the upper surface by S_{2}. Let the equation of S_{1} be z
= f_{1}(x, y) and the equation of S_{2} be z = f_{2}(x, y). See Fig. 1. Now

Now by the Fundamental Theorem of Integral Calculus

Substituting 3) into the right member of 2) we get

For the upper surface S_{2}, dy dx = cos γ_{2} dS_{2} = k•n_{2} dS_{2} since the normal n_{2} to S_{2} makes an
acute angle γ_{2} with k.

For the lower surface S_{1}, dy dx = -cos γ_{1} dS_{1} = -k•n_{1} dS_{1} since the normal n_{1} to S_{1} makes an
obtuse angle γ_{1} with k.

Consequently

Substituting 5) and 6) into 4) we get

or

which is what we wished to prove.

In the same way, by projecting S on the other coordinate planes, we can obtain

Adding 8), 9) and 10) we get

or

The theorem can be extended to surfaces which are such that lines parallel to the coordinate axes meet them in more than two points by subdividing the region into subregions whose surfaces do satisfy this condition. The procedure is similar to the one used in Green’s theorem in the plane.

References.

Spiegel. Vector Analysis.

Taylor. Advanced Calculus.

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