The divergence theorem states that for a compact domain in with piecewise smooth boundary , then for a smooth vector field on , we have
where is the unit outward normal and is the divergence of .
In most textbooks, the divergence theorem is proved by the following strategy:
- Prove that
where is a so called type domain, i.e. for some region and . i.e. is bounded by the graphs of two functions with variables . This can be quite easily done using the fundamental theorem of calculus.
- Similarly prove
for the so called type domain and type domain respectively.
- Argue that for sufficiently nice , it can be cut into finitely many pieces , each of which are of both type , , and (the so called type domain). Thus we can compute the divergence integral by adding the integrals of the pieces and apply the previous result (using linearity). Argue that the orientation of the common face of these (if any) is opposite to each other and thus a cancellation argument will give the surface integral .
I am slightly disappointed by this approach because even for the so called type domain, we still cannot prove the divergence theorem very directly (say, by reducing to a double integral on some planar domain and apply Green’s theorem), but have to further cut it into even smaller type domains.
- We first prove (1). Let , .
Let and be the top and bottom face respectively.
As an outward normal of and is given by and respectively, it is easy to see that on the lateral face , and so
- Now we prove the remaining cases. For simplicity let us prove only (2), as (3) is similar. With the same setting as in the previous part except that , we compute
On the other hand, let be parametrized by for (with positive orientation), then the lateral face is given by
and an outward normal is then given by . From this we compute
Then by Green’s theorem,
Exercise 1 Show that
Hint: let and use the chain rule and the fundamental theorem of calculus.