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Calculus Volume 3

6.5 Divergence and Curl

Calculus Volume 36.5 Divergence and Curl
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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

  • 6.5.1. Determine divergence from the formula for a given vector field.
  • 6.5.2. Determine curl from the formula for a given vector field.
  • 6.5.3. Use the properties of curl and divergence to determine whether a vector field is conservative.

In this section, we examine two important operations on a vector field: divergence and curl. They are important to the field of calculus for several reasons, including the use of curl and divergence to develop some higher-dimensional versions of the Fundamental Theorem of Calculus. In addition, curl and divergence appear in mathematical descriptions of fluid mechanics, electromagnetism, and elasticity theory, which are important concepts in physics and engineering. We can also apply curl and divergence to other concepts we already explored. For example, under certain conditions, a vector field is conservative if and only if its curl is zero.

In addition to defining curl and divergence, we look at some physical interpretations of them, and show their relationship to conservative and source-free vector fields.

Divergence

Divergence is an operation on a vector field that tells us how the field behaves toward or away from a point. Locally, the divergence of a vector field F in 22 or 33 at a particular point P is a measure of the “outflowing-ness” of the vector field at P. If F represents the velocity of a fluid, then the divergence of F at P measures the net rate of change with respect to time of the amount of fluid flowing away from P (the tendency of the fluid to flow “out of” P). In particular, if the amount of fluid flowing into P is the same as the amount flowing out, then the divergence at P is zero.

Definition

If F=P,Q,RF=P,Q,R is a vector field in 33 and Px,Qy,Px,Qy, and RzRz all exist, then the divergence of F is defined by

divF=Px+Qy+Rz=Px+Qy+Rz.divF=Px+Qy+Rz=Px+Qy+Rz.
6.16

Note the divergence of a vector field is not a vector field, but a scalar function. In terms of the gradient operator =x,y,z,=x,y,z, divergence can be written symbolically as the dot product

divF=·F.divF=·F.

Note this is merely helpful notation, because the dot product of a vector of operators and a vector of functions is not meaningfully defined given our current definition of dot product.

If F=P,QF=P,Q is a vector field in 2,2, and PxPx and QyQy both exist, then the divergence of F is defined similarly as

divF=Px+Qy=Px+Qy=·F.divF=Px+Qy=Px+Qy=·F.

To illustrate this point, consider the two vector fields in Figure 6.50. At any particular point, the amount flowing in is the same as the amount flowing out, so at every point the “outflowing-ness” of the field is zero. Therefore, we expect the divergence of both fields to be zero, and this is indeed the case, as

div(1,2)=x(1)+y(2)=0anddiv(y,x)=x(y)+y(x)=0.div(1,2)=x(1)+y(2)=0anddiv(y,x)=x(y)+y(x)=0.
Two images of vector fields A and B in two dimensions. Vector field A has arrows pointing up and to the right. They do not change in size or direction. It has zero divergence. Vector field B has arrows surrounding the origin in a counterclockwise direction. The arrows are larger the closer they are to the origin. It also has zero divergence.
Figure 6.50 (a) Vector field 1,21,2 has zero divergence. (b) Vector field y,xy,x also has zero divergence.

By contrast, consider radial vector field R(x,y)=x,yR(x,y)=x,y in Figure 6.51. At any given point, more fluid is flowing in than is flowing out, and therefore the “outgoingness” of the field is negative. We expect the divergence of this field to be negative, and this is indeed the case, as div(R)=x(x)+y(y)=−2.div(R)=x(x)+y(y)=−2.

A vector field in two dimensions with negative divergence. The arrows point in towards the origin in a radial pattern. The closer the arrows are to the origin, the larger they are.
Figure 6.51 This vector field has negative divergence.

To get a global sense of what divergence is telling us, suppose that a vector field in 22 represents the velocity of a fluid. Imagine taking an elastic circle (a circle with a shape that can be changed by the vector field) and dropping it into a fluid. If the circle maintains its exact area as it flows through the fluid, then the divergence is zero. This would occur for both vector fields in Figure 6.50. On the other hand, if the circle’s shape is distorted so that its area shrinks or expands, then the divergence is not zero. Imagine dropping such an elastic circle into the radial vector field in Figure 6.51 so that the center of the circle lands at point (3, 3). The circle would flow toward the origin, and as it did so the front of the circle would travel more slowly than the back, causing the circle to “scrunch” and lose area. This is how you can see a negative divergence.

Example 6.48

Calculating Divergence at a Point

If F(x,y,z)=exi+yzjy2k,F(x,y,z)=exi+yzjy2k, then find the divergence of F at (0,2,−1).(0,2,−1).

Solution

The divergence of F is

x(ex)+y(yz)z(yz2)=ex+z2yz.x(ex)+y(yz)z(yz2)=ex+z2yz.

Therefore, the divergence at (0,2,−1)(0,2,−1) is e01+4=4.e01+4=4. If F represents the velocity of a fluid, then more fluid is flowing out than flowing in at point (0,2,−1).(0,2,−1).

Checkpoint 6.40

Find divFdivF for F(x,y,z)=xy,5z2y,x2+y2.F(x,y,z)=xy,5z2y,x2+y2.

One application for divergence occurs in physics, when working with magnetic fields. A magnetic field is a vector field that models the influence of electric currents and magnetic materials. Physicists use divergence in Gauss’s law for magnetism, which states that if B is a magnetic field, then ·B=0;·B=0; in other words, the divergence of a magnetic field is zero.

Example 6.49

Determining Whether a Field Is Magnetic

Is it possible for F(x,y)=x2y,yxy2F(x,y)=x2y,yxy2 to be a magnetic field?

Solution

If F were magnetic, then its divergence would be zero. The divergence of F is

x(x2y)+y(yxy2)=2xy+12xy=1x(x2y)+y(yxy2)=2xy+12xy=1

and therefore F cannot model a magnetic field (Figure 6.52).

A vector field in two dimensions with divergence equal to 1. The arrows are quite flat near the x axis and vertical near the y axis. They seem to asymptotically approach the axes in quadrants 2 and 4, pointing up and to the right in quadrant 2 and down and to the left in quadrant 4. In quadrant 1, they start by pointing up and to the right close to the y axis, but they soon shift to pointing down and to the right. In quadrant 3, they start by pointing down and to the left close to the y axis, bu they soon shift to pointing up and to the left. The closer the arrows are to the origin, the shorter they are.
Figure 6.52 The divergence of vector field F(x,y)=x2y,yxy2F(x,y)=x2y,yxy2 is one, so it cannot model a magnetic field.

Another application for divergence is detecting whether a field is source free. Recall that a source-free field is a vector field that has a stream function; equivalently, a source-free field is a field with a flux that is zero along any closed curve. The next two theorems say that, under certain conditions, source-free vector fields are precisely the vector fields with zero divergence.

Theorem 6.14

Divergence of a Source-Free Vector Field

If F=P,QF=P,Q is a source-free continuous vector field with differentiable component functions, then divF=0.divF=0.

Proof

Since F is source free, there is a function g(x,y)g(x,y) with gy=Pgy=P and gx=Q.gx=Q. Therefore, F=gy,gxF=gy,gx and divF=gyxgxy=0divF=gyxgxy=0 by Clairaut’s theorem.

The converse of Divergence of a Source-Free Vector Field is true on simply connected regions, but the proof is too technical to include here. Thus, we have the following theorem, which can test whether a vector field in 22 is source free.

Theorem 6.15

Divergence Test for Source-Free Vector Fields

Let F=P,QF=P,Q be a continuous vector field with differentiable component functions with a domain that is simply connected. Then, divF=0divF=0 if and only if F is source free.

Example 6.50

Determining Whether a Field Is Source Free

Is field F(x,y)=x2y,5xy2F(x,y)=x2y,5xy2 source free?

Solution

Note the domain of F is 2,2, which is simply connected. Furthermore, F is continuous with differentiable component functions. Therefore, we can use Divergence Test for Source-Free Vector Fields to analyze F. The divergence of F is

x(x2y)+y(5xy2)=2xy2xy=0.x(x2y)+y(5xy2)=2xy2xy=0.

Therefore, F is source free by Divergence Test for Source-Free Vector Fields.

Checkpoint 6.41

Let F(x,y)=ay,bxF(x,y)=ay,bx be a rotational field where a and b are positive constants. Is F source free?

Recall that the flux form of Green’s theorem says that

CF·Nds=DPx+QydA,CF·Nds=DPx+QydA,

where C is a simple closed curve and D is the region enclosed by C. Since Px+Qy=divF,Px+Qy=divF, Green’s theorem is sometimes written as

CF·Nds=DdivFdA.CF·Nds=DdivFdA.

Therefore, Green’s theorem can be written in terms of divergence. If we think of divergence as a derivative of sorts, then Green’s theorem says the “derivative” of F on a region can be translated into a line integral of F along the boundary of the region. This is analogous to the Fundamental Theorem of Calculus, in which the derivative of a function ff on a line segment [a,b][a,b] can be translated into a statement about ff on the boundary of [a,b].[a,b]. Using divergence, we can see that Green’s theorem is a higher-dimensional analog of the Fundamental Theorem of Calculus.

We can use all of what we have learned in the application of divergence. Let v be a vector field modeling the velocity of a fluid. Since the divergence of v at point P measures the “outflowing-ness” of the fluid at P, divv(P)>0divv(P)>0 implies that more fluid is flowing out of P than flowing in. Similarly, divv(P)<0divv(P)<0 implies the more fluid is flowing in to P than is flowing out, and divv(P)=0divv(P)=0 implies the same amount of fluid is flowing in as flowing out.

Example 6.51

Determining Flow of a Fluid

Suppose v(x,y)=xy,y,y>0v(x,y)=xy,y,y>0 models the flow of a fluid. Is more fluid flowing into point (1,4)(1,4) than flowing out?

Solution

To determine whether more fluid is flowing into (1,4)(1,4) than is flowing out, we calculate the divergence of v at (1,4):(1,4):

div(v)=x(xy)+y(y)=y+1.div(v)=x(xy)+y(y)=y+1.

To find the divergence at (1,4),(1,4), substitute the point into the divergence: −4+1=−3.−4+1=−3. Since the divergence of v at (1,4)(1,4) is negative, more fluid is flowing in than flowing out (Figure 6.53).

A vector field in two dimensions with negative divergence at (1,4). The arrows are very flat but become more vertical closer to the y axis. Above the x axis, the arrows point up and towards the y axis on either side of it. Below the x axis, the arrows point down and away from the y axis on either side of it.
Figure 6.53 Vector field v(x,y)=xy,yv(x,y)=xy,y has negative divergence at (1,4).(1,4).

Checkpoint 6.42

For vector field v(x,y)=xy,y,y>0,v(x,y)=xy,y,y>0, find all points P such that the amount of fluid flowing in to P equals the amount of fluid flowing out of P.

Curl

The second operation on a vector field that we examine is the curl, which measures the extent of rotation of the field about a point. Suppose that F represents the velocity field of a fluid. Then, the curl of F at point P is a vector that measures the tendency of particles near P to rotate about the axis that points in the direction of this vector. The magnitude of the curl vector at P measures how quickly the particles rotate around this axis. In other words, the curl at a point is a measure of the vector field’s “spin” at that point. Visually, imagine placing a paddlewheel into a fluid at P, with the axis of the paddlewheel aligned with the curl vector (Figure 6.54). The curl measures the tendency of the paddlewheel to rotate.

A diagram of a small paddlewheel in water. Arrows are drawn surrounding the center in a counterclockwise circle. At the center, the height is labeled n.
Figure 6.54 To visualize curl at a point, imagine placing a small paddlewheel into the vector field at a point.

Consider the vector fields in Figure 6.50. In part (a), the vector field is constant and there is no spin at any point. Therefore, we expect the curl of the field to be zero, and this is indeed the case. Part (b) shows a rotational field, so the field has spin. In particular, if you place a paddlewheel into a field at any point so that the axis of the wheel is perpendicular to a plane, the wheel rotates counterclockwise. Therefore, we expect the curl of the field to be nonzero, and this is indeed the case (the curl is 2k).2k).

To see what curl is measuring globally, imagine dropping a leaf into the fluid. As the leaf moves along with the fluid flow, the curl measures the tendency of the leaf to rotate. If the curl is zero, then the leaf doesn’t rotate as it moves through the fluid.

Definition

If F=P,Q,RF=P,Q,R is a vector field in 3,3, and Px,Qy,Px,Qy, and RzRz all exist, then the curl of F is defined by

curlF=(RyQz)i+(PzRx)j+(QxPy)k=(RyQz)i+(PzRx)j+(QxPy)k.curlF=(RyQz)i+(PzRx)j+(QxPy)k=(RyQz)i+(PzRx)j+(QxPy)k.
6.17

Note that the curl of a vector field is a vector field, in contrast to divergence.

The definition of curl can be difficult to remember. To help with remembering, we use the notation ×F×F to stand for a “determinant” that gives the curl formula:

|ijkxyzPQR|.|ijkxyzPQR|.

The determinant of this matrix is

(RyQz)i(RxPz)j+(QxPy)k=(RyQz)i+(PzRx)j+(QxPy)k=curlF.(RyQz)i(RxPz)j+(QxPy)k=(RyQz)i+(PzRx)j+(QxPy)k=curlF.

Thus, this matrix is a way to help remember the formula for curl. Keep in mind, though, that the word determinant is used very loosely. A determinant is not really defined on a matrix with entries that are three vectors, three operators, and three functions.

If F=P,QF=P,Q is a vector field in 2,2, then the curl of F, by definition, is

curlF=(QxPy)k=(QxPy)k.curlF=(QxPy)k=(QxPy)k.

Example 6.52

Finding the Curl of a Three-Dimensional Vector Field

Find the curl of F(P,Q,R)=x2z,ey+xz,xyz.F(P,Q,R)=x2z,ey+xz,xyz.

Solution

The curl is

curlF=×F=|ijk/x/y/zPQR|=(RyQz)i+(PzRx)j+(QxPy)k=(xzx)i+(x2yz)j+zk.curlF=×F=|ijk/x/y/zPQR|=(RyQz)i+(PzRx)j+(QxPy)k=(xzx)i+(x2yz)j+zk.
Checkpoint 6.43

Find the curl of F=sinxcosz,sinysinz,cosxcosyF=sinxcosz,sinysinz,cosxcosy at point (0,π2,π2).(0,π2,π2).

Example 6.53

Finding the Curl of a Two-Dimensional Vector Field

Find the curl of F=P,Q=y,0.F=P,Q=y,0.

Solution

Notice that this vector field consists of vectors that are all parallel. In fact, each vector in the field is parallel to the x-axis. This fact might lead us to the conclusion that the field has no spin and that the curl is zero. To test this theory, note that

curlF=(QxPy)k=k0.curlF=(QxPy)k=k0.

Therefore, this vector field does have spin. To see why, imagine placing a paddlewheel at any point in the first quadrant (Figure 6.55). The larger magnitudes of the vectors at the top of the wheel cause the wheel to rotate. The wheel rotates in the clockwise (negative) direction, causing the coefficient of the curl to be negative.

Two vector field diagrams consisting of vectors that are all parallel. The closer they are to the x axis, the shorter the arrows are. Above the x axis, the arrows point to the right, and below the x axis, the arrows point to the left.
Figure 6.55 Vector field F(x,y)=y,0F(x,y)=y,0 consists of vectors that are all parallel.

Note that if F=P,QF=P,Q is a vector field in a plane, then curlF·k=(QxPy)k·k=QxPy.curlF·k=(QxPy)k·k=QxPy. Therefore, the circulation form of Green’s theorem is sometimes written as

CF·dr=DcurlF·kdA,CF·dr=DcurlF·kdA,

where C is a simple closed curve and D is the region enclosed by C. Therefore, the circulation form of Green’s theorem can be written in terms of the curl. If we think of curl as a derivative of sorts, then Green’s theorem says that the “derivative” of F on a region can be translated into a line integral of F along the boundary of the region. This is analogous to the Fundamental Theorem of Calculus, in which the derivative of a function ff on line segment [a,b][a,b] can be translated into a statement about ff on the boundary of [a,b].[a,b]. Using curl, we can see the circulation form of Green’s theorem is a higher-dimensional analog of the Fundamental Theorem of Calculus.

We can now use what we have learned about curl to show that gravitational fields have no “spin.” Suppose there is an object at the origin with mass m1m1 at the origin and an object with mass m2.m2. Recall that the gravitational force that object 1 exerts on object 2 is given by field

F(x,y,z)=Gm2m2x(x2+y2+z2)3/2,y(x2+y2+z2)3/2,z(x2+y2+z2)3/2.F(x,y,z)=Gm2m2x(x2+y2+z2)3/2,y(x2+y2+z2)3/2,z(x2+y2+z2)3/2.

Example 6.54

Determining the Spin of a Gravitational Field

Show that a gravitational field has no spin.

Solution

To show that F has no spin, we calculate its curl. Let P(x,y,z)=x(x2+y2+z2)3/2,P(x,y,z)=x(x2+y2+z2)3/2, Q(x,y,z)=y(x2+y2+z2)3/2,Q(x,y,z)=y(x2+y2+z2)3/2, and R(x,y,z)=z(x2+y2+z2)3/2.R(x,y,z)=z(x2+y2+z2)3/2. Then,

curlF=Gm1m2[(RyQz)i+(PzRx)j+(QxPy)k]=Gm1m2[(−3yz(x2+y2+z2)5/2(−3yz(x2+y2+z2)5/2))i+(−3xz(x2+y2+z2)5/2(−3xz(x2+y2+z2)5/2))j+(−3xy(x2+y2+z2)5/2(−3xy(x2+y2+z2)5/2))k]=0.curlF=Gm1m2[(RyQz)i+(PzRx)j+(QxPy)k]=Gm1m2[(−3yz(x2+y2+z2)5/2(−3yz(x2+y2+z2)5/2))i+(−3xz(x2+y2+z2)5/2(−3xz(x2+y2+z2)5/2))j+(−3xy(x2+y2+z2)5/2(−3xy(x2+y2+z2)5/2))k]=0.

Since the curl of the gravitational field is zero, the field has no spin.

Checkpoint 6.44

Field v(x,y)=yx2+y2,xx2+y2v(x,y)=yx2+y2,xx2+y2 models the flow of a fluid. Show that if you drop a leaf into this fluid, as the leaf moves over time, the leaf does not rotate.

Using Divergence and Curl

Now that we understand the basic concepts of divergence and curl, we can discuss their properties and establish relationships between them and conservative vector fields.

If F is a vector field in 3,3, then the curl of F is also a vector field in 3.3. Therefore, we can take the divergence of a curl. The next theorem says that the result is always zero. This result is useful because it gives us a way to show that some vector fields are not the curl of any other field. To give this result a physical interpretation, recall that divergence of a velocity field v at point P measures the tendency of the corresponding fluid to flow out of P. Since divcurl(v)=0,divcurl(v)=0, the net rate of flow in vector field curl(v) at any point is zero. Taking the curl of vector field F eliminates whatever divergence was present in F.

Theorem 6.16

Divergence of the Curl

Let F=P,Q,RF=P,Q,R be a vector field in 33 such that the component functions all have continuous second-order partial derivatives. Then, divcurl(F)=·(×F)=0.divcurl(F)=·(×F)=0.

Proof

By the definitions of divergence and curl, and by Clairaut’s theorem,

div curlF=div[(RyQz)i+(PzRx)j+(QxPy)k]=RyxQxz+PyzRyx+QzxPzy=0.div curlF=div[(RyQz)i+(PzRx)j+(QxPy)k]=RyxQxz+PyzRyx+QzxPzy=0.

Example 6.55

Showing That a Vector Field Is Not the Curl of Another

Show that F(x,y,z)=exi+yzj+xz2kF(x,y,z)=exi+yzj+xz2k is not the curl of another vector field. That is, show that there is no other vector G with curlG=F.curlG=F.

Solution

Notice that the domain of F is all of 33 and the second-order partials of F are all continuous. Therefore, we can apply the previous theorem to F.

The divergence of F is ex+z+2xz.ex+z+2xz. If F were the curl of vector field G, then divF=div curlG=0.divF=div curlG=0. But, the divergence of F is not zero, and therefore F is not the curl of any other vector field.

Checkpoint 6.45

Is it possible for G(x,y,z)=sinx,cosy,sin(xyz)G(x,y,z)=sinx,cosy,sin(xyz) to be the curl of a vector field?

With the next two theorems, we show that if F is a conservative vector field then its curl is zero, and if the domain of F is simply connected then the converse is also true. This gives us another way to test whether a vector field is conservative.

Theorem 6.17

Curl of a Conservative Vector Field

If F=P,Q,RF=P,Q,R is conservative, then curlF=0.curlF=0.

Proof

Since conservative vector fields satisfy the cross-partials property, all the cross-partials of F are equal. Therefore,

curlF=(RyQz)i+(PzRx)j+(QxPy)k=0.curlF=(RyQz)i+(PzRx)j+(QxPy)k=0.

The same theorem is true for vector fields in a plane.

Since a conservative vector field is the gradient of a scalar function, the previous theorem says that curl(f)=0curl(f)=0 for any scalar function f.f. In terms of our curl notation, ×(f)=0.×(f)=0. This equation makes sense because the cross product of a vector with itself is always the zero vector. Sometimes equation ×(f)=0×(f)=0 is simplified as ×=0.×=0.

Theorem 6.18

Curl Test for a Conservative Field

Let F=P,Q,RF=P,Q,R be a vector field in space on a simply connected domain. If curlF=0,curlF=0, then F is conservative.

Proof

Since curlF=0,curlF=0, we have that Ry=Qz,Pz=Rx,Ry=Qz,Pz=Rx, and Qx=Py.Qx=Py. Therefore, F satisfies the cross-partials property on a simply connected domain, and Cross-Partial Property of Conservative Fields implies that F is conservative.

The same theorem is also true in a plane. Therefore, if F is a vector field in a plane or in space and the domain is simply connected, then F is conservative if and only if curlF=0.curlF=0.

Example 6.56

Testing Whether a Vector Field Is Conservative

Use the curl to determine whether F(x,y,z)=yz,xz,xyF(x,y,z)=yz,xz,xy is conservative.

Solution

Note that the domain of F is all of 3,3, which is simply connected (Figure 6.56). Therefore, we can test whether F is conservative by calculating its curl.

A diagram showing the curl of a vector field in two dimensions. The curl is zero. The arrows seem to be pointing up and over into the yz plane.
Figure 6.56 The curl of vector field F(x,y,z)=yz,xz,xyF(x,y,z)=yz,xz,xy is zero.

The curl of F is

(yxyzxz)i+(yyzzxy)j+(yxzzyz)k=(xx)i+(yy)j+(zz)k=0.(yxyzxz)i+(yyzzxy)j+(yxzzyz)k=(xx)i+(yy)j+(zz)k=0.

Thus, F is conservative.

We have seen that the curl of a gradient is zero. What is the divergence of a gradient? If ff is a function of two variables, then div(f)=·(f)=fxx+fyy.div(f)=·(f)=fxx+fyy. We abbreviate this “double dot product” as 2.2. This operator is called the Laplace operator, and in this notation Laplace’s equation becomes 2f=0.2f=0. Therefore, a harmonic function is a function that becomes zero after taking the divergence of a gradient.

Similarly, if ff is a function of three variables then

div(f)=·(f)=fxx+fyy+fzz.div(f)=·(f)=fxx+fyy+fzz.

Using this notation we get Laplace’s equation for harmonic functions of three variables:

2f=0.2f=0.

Harmonic functions arise in many applications. For example, the potential function of an electrostatic field in a region of space that has no static charge is harmonic.

Example 6.57

Analyzing a Function

Is it possible for f(x,y)=x2+xyf(x,y)=x2+xy to be the potential function of an electrostatic field that is located in a region of 22 free of static charge?

Solution

If ff were such a potential function, then ff would be harmonic. Note that fxx=2fxx=2 and fyy=0,fyy=0, and so fxx+fyy0.fxx+fyy0. Therefore, ff is not harmonic and ff cannot represent an electrostatic potential.

Checkpoint 6.46

Is it possible for function f(x,y)=x2y2+xf(x,y)=x2y2+x to be the potential function of an electrostatic field located in a region of 22 free of static charge?

Section 6.5 Exercises

For the following exercises, determine whether the statement is true or false.

206.

If the coordinate functions of F:33F:33 have continuous second partial derivatives, then curl(div(F))curl(div(F)) equals zero.

207.

·(xi+yj+zk)=1.·(xi+yj+zk)=1.

208.

All vector fields of the form F(x,y,z)=f(x)i+g(y)j+h(z)kF(x,y,z)=f(x)i+g(y)j+h(z)k are conservative.

209.

If curlF=0,curlF=0, then F is conservative.

210.

If F is a constant vector field then divF=0.divF=0.

211.

If F is a constant vector field then curlF=0.curlF=0.

For the following exercises, find the curl of F.

212.

F(x,y,z)=xy2z4i+(2x2y+z)j+y3z2kF(x,y,z)=xy2z4i+(2x2y+z)j+y3z2k

213.

F(x,y,z)=x2zi+y2xj+(y+2z)kF(x,y,z)=x2zi+y2xj+(y+2z)k

214.

F(x,y,z)=3xyz2i+y2sinzj+xe2zkF(x,y,z)=3xyz2i+y2sinzj+xe2zk

215.

F(x,y,z)=x2yzi+xy2zj+xyz2kF(x,y,z)=x2yzi+xy2zj+xyz2k

216.

F(x,y,z)=(xcosy)i+xy2jF(x,y,z)=(xcosy)i+xy2j

217.

F(x,y,z)=(xy)i+(yz)j+(zx)kF(x,y,z)=(xy)i+(yz)j+(zx)k

218.

F(x,y,z)=xyzi+x2y2z2j+y2z3kF(x,y,z)=xyzi+x2y2z2j+y2z3k

219.

F(x,y,z)=xyi+yzj+xzkF(x,y,z)=xyi+yzj+xzk

220.

F(x,y,z)=x2i+y2j+z2kF(x,y,z)=x2i+y2j+z2k

221.

F(x,y,z)=axi+byj+ckF(x,y,z)=axi+byj+ck for constants a, b, c

For the following exercises, find the divergence of F.

222.

F(x,y,z)=x2zi+y2xj+(y+2z)kF(x,y,z)=x2zi+y2xj+(y+2z)k

223.

F(x,y,z)=3xyz2i+y2sinzj+xe2zkF(x,y,z)=3xyz2i+y2sinzj+xe2zk

224.

F(x,y)=(sinx)i+(cosy)jF(x,y)=(sinx)i+(cosy)j

225.

F(x,y,z)=x2i+y2j+z2kF(x,y,z)=x2i+y2j+z2k

226.

F(x,y,z)=(xy)i+(yz)j+(zx)kF(x,y,z)=(xy)i+(yz)j+(zx)k

227.

F(x,y)=xx2+y2i+yx2+y2jF(x,y)=xx2+y2i+yx2+y2j

228.

F(x,y)=xiyjF(x,y)=xiyj

229.

F(x,y,z)=axi+byj+ckF(x,y,z)=axi+byj+ck for constants a, b, c

230.

F(x,y,z)=xyzi+x2y2z2j+y2z3kF(x,y,z)=xyzi+x2y2z2j+y2z3k

231.

F(x,y,z)=xyi+yzj+xzkF(x,y,z)=xyi+yzj+xzk

For the following exercises, determine whether each of the given scalar functions is harmonic.

232.

u(x,y,z)=ex(cosysiny)u(x,y,z)=ex(cosysiny)

233.

w(x,y,z)=(x2+y2+z2)1/2w(x,y,z)=(x2+y2+z2)1/2

234.

If F(x,y,z)=2i+2xj+3ykF(x,y,z)=2i+2xj+3yk and G(x,y,z)=xiyj+zk,G(x,y,z)=xiyj+zk, find curl(F×G).curl(F×G).

235.

If F(x,y,z)=2i+2xj+3ykF(x,y,z)=2i+2xj+3yk and G(x,y,z)=xiyj+zk,G(x,y,z)=xiyj+zk, find div(F×G).div(F×G).

236.

Find divF,divF, given that F=f,F=f, where f(x,y,z)=xy3z2.f(x,y,z)=xy3z2.

237.

Find the divergence of F for vector field F(x,y,z)=(y2+z2)(x+y)i+(z2+x2)(y+z)j+(x2+y2)(z+x)k.F(x,y,z)=(y2+z2)(x+y)i+(z2+x2)(y+z)j+(x2+y2)(z+x)k.

238.

Find the divergence of F for vector field F(x,y,z)=f1(y,z)i+f2(x,z)j+f3(x,y)k.F(x,y,z)=f1(y,z)i+f2(x,z)j+f3(x,y)k.

For the following exercises, use r=|r|r=|r| and r=(x,y,z).r=(x,y,z).

239.

Find the curlr.curlr.

240.

Find the curlrr.curlrr.

241.

Find the curlrr3.curlrr3.

242.

Let F(x,y)=yi+xjx2+y2,F(x,y)=yi+xjx2+y2, where F is defined on {(x,y)|(x,y)(0,0)}.{(x,y)|(x,y)(0,0)}. Find curlF.curlF.

For the following exercises, use a computer algebra system to find the curl of the given vector fields.

243.

[T] F(x,y,z)=arctan(xy)i+lnx2+y2j+kF(x,y,z)=arctan(xy)i+lnx2+y2j+k

244.

[T] F(x,y,z)=sin(xy)i+sin(yz)j+sin(zx)kF(x,y,z)=sin(xy)i+sin(yz)j+sin(zx)k

For the following exercises, find the divergence of F at the given point.

245.

F(x,y,z)=i+j+kF(x,y,z)=i+j+k at (2,−1,3)(2,−1,3)

246.

F(x,y,z)=xyzi+yj+zkF(x,y,z)=xyzi+yj+zk at (1,2,3)(1,2,3)

247.

F(x,y,z)=exyi+exzj+eyzkF(x,y,z)=exyi+exzj+eyzk at (3,2,0)(3,2,0)

248.

F(x,y,z)=xyzi+yj+zkF(x,y,z)=xyzi+yj+zk at (1, 2, 1)

249.

F(x,y,z)=exsinyiexcosyjF(x,y,z)=exsinyiexcosyj at (0, 0, 3)

For the following exercises, find the curl of F at the given point.

250.

F(x,y,z)=i+j+kF(x,y,z)=i+j+k at (2,−1,3)(2,−1,3)

251.

F(x,y,z)=xyzi+yj+xkF(x,y,z)=xyzi+yj+xk at (1,2,3)(1,2,3)

252.

F(x,y,z)=exyi+exzj+eyzkF(x,y,z)=exyi+exzj+eyzk at (3, 2, 0)

253.

F(x,y,z)=xyzi+yj+zkF(x,y,z)=xyzi+yj+zk at (1, 2, 1)

254.

F(x,y,z)=exsinyiexcosyjF(x,y,z)=exsinyiexcosyj at (0, 0, 3)

255.

Let F(x,y,z)=(3x2y+az)i+x3j+(3x+3z2)k.F(x,y,z)=(3x2y+az)i+x3j+(3x+3z2)k. For what value of a is F conservative?

256.

Given vector field F(x,y)=1x2+y2(y,x)F(x,y)=1x2+y2(y,x) on domain D=2{(0,0)}={(x,y)2|(x,y)(0,0)},D=2{(0,0)}={(x,y)2|(x,y)(0,0)}, is F conservative?

257.

Given vector field F(x,y)=1x2+y2(x,y)F(x,y)=1x2+y2(x,y) on domain D=2{(0,0)},D=2{(0,0)}, is F conservative?

258.

Find the work done by force field F(x,y)=eyixeyjF(x,y)=eyixeyj in moving an object from P(0, 1) to Q(2, 0). Is the force field conservative?

259.

Compute divergence F=(sinhx)i+(coshy)jxyzk.F=(sinhx)i+(coshy)jxyzk.

260.

Compute curl F=(sinhx)i+(coshy)jxyzk.F=(sinhx)i+(coshy)jxyzk.

For the following exercises, consider a rigid body that is rotating about the x-axis counterclockwise with constant angular velocity ω=a,b,c.ω=a,b,c. If P is a point in the body located at r=xi+yj+zk,r=xi+yj+zk, the velocity at P is given by vector field F=ω×r.F=ω×r.

A three dimensional diagram of an object rotating about the x axis in a counterclockwise manner with constant angular velocity w = <a,b,c>. The object is roughly a sphere with pointed ends on the x axis, which cuts it in half. An arrow r is drawn from (0,0,0) to P(x,y,z) and down from P(x,y,z) to the x axis.
261.

Express F in terms of i, j, and k vectors.

262.

Find divF.divF.

263.

Find curlFcurlF

In the following exercises, suppose that ·F=0·F=0 and ·G=0.·G=0.

264.

Does F+GF+G necessarily have zero divergence?

265.

Does F×GF×G necessarily have zero divergence?

In the following exercises, suppose a solid object in 33 has a temperature distribution given by T(x,y,z).T(x,y,z). The heat flow vector field in the object is F=kT,F=kT, where k>0k>0 is a property of the material. The heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is ·F=k·T=k2T.·F=k·T=k2T.

266.

Compute the heat flow vector field.

267.

Compute the divergence.

268.

[T] Consider rotational velocity field v=0,10z,−10y.v=0,10z,−10y. If a paddlewheel is placed in plane x+y+z=1x+y+z=1 with its axis normal to this plane, using a computer algebra system, calculate how fast the paddlewheel spins in revolutions per unit time.

A three dimensional diagram of a rotational velocity field. The arrows are showing a rotation in a clockwise manner. A paddlewheel is shown in plan x + y + z = 1 with n extended out perpendicular to the plane.
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