The divergence of a vector field F = M (x, y, z) i + N (x, y, z) j + P (x, y, z) k is the scalar function

The div F has the same physical interpretation in three dimensions that it does in two. If F is the velocity field of a fluid flow, the value of div F at a point (x, y, z) is the rate at which fluid is being piped in or drained away at (x, y, z).

The flux of a vector field F = M i + N j + P k across a closed oriented surface S in the direction of the surface's outward unit normal field n equals the integral Ñ · F over the region D enclosed by the surface:

EXAMPLE 1:

Use the Divergence theorem to find the outward flux of F = (y - x) i + (x - y) j + (y - x) k across the boundary of the cube bounded by the planes x = ± 1, y = ± 1, and z = ± 1.

SOLUTION: Let M = y - x, N = z - y, and P = y - x.

EXAMPLE 2:

Use the Divergence theorem to find the outward flux of F = x ² i - 2xy j + 3xz k across the boundary cut from the first octant by the sphere x ² + y ² + z ² = 4.

SOLUTION: Let M = x ², N = -2xy, and P = 3xz.

I am going to use spherical coordinates to work this integral. Remember that the surface is in the first octant.

EXAMPLE 3:

Use the Divergence theorem to find the outward flux of

across the boundary of the region 1 £ x ² + y ² + z ² £ 4.

SOLUTION:

Now convert everything to spherical coordinates.

EXAMPLE 4:

Use the Divergence theorem to find the outward flux of

across the boundary of the region D which is the thick-walled cylinder 1 £ x ² + y ² £ 2, -1 £ z £ 2.

SOLUTION:

Now convert everything to cylindrical coordinates. (The region that we will be integrating over is a cylinder.)

The Divergence theorem can be extended to regions that can be partitioned into a finite number of simple regions and to regions that can be defined as limits of simpler regions in certain ways.

Suppose that D is the region between two concentric spheres and F has continuously differentiable components throughout D and on the boundary surfaces. Now split D by an equatorial plane and apply the Divergence theorem to each half separately. The bottom half, D₁, surface consists of an outer hemisphere, a plane washer-shaped base, and an inner hemisphere. Now the Divergence theorem states that

The unit normal n ₁ for S₁ that points outward from D₁ points away from the origin along the outer surface. It is equal to k along the flat base, and points toward the origin on the inner surface. Now for the top half of the sphere.

Now as we follow n ₂ over S₂, it is pointing outward from D₂. On the outer surface, n ₂ points away from the origin. On the inner surface, n ₂ points toward the origin, and on the flat base, n ₂ = -k. Therefore when I sum up all of these integrals, the integrals over the flat bases cancel out. Therefore, we can say that