9.16 Divergence Theorem

INTRODUCTION

In Section 9.14 we saw that Stokes’ theorem was a three-dimensional generalization of a vector form of Green’s theorem. In this section we present a second vector form of Green’s theorem and its three-dimensional analogue.

Another Vector Form of Green’s Theorem

Let F(x, y) = P(x, y)i + Q(x, y)j be a two-dimensional vector field, and let T = (dx/ds)i + (dy/ds)j be a unit tangent to a simple closed plane curve C. In (1) of Section 9.14 we saw that (F · T) ds can be evaluated by a double integral involving curl F. Similarly, if n = (dy/ds)i – (dx/ds)j is a unit normal to C (check T · n), then (F · n) ds can be expressed in terms of a double integral of div F. From Green’s theorem,

that is, (1)

The result in (1) is a special case of the divergence or Gauss’ theorem. The following is a generalization of (1) to 3-space:

THEOREM 9.16.1 Divergence Theorem

Let D be a closed and bounded region in 3-space with a piecewise-smooth boundary S that is oriented outward. Let F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k be a vector field for which P, Q, and R are continuous and have continuous first partial derivatives in a region of 3-space containing D. Then

(2)

PARTIAL PROOF:

We will prove (2) for the special region D shown in FIGURE 9.16.1 whose surface S consists of three pieces:

where R is the projection of D onto the xy-plane and C is the boundary of R. Since

we can write

and

A graph. A region labeled D is graphed on a three dimensional x y z coordinate system. The shaded region is bound by a horizontal surface labeled S subscript 1 below and a similar horizontal surface labeled S subscript 2 above. A third vertical surface labeled S subscript 3 joins the 2 surfaces along their boundaries. The region D is projected vertically down on the x y plane as a closed curve labeled C. The region inside the closed curve is shaded and labeled R. A vector labeled n points upward on the surface S subscript 2. A second vector labeled n points downward on the surface S subscript 1. A third vector labeled n points to the right on the surface S subscript 3.

FIGURE 9.16.1 Region D used in proof of Theorem 9.16.1

To prove (2) we need only establish that

(3)

(4)

(5)

Indeed, we shall prove only (5), since the proofs of (3) and (4) follow in a similar manner. Now,

(6)

Next we write

On S1: Since the outward normal points downward, we describe the surface as g(x, y, z) = f1(x, y) – z = 0. Thus,

From the definition of dS we then have

(7)

On S2: The outward normal points upward, so

from which we find

(8)

On S3: Since this side is vertical, k is perpendicular to n. Consequently, k · n = 0 and

(9)

Finally, adding (7), (8), and (9), we get

which is the same as (6). ≡

Although we proved (2) for a special region D that has a vertical side, we note that this type of region is not required in Theorem 9.16.1. A region D with no vertical side is illustrated in FIGURE 9.16.2; a region bounded by a sphere or an ellipsoid also does not have a vertical side. The divergence theorem also holds for the region D bounded between two closed surfaces, such as the concentric spheres Sa and Sb shown in FIGURE 9.16.3; the boundary surface S of D is the union of Sa and Sb. In this case (F · n) dS = D div F dV becomes

where n points outward from D; that is, n points away from the origin on Sb and n points toward the origin on Sa.

A graph. A region labeled D with no vertical side is graphed on a three dimensional x y z coordinate system. A closed smooth curve labeled S goes horizontally around the region D. A vector labeled n points up to the left at the top of the region D, and a second vector labeled n points down to the left at the bottom of the region D.

FIGURE 9.16.2 Region D with no vertical side

A graph. 2 concentric spheres are graphed on a three dimensional x y z coordinate system. The smaller sphere is labeled S subscript a and the larger sphere is labeled S subscript b. The region inside the larger sphere, excluding the region insdie the smaller sphere, is shaded and labeld D. A vector labeled n points outward on the larger sphere S subscript b. A second vector, on the smaller sphere and on the same radius as the vector n, points toward the center.

FIGURE 9.16.3 Region D is bounded between two concentric spheres

EXAMPLE 1 Verifying Divergence Theorem

Let D be the region bounded by the hemisphere x2 + y2 + (z – 1)2 = 9, 1 ≤ z ≤ 4, and the plane z = 1. Verify the divergence theorem if F = xi + yj + (z – 1)k.

SOLUTION

The closed region is shown in FIGURE 9.16.4.

A graph. A hemispherical region labeled D is graphed on a three dimensional x y z coordinate system. The hemispherical region is bound by the surface labeled S subscript 1: x^2 + y^2 + (z minus 1)^2 = 9, 1 <= z <= 4 above, and the circular plane labeled S subscript 2: z = 1 below. A vector labeled n on the surface S subscript 1 points up to the right. A second vector labeled n on the surface S subscript 2 points vertically down. The region D is projected vertically down on the x y plane as a circle labeled x^2 + y^2 = 9. The region inside the projected circle is shaded and labeled R.

FIGURE 9.16.4 Hemispherical region D in Example 1

Triple Integral: Since F = xi + yj + (z – 1)k, we see div F = 3. Hence,

(10)

In the last calculation, we used the fact that DdV gives the volume of the hemisphere ( π33).

Surface Integral: We write where S1 is the hemisphere and S2 is the plane z = 1. If S1 is a level surface of g(x, y, z) = x2 + y2 + (z – 1)2, then a unit outer normal is

Now

and so

On S2, we take n = −k so that F · n = –z + 1. But since z = 1, 2 (–z + 1) dS = 0.

Hence, we see that S (F · n) dS = 54π + 0 = 54π agrees with (10). ≡

EXAMPLE 2 Using Divergence Theorem

If F = xyi + y2zj + z3k, evaluate (F · n) dS, where S is the unit cube defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1.

SOLUTION

See Figure 9.13.14 and Problem 38 in Exercises 9.13. Rather than evaluate six surface integrals, we apply the divergence theorem. Since div F = ∇ · F = y + 2yz + 3z2, we have from (2)

≡

Physical Interpretation of Divergence

In Section 9.14 we saw that we could express the normal component of the curl of a vector field F at a point as a limit involving the circulation of F. In view of (2), it is possible to interpret the divergence of F at a point as a limit involving the flux of F. Recall from Section 9.7 that the flux of the velocity field F of a fluid is the rate of fluid flow—that is, the volume of fluid flowing through a surface per unit time. In Section 9.7 we saw that the divergence of F is the flux per unit volume. To reinforce this last idea let us suppose P0(x0, y0, z0) is any point in the fluid and Sr is a small sphere of radius r centered at P0. See FIGURE 9.16.5. If Dr is the sphere Sr and its interior, then the divergence theorem gives

(11)

If we take the approximation div F(P) ≈ div F(P0) at all points P(x, y, z) within the small sphere, then (11) gives

(12)

where Vr is the volume ( πr3) of the spherical region Dr. By letting r → 0, we see from (12) that the divergence of F is the limiting value of the ratio of the flux of F to the volume of the spherical region:

.

Hence, divergence F is flux per unit volume.

A spherical region is labeled D subscript r. The center of the sphere is labeled P subscript 0. The surface of the sphere is labeled S subscript r. 2 vectors labeled n on the surface S subscript r, point outward.

FIGURE 9.16.5 Region Dr in (11)

The divergence theorem is extremely useful in the derivation of some of the famous equations in electricity and magnetism and hydrodynamics. In the discussion that follows we shall consider an example from the study of fluids.

Continuity Equation

At the end of Section 9.7 we mentioned that one interpretation of div F was a measure of the rate of change of the density of a fluid at a point. To see why this is so, let us suppose that F is a velocity field of a fluid and that ρ(x, y, z, t) is the density of the fluid at a point P(x, y, z) at a time t. Let D be the closed region consisting of a sphere S and its interior. We know from Section 9.15 that the total mass m of the fluid in D is given by m = D ρ(x, y, z, t)dV. The rate at which the mass increases in D is given by

(13)

Now from Figure 9.7.3 we saw that the volume of fluid flowing through an element of surface area ΔS per unit time is approximated by (F · n) ΔS. The mass of the fluid flowing through an element of surface area ΔS per unit time is then (ρF · n) ΔS. If we assume that the change in mass in D is due only to the flow in and out of D, then the volume of fluid flowing out of D per unit time is given by (10) of Section 9.13, (F · n) dS, whereas the mass of the fluid flowing out of D per unit time is (ρF · n) dS. Hence, an alternative expression for the rate at which the mass increases in D is

(14)

By the divergence theorem, (14) is the same as

(15)

Equating (13) and (15) then yields

Since this last result is to hold for every sphere, we obtain the equation of continuity for fluid flows:

(16)

On page 521 we stated that if div F = ∇ · F = 0, then a fluid is incompressible. This fact follows immediately from (16). If a fluid is incompressible (such as water), then ρ is constant, so consequently ∇ · (ρ F) = ρ∇ · F. But in addition ∂ρ/∂t = 0 and so (16) implies ∇ · F = 0.

9.16 Exercises Answers to selected odd-numbered problems begin on page ANS-26.

In Problems 1 and 2, verify the divergence theorem.

  1. F = xyi + yzj + xzk; D the region bounded by the unit cube defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1
  2. F = 6xyi + 4yzj + xe−yk; D the region bounded by the three coordinate planes and the plane x + y + z = 1

In Problems 3–14, use the divergence theorem to find the outward flux (F · n) dS of the given vector field F.

  1. F = x3i + y3j + z3k; D the region bounded by the sphere x2 + y2 + z2 = a2
  2. F = 4xi + yj + 4zk; D the region bounded by the sphere x2 + y2 + z2 = 4
  3. F = y2i + xz3j + (z – 1)2k; D the region bounded by the cylinder x2 + y2 = 16 and the planes z = 1, z = 5
  4. F = x2i + 2yzj + 4z3k; D the region bounded by the parallelepiped defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3
  5. F = y3i + x3j + z3k; D the region bounded within by z = , x2 + y2 = 3, z = 0
  6. F = (x2 + sin y)i + z2j + xy3k; D the region bounded by y = x2, z = 9 – y, z = 0
  7. F = (xi + yj + zk)/(x2 + y2 + z2); D the region bounded by the concentric spheres x2 + y2 + z2 = a2, x2 + y2 + z2 = b2, b > a
  8. F = 2yzi + x3j + xy2k; D the region bounded by the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1
  9. F = 2xzi + 5y2j – z2k; D the region bounded by z = y, z = 4 – y, z = 2 – x2, x = 0, z = 0. See FIGURE 9.16.6.
    A graph. A region is graphed on a three dimensional x y z coordinate system. The region, graphed in the first octant, is bound by the plane labeled x = 0 at the back, the plane labeled z = y on the left, the surface labeled z = 2 minus 1 over 2 x^2 in the front, and the plane labeled z = 4 minus y on the right.

    FIGURE 9.16.6 Region D for Problem 11

  10. F = 15x2yi + x2zj + y4k; D the region bounded by x + y = 2, z = x + y, z = 3, x = 0, y = 0
  11. F = 3x2y2i + yj – 6zxy2k; D the region bounded by the paraboloid z = x2 + y2 and the plane z = 2y
  12. F = xy2i + x2yj + 6 sin xk; D the region bounded by the cone z = and the planes z = 2, z = 4
  13. The electric field at a point P(x, y, z) due to a point charge q located at the origin is given by the inverse square field

    where r = xi + yj + zk.

    1. Suppose S is a closed surface, Sa is a sphere x2 + y2 + z2 = a2 lying completely within S, and D is the region bounded between S and Sa. See FIGURE 9.16.7. Show that the outward flux of E for the region D is zero.
      A graph. A region labeled D in graphed on a three dimensional x y z coordinate system. The region D contains a sphere labeled S subscript a. The center of the sphere is at the origin.

      FIGURE 9.16.7 Region D for Problem 15(a)

    2. Use the result of part (a) to prove Gauss’ law:

      that is, the outward flux of the electric field E through any closed surface (for which the divergence theorem applies) containing the origin is 4πq.

  14. Suppose there is a continuous distribution of charge throughout a closed and bounded region D enclosed by a surface S. Then the natural extension of Gauss’ law is given by

    where ρ(x, y, z) is the charge density or charge per unit volume.

    1. Proceed as in the derivation of the continuity equation (16) to show that div E = 4πρ.
    2. Given that E is an irrotational vector field, show that the potential ϕ for E satisfies Poisson’s equation ∇2ϕ = 4πρ.

In Problems 17–21, assume that S forms the boundary of a closed and bounded region D.

  1. If a is a constant vector, show that (a · n) dS = 0.
  2. If F = Pi + Qj + Rk and P, Q, and R have continuous second partial derivatives, prove that

In Problems 19 and 20, assume that f and g are scalar functions with continuous second partial derivatives. Use the divergence theorem to establish Green’s identities.

  3. If f is a scalar function with continuous first partial derivatives, prove that

    [Hint: Use (2) on fa, where a is a constant vector, and Problem 27 in Exercises 9.7.]
  4. The buoyancy force on a floating object is B = –pn dS, where p is the fluid pressure. The pressure p is related to the density of the fluid ρ(x, y, z) by a law of hydrostatics: ∇p = ρ(x, y, z)g, where g is the constant acceleration of gravity. If the weight of the object is W = mg, use the result of Problem 21 to prove Archimedes’ principle, B + W = 0. See FIGURE 9.16.8.
    A floating object is half submerged in water. A vertical up arrow at the water surface is labeled B, and a vertical down arrow of similar magnitude and at the same point, is labeled W.

    FIGURE 9.16.8 Floating object in Problem 22