Divergence theorem
A novice might find a proof easier to follow if we greatly restrict the conditions of the theorem, but carefully explain each step. For that reason, we prove the divergence theorem for a rectangular box, using a vector field that depends on only one variable.
 See Wikipedia:Divergence theorem for more rigorous proofs.
 See the subpage, Divergence theorem/Proof for another proof.
The Divergence (GaussOstrogradsky) theorem relates the integral over a volume, , of the divergence of a vector function, , and the integral of that same function over the the volume's surface:

(1)

Special case of a field that depends on only one variable:
[edit  edit source]Figure 2 shows a vector field using red arrows to denote the field's direction and magnitude. The field occupies all three dimensions, but is only directed in the x or y directions. And, the field varies only along the xaxis. In other words, is of the form,

(2)

where and are arbitrary "smooth" functions of only the variable A zcomponent, , could be included in (2) without adding any complexity to the proof, except that the third component could not easily be depicted in Figure 2.
Volume integral
[edit  edit source]Our first task is to use the fundamental theorem of calculus to perform this integral over the volume of the box. Despite the fact that the vector field points in two directions, the divergence operator removes any involvement of the ycomponent, since the field does not depend on :

(3)

Integrals over threedimensional boxes are easy to perform:

(4)

Since does not depend on or the partial derivative may be treated as a constant with regard to those to variables. The integrations in (4) are structured so that they may be performed in any order, which permits us to write:

,
(5)

where is the area of the two sides at and as shown in Figure 2.
Surface integrals
[edit  edit source]Now we calculate the surface integral and verify that it yields the same result as (5). The surface integral over the box involves six rectangles. One way to define a twodimensional surface in threedimensions is to state one equation involving the three variables.^{[1]} The six surfaces of the box in Figure 2 can be labeled by the set of six (infinite) planes: . In contrast with onedimensional integrals, surface and volume integrals require much more than simply stating the two endpoints associated with a one dimensional integral. Each of the six surfaces must be integrated separately, and the integral over the entire surface equals the sum of the six individual integrals. A common shorthand for surface integrals is, , where the outward unit normal, and

(6)

Integration over xpair surfaces
[edit  edit source]The two most important surfaces are the pair already discussed, with area , and outward unit normals in the xdirection, shown as the left and right ends of the box shown in Figure 2. Since is constant over each of these two surfaces, the the surface integral of anything that depends only on is trivial, and depends only on the area, which is the same area, , shown in Figure 2 and also at (5). The algebra for the surface at is:

.
(7)

To understand how was evaluated in (7), recall that the outward unit normal points in the negative x direction on that side. Also, when a vector is dotted with a unit vector, only one component of the vector survives:^{[2]}

,
(8)

since, , at On the other side, points in the opposite direction, so that combining (5) with (7) establishes that (1) is true, but only if we can establish that the other four surface integrals vanish.
Integration over the ypair and zpair surfaces
[edit  edit source]We begin with the ypair of surfaces, i.e., those that occupy a portion of the or the planes. Since the vector field depends on the variable, the integral is not trivial. By the reasoning used at (7) and (8), we only need integrate the component, , because the unit vector points in the direction. Fortunately, the absence of a dependence of our vector field ensures that the following two integrals are equal:

(9)

Since the outward unit normals, are equal and opposite on both sides, the pair of surface integrals at (9) cancel. The same argument can be the pair of surfaces.
Conclusion
[edit  edit source]Although we have proven the divergence theorem on a rectangular box for a small subset of all possible differentiable vector fields , we have established the essential role played by the fundamental theorem of calculus in onedimension. Essentially, we have placed the two endpoints of segment along the xaxis by the six surfaces that form the boundary of a threedimensional box. Also, the reader who can grasp this proof is probably capable of developing a proof for any vector field for which the divergence is well defined. It just takes a lot of algebra.
Footnotes
[edit  edit source]Subpages: