The Divergence Theorem relates relates volume
integrals to surface
integrals of vector fields. Let R be a region in xyz space with surface
S. Let **n** denote the unit normal vector to S pointing in the outward
direction. Let **F**(x,y,z)=<P(x,y,z),Q(x,y,z),R(x,y,z)> be a vector
field whose components P, Q, and R have continuous partial derivatives. The
Divergence Theorem states:

Here div **F** is the divergence
of **F**. There are various
technical restrictions on the region R and the surface S; see the
references for the details. The theorem
is valid for regions bounded
by ellipsoids, spheres, and rectangular boxes, for example.

**Example**

Verify the Divergence Theorem in the case that R is the region
satisfying 0<=z<=16-x^2-y^2 and **F**=<y,x,z>.

A plot of the paraboloid is z=g(x,y)=16-x^2-y^2 for z>=0 is shown on the left in the figure above. The surface of the region R consists of two pieces. Let us denote the paraboloid by S_1. The intersection of the parabaloid with the z plane is the circle x^2+y^2=16. It follows that that the bottom of R, which we denote by S_2, is the disk x^2+y^2<=16. The surface integral of the vector field is

where the unit normal vector **n** points away from the region
R.

Let us first consider the surface S_1. The outward normal vector points in the positive z direction. Our previous computation in the flux web page shows that if the normal vector points in the negative direction, then the surface integral equals -128*pi. For a normal vector pointing in the positive z direction the surface integral equals 128*pi.

On the surface S_2, the normal vector points in the negative
z direction. In fact, since S_2 lies in the plane z=0, the unit
normal vector has x component and y component equal to 0. Hence,
<0,0,-1>. On S_2, **F**=<y,x,z>=<y,x,0>, since S_2 is
in the
xy plane. It follows that

Hence, the surface integral on S_2 is 0. The surface integral of **F**
on the entire surface is
128*pi.

Now, let us compute the volume integral. The divergence of
**F**=<P,Q,R >=<y,x,z> is

It follows that the relevant volume integral is

The circular symmetry of the region R suggests we convert to cylindrical coordinates. In cartesian coordinates the region R is described by the inequalities 0<=z<=16-x^2-y^2 and 0<=x^2+y^2<=16. In cylindrical coordinates, we have 0<=z<=16-r^2, 0<=r<=4, and 0<=theta<= 2*pi. The above volume integral becomes:

(There are several ways in the which the integrals can be ordered. However, the z integral must be done before the r integral.) The inner integral is

(Note that r is held constant). The middle integral with respect to r is

The outer integral is

This is the same as the surface integral!

**Copyright **© **1996 Department
of Mathematics, Oregon State
University**

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