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16.9 The Divergence Theorem
The Divergence Theorem
We write Green’s Theorem in a vector version as
where C is the positively oriented boundary curve of the plane region D.
If we were seeking to extend this theorem to vector fields on we might make the guess that
where S is the boundary surface of the solid region E.
The Divergence Theorem
It turns out that Equation 1 is true, under appropriate hypotheses, and is called the Divergence Theorem.
Notice its similarity to Green’s Theorem and Stokes’
Theorem in that it relates the integral of a derivative of a function (div F in this case) over a region to the integral of the original function F over the boundary of the region.
We state the Divergence Theorem for regions E that are simultaneously of types 1, 2, and 3 and we call such
regions simple solid regions. (For instance, regions
bounded by ellipsoids or rectangular boxes are simple solid regions.)
The Divergence Theorem
The boundary of E is a closed surface, and we use the
convention, that the positive orientation is outward; that is, the unit normal vector n is directed outward from E.
Thus the Divergence Theorem states that, under the given conditions, the flux of F across the boundary surface of E is equal to the triple integral of the divergence of F over E.
Example 1
Find the flux of the vector field F(x, y, z) = z i + y j + x k over the unit sphere x2 + y2 + z2 = 1.
Solution:
First we compute the divergence of F:
The unit sphere S is the boundary of the unit ball B given by x2 + y2 + z2 ≤ 1.
Example 1 – Solution
Thus the Divergence Theorem gives the flux as
cont’d
The Divergence Theorem
Let’s consider the region E that lies between the closed surfaces S1 and S2, where S1 lies inside S2. Let n1 and n2 be outward normals of S1 and S2.
Then the boundary surface of E is S = S1 U S2 and its normal n is given by n = –n1 on S1 and n = n2 on S2. (See Figure 3.)
Figure 3
The Divergence Theorem
Applying the Divergence Theorem to S, we get
Example 3
We consider the electric field
where the electric charge Q is located at the origin and is a position vector.
Use the Divergence Theorem to show that the electric flux of E through any closed surface S2 that encloses the origin is
Example 3 – Solution
The difficulty is that we don’t have an explicit equation for S2 because it is any closed surface enclosing the origin.
The simplest such surface would be a sphere, so we let S1 be a small sphere with radius a and center the origin. You can verify that div E = 0.
Therefore Equation 7 gives
Example 3 – Solution
The point of this calculation is that we can compute the surface integral over S1 because S1 is a sphere. The normal vector at x is x/| x |.
Therefore
since the equation of S1 is |x | = a.
cont’d
Example 3 – Solution
Thus we have
This shows that the electric flux of E is 4πεQ through any closed surface S2 that contains the origin. [This is a special case of Gauss’s Law for a single charge. The relationship between ε and ε0 is ε =1/(4πε0).]
cont’d
The Divergence Theorem
Another application of the Divergence Theorem occurs in fluid flow. Let v(x, y, z) be the velocity field of a fluid with constant density ρ. Then F = ρv is the rate of flow per unit area.
The Divergence Theorem
If P0(x0, y0, z0) is a point in the fluid and Ba is a ball with
center P0 and very small radius a, then div F(P) ≈ div F(P0) for all points P in Ba since div F is continuous. We
approximate the flux over the boundary sphere Sa as follows:
This approximation becomes better as a → 0 and suggests that
The Divergence Theorem
Equation 8 says that div F(P0) is the net rate of outward flux per unit volume at P0. (This is the reason for the name
divergence.)
If div F(P) > 0, the net flow is outward near P and P is called a source.
If div F(P) < 0, the net flow is inward near P and P is called a sink.
The Divergence Theorem
For the vector field in Figure 4, it appears that the vectors that end near P1 are shorter than the vectors that start
near P1.
Figure 4
The vector field F = x2i + y2j
The Divergence Theorem
Thus the net flow is outward near P1, so div F(P1) > 0 and P1 is a source. Near P2, on the other hand, the incoming arrows are longer than the outgoing arrows.
Here the net flow is inward, so div F(P2) < 0 and P2 is a sink.
We can use the formula for F to confirm this impression.
Since F = x2 i + y2 j, we have div F = 2x + 2y, which is positive when y > –x. So the points above the line y = –x are sources and those below are sinks.
