The Divergence Theorem
Here’s a really cool idea relating the flux of a vector field through a surface to the
integral of that field throughout the space enclosed by the surface. Start by considering
an arbitrary surface S. Divide up the interior into N little cuboids. The flux of F through
the surface S equals the sum of the fluxes through the surfaces of each of the N little
cuboids. In other words,

N
S
F  n dS    F  n dS
i 1
Si
If you think about it, this makes sense because anywhere one of the little cuboids touches
another little cuboid, the flux through the face of the first contributes some amount to the
total flux, but the flux through the face of the second that’s touching it is just the negative
of this amount because the unit vector of the first face is just the negative of the unit
vector of the face on the second one. Think of two little boxes touching on one face: the
normal vectors to the common face are negatives of one another, so whatever one of them
contributes to the flux the other one cancels out. This will be true for all faces of all the
little cuboids except for ones that touch the surface of S. These have no “corresponding
face” on a neighboring cuboid to cancel out the flux through the exterior face, so that
little bit of flux contributes to the total.
Since we want to see if there’s a relationship between the flux through the surface and the
vector field in the interior, we want in some way to do something with our surface
integral so that it represents a volume. The simplest way to do that is to introduce a
volume element, Vi that corresponds to each one of our little cuboids. But what the
Lord giveth, the Lord taketh away, so if we multiply part of the integral by the volume
element, we have to divide another part by the same thing. So, we do:
N
 1

F

n
dS

F  n dS  Vi


 S

Si
i 1  Vi

However, by the discussion we just went through above, you should be able to recognize
that if we let N approach infinity, the quantity in square brackets simply becomes div F
through the little piece of volume Vi . Furthermore, in the limit of N’s approaching
infinity, the sum of all the little volume elements Vi is simply the volume of the region
V, which can be found as simply the triple integral

V
dV
Of course, we don’t have just the volume; we have a scalar field permeating that volume
and hence being integrated over that volume, so we have on the right instead of just the
simple volume integral above, the triple integral

V
  F dV
Putting all this together, we’ve now found a relationship between the flux of a vector
function through a surface and the integral of the divergence of that field over the volume
enclosed by the surface:

S
F  n dS  
V
  F dV
This result, important enough to have its own name—The Divergence Theorem—is
either reasonable or amazing, depending on how you look at it. If you think just about
the math, it’s amazing that a vector field integrated over a surface can be equal to a kind
of derivative of the field integrated over a volume.
On the other hand, if you think about it in physics terms, it makes more sense: the flux
through a surface (i.e., the amount of “stuff” streaming out through the surface of an
object) is probably related to the divergence of the field producing the flux in the interior
(i.e., the amount of “stuff” being generated in the volume inside the surface).
One of the reasons we’ve been defining quantities like divergence and flux in physics
terms and then deriving the math is so that the results would seem more intuitive. If they
haven’t, you can be like all the math majors and memorize these formulas and think,
“Wow! What a coincidence that these formulas apply to the real world!” 
Another reason for doing things this way is so that if you follow through the same
procedure, you can derive the results in other coordinate systems should you ever need to
do so.
A simple illustration
For a simple illustration of the divergence theorem, take the field F = {x, y, z} over the
region of space represented by the surface that’s composed of the hemisphere of radius 1
and z > 0 centered at the origin and the unit circle in the xy-plane. Together, those two
surfaces enclose a volume of space. On the hemisphere, the normal unit vector simply
points radially outward and can be represented in Cartesian coordinates by {x, y, z}.
Hence F  n = {x,y,z}{x,y,z}= x2
+ y2 + z2 , which in turn equals 1 since it’s a unit hemisphere. Thus, F  n = 1. Hence,
the flux through the hemisphere is simply

F  n dS   dS  2
hemisphere
because the surface area of the hemisphere is 2. On the bottom of the hemisphere
(a.k.a. the unit circle), n = -k and hence F  n = - z. Hence,

circle
F  n dS    z dydx  0
because z = 0 everywhere in the xy-plane. Thus, there is no contribution to the flux from
the bottom of the hemisphere. Hence, the total flux is 2. Now, the divergence of F is
F
Fx
x

F y
y

Fz
z

x y z


111 3
x y z
Hence,

hemisphere
  F dV  3
hemisphere
dV  3
2
 2
3
since the volume of the hemisphere is half that of the unit sphere, or 2/3. Hence, we’ve
just illustrated that for the given vector field F over the given hemisphere, the divergence
theorem holds. You’ll get to use it with harder examples. As you should have noticed by
now, it’s much easier to teach math than to take it—the teachers take all the easy
examples for themselves and give the hard ones as exercises for students. No doubt,
that’s why Kenny enjoyed teaching the ninth-graders so much.
There are really only two things that make teaching math work (or three, if you count the
huge amount of work making up these labs): 1) sometimes students insist on asking
questions and depending on your level of knowledge it’s either very hard to answer them
(if you know nothing) or very hard to answer them in such a way that the students are led
to the answer rather than just having it delivered to them; 2) it’s often very timeconsuming to make problems that are either realistic or unify several disparate topics
you’ve studied (or both) without being so hard that they can’t be done by students just
starting their study of the material (or can’t be done in a reasonable time frame). I find
high-school students usually complain if you give them take-home tests that require more
than three or four hours to finish….