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Calculus Volume 3

6.8 The Divergence Theorem

Calculus Volume 36.8 The Divergence Theorem
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 6.8.1. Explain the meaning of the divergence theorem.
  • 6.8.2. Use the divergence theorem to calculate the flux of a vector field.
  • 6.8.3. Apply the divergence theorem to an electrostatic field.

We have examined several versions of the Fundamental Theorem of Calculus in higher dimensions that relate the integral around an oriented boundary of a domain to a “derivative” of that entity on the oriented domain. In this section, we state the divergence theorem, which is the final theorem of this type that we will study. The divergence theorem has many uses in physics; in particular, the divergence theorem is used in the field of partial differential equations to derive equations modeling heat flow and conservation of mass. We use the theorem to calculate flux integrals and apply it to electrostatic fields.

Overview of Theorems

Before examining the divergence theorem, it is helpful to begin with an overview of the versions of the Fundamental Theorem of Calculus we have discussed:

  1. The Fundamental Theorem of Calculus:
    abf(x)dx=f(b)f(a).abf(x)dx=f(b)f(a).

    This theorem relates the integral of derivative ff over line segment [a,b][a,b] along the x-axis to a difference of ff evaluated on the boundary.
  2. The Fundamental Theorem for Line Integrals:
    Cf·dr=f(P1)f(P0),Cf·dr=f(P1)f(P0),

    where P0P0 is the initial point of C and P1P1 is the terminal point of C. The Fundamental Theorem for Line Integrals allows path C to be a path in a plane or in space, not just a line segment on the x-axis. If we think of the gradient as a derivative, then this theorem relates an integral of derivative ff over path C to a difference of ff evaluated on the boundary of C.
  3. Green’s theorem, circulation form:
    D(QxPy)dA=CF·dr.D(QxPy)dA=CF·dr.

    Since QxPy=curlF·kQxPy=curlF·k and curl is a derivative of sorts, Green’s theorem relates the integral of derivative curlF over planar region D to an integral of F over the boundary of D.
  4. Green’s theorem, flux form:
    D(Px+Qy)dA=CF·Nds.D(Px+Qy)dA=CF·Nds.

    Since Px+Qy=divFPx+Qy=divF and divergence is a derivative of sorts, the flux form of Green’s theorem relates the integral of derivative divF over planar region D to an integral of F over the boundary of D.
  5. Stokes’ theorem:
    ScurlF·dS=CF·dr.ScurlF·dS=CF·dr.

    If we think of the curl as a derivative of sorts, then Stokes’ theorem relates the integral of derivative curlF over surface S (not necessarily planar) to an integral of F over the boundary of S.

Stating the Divergence Theorem

The divergence theorem follows the general pattern of these other theorems. If we think of divergence as a derivative of sorts, then the divergence theorem relates a triple integral of derivative divF over a solid to a flux integral of F over the boundary of the solid. More specifically, the divergence theorem relates a flux integral of vector field F over a closed surface S to a triple integral of the divergence of F over the solid enclosed by S.

Theorem 6.20

The Divergence Theorem

Let S be a piecewise, smooth closed surface that encloses solid E in space. Assume that S is oriented outward, and let F be a vector field with continuous partial derivatives on an open region containing E (Figure 6.87). Then

EdivFdV=SF·dS.EdivFdV=SF·dS.
6.24
A diagram of a closed surface S, vector field, and solid E enclosed by the surface in three dimensions. The surface is a roughly rectangular prism with curved sides. The normal vectors stretch out and away from the surface. The arrows have negative x components and positive y and z components.
Figure 6.87 The divergence theorem relates a flux integral across a closed surface S to a triple integral over solid E enclosed by the surface.

Recall that the flux form of Green’s theorem states that DdivFdA=CF·Nds.DdivFdA=CF·Nds. Therefore, the divergence theorem is a version of Green’s theorem in one higher dimension.

The proof of the divergence theorem is beyond the scope of this text. However, we look at an informal proof that gives a general feel for why the theorem is true, but does not prove the theorem with full rigor. This explanation follows the informal explanation given for why Stokes’ theorem is true.

Proof

Let B be a small box with sides parallel to the coordinate planes inside E (Figure 6.88). Let the center of B have coordinates (x,y,z)(x,y,z) and suppose the edge lengths are Δx,Δy,Δx,Δy, and ΔzΔz (Figure 6.88(b)). The normal vector out of the top of the box is k and the normal vector out of the bottom of the box is k.k. The dot product of F=P,Q,RF=P,Q,R with k is R and the dot product with kk is R.R. The area of the top of the box (and the bottom of the box) ΔSΔS is ΔxΔy.ΔxΔy.

This figure has three diagrams. The first is a surface E in three dimensions with a small box B inside it. The second just has box B. The height is labeled as delta z, the width is labeled as delta x, and the width is labeled as delta y. An arrow perpendicular to the top points up and away from the box and is labeled k. An arrow perpendicular to the bottom points down and away from the box and is labeled –k. The third diagram is a side views of box B. The center is (x, y, z), the midpoint of the side below it is (x, y, z – delta z / 2), and the midpoint of the size above it is (x, y, z + delta z / 2). The height is delta z.
Figure 6.88 (a) A small box B inside surface E has sides parallel to the coordinate planes. (b) Box B has side lengths Δx,Δy,Δx,Δy, and ΔzΔz (c) If we look at the side view of B, we see that, since (x,y,z)(x,y,z) is the center of the box, to get to the top of the box we must travel a vertical distance of Δz/2Δz/2 up from (x,y,z).(x,y,z). Similarly, to get to the bottom of the box we must travel a distance Δz/2Δz/2 down from (x,y,z).(x,y,z).

The flux out of the top of the box can be approximated by R(x,y,z+Δz2)ΔxΔyR(x,y,z+Δz2)ΔxΔy (Figure 6.88(c)) and the flux out of the bottom of the box is R(x,y,zΔz2)ΔxΔy.R(x,y,zΔz2)ΔxΔy. If we denote the difference between these values as ΔR,ΔR, then the net flux in the vertical direction can be approximated by ΔRΔxΔy.ΔRΔxΔy. However,

ΔRΔxΔy=(ΔRΔz)ΔxΔyΔz(Rz)ΔV.ΔRΔxΔy=(ΔRΔz)ΔxΔyΔz(Rz)ΔV.

Therefore, the net flux in the vertical direction can be approximated by (Rz)ΔV.(Rz)ΔV. Similarly, the net flux in the x-direction can be approximated by (Px)ΔV(Px)ΔV and the net flux in the y-direction can be approximated by (Qy)ΔV.(Qy)ΔV. Adding the fluxes in all three directions gives an approximation of the total flux out of the box:

Total flux(Px+Qy+Rz)ΔV=divFΔV.Total flux(Px+Qy+Rz)ΔV=divFΔV.

This approximation becomes arbitrarily close to the value of the total flux as the volume of the box shrinks to zero.

The sum of divFΔVdivFΔV over all the small boxes approximating E is approximately EdivFdV.EdivFdV. On the other hand, the sum of divFΔVdivFΔV over all the small boxes approximating E is the sum of the fluxes over all these boxes. Just as in the informal proof of Stokes’ theorem, adding these fluxes over all the boxes results in the cancelation of a lot of the terms. If an approximating box shares a face with another approximating box, then the flux over one face is the negative of the flux over the shared face of the adjacent box. These two integrals cancel out. When adding up all the fluxes, the only flux integrals that survive are the integrals over the faces approximating the boundary of E. As the volumes of the approximating boxes shrink to zero, this approximation becomes arbitrarily close to the flux over S.

Example 6.77

Verifying the Divergence Theorem

Verify the divergence theorem for vector field F=xy,x+z,zyF=xy,x+z,zy and surface S that consists of cone x2+y2=z2,0z1,x2+y2=z2,0z1, and the circular top of the cone (see the following figure). Assume this surface is positively oriented.

This figure is a vector diagram in three dimensions. The cone x^2 + y^2 = z^2 is shown. Its point is at the origin, and it opens up. There is a cover across the top. The arrows seem to be following the shape of the cone.

Solution

Let E be the solid cone enclosed by S. To verify the theorem for this example, we show that EdivFdV=SF·dSEdivFdV=SF·dS by calculating each integral separately.

To compute the triple integral, note that divF=Px+Qy+Rz=2,divF=Px+Qy+Rz=2, and therefore the triple integral is

EdivFdV=2EdV=2(volume ofE).EdivFdV=2EdV=2(volume ofE).

The volume of a right circular cone is given by πr2h3.πr2h3. In this case, h=r=1.h=r=1. Therefore,

EdivFdV=2(volume ofE)=2π3.EdivFdV=2(volume ofE)=2π3.

To compute the flux integral, first note that S is piecewise smooth; S can be written as a union of smooth surfaces. Therefore, we break the flux integral into two pieces: one flux integral across the circular top of the cone and one flux integral across the remaining portion of the cone. Call the circular top S1S1 and the portion under the top S2.S2. We start by calculating the flux across the circular top of the cone. Notice that S1S1 has parameterization

r(u,v)=ucosv,usinv,1,0u1,0v2π.r(u,v)=ucosv,usinv,1,0u1,0v2π.

Then, the tangent vectors are tu=cosv,sinv,0tu=cosv,sinv,0 and tv=ucosv,usinv,0.tv=ucosv,usinv,0. Therefore, the flux across S1S1 is

S1F·dS=0102πF(r(u,v))·(tu×tv)dA=0102πucosvusinv,ucosv+1,1usinv·0,0,udvdu=0102πuu2sinvdvdu=π.S1F·dS=0102πF(r(u,v))·(tu×tv)dA=0102πucosvusinv,ucosv+1,1usinv·0,0,udvdu=0102πuu2sinvdvdu=π.

We now calculate the flux over S2.S2. A parameterization of this surface is

r(u,v)=ucosv,usinv,u,0u1,0v2π.r(u,v)=ucosv,usinv,u,0u1,0v2π.

The tangent vectors are tu=cosv,sinv,1tu=cosv,sinv,1 and tv=usinv,ucosv,0,tv=usinv,ucosv,0, so the cross product is

tu×tv=ucosv,usinv,u.tu×tv=ucosv,usinv,u.

Notice that the negative signs on the x and y components induce the negative (or inward) orientation of the cone. Since the surface is positively oriented, we use vector tv×tu=ucosv,usinv,utv×tu=ucosv,usinv,u in the flux integral. The flux across S2S2 is then

S2F·dS=0102πF(r(u,v))·(tv×tu)dA=0102πucosvusinv,ucosv+u,usinv·ucosv,usinv,u=0102πu2cos2v+2u2sinvu2dvdu=π3.S2F·dS=0102πF(r(u,v))·(tv×tu)dA=0102πucosvusinv,ucosv+u,usinv·ucosv,usinv,u=0102πu2cos2v+2u2sinvu2dvdu=π3.

The total flux across S is

S2F·dS=S1F·dS+S2F·dS=2π3=EdivFdV,S2F·dS=S1F·dS+S2F·dS=2π3=EdivFdV,

and we have verified the divergence theorem for this example.

Checkpoint 6.65

Verify the divergence theorem for vector field F(x,y,z)=x+y+z,y,2xyF(x,y,z)=x+y+z,y,2xy and surface S given by the cylinder x2+y2=1,0z3x2+y2=1,0z3 plus the circular top and bottom of the cylinder. Assume that S is positively oriented.

Recall that the divergence of continuous field F at point P is a measure of the “outflowing-ness” of the field at P. If F represents the velocity field of a fluid, then the divergence can be thought of as the rate per unit volume of the fluid flowing out less the rate per unit volume flowing in. The divergence theorem confirms this interpretation. To see this, let P be a point and let BrBr be a ball of small radius r centered at P (Figure 6.89). Let SrSr be the boundary sphere of Br.Br. Since the radius is small and F is continuous, divF(Q)divF(P)divF(Q)divF(P) for all other points Q in the ball. Therefore, the flux across SrSr can be approximated using the divergence theorem:

SrF·dS=BrdivFdVBrdivF(P)dV.SrF·dS=BrdivFdVBrdivF(P)dV.

Since divF(P)divF(P) is a constant,

BrdivF(P)dV=divF(P)V(Br).BrdivF(P)dV=divF(P)V(Br).

Therefore, flux SrF·dSSrF·dS can be approximated by divF(P)V(Br).divF(P)V(Br). This approximation gets better as the radius shrinks to zero, and therefore

divF(P)=limr01V(Br)SrF·dS.divF(P)=limr01V(Br)SrF·dS.

This equation says that the divergence at P is the net rate of outward flux of the fluid per unit volume.

This figure is a diagram of ball B_r, with small radius r centered at P. Arrows are drawn pointing up and to the right across the ball.
Figure 6.89 Ball BrBr of small radius r centered at P.

Using the Divergence Theorem

The divergence theorem translates between the flux integral of closed surface S and a triple integral over the solid enclosed by S. Therefore, the theorem allows us to compute flux integrals or triple integrals that would ordinarily be difficult to compute by translating the flux integral into a triple integral and vice versa.

Example 6.78

Applying the Divergence Theorem

Calculate the surface integral SF·dS,SF·dS, where S is cylinder x2+y2=1,0z2,x2+y2=1,0z2, including the circular top and bottom, and F=x33+yz,y33sin(xz),zxy.F=x33+yz,y33sin(xz),zxy.

Solution

We could calculate this integral without the divergence theorem, but the calculation is not straightforward because we would have to break the flux integral into three separate integrals: one for the top of the cylinder, one for the bottom, and one for the side. Furthermore, each integral would require parameterizing the corresponding surface, calculating tangent vectors and their cross product, and using Equation 6.19.

By contrast, the divergence theorem allows us to calculate the single triple integral EdivFdV,EdivFdV, where E is the solid enclosed by the cylinder. Using the divergence theorem and converting to cylindrical coordinates, we have

sF·dS=EdivFdV=E(x2+y2+1)dV=02π0102(r2+1)rdzdrdθ=3202πdθ=3π.sF·dS=EdivFdV=E(x2+y2+1)dV=02π0102(r2+1)rdzdrdθ=3202πdθ=3π.
Checkpoint 6.66

Use the divergence theorem to calculate flux integral SF·dS,SF·dS, where S is the boundary of the box given by 0x2,1y4,0z1,0x2,1y4,0z1, and F=x2+yz,y-z,2x+2y+2zF=x2+yz,y-z,2x+2y+2z (see the following figure).

This figure is a vector diagram in three dimensions. The box of the figure spans x from 0 to 2; y from 0 to 4; and z from 0 to 1. The vectors point up increasingly with distance from the origin; toward larger x with increasing distance from the origin; and toward smaller y values with increasing height.

Example 6.79

Applying the Divergence Theorem

Let v=yz,xz,0v=yz,xz,0 be the velocity field of a fluid. Let C be the solid cube given by 1x4,2y5,1z4,1x4,2y5,1z4, and let S be the boundary of this cube (see the following figure). Find the flow rate of the fluid across S.

This is a figure of a diagram of the given vector field in three dimensions. The x components are –y/z, the y components are x/z, and the z components are 0.
Figure 6.90 Vector field v=yz,xz,0.v=yz,xz,0.

Solution

The flow rate of the fluid across S is Sv·dS.Sv·dS. Before calculating this flux integral, let’s discuss what the value of the integral should be. Based on Figure 6.90, we see that if we place this cube in the fluid (as long as the cube doesn’t encompass the origin), then the rate of fluid entering the cube is the same as the rate of fluid exiting the cube. The field is rotational in nature and, for a given circle parallel to the xy-plane that has a center on the z-axis, the vectors along that circle are all the same magnitude. That is how we can see that the flow rate is the same entering and exiting the cube. The flow into the cube cancels with the flow out of the cube, and therefore the flow rate of the fluid across the cube should be zero.

To verify this intuition, we need to calculate the flux integral. Calculating the flux integral directly requires breaking the flux integral into six separate flux integrals, one for each face of the cube. We also need to find tangent vectors, compute their cross product, and use Equation 6.19. However, using the divergence theorem makes this calculation go much more quickly:

Sv·dS=Cdiv(v)dV=C0dV=0.Sv·dS=Cdiv(v)dV=C0dV=0.

Therefore the flux is zero, as expected.

Checkpoint 6.67

Let v=xz,yz,0v=xz,yz,0 be the velocity field of a fluid. Let C be the solid cube given by 1x4,2y5,1z4,1x4,2y5,1z4, and let S be the boundary of this cube (see the following figure). Find the flow rate of the fluid across S.

This is a figure of a diagram of the given vector field in three dimensions. The x components are x/z, the y components are y/z, and the z components are 0.

Example 6.79 illustrates a remarkable consequence of the divergence theorem. Let S be a piecewise, smooth closed surface and let F be a vector field defined on an open region containing the surface enclosed by S. If F has the form F=f(y,z),g(x,z),h(x,y),F=f(y,z),g(x,z),h(x,y), then the divergence of F is zero. By the divergence theorem, the flux of F across S is also zero. This makes certain flux integrals incredibly easy to calculate. For example, suppose we wanted to calculate the flux integral SF·dSSF·dS where S is a cube and

F=sin(y)eyz,x2z2,cos(xy)esinx.F=sin(y)eyz,x2z2,cos(xy)esinx.

Calculating the flux integral directly would be difficult, if not impossible, using techniques we studied previously. At the very least, we would have to break the flux integral into six integrals, one for each face of the cube. But, because the divergence of this field is zero, the divergence theorem immediately shows that the flux integral is zero.

We can now use the divergence theorem to justify the physical interpretation of divergence that we discussed earlier. Recall that if F is a continuous three-dimensional vector field and P is a point in the domain of F, then the divergence of F at P is a measure of the “outflowing-ness” of F at P. If F represents the velocity field of a fluid, then the divergence of F at P is a measure of the net flow rate out of point P (the flow of fluid out of P less the flow of fluid in to P). To see how the divergence theorem justifies this interpretation, let BrBr be a ball of very small radius r with center P, and assume that BrBr is in the domain of F. Furthermore, assume that BrBr has a positive, outward orientation. Since the radius of BrBr is small and F is continuous, the divergence of F is approximately constant on Br.Br. That is, if PP is any point in Br,Br, then divF(P)divF(P).divF(P)divF(P). Let SrSr denote the boundary sphere of Br.Br. We can approximate the flux across SrSr using the divergence theorem as follows:

SrF·dS=BrdivFdVBrdivF(P)dV=divF(P)V(Br).SrF·dS=BrdivFdVBrdivF(P)dV=divF(P)V(Br).

As we shrink the radius r to zero via a limit, the quantity divF(P)V(Br)divF(P)V(Br) gets arbitrarily close to the flux. Therefore,

divF(P)=limr01V(Br)SrF·dSdivF(P)=limr01V(Br)SrF·dS

and we can consider the divergence at P as measuring the net rate of outward flux per unit volume at P. Since “outflowing-ness” is an informal term for the net rate of outward flux per unit volume, we have justified the physical interpretation of divergence we discussed earlier, and we have used the divergence theorem to give this justification.

Application to Electrostatic Fields

The divergence theorem has many applications in physics and engineering. It allows us to write many physical laws in both an integral form and a differential form (in much the same way that Stokes’ theorem allowed us to translate between an integral and differential form of Faraday’s law). Areas of study such as fluid dynamics, electromagnetism, and quantum mechanics have equations that describe the conservation of mass, momentum, or energy, and the divergence theorem allows us to give these equations in both integral and differential forms.

One of the most common applications of the divergence theorem is to electrostatic fields. An important result in this subject is Gauss’ law. This law states that if S is a closed surface in electrostatic field E, then the flux of E across S is the total charge enclosed by S (divided by an electric constant). We now use the divergence theorem to justify the special case of this law in which the electrostatic field is generated by a stationary point charge at the origin.

If (x,y,z)(x,y,z) is a point in space, then the distance from the point to the origin is r=x2+y2+z2.r=x2+y2+z2. Let FrFr denote radial vector field Fr=1r2xr,yr,zr.Fr=1r2xr,yr,zr. The vector at a given position in space points in the direction of unit radial vector xr,yr,zrxr,yr,zr and is scaled by the quantity 1/r2.1/r2. Therefore, the magnitude of a vector at a given point is inversely proportional to the square of the vector’s distance from the origin. Suppose we have a stationary charge of q Coulombs at the origin, existing in a vacuum. The charge generates electrostatic field E given by

E=q4πε0Fr,E=q4πε0Fr,

where the approximation ε0=8.854×10−12ε0=8.854×10−12 farad (F)/m is an electric constant. (The constant ε0ε0 is a measure of the resistance encountered when forming an electric field in a vacuum.) Notice that E is a radial vector field similar to the gravitational field described in Example 6.6. The difference is that this field points outward whereas the gravitational field points inward. Because

E=q4πε0Fr=q4πε0(1r2xr,yr,zr),E=q4πε0Fr=q4πε0(1r2xr,yr,zr),

we say that electrostatic fields obey an inverse-square law. That is, the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge (which in this case is at the origin). Given this vector field, we show that the flux across closed surface S is zero if the charge is outside of S, and that the flux is q/ε0q/ε0 if the charge is inside of S. In other words, the flux across S is the charge inside the surface divided by constant ε0.ε0. This is a special case of Gauss’ law, and here we use the divergence theorem to justify this special case.

To show that the flux across S is the charge inside the surface divided by constant ε0,ε0, we need two intermediate steps. First we show that the divergence of FrFr is zero and then we show that the flux of FrFr across any smooth surface S is either zero or 4π.4π. We can then justify this special case of Gauss’ law.

Example 6.80

The Divergence of FrFr Is Zero

Verify that the divergence of FrFr is zero where FrFr is defined (away from the origin).

Solution

Since r=x2+y2+z2,r=x2+y2+z2, the quotient rule gives us

x(xr3)=x(x(x2+y2+z2)3/2)=(x2+y2+z2)3/2x[32(x2+y2+z2)1/22x](x2+y2+z2)3=r33x2rr6=r23x2r5.x(xr3)=x(x(x2+y2+z2)3/2)=(x2+y2+z2)3/2x[32(x2+y2+z2)1/22x](x2+y2+z2)3=r33x2rr6=r23x2r5.

Similarly,

y(yr3)=r23y2r5andz(zr3)=r23z2r5.y(yr3)=r23y2r5andz(zr3)=r23z2r5.

Therefore,

divFr=r23x2r5+r23y2r5+r23z2r5=3r23(x2+y2+z2)r5=3r23r2r5=0.divFr=r23x2r5+r23y2r5+r23z2r5=3r23(x2+y2+z2)r5=3r23r2r5=0.

Notice that since the divergence of FrFr is zero and E is FrFr scaled by a constant, the divergence of electrostatic field E is also zero (except at the origin).

Theorem 6.21

Flux across a Smooth Surface

Let S be a connected, piecewise smooth closed surface and let Fr=1r2xr,yr,zr.Fr=1r2xr,yr,zr. Then,

SFr·dS={0ifSdoes not encompass the origin4πifSencompasses the origin.SFr·dS={0ifSdoes not encompass the origin4πifSencompasses the origin.

In other words, this theorem says that the flux of FrFr across any piecewise smooth closed surface S depends only on whether the origin is inside of S.

Proof

The logic of this proof follows the logic of Example 6.46, only we use the divergence theorem rather than Green’s theorem.

First, suppose that S does not encompass the origin. In this case, the solid enclosed by S is in the domain of Fr,Fr, and since the divergence of FrFr is zero, we can immediately apply the divergence theorem and find that SF·dSSF·dS is zero.

Now suppose that S does encompass the origin. We cannot just use the divergence theorem to calculate the flux, because the field is not defined at the origin. Let SaSa be a sphere of radius a inside of S centered at the origin. The outward normal vector field on the sphere, in spherical coordinates, is

tϕ×tθ=a2cosθsin2ϕ,a2sinθsin2ϕ,a2sinϕcosϕtϕ×tθ=a2cosθsin2ϕ,a2sinθsin2ϕ,a2sinϕcosϕ

(see Example 6.64). Therefore, on the surface of the sphere, the dot product Fr·NFr·N (in spherical coordinates) is

Fr·N=sinϕcosθa2,sinϕsinθa2,cosϕa2·a2cosθsin2ϕ,a2sinθsin2ϕ,a2sinϕcosϕ=sinϕ(sinϕcosθ,sinϕsinθ,cosϕ·sinϕcosθ,sinϕsinθ,cosϕ)=sinϕ.Fr·N=sinϕcosθa2,sinϕsinθa2,cosϕa2·a2cosθsin2ϕ,a2sinθsin2ϕ,a2sinϕcosϕ=sinϕ(sinϕcosθ,sinϕsinθ,cosϕ·sinϕcosθ,sinϕsinθ,cosϕ)=sinϕ.

The flux of FrFr across SaSa is

SaFr·NdS=02π0πsinϕdϕdθ=4π.SaFr·NdS=02π0πsinϕdϕdθ=4π.

Now, remember that we are interested in the flux across S, not necessarily the flux across Sa.Sa. To calculate the flux across S, let E be the solid between surfaces SaSa and S. Then, the boundary of E consists of SaSa and S. Denote this boundary by SSaSSa to indicate that S is oriented outward but now SaSa is oriented inward. We would like to apply the divergence theorem to solid E. Notice that the divergence theorem, as stated, can’t handle a solid such as E because E has a hole. However, the divergence theorem can be extended to handle solids with holes, just as Green’s theorem can be extended to handle regions with holes. This allows us to use the divergence theorem in the following way. By the divergence theorem,

SSaFr·dS=SFr·dSSaFr·dS=EdivFrdV=E0dV=0.SSaFr·dS=SFr·dSSaFr·dS=EdivFrdV=E0dV=0.

Therefore,

SFr·dS=SaFr·dS=4π,SFr·dS=SaFr·dS=4π,

and we have our desired result.

Now we return to calculating the flux across a smooth surface in the context of electrostatic field E=q4πε0FrE=q4πε0Fr of a point charge at the origin. Let S be a piecewise smooth closed surface that encompasses the origin. Then

SE·dS=Sq4πε0Fr·dS=q4πε0SFr·dS=qε0.SE·dS=Sq4πε0Fr·dS=q4πε0SFr·dS=qε0.

If S does not encompass the origin, then

SE·dS=q4πε0SFr·dS=0.SE·dS=q4πε0SFr·dS=0.

Therefore, we have justified the claim that we set out to justify: the flux across closed surface S is zero if the charge is outside of S, and the flux is q/ε0q/ε0 if the charge is inside of S.

This analysis works only if there is a single point charge at the origin. In this case, Gauss’ law says that the flux of E across S is the total charge enclosed by S. Gauss’ law can be extended to handle multiple charged solids in space, not just a single point charge at the origin. The logic is similar to the previous analysis, but beyond the scope of this text. In full generality, Gauss’ law states that if S is a piecewise smooth closed surface and Q is the total amount of charge inside of S, then the flux of E across S is Q/ε0.Q/ε0.

Example 6.81

Using Gauss’ law

Suppose we have four stationary point charges in space, all with a charge of 0.002 Coulombs (C). The charges are located at (0,1,1),(1,1,4),(−1,0,0),and(−2,−2,2).(0,1,1),(1,1,4),(−1,0,0),and(−2,−2,2). Let E denote the electrostatic field generated by these point charges. If S is the sphere of radius 2 oriented outward and centered at the origin, then find SE·dS.SE·dS.

Solution

According to Gauss’ law, the flux of E across S is the total charge inside of S divided by the electric constant. Since S has radius 2, notice that only two of the charges are inside of S: the charge at (0,1,1)(0,1,1) and the charge at (−1,0,0).(−1,0,0). Therefore, the total charge encompassed by S is 0.004 and, by Gauss’ law,

SE·dS=0.0048.854×10−124.518×109V-m.SE·dS=0.0048.854×10−124.518×109V-m.

Checkpoint 6.68

Work the previous example for surface S that is a sphere of radius 4 centered at the origin, oriented outward.

Section 6.8 Exercises

For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral SF·ndsSF·nds for the given choice of F and the boundary surface S. For each closed surface, assume N is the outward unit normal vector.

376.

[T] F(x,y,z)=xi+yj+zk;F(x,y,z)=xi+yj+zk; S is the surface of cube 0x1,0y1,0<z1.0x1,0y1,0<z1.

377.

[T] F(x,y,z)=(cosyz)i+exzj+3z2k;F(x,y,z)=(cosyz)i+exzj+3z2k; S is the surface of hemisphere z=4x2y2z=4x2y2 together with disk x2+y24x2+y24 in the xy-plane.

378.

[T] F(x,y,z)=(x2+y2x2)i+x2yj+3zk;F(x,y,z)=(x2+y2x2)i+x2yj+3zk; S is the surface of the five faces of unit cube 0x1,0y1,0<z1.0x1,0y1,0<z1.

379.

[T] F(x,y,z)=xi+yj+zk;F(x,y,z)=xi+yj+zk; S is the surface of paraboloid z=x2+y2for0z9.z=x2+y2for0z9.

380.

[T] F(x,y,z)=x2i+y2j+z2k;F(x,y,z)=x2i+y2j+z2k; S is the surface of sphere x2+y2+z2=4.x2+y2+z2=4.

381.

[T] F(x,y,z)=xi+yj+(z21)k;F(x,y,z)=xi+yj+(z21)k; S is the surface of the solid bounded by cylinder x2+y2=4x2+y2=4 and planes z=0andz=1.z=0andz=1.

382.

[T] F(x,y,z)=xy2i+yz2j+x2zk;F(x,y,z)=xy2i+yz2j+x2zk; S is the surface bounded above by sphere ρ=2ρ=2 and below by cone φ=π4φ=π4 in spherical coordinates. (Think of S as the surface of an “ice cream cone.”)

383.

[T] F(x,y,z)=x3i+y3j+3a2zk(constanta>0);F(x,y,z)=x3i+y3j+3a2zk(constanta>0); S is the surface bounded by cylinder x2+y2=a2x2+y2=a2 and planes z=0andz=1.z=0andz=1.

384.

[T] Surface integral SF·dS,SF·dS, where S is the solid bounded by paraboloid z=x2+y2z=x2+y2 and plane z=4,z=4, and F(x,y,z)=(x+y2z2)i+(y+z2x2)j+(z+x2y2)kF(x,y,z)=(x+y2z2)i+(y+z2x2)j+(z+x2y2)k

385.

Use the divergence theorem to calculate surface integral SF·dS,SF·dS, where F(x,y,z)=(ey2)i+(y+sin(z2))j+(z1)kF(x,y,z)=(ey2)i+(y+sin(z2))j+(z1)k and S is upper hemisphere x2+y2+z2=1,z0,x2+y2+z2=1,z0, oriented upward.

386.

Use the divergence theorem to calculate surface integral SF·dS,SF·dS, where F(x,y,z)=x4ix3z2j+4xy2zkF(x,y,z)=x4ix3z2j+4xy2zk and S is the surface bounded by cylinder x2+y2=1x2+y2=1 and planes z=x+2z=x+2 and z=0.z=0.

387.

Use the divergence theorem to calculate surface integral SF·dSSF·dS when F(x,y,z)=x2z3i+2xyz3j+xz4kF(x,y,z)=x2z3i+2xyz3j+xz4k and S is the surface of the box with vertices (±1,±2,±3).(±1,±2,±3).

388.

Use the divergence theorem to calculate surface integral SF·dSSF·dS when F(x,y,z)=ztan−1(y2)i+z3ln(x2+1)j+zkF(x,y,z)=ztan−1(y2)i+z3ln(x2+1)j+zk and S is a part of paraboloid x2+y2+z=2x2+y2+z=2 that lies above plane z=1z=1 and is oriented upward.

389.

[T] Use a CAS and the divergence theorem to calculate flux SF·dS,SF·dS, where F(x,y,z)=(x3+y3)i+(y3+z3)j+(z3+x3)kF(x,y,z)=(x3+y3)i+(y3+z3)j+(z3+x3)k and S is a sphere with center (0, 0) and radius 2.

390.

Use the divergence theorem to compute the value of flux integral SF·dS,SF·dS, where F(x,y,z)=(y3+3x)i+(xz+y)j+[z+x4cos(x2y)]kF(x,y,z)=(y3+3x)i+(xz+y)j+[z+x4cos(x2y)]k and S is the area of the region bounded by x2+y2=1,x0,y0,and0z1.x2+y2=1,x0,y0,and0z1.

A vector field in three dimensions, with focus on the area with x > 0, y>0, and z>0. A quarter of a cylinder is drawn with center on the z axis. The arrows have positive x, y, and z components; they point away from the origin.
391.

Use the divergence theorem to compute flux integral SF·dS,SF·dS, where F(x,y,z)=yjzkF(x,y,z)=yjzk and S consists of the union of paraboloid y=x2+z2,0y1,y=x2+z2,0y1, and disk x2+z21,y=1,x2+z21,y=1, oriented outward. What is the flux through just the paraboloid?

392.

Use the divergence theorem to compute flux integral SF·dS,SF·dS, where F(x,y,z)=x+yj+z4kF(x,y,z)=x+yj+z4k and S is a part of cone z=x2+y2z=x2+y2 beneath top plane z=1,z=1, oriented downward.

393.

Use the divergence theorem to calculate surface integral SF·dSSF·dS for F(x,y,z)=x4ix3z2j+4xy2zk,F(x,y,z)=x4ix3z2j+4xy2zk, where S is the surface bounded by cylinder x2+y2=1x2+y2=1 and planes z=x+2andz=0.z=x+2andz=0.

394.

Consider F(x,y,z)=x2i+xyj+(z+1)k.F(x,y,z)=x2i+xyj+(z+1)k. Let E be the solid enclosed by paraboloid z=4x2y2z=4x2y2 and plane z=0z=0 with normal vectors pointing outside E. Compute flux F across the boundary of E using the divergence theorem.

For the following exercises, use a CAS along with the divergence theorem to compute the net outward flux for the fields across the given surfaces S.

395.

[T] F=x,−2y,3z;F=x,−2y,3z; S is sphere {(x,y,z):x2+y2+z2=6}.{(x,y,z):x2+y2+z2=6}.

396.

[T] F=x,2y,z;F=x,2y,z; S is the boundary of the tetrahedron in the first octant formed by plane x+y+z=1.x+y+z=1.

397.

[T] F=y2x,x3y,y2z;F=y2x,x3y,y2z; S is sphere {(x,y,z):x2+y2+z2=4}.{(x,y,z):x2+y2+z2=4}.

398.

[T] F=x,y,z;F=x,y,z; S is the surface of paraboloid z=4x2y2,z=4x2y2, for z0,z0, plus its base in the xy-plane.

For the following exercises, use a CAS and the divergence theorem to compute the net outward flux for the vector fields across the boundary of the given regions D.

399.

[T] F=zx,xy,2yz;F=zx,xy,2yz; D is the region between spheres of radius 2 and 4 centered at the origin.

400.

[T] F=r|r|=x,y,zx2+y2+z2;F=r|r|=x,y,zx2+y2+z2; D is the region between spheres of radius 1 and 2 centered at the origin.

401.

[T] F=x2,y2,z2;F=x2,y2,z2; D is the region in the first octant between planes z=4xyz=4xy and z=2xy.z=2xy.

402.

Let F(x,y,z)=2xi3xyj+xz2k.F(x,y,z)=2xi3xyj+xz2k. Use the divergence theorem to calculate SF·dS,SF·dS, where S is the surface of the cube with corners at (0,0,0),(1,0,0),(0,1,0),(0,0,0),(1,0,0),(0,1,0), (1,1,0),(0,0,1),(1,0,1),(0,1,1),and(1,1,1),(1,1,0),(0,0,1),(1,0,1),(0,1,1),and(1,1,1), oriented outward.

403.

Use the divergence theorem to find the outward flux of field F(x,y,z)=(x33y)i+(2yz+1)j+xyzkF(x,y,z)=(x33y)i+(2yz+1)j+xyzk through the cube bounded by planes x=±1,y=±1,andz=±1.x=±1,y=±1,andz=±1.

404.

Let F(x,y,z)=2xi3yj+5zkF(x,y,z)=2xi3yj+5zk and let S be hemisphere z=9x2y2z=9x2y2 together with disk x2+y29x2+y29 in the xy-plane. Use the divergence theorem.

405.

Evaluate SF·NdS,SF·NdS, where F(x,y,z)=x2i+xyj+x3y3kF(x,y,z)=x2i+xyj+x3y3k and S is the surface consisting of all faces except the tetrahedron bounded by plane x+y+z=1x+y+z=1 and the coordinate planes, with outward unit normal vector N.

A vector field in three dimensions, with arrows becoming larger the further away from the origin they are, especially in their x components. S is the surface consisting of all faces except the tetrahedron bounded by the plane x + y + z = 1. As such, a portion of the given plane, the (x, y) plane, the (x, z) plane, and the (y, z) plane are shown. The arrows point towards the origin for negative x components, away from the origin for positive x components, down for positive x and negative y components, as well as positive y and negative x components, and for positive x and y components, as well as negative x and negative y components.
406.

Find the net outward flux of field F=bzcy,cxaz,aybxF=bzcy,cxaz,aybx across any smooth closed surface in R3,R3, where a, b, and c are constants.

407.

Use the divergence theorem to evaluate SRR·nds,SRR·nds, where R(x,y,z)=xi+yj+zkR(x,y,z)=xi+yj+zk and S is sphere x2+y2+z2=a2,x2+y2+z2=a2, with constant a>0.a>0.

408.

Use the divergence theorem to evaluate SF·dS,SF·dS, where F(x,y,z)=y2zi+y3j+xzkF(x,y,z)=y2zi+y3j+xzk and S is the boundary of the cube defined by −1x1,−1y1,and0z2.−1x1,−1y1,and0z2.

409.

Let R be the region defined by x2+y2+z21.x2+y2+z21. Use the divergence theorem to find Rz2dV.Rz2dV.

410.

Let E be the solid bounded by the xy-plane and paraboloid z=4x2y2z=4x2y2 so that S is the surface of the paraboloid piece together with the disk in the xy-plane that forms its bottom. If F(x,y,z)=(xzsin(yz)+x3)i+cos(yz)j+(3zy2ex2+y2)k,F(x,y,z)=(xzsin(yz)+x3)i+cos(yz)j+(3zy2ex2+y2)k, find SF·dSSF·dS using the divergence theorem.

A vector field in three dimensions with all of the arrows pointing down. They seem to follow the path of the paraboloid drawn opening down with vertex at the origin. S is the surface of this paraboloid and the disk in the (x, y) plane that forms its bottom.
411.

Let E be the solid unit cube with diagonally opposite corners at the origin and (1, 1, 1), and faces parallel to the coordinate planes. Let S be the surface of E, oriented with the outward-pointing normal. Use a CAS to find SF·dSSF·dS using the divergence theorem if F(x,y,z)=2xyi+3yezj+xsinzk.F(x,y,z)=2xyi+3yezj+xsinzk.

412.

Use the divergence theorem to calculate the flux of F(x,y,z)=x3i+y3j+z3kF(x,y,z)=x3i+y3j+z3k through sphere x2+y2+z2=1.x2+y2+z2=1.

413.

Find SF·dS,SF·dS, where F(x,y,z)=xi+yj+zkF(x,y,z)=xi+yj+zk and S is the outwardly oriented surface obtained by removing cube [1,2]×[1,2]×[1,2][1,2]×[1,2]×[1,2] from cube [0,2]×[0,2]×[0,2].[0,2]×[0,2]×[0,2].

414.

Consider radial vector field F=r|r|=x,y,z(x2+y2+z2)1/2.F=r|r|=x,y,z(x2+y2+z2)1/2. Compute the surface integral, where S is the surface of a sphere of radius a centered at the origin.

415.

Compute the flux of water through parabolic cylinder S:y=x2,S:y=x2, from 0x2,0z3,0x2,0z3, if the velocity vector is F(x,y,z)=3z2i+6j+6xzk.F(x,y,z)=3z2i+6j+6xzk.

416.

[T] Use a CAS to find the flux of vector field F(x,y,z)=zi+zj+x2+y2kF(x,y,z)=zi+zj+x2+y2k across the portion of hyperboloid x2+y2=z2+1x2+y2=z2+1 between planes z=0z=0 and z=33,z=33, oriented so the unit normal vector points away from the z-axis.

417.

[T] Use a CAS to find the flux of vector field F(x,y,z)=(ey+x)i+(3cos(xz)y)j+zkF(x,y,z)=(ey+x)i+(3cos(xz)y)j+zk through surface S, where S is given by z2=4x2+4y2z2=4x2+4y2 from 0z4,0z4, oriented so the unit normal vector points downward.

418.

[T] Use a CAS to compute SF·dS,SF·dS, where F(x,y,z)=xi+yj+2zkF(x,y,z)=xi+yj+2zk and S is a part of sphere x2+y2+z2=2x2+y2+z2=2 with 0z1.0z1.

419.

Evaluate SF·dS,SF·dS, where F(x,y,z)=bxy2i+bx2yj+(x2+y2)z2kF(x,y,z)=bxy2i+bx2yj+(x2+y2)z2k and S is a closed surface bounding the region and consisting of solid cylinder x2+y2a2x2+y2a2 and 0zb.0zb.

420.

[T] Use a CAS to calculate the flux of F(x,y,z)=(x3+ysinz)i+(y3+zsinx)j+3zkF(x,y,z)=(x3+ysinz)i+(y3+zsinx)j+3zk across surface S, where S is the boundary of the solid bounded by hemispheres z=4x2y2z=4x2y2 and z=1x2y2,z=1x2y2, and plane z=0.z=0.

421.

Use the divergence theorem to evaluate SF·dS,SF·dS, where F(x,y,z)=xyi12y2j+zkF(x,y,z)=xyi12y2j+zk and S is the surface consisting of three pieces: z=43x23y2,1z4z=43x23y2,1z4 on the top; x2+y2=1,0z1x2+y2=1,0z1 on the sides; and z=0z=0 on the bottom.

422.

[T] Use a CAS and the divergence theorem to evaluate SF·dS,SF·dS, where F(x,y,z)=(2x+ycosz)i+(x2y)j+y2zkF(x,y,z)=(2x+ycosz)i+(x2y)j+y2zk and S is sphere x2+y2+z2=4x2+y2+z2=4 orientated outward.

423.

Use the divergence theorem to evaluate SF·dS,SF·dS, where F(x,y,z)=xi+yj+zkF(x,y,z)=xi+yj+zk and S is the boundary of the solid enclosed by paraboloid y=x2+z22,y=x2+z22, cylinder x2+z2=1,x2+z2=1, and plane x+y=2,x+y=2, and S is oriented outward.

For the following exercises, Fourier’s law of heat transfer states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, F=kT,F=kT, which means that heat energy flows hot regions to cold regions. The constant k>0k>0 is called the conductivity, which has metric units of joules per meter per second-kelvin or watts per meter-kelvin. A temperature function for region D is given. Use the divergence theorem to find net outward heat flux SF·NdS=kST·NdSSF·NdS=kST·NdS across the boundary S of D, where k=1.k=1.

424.

T(x,y,z)=100+x+2y+z;T(x,y,z)=100+x+2y+z; D={(x,y,z):0x1,0y1,0z1}D={(x,y,z):0x1,0y1,0z1}

425.

T(x,y,z)=100+ez;T(x,y,z)=100+ez; D={(x,y,z):0x1,0y1,0z1}D={(x,y,z):0x1,0y1,0z1}

426.

T(x,y,z)=100ex2y2z2;T(x,y,z)=100ex2y2z2; D is the sphere of radius a centered at the origin.

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