Let
be a smooth (differentiable) three-component vector field on the three dimensional space and
is its divergence then the field divergence integral over the arbitrary three dimensional volume
equals to the integral over the surface
of the field itself projected onto the unite length vector field
always perpendicular to the surface and pointing outside the surface which contains this volume or otherwise the inner values of the field divergence make virtually no contributions to the integral over the volume i.e,
where
and the
wraps the
.
We can approximate the integral of the divergence over the volume by the finite sum by dividing densely the space inside the volume
into small cubes with the edges
and the corners
as well as approximating three of the coordinate derivatives
by their difference quotients. We will keep the edges coordinate names for the convenience even if they are equal.
We get
Let us focus on a single contribution to this sum related to the derivative with respect to a chosen coordinate for example
i.e.
for example
. For a fixed
we have
Note now that because of the alternating signs the vast majority of terms in the right sum vanish and we have
which reduces only to two terms or
where the bordering
and
with the first coordinate obviously depending on the choice of
and
are such that those points are the closed to the
surface
containing the volume
.
Also note that while
is an infinitesimal (small) element of the surface parallel to the
plane and for the unite vector
perpendicular to it
and so for the second point the right side is an approximate to the growth
of the surface integral
i.e.
,
,
Repeating the estimate for the two other dimensions and coming back to the original sum we get
so the right side is the approximate surface integral (sum over the surfaces of the cubes closest to the surface
) of the field itself projected on the outward unit vector field which proves the therem i.e.
.