Section 6.2 Divergence theorem in
Objectives
You should be able to:
Define integration of three-forms over closed bounded regions in
For exact three-forms, rephrase the generalized Stokes' theorem as the divergence theorem in
Summarize all the integral theorems of vector calculus as particular cases of the generalized Stokes' theorem.
Subsection 6.2.1 Integrating a three-form over a region in
Our goal in this section is to study Stokes' theorem for two-/three-forms. As a first step we need to define integration of three-forms over regions in
Definition 6.2.1. Orientation of a region in and induced orientation on the boundary.
Let
If
Example 6.2.2. Solid region bounded by a sphere in .
Let
Definition 6.2.3. Integral of a three-form over a closed bounded region in .
Let
where on the right-hand-side we mean the standard triple integral from calculus. If
Note that, as for the integral of two-forms over regions, the choice of basic three-form
A region
is -supported (also called type 1) if there exists a region in the -plane and two continuous functions such thatA region
is -supported (also called type 2) if there exists a region in the -plane and two continuous functions such thatA region
is -supported (also called type 3) if there exists a region in the -plane and two continuous functions such that
Example 6.2.4. Integral of a three-form over a recursively supported region.
Let
over
By definition, we know that
We want to evaluate the triple integral on the right-hand-side.
We can express the region
In fact, using the definition of the rectangular region
which shows that
Lemma 6.2.5. Integrals of three-forms over regions in are oriented and reparametrization-invariant.
Let
If
in the interior ofIf
in the interior of
Subsection 6.2.2 The divergence theorem in
Now that we understand integration of three-forms over regions in
Theorem 6.2.6. The divergence theorem in .
Let
where the integral on the right-hand-side is a surface integral of
In vector calculus language, if
where the integral on the right-hand-side is the surface integral of the vector field
Remark 6.2.7.
Just as for Green's theorem, we can read the divergence theorem in two different ways, starting from the left-hand-side or the right-hand-side. This results in two potential applications: either to evaluate the volume integral of the divergence of a vector field (or an exact three-form), or to evaluate the surface integral of a vector field (a two-form) over a closed surface. However, as was the case for Green's theorem, the divergence theorem is mostly useful to evaluate surface integrals over closed surfaces by transforming them into volume integrals over the interior of the region.
Example 6.2.8. Using the divergence theorem to evaluate the flux of a vector field over a closed surface in .
Find the flux of the vector field
in the outward direction over the surface of the solid region
We could try to evaluate the surface integral directly, but given how complicated the vector field is, this would probably be a nightmare. Or we can use the divergence theorem, which tells us that
The divergence of the vector field is
which is of course much simpler, so using the divergence theorem looks like a good strategy.
To evaluate the volume integral we need to write the solid region

The solid region
The volume integral can then be evaluated:
Therefore, by the divergence theorem, the flux of
Exercises 6.2.3 Exercises
1.
Use the divergence theorem to find the surface integral of the two-form
over the surface of the solid bounded by the cylinder
Let
where the integral on the right-hand-side is over the solid region
We calculate the exterior derivative:
To evaluate the integral of
The pullback of
We note that the determinant of the Jacobian is
The region
The volume integral can finally be evaluated:
Therefore, the integral of
2.
Use the divergence theorem to find the flux of the vector field
across (in the outward direction) the sphere
Let
where the integral on the right-hand-side is with canonical orientation.
We first calculate the divergence of
Thus the integral that we are interested in is
which is simply the volume of the solid region
so we could conclude right away that the flux of the vector field across the sphere
3.
Use the divergence theorem to evaluate the surface integral of the two-form
over the surface of the solid bounded by the paraboloid
First, we note that the problem is asking to evaluate the surface integral with a normal vector pointing inward. Thus, if we denote the solid region by
where on the left-hand-side we mean the surface integral with normal vector pointing inward, while on the right-hand-side we mean the volume integral with canonical orientation.
We calculate the exterior derivative:
We can either work in Cartesian or cylindrical coordinates here. But cylindrical coordinates will make the calculation much easier (believe me: I first did it in Cartesian coordinates, and it is no fun! See below. :-). So we do the change of coordinates
The pullback of
We note that the determinant of the Jacobian is
The region
The volume integral becomes
Therefore, by the divergence theorem the surface integral of the two-form that we are asked to evaluate is equal to minus this result (the surface integral is with respect to a normal vector pointing inward, as mentioned above), and thus is equal to
For completeness, let me do the calculation in Cartesian coordinates as well -- you will see how much uglier it is. The solid is given by the region enclosed by the parabola
The volume integral becomes:
Here I used the trigonometric substitution
4.
Show that the volume
where the surface integral is evaluated with the orientation given by an outward pointing normal vector.
We know that the volume of the region
But
where the integral on the right-hand-side is the surface integral with orientation given by an upward pointing normal vector.
5.
We studied the two-form
in Remark 4.6.6 and Exercise 5.8.3.5.
Consider an arbitrary closed surface
with outward pointing normal vector that does not contain or pass through the origin. Use the divergence theorem to show thatLet
be an arbitrary closed surface with outward pointing normal vector that contains the origin. Explain why the argument of (a) does not apply here.Let
be an arbitrary closed surface with outward pointing normal vector that contains the origin. Let be a sphere centered at the origin, with a radius small enough that it is contained completely inside Give the orientation of a normal vector pointing outward (outward of the sphere ). Use the divergence theorem to show thatUsing part (c), show that it implies that
You have shown that the surface integral of
(a) We know that the two-form
(b) If
(c) We stated the divergence theorem only for solid regions that consisted of a single closed surface and its interior, but it in fact applies to any closed bounded region. One simply needs to make sure that each bounding surface is oriented with an outward pointing normal vector, where βoutwardβ means away from the solid region bounded by the surfaces.
In particular, we can consider the solid region
We conclude that
To get rid of the minus sign, we can reverse the orientation on
as stated in the question.
(d) We already calculated the surface integral of