From lecture 21 we introduced the notion of divergence of a vector field. We
wanted a measurement of how compressed a given vector field F was within a simply
connected region D with smooth boundary C. We discovered that the overall rate
of flow (in units of area squared per unit of time) across C, or in other words the
overall rate of flow into or out of D, is determined by
Z
F · n ds,
C
where n is, say, outwardly oriented. If the above is positive, then there is an overall
outflow of F from D, and if the above is negative, then there is an overall inflow of
F into D. In order to measure the magnitude of the compression, it made sense to
divide by the area of D. More explicitly, the quantity
Z
1
F · n ds
A(D) C
tells us the rate of flow per unit of area (in units of area per unit of time per units
of area). This quantity gives us an idea of how compressed F is within D.
Now, in order to understand how F was compressed about a given point (x, y),
it made sense to compute
lim+
r→0
1
πr2
Z
F · n ds.
Cr
where Cr is a circle of radius r centered about (x, y). By brute force computation,
we showed that this quantity was equal to
div F := Px (x, y) + Qy (x, y).
As it turns out, all of the above works just as well for 3-d vector fields F =
hP, Q, Ri. Specifically, let E be a region with smooth boundary surface S. Let n
orient S outwardly. Then the quantity
ZZ
ZZ
F · dS
F · n dS =
1
V(E)
ZZ
F · n dS =
S
1
V(E)
F · dr,
is analagous to
C
S
S
and
Z
ZZ
F · dS
1
A(D)
is analagous to
S
where V(E) is the volume of E. Moreover,
ZZ
1
lim+
F · dS
is analogous to
r→0 4/3πr3
Br
lim+
r→0
1
πr2
Z
F · dr,
C
Z
F · n ds,
Cr
where Br is the sphere of radius r centered at a point, say (x, y, z). A very similar
type of brute force computation to that which was used in lecture 21 to determine
the limit on the right hand side, shows that
ZZ
1
lim+
F · dS = div F = Px (x, y, z) + Qy (x, y, z) + Ry (x, y, z).
r→0 4/3πr3
Br
1
2
Now, in lecture 21, we also discovered that given a vector field F with continuous
first partial derivatives, we have
Z
ZZ
ZZ
F · n ds =
Px + Qy dA =
div F dA1
C
D
D
Along completely analogous lines, we have the following.
Theorem 1. (Gauss’/Divergence theorem) Let E be a region in R3 with smooth
boundary surface S with outward orientation. Suppose F has continuous first partial
derivatives. Then
ZZ
ZZZ
F · dS =
S
ZZZ
div F dV.
Px + Qy + Rz dV =
E
E
One can loosely interpret the above theorem in the following way: the tendency
for F to flow through the boundary of E is equal to the average tendency for F to
flow away from or toward the interior points of E.
Gauss’ theorem is not only very interesting, but it’s also a very useful computational tool.
RR
Example 1. Compute
2
S
F · dS where S is the boundary of the cube E = [0, 1]3
and F = hx + (y + 1) , y + (z2 + 1)x , z + (x2 + 1)x i.
y
Solution. We apply Gauss’ theorem. Note that
div F = Px + Qy + Rz = 1 + 1 + 1 = 3.
ZZZ
ZZ
So
F · dS =
S
3 dV = 3V(E) = 3.
[0,1]3
Remark. Notice how difficult it would have been to actually compute the above
surface integral. The sides of the cube would have to be parameterized separately.
This means it would have taken 6 surface integrals. What’s more, the vector field F
is quite complicated perhaps even unintegrable. On the other hand, div F is quite
simple.
Problem 1. Find a region E in R3 with a boundary surface that would be difficult to
integrate over, but that E itself is easy to integrate over (as a triple integral). Then
find a vector field F such that F is ridiculously complicated but div F is ridiculously
RR
simple. Finally, ask a calculus student to compute S F · dS.
1NOT to be confused with Green’s theorem.