Note that there is no assumption that r is one to one or that Dr has full rank. Thus the
image of r can have all sorts of creases and points and such things. Also, it is not necessary to
assume that r is C^{2}. It suffices to assume it is only C^{1}. This can be concluded by
approximating such a C^{1} function with a sequence of C^{2} functions such that they and their
first derivatives converge uniformly to r. Then you apply the theorem to each of the
approximating sequence. Passing to the limit gives the same thing as above. In fact, you can
generalize beyond C^{1} to Lipschitz continuous functions but this requires a little more analysis
so these considerations will not be developed here. In addition, it may be possible to
generalize even further but then you would be getting in to things which are very
specialized.
The explicit description of the boundary integral is
There would have been no change in any of the arguments if the unit cube
[0,1]
^{n} were
replaced with an arbitrary box ∏_{i=1}^{n}
[a ,b ]
i i
. In this case, the boundary integral would take
the form
∑n ∫ ∑ ( )
∏ aI (r (u ))(− 1)1+l-∂-xi1,⋅⋅⋅,xin−1--dul|blal
l=1 j⁄=l[aj,bj]I=(xi,⋅⋅⋅,xi ) ∂ (u1,⋅⋅⋅,ˆul⋅⋅⋅,un)
1 n−1
Also note that the boundary integral is represented as an integral over a side of a box. What
about the edges of this side? Do they matter? In fact, they don’t. You could integrate over
∏_{j≠l}
(aj,bj)
just as well.
Another thing to notice is that you could consider a chain of these boxes such that each
box intersects another along a complete face. Then you could do the above for each box
in the chain and note that the boundary integrals will cancel along the common
faces. Thus, in place of a single box you could have a much more complicated shape
and the boundary integral would take place over exactly those faces which do not
have intersection with faces of other boxes. The following picture illustrates what is
meant.