Instructor: Abdullah Hamadeh
Part IV
Vector calculus
1 Vector fields
Historically, vector calculus was developed in order to understand a number of physical phenomena occurring in nature, e.g., gravitation, electromagnetism, fluid flow and heat transfer. Note that these phenomena require the concept of a field which exists in space and which can be used to explain their behavior. You are already familiar with the idea that forces are vector quantities: In order to specify a force F, you need to specify both its magnitude, or strength, as well as its direction of action.
In this final part of the course, we shall be concerned with vector-valued functions of several variables that will have the form
F :Rm→Rn, m,n>1
Usually, the input to F will be a point(x,y,z)in physical space. This point will also be denoted as r, which is also used to represent the position vector of the point. The output of F will most often be one of the following:
1. The velocity v= (v1,· · ·,vn)of a particle moving inRn, or 2. The force F= (F1,· · ·,Fn)exerted on a particle at the point r.
Therefore, for a vector field F(r), where the input to the vector fields is r=xi+yj+zk, there is a vector F associated with each point(x,y,z) in space. In other words, each of the components Fi of the vector F is a multivariable function Fi=Fi(r) =Fi(x,y,z), of the form we saw in Part 2.
It is useful to be able to visualize these vector fields. For example, a velocity vector field could give an idea of how a fluid is travelling in space. The vector field F(x,y) = (F1(x,y),F2(x,y))inR2is graphically represented by drawing an arrow representing the vector
F1(x,y)i+F2(x,y)j that starts at the point(x,y). Let us look at some examples.
1. The vector field F(x,y) =i. This is a quite trivial field – at each point(x,y)we simply draw the vector i, a unit vector pointing in the positive x-direction.
One possible physical interpretation of this vector field is the velocity field v(x,y)of a thin layer of fluid that is moving horizontally over the surface of a table with constant velocity.
2. The vector field F(x,y) =xi+yj. At each point P with coordinates(x,y), we see that F(x,y)is the position vector of P. If we draw such arrows at points P (see below), we obtain a set of arrows which point away from the origin and which get larger and larger as we move away from the origin.
x y
0 1 2 3
-1 -2 -3
The vector field F(x,y) =i.
x y
The vector field F(x,y) =xi+yj.
Here is another planar vector field (i.e., a vector field in the planeR2) that will be important in future applications:
F(x,y) =−yi+xj.
A sketch of the field is given below. You should compute a few values of F(x,y)for convenient values of(x,y)in order to convince yourself that the vector field behaves in this way.
Obviously, the vector field suggests some kind of rotational motion around the origin. But the magnitude of the field vectors, as represented by the lengths of the arrows, appears to increase. In fact,
kF(x,y)k=p x2+y2.
One might wonder where such a vector field would correspond to a realistic situation: How can you have rotation, in which the magnitude of the rotation, i.e., the velocity, increases as you move away from the origin. It is not too hard to come up with some situations, for example, the rotation of a turntable (or a merry-go-round platform) at constant angular velocity. Each point on the turntable travels along a circular trajectory around the origin. And all points take the same time T to execute one cycle of rotation. This means that points that are farther away from the origin must travel faster. The vector field pictured above describes, up to a constant factor, the velocity field of particles on a turntable that rotates about the origin with constant angular speed. More on this later.
Let us conclude with a look at a vector field of major importance in physics: the gravitational force.
1.1 Gravitational force fields
Suppose that a mass M is located at the origin O inR3. And suppose that there is another mass m situated at point P with coordinates r= (x,y,z). (You can refer to the relevant figure in the set of illustrations of vector fields.) What do we know about gravitation?
y
x
The vector field F=−yi+xj.
1. First, we know that the force exerted by M on m points in the direction from m to M. In this case, it points in the direction of−r, where r is the position vector of m. (For convenience, we have placed M at the origin.) Therefore, the gravitational force exerted by M on m will have the form
FMm=−g r (1)
where g(x,y,z)will be a positive quantity that is not necessarily constant.
2. Secondly, we know that the magnitude of FMmis inversely proportional to the square of the distance between M and m. In other words,
kFMmk= K krk2 =K
r2, (2)
where K is the constant of proportionality and, for convenience, we use the notation r=krk.
3. The magnitude of FMmis also proportional to each of M and m.
We shall first consider the effect of distance on the force, for fixed M and m. From Eq. (1), we have
kFMmk=gr (3)
Comparing (3) and (2), it follows that
g=K
r3. (4)
Therefore, from (1), we have that
FMm=−K
r3r. (5)
We now use Fact No. 3 to deduce that K=GMm, where G is a proportionality constant. The final result is well known to you:
FMm(r) =−GMm
r3 r, (6)
where G is the so-called gravitational constant. Note that this result can also be written as FMm(r) =−GMm
r2 ˆr, (7)
where ˆr=r
r denotes the unit vector pointing in the direction of r.
Eq. (6) is a very compact notation for the vector force field. If we express it in Cartesian coordinates x, y and z, the result is rather complicated looking. First of all, acknowledging that r=xi+yj+zk, we have
r=krk=p
x2+y2+z2 (8)
so that
FMm(x,y,z) =− GMm
(x2+y2+z2)3/2[xi+yj+zk] (9) This expression will be useful later.
What does this vector field look like? As we reasoned above, at each point P with coordinates r= (x,y,z), the force exerted by mass M on mass m will point toward the origin. So all arrows point toward the origin. As we move away from the origin, however, the lengths of the arrows get smaller. Here is a rough sketch:
x
y z
The gravitational force field vector F=−GMm r3 r.
It is often convenient to define the gravitational field f that exists due to the presence of the mass M at the origin:
It is the gravitational force per unit mass exerted by M, i.e.
f= 1
mFMm=−GM
r3 r. (10)
Then the force exerted on a mass m at point P with position r is given by
F=mf. (11)
1.1.1 Electrostatic force fields
The electrostatic force on a “test point charge” q due to the presence of a point charge Q at the origin is F(r) = Qq
4πε0r3r, (12)
whereε0is known as the permittivity of free space. Note that F is (1) repulsive when Q and q have the same sign and (2) attractive when they have opposite signs, as expected.
The electric field due to the presence of charge Q at the origin is the electrostatic force per unit charge, i.e., E(r) = Q
4πε0r3r, (13)
so that the force exerted on a charge q at position r is given by
F=qE. (14)
In the case Q>0, the arrows representing the electric field E will point outward, as sketched below.
x
y z
The electric field vector E= Q
4πεr3r, for Q>0.
2 Integration along curves in R
nIn this section, we shall be integrating scalar-valued functions f(x,y,z)and vector-valued functions F(x,y,z)along curves inR3. The motivation for doing these things will become clear as the discussion progresses.
In what follows, it will be important to keep in mind that we shall be working on curves C inR3 that will be parameterized in terms of a single variable as follows
r(t) = (x(t),y(t),z(t)), a≤t≤b. (15)
Let us briefly recall the working definition of the Riemann integral of the single-variable function f(x)over the interval[a,b]:
1. We partition the interval I= [a,b]into n subintervals of length∆x= b−an using the partition points xk= a+k∆x, k=0,1,2,· · ·,n.
2. From each subinterval Ik= [xk−1,xk], pick a sample point x∗kand evaluate f(x∗k).
3. Form the Riemann sum
Sn=
∑
n k=1f(x∗k)∆x
Now let n→∞, implying that∆x→0. If f(x)is a sufficiently “smooth” function (i.e. piecewise continuous), then
n→∞limSn=S= Z b
a
f(x)dx
2.1 Line integrals of vector-valued functions of multiple variables
The integration of a vector field F over a curve C is a commonly encountered application of vector calculus. In addition to other applications that will be encountered later in the course, the line integral can be used to compute the work done by a nonconstant force acting on a particle that moves along a curve. It can also be used to measure the circulation of a rotating fluid or a magnetic field.
In what follows, we shall motivate our treatment with reference to the computation of work done by a force. As such, it will be useful to recall a fundamental formula from physics. Suppose that a constant force F=F1i+F2j is applied to a mass m that is constrained to move only in the x-direction. The mass is moved from position x=a
to x=b. In this case, it is easy to see that the total work done by the force is W=F1(b−a). It is the component of the force F in the direction of motion that “does the work.” This is an example of the more general formula that can be used inRn:
W =F·d, (16)
where F is the constant force acting on an object and d is the displacement vector−→
PQ of the mass, where P and Q are, respectively, the initial and final positions of the mass. There is another assumption behind this formula: that the actual motion of the mass was along the straight line PQ.
Now let us consider a more general case: A force F(r), which is not necessarily constant in space, is acting on a mass m, as the mass moves along a curve C from point P to point Q as shown in the diagram below. The goal is to
m
x y
z
P
Q
F(x(t)) x(t)
compute the total amount of work W done by the force. Clearly Eq. (16) does not apply here. But the fundamental idea is to break up the motion into tiny pieces over which we can use (16) as an approximation. We then “sum up,”
i.e., integrate, over all contributions to obtain W . We assume that the curve C can be parametrized, i.e.,
r(t) = (x(t),y(t),z(t)), t∈[a,b], (17)
so that r(a)is point P and r(b)is point Q. Now divide the parameter interval[a,b]into n subintervals of length
∆t= (b−a)/n: We do this by defining the partition points tk=a+k∆t, k=0,1,2,· · ·,n. These points define a set of n+1 points Pk=r(tk) = (x(tk),y(tk),z(tk)), k=0,1,· · ·,n that lie on the curve C, with P0=P and Pn=Q.
We now come to the major point of this procedure. We shall make two approximations in the computation of the work∆Wkdone by the force F in moving mass m from point Pk−1to point Pk:
1. We shall assume that the force acting on the mass during this time interval is constant. From each parameter subinterval[tk−1,tk], pick a sample point tk∗so that
(x∗k,y∗k,z∗k) =r(tk∗) = (x(tk∗),y(tk∗),z(tk∗))
is a sample point on the curve extending from Pk−1 to Pk Now evaluate the force vector F at this sample point, i.e., F(r(tk)). This is the constant force that we shall assume acts on the mass from Pk−1to Pk. (For n very large, we don’t expect the force F to change appreciably over this subcurve.)
2. We shall also assume that the motion of the mass is along the straight line segment Pk−1Pk.
These two approximations now allow us to use the “F·d” formula in (16): the work∆Wkdone by the force over the subcurve Pk−1Pkis approximated by
∆Wk≈F(r(tk∗))·−−−−→
Pk−1Pk. (18)
We still need to express the displacement vector−−−−→
Pk−1Pkin terms of the parameter t, in order to be able to produce
an integration over t. Note that
−−−−→
Pk−1Pk = ∆r(tk) (19)
= r(tk∗)−r(tk−1∗ )
= r(tk∗)−r(tk−1∗ )
∆t ∆t
≈ dr(tk∗) dt ∆t.
The final line is another approximation, but as n is increased, the length of the interval[tk−1,tk]decreases, and the approximation gets better.
Thus the work∆Wkis now approximated by
∆Wk≈F(r(tk∗))·dr(tk∗)
dt ∆t. (20)
The total work is now approximated as a sum over all subcurves:
W =
∑
n k=1∆Wk≈
∑
n k−1F(r(tk∗))·dr(tk∗)
dt ∆t. (21)
This is now a Riemann sum involving the scalar-valued function f(r(t)) =F(r(t))·dr(t)
dt (22)
evaluated at the sample points tk∗. In the limit n→∞, this Riemann sum converges to a definite integral:
W= Z b
a
F(r(t))·dr(t) dt dt=“
Z C
F· dr.” (23)
The term
dr=dr(t)
dt dt (24)
represents the infinitesimal displacement vector associated with the velocity vectordr(t)dt . The expression “ Z
C
F·dr”
denotes the “line integral of the vector field F over the curve C”. But it will be the expression W =
Z b
a F(r(t))·dr(t)
dt dt (25)
that will be useful for the practical computation of these line integrals.
Examples:
1. Evaluate the line integralRCF·dr where F=xyzi+y2j+zk along the curve r(t) = (t,t,t)for 0≤t≤1.
This parametrization produces a straight line that starts at (0,0,0) and ends at (1,1,1).
Step 1: Evaluate the velocity vector: dr(t)dt = (1,1,1).
Step 2: Evaluate F at points on the curve, using the parametrization. Dropping the unit vectors i, j, and k for convenience, we simply write F= (xyz,y2,z)so that
F(r(t)) = (x(t)y(t)z(t),y(t)2,z(t)) = (t3,t2,t). (26)
Step 3: Now construct the dot product that will appear in the integrand:
F(r(t))·dr(t)
dt = (t3,t2,t)·(1,1,1) =t3+t2+t. (27) We may now evaluate the line integral:
Z C
F·dr= Z 1
0 (t3+t2+t)dt=1 4+1
3+1 2 =13
12. (28)
2. Evaluate the integralRCF·dr where F=xyzi+y2j+zk as in Example 1, but the curve is now r(t) = (t,t2,t2), 0≤t≤1
This parametrization produces a parabolic curve line that also starts at (0,0,0) and ends at (1,1,1).
Step 1: Evaluate the velocity vector: dr(t)dt = (1,2t,2t).
Step 2: Evaluate F at points on the curve, using the parametrization. Here,
F(r(t)) = (x(t)y(t)z(t),y(t)2,z(t)) = (t5,t4,t2). (29) Step 3: Now construct the dot product that will appear in the integrand:
F(r(t))·dr(t)
dt = (t5,t4,t2)·(1,2t,2t) =t5+2t5+2t3=3t5+2t3. (30) We may now evaluate the line integral:
Z C
F·dr= Z 1
0
(3t5+2t3)dt=1 2+1
2=1. (31)
Note that this result differs from that of Example 1. The line integrals had the same endpoints but different paths. There is no guarantee that the results will be the same.
2.1.1 A comment on the notations used for line integrals of vector fields
There is another standard notation used for such line integrals. Firstly, the vector field F is written explicitly in component form as follows,
F(x,y,z) =P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k. (32) The infinitesimal vector element of displacement dr is also written in component form as
dr=dxi+dyj+dzk. (33)
Formally, the dot product of these two vectors becomes
F·dr= (P,Q,R)·(dx,dy,dz) =Pdx+Qdy+Rdz, (34) which is then inserted into the integral to give
Z C
F·dr= Z
C
Pdx+Qdy+Rdz. (35)
This “P,Q,R” notation is often employed in calculus texts, possibly because of the use of line integrals in Physical Chemistry, where changes to a system are often applied to each variable separately: For example, a container of gas may first be subjected to a change in pressure, keeping temperature and volume constant – this could be the
“dx” integration, then a change in temperature, keeping pressure and volume constant – the “dy” integration, etc..
We shall largely continue to use only the more general “F·dr” notation.
2.2 Line integrals of vector-valued functions over closed curves
Consider the situation sketched in Figure 1. The vector field F could represent the velocity field of fluid particles moving in a region of the plane. A few curves are also sketched, and we are interested the line integral of F over these curves.
A B
C D
E F
F(r)
Figure 1: The vector field F and curves in space.
Without even calculating these line integrals explicitly, we can get an idea of their values, keeping in mind that the line integral of a vector field F sums over the projection of F in the direction of the tangent vector v at each point on the curve. For example, regarding curve CABthat starts at A and ends at B, its tangent vectors appear to be pointing in roughly the same direction as F. As such, we would expect the line integral over CABto be a positive number.
On the other hand, the tangent vectors of curve CCD, that starts at C and ends at D, seem to be pointing in directions opposite to F. As such, we would expect the line integral over CCDto be a negative number.
And for the curve CEF starting at E and ending at F, its tangent vectors seem to be roughly perpendicular to F, so we would expect the line integral over CEF to be very small in value, i.e., close to zero.
Of greater importance will be the case when the curves are closed, which will lead to the concepts of circulation and outward flux.
2.2.1 Circulation of a vector field around a closed curve C inR2
We have seen that if F is a vector field and C is a curve that travels from point A to point B, then we can define the line integral
Z C
F·dr (36)
What would happen if curve C started at A, travelled around for a while and then ended back at A – in other words, A=B? In physics and other applications, one is often concerned with line integrals of vector fields over simple, closed curves C. “Simple” means nonintersecting. “Closed” means that the curve has no endpoints – you pick any starting point on the curve and you’ll eventually arrive back at it. Such line integrals are often denoted as follows
I C
F·dr. (37)
In what follows, we consider the line integral of a planar vector field F around a simple closed curve C inR2. The convention is that the integration along C is performed in the counterclockwise direction so that the region D enclosed by C lies always to the left of C as we move along the curve. Let’s recall that this line integral sums up the projection of the vector field onto the unit tangent vector ˆT to the curve. Assuming that we can parametrize
this curve as r(t), a≤t≤b, I
C
F·dr = Z b
a
F(r)(t))·dr(t)
dt dt (38)
= Z b
a
F(r)(t))·
dr(t)
dt
dr(t) dt
dr(t) dt
dt
= Z b
a
F(r)(t))·T(tˆ )ds
= I
f ds
where f(r(t)) =F(r(t))·T(t)ˆ is the projection of F in the direction of the unit tangent vector to the curve C at r(t):
T(tˆ 1) T(tˆ 2)
T(tˆ 3)
F(r(t3))
T(tˆ 4)
F(r(t4))
F(r(t1)) F(r(t2))
C
Starting at any point P on the curve C, the orientation of the tangent vector ˆT will change as we travel along C. In one traversal of C, the net rotation of the tangent vector is 2π. This is quite clear when C is a circle. As such, we say that the line integral in (38) is the circulation of the vector field F around the closed curve C.
If the vector field F is roughly parallel over the region D enclosed by curve C, then we expect the line integral to be small in value – in some regions of the curve, F points in the same direction as ˆT and in others, it points in the opposite direction. In other words, the vector field exhibits very little circulation. Let’s consider a very simple case:
F=Ki (39)
and C=CRis the circle of radius R centered at the origin:
y
x
CR R F=Kˆi
1. Parametrization of curve CR: r(t) = (R cost,R sint), 0≤t≤2π. 2. Velocity: dr(t)dt = (−R sint,R cost).
3. Evaluate F on curve: F(r(t)) = (K,0).
4. Integrand: F(r(t))·dr(t)dt = (K,0)·(−R sint,R cost) =−KR sint.
Thus I
F·dr=−KR Z 2π
0
sint dt=0. (40)
As expected, the line integral is zero.
Here is another situation that we expect will produce a zero result: C=CRas before and
F=Kr=Kxi+Kyj (41)
In this case, the vector F on the curve is perpendicular to the tangent vector:
x y
R
CR
F=Kr
1. Parametrization of curve CR: r(t) = (R cost,R sint), 0≤t≤2π. 2. Velocity: dr(t)dt = (−R sint,R cost).
3. Evaluate F on curve: F(r(t)) =K(x,y) =K(R cost,R sint).
4. Integrand: F(r(t))·dr(t)dt =K(R cost,R sint)·(−R sint,R cost) =−KR cost sint+KR cost sint=0.
Thus I
F·dr= Z 2π
0
0 dt=0. (42)
As expected, the line integral is zero.
Finally, let’s consider the vector field
F=K(−yi+xj) (43)
as sketched below.
You will recall that this is the velocity vector field of a disk that is revolving about the origin with angular frequency ω=K In this case, F points in the direction of the unit tangent vector ˆT at every point on the curve CR. As such, we expect a nonzero result:
1. Parametrization of curve CR: r(t) = (R cost,R sint), 0≤t≤2π. 2. Velocity: dr(t)dt = (−R sint,R cost).
3. Evaluate F on curve: F(r(t)) =K(−y,x) =K(−R sint,R cost).
4. Integrand: F(r(t))·dr(t)dt =K(−R sint,R cost)·(−R sint,R cost) =KR2sin2t+KR2cos2t=KR2.
Thus I
F·dr= Z 2π
0
KR2dt=2πKR2. (44)
As expected, the line integral is nonzero.
x y
CR R
2.2.2 An important application to physics
We now consider an important application of line integrals of vector fields in physics.
Proposition: A mass m moves inR3under the influence of a force F according to Newton’s Law F=ma. The mass moves from A to point B along a trajectory r(t)which we shall denote as curve CAB. Then the work W done by the force F along CABis equal to the change in kinetic energy of the mass, i.e.
W =EK(B)−EK(A) =∆EK, (45)
where EKdenotes the kinetic energy of the particle.
Proof: The total work done by the force is
W =
Z CAB
F·dr (46)
= Z b
a
F(r(t))·dr(t) dt dt,
where a and b denote the times that the particle is at A and B, respectively. But dr(t)dt =v(t), the velocity of the mass. At all points on the trajectory, Newton’s Law is obeyed, implying that
F(r(t)) =ma(t) =mdr(t)
dt . (47)
We substitute this result into the work integral:
W= Z b
a
mdv(t)
dt ·v(t)dt (48)
The product rule applied to the dot product yields d
dt(v(t)·v(t)) =dv(t)
dt ·v(t) +v(t)·dv(t)
dt =2dv(t)
dt ·v(t) (49)
or dv(t)
dt ·v(t) =1 2
d
dt(v(t)·v(t)) =1 2
d
dt|v(t)|2. (50)
Therefore
W = 1
2m Z b
a
d
dt|v(t)|2dt (51)
= 1
2m|v(b)|2−1
2m|v(a)|2
= EK(b)−EK(a)
= ∆EK.
3 Integrals of flux across curves and surfaces
3.1 Outward flux of a vector field across a closed curve C in R
2Let us now return to the line integral in (38), replacing the unit tangent vector ˆT with the unit outward normal ˆn to the curve C. This is the unit vector at a point r(t)on the curve which is perpendicular to the unit tangent vector T(t)ˆ and which points outward from the region D enclosed by C, as shown below:
The result is the line integral that we denote as I
C
F·ˆn ds= Z b
a
F(r(t))·ˆn(t)
dr(t) dt
dt. (52) Clearly, this line integral adds up the projection of the vector field F onto the outward normal along the curve C.
The result is the total outward flux of F across the closed curve C. If F represents the velocity field of a fluid, then the total outward flux measures the net rate of fluid escape from the region D through the curve C per unit time.
The practical calculation of the outward flux integral is not difficult – the only complication is that you have to determine the outward unit normal vector ˆn(t)to the curve C. This is easily done from a knowledge of the velocity vectordr(t)dt . You first construct the unit tangent vector as follows:
T(tˆ ) = 1 dr(t)dt
dr(t)
dt = (T1(t),T2(t)). (53)
Note that T12+T22=1. There are two unit vectors that are perpendicular to ˆT(t)– we’ll define them as
ˆn1= (T2(t),−T1(t)), ˆn2=−(T2(t),−T1(t)). (54) You choose the vector that points outward.
ˆn(t4) F(x(t3)
C
ˆn(t3)
F(x(t4))
F(x(t1)) F(x(t2))
ˆn(t1) ˆn(t2)
Example 1. Find the flux integrals through the circle CRof radius R, centered on the origin for the vector fields 1. F=Kr=Kxi+Kyj
2. F=−Kyi+Kxj where K is a positive constant.
1. In this case, the vector F on the curve is normal to the curve, as shown in Figure 2.
x y
R
CR
F=Kr
Figure 2: The vector field F=Kxi+Kyj (a) Parametrization of curve CR: r(t) = (R cost,R sint), 0≤t≤2π. (b) Unit tangent to curve:
dr(t)
dt
dr(tdt)
= (−sint,cost).
(c) Normal to curve: We have two choices for the normal, ˆn= (cost,sint)and ˆn= (−cost,−sint). By convention, we choose the first of these vectors since it points ‘outwards’ with respect to the closed curve CR.
(d) Evaluate F on curve: F(r(t)) =K(x,y) =K(R cost,R sint).
(e) Integrand: F(r(t))·ˆn=K(R cost,R sint)·(cost,sint) =KR cos2t+KR sin2t=KR.
(f) Elemental segments ds: Since dsdt = drdt
, then ds= drdt
dt=|(−R sint,R cost)|dt=Rdt
Thus I
F·ˆnds= Z 2π
0
KR2dt=2πKR2 (55)
2. The vector field
F=K(−yi+xj) (56)
which is sketched in Figure 3
is tangential to the curve CR. In other words, it does not cross CRat any point, and we may therefore expect its flux integral across CRto be zero. To see this, we proceed as follows
(a) Parametrization of curve CR: r(t) = (R cost,R sint), 0≤t≤2π. (b) Unit tangent to curve:
dr(t)
dt
dr(t) dt
= (−sint,cost).
(c) Normal to curve: We have two choices for the normal, ˆn= (cost,sint)and ˆn= (−cost,−sint). By convention, we choose the first of these vectors since it points ‘outwards’ with respect to the closed curve CR.
x y
CR R
Figure 3: The vector field F=−Kyi+Kxj (d) Evaluate F on curve: F(r(t)) =K(x,y) =K(−R sint,R cost).
(e) Integrand: F·ˆn=K(−R sint,R cost)·(cost,sint) =0.
(f) Elemental segments ds: Since dsdt = dr
dt
, then ds= dr
dt
dt=|(−R sint,R cost)|dt=Rdt
Thus I
F·ˆnds= Z 2π
0
0 dt=0 (57)
3.2 Surface integrals of vector functions
In the previous section, we examined the outward flux of a vector field through a closed curve C that enclosed a region D in the plane. In this section, we extend this idea toR3, namely, the flux of a vector field F through a surface S.
We’ll develop this idea by means of some simple steps. First, consider a surface D that lies parallel to the xy-plane as sketched in Figure 4, left. Now suppose that the constant vector field F=vk, with v>0 a constant, is also defined at all points inR3. We shall suppose that this vector field represents the constant velocity of fluid particles that are moving in the positive z-direction. A subset of these particles move through region D – those that lie directly below region D.
v∆t
∆S
∆V=v∆S∆t
D z
x y
∆A D
z
x y
F=vk
∆x
∆y
Figure 4: A surface D, parallel to the xy-plane. Left: flow of vector field F through D, representing the velocity of a fluid. Right: the small volume∆V of fluid that passes vertically through D in time∆t.
We first ask the question: How much fluid flows through surface D during a time interval∆t? Consider a tiny rectangular element of the surface area of D, labelled∆S in Figure 4, right. Since D is parallel to the xy-plane,
∆S=∆A=∆x∆y centered at a point(x,y)in D. After a time∆t, the fluid particles situated in this element will have moved a distance v∆t upward. The volume of fluid that has passed through the element∆S on D is the volume of the box of base area∆S and height v∆t:
v∆t∆A. (58)
This box is sketched in Figure 4, right.
The total volume∆V of fluid that has passed through region D over the time interval∆t is obtained by summing up over all area elements∆S in D:
∆V =v∆tx
D
dS=v∆tS(D), (59)
where S(D)denotes the area of D. Of course, this is a rather trivial result: the volume of fluid passing through D is simply the volume of the solid of base area S(D)and height v∆t. Dividing both sides by∆t, we have
∆V
∆t =vS(D). (60)
In the limit∆t→0, we have the instantaneous rate of change of the volume of fluid passing through region D, or simply the rate of fluid flow through region D:
dV(t)
dt =vS(D). (61)
This quantity is the flux of the vector field F through region D.
Suppose now that the fluid is moving at a constant speed v through region D but not necessarily at right angles to it, i.e., not necessarily parallel to its normal vector k. We shall suppose that
F=v1i+v2j+v3k, |F|=v (62)
and letγdenote the angle between F and the normal vector k.
In this case, the fluid particles that pass through the tiny element∆S after a time interval∆t form a parallelepiped of base area∆S and height v∆t cosγ, as sketched in Figure 5.
v∆t cos γ
∆S D
z
x y
F γ
∆V=v∆t∆S cos γ
Figure 5: Flow of vector field F through surface D, representing the velocity of a fluid. The small volume∆V of fluid passes through D at an angleγin time∆t.
The volume of this box is
v cosγ∆t∆S (63)
The total volume∆V of fluid that has passed through region D over the time interval∆t is obtained by summing up over all area elements∆S in D:
∆V=v cosγ∆tx
D
dS=v cosγ∆tS(D), (64)
Dividing both sides by∆t, we have
∆V
∆t =v cosγS(D). (65)
In the limit∆t→0, we obtain the flux of the vector field F through region D:
V′(t) =v cosγS(D). (66)
We shall rewrite this flux as follows,
V′(t) =F·ˆn S(D), (67)
where ˆn=k is the normal vector to surface D. Note that this general case includes the first case,γ=0. And in the case thatγ=π/2, there is no flow through the region D, so the flux is zero.
Of course, the above results have been rather trivially obtained since (i) the vector fields are constant and (ii) the region D is flat. Let us now generalize the first case, i.e., the vector field F is assumed to be nonconstant over region D, i.e.,
F(x,y) =v1(x,y)i+v2(x,y)j+v3(x,y)k. (68) In this case, the total volume∆V of fluid that has passed through region D over the time interval∆t is obtained by summing up over all area elements∆S in D:
∆V=∆tx
D
F(x,y)·ˆn dS=∆tZ
D
v3(x,y)dS (69)
Once again dividing by∆t and taking the limit∆t→0, we obtain the total flux of F through region D:
V′(t) =x
D
F(x,y)·ˆn dS (70)
We now arrive at the final generalization which will cover all surfaces inR3: we do not require the surface to be flat, as was region D in the plane, but rather a general surface S inR3– for example, a portion of a sphere, or perhaps the entire sphere. Very briefly, we first divide S into tiny infinitesimal pieces dS. We then construct a normal vector ˆn to each surface element dS at a point in dS, as sketched in Figure 6.
We then form the dot product of the vector field F at that point with the normal vector ˆn. This will represent the local flux of F through the surface element dS. To obtain the total flux through the surface S, we add up the fluxes of all elements dS – an integration over S that is denoted as
x
S
F·ˆn dS (71)
As a matter of convention
• often, the vector quantity ˆndS is combined into one term, dS, so the the flux integral of a vector field F through a surface S, which has norma vector ˆn is written as either one of
x
S
F·ˆndS orx
S
F·dS
• when the flux integral is through a closed surface S, the notation {
S
F·ˆndS is often used
dS
F
n
F F
F
F
F ˆ
nˆ
nˆ nˆ nˆ
nˆ nˆ nˆ
nˆ nˆ
nˆ
nˆ nˆ
Figure 6: The vector field F, crossing a surface with elemental surface area dS and normal vectors ˆn.
• for surface integrals across closed surfaces, we always measure the flux out of the closed surface. For this reason, the normal to the surface is taken to point outwards with respect to the closed surface.
Example 2. Evaluate the flux integral of the vector field F=r=
x y z T
through the sphere S, which is of radius R and centered on the origin.
The sphere of interest is illustrated in Figure 7, left. A vector that is normal to this sphere is the position vector r. A unit vector normal to the sphere is therefore ˆn= |r|r. Therefore the integrand of the surface integral is F·ˆn=r·|r|r =|r|. On the surface S, this integrand is simply R.
To find the elemental surface area dS, we use spherical polar coordinates to divide the surface of the sphere.
As illustrated in Figure 7, right, a small ‘rectangular’ elementδS of the surface area of the sphere has ‘height’
Rδφ and ‘width’ R sinφδθ. Therefore the area of this element isδS=R2sinφδφδθ (compare this with the elemental volume δV in spherical polar coordinates), and the flux of F through just one of these elements is F·ˆnδS=R3sinφδφδθ. To obtain the flux through the entire volume, we sum together the flux through each of the elementsδS that make up the surface. In the limit asδS→0, this summation becomes the surface integral
{
S
F·ˆndS= Z 2π
0 Z π
0
R3sinφdφdθ
= Z 2π
0
2R3dθ
=4πR3
δθ
δφ z
y
x
Rsin φ δS
φ
R Rsin φ δθ δS
R z
y
x
Rδφ
Figure 7: Finding the elemental surface areaδS in terms of polar coordinates.
4 The divergence and the curl of a vector field
4.1 Divergence of a vector field
It is convenient to define the following operator, called the “del” operator,
∇= ∂
∂xi+
∂
∂yi+
∂
∂zk (72)
This operator could also be written as an ordered triple of operators, i.e.
∇= ∂
∂x,
∂
∂y,
∂
∂z
. (73)
Now, an operator acts on suitable mathematical objects. We have already seen an example of the action of the del operator – it can act on a scalar-valued function f(x,y,z)to produce the associated gradient vector∇f :
∇f =∂f
∂xi+
∂f
∂yj+
∂f
∂zk (74)
But operators can also act on vectors. For example, we could form the scalar product of the del operator with a vector field F:
∇·F = ∂
∂xi+ ∂
∂yi+ ∂
∂zk
·(F1i+F2j+F3k) (75)
= ∂F1
∂x +
∂F2
∂y +
∂F3
∂z .
This is known as the divergence of the vector field F. In some books, it is also written as “div F”. It is a scalar quantity, as should be the case when the dot product of two vectors is taken. Note that we can also define the divergence of vector fields inR2by omitting the k terms.
Examples:
1. ∇·C=0 for any constant vector C(x,y,z) =C1i+C2j+C3k.
2. If F=x2yzi+xz5sin yj+x2y sin 4zk, then
∇·F= ∂
∂x(x2yz) + ∂
∂y(xz5sin y) + ∂
∂z(x2y sin 4z) =2xyz+xz5cos y+4x2y cos 4z. (76) 3. If F=r=xi+yj+zk, then F1=x, F2=y and F3=z so that
∇·r=∂x
∂x+
∂y
∂y+
∂z
∂z=3. (77)
Recall the importance of the class of vector fields K
r3r in physics. When K=−GMm, we have the gravitational force exerted on a mass m at r by a point mass M at the origin of a coordinate system. When K=Qq/(4πε0), we have the electrostatic force exerted on a charge q at r due to a point mass Q at the origin. For convenience, we shall omit the multiplicative factor K. Let us first express this field in terms of Cartesian coordinates:
1
r3r= 1
(x2+y2+z2)3/2[xi+yj+zk] (78) We have the final result (Exercise),
∇· 1
r3r=div1
r3r=0, (x,y,z)6= (0,0,0) (79) This is an extremely important result! As we’ll see later, it reflects how the electrostatic and gravitational fields inR3naturally behave in the presence and absence of charges/masses, respectively.
Definition: A vector field F for which∇·F=0 for all points(x,y,z)∈D⊆Ris said to be incompressible over D. (Some books also use the term solenoidal.)
The vector field F= K
r3r examined above is incompressible over the setR− {(0,0,0)}.
4.1.1 Physical interpretation of the divergence
Consider a velocity field u(x,y,z) =uxi+uyj+uzk and a small rectangular volumeδV=δxδyδz, illustrated in Figure 8. Let us calculate the net flux of u out ofδV , given by
I u·dA
Φ1 Φ2
Surface S1, area δA1
δx
δy δz
x0
Surface S2, area δA2
Figure 8: A small volumeδV .
Consider first the fluxesΦ1andΦ2through the surfaces S1and S2. These are Φ1=
Z S1
u·dA=−ux(x0)δA1=−uxδyδz Φ2=
Z S2
u·dA=ux(x0+δx)δA2=ux(x0+δx)δyδz This gives the total flux out ofδV from via surfaces S1and S2as
Φ1+Φ2= [ux(x0+δx)−ux(x0)]δyδz=∂ux
∂xδV
Similarly, the pair of surfaces with normals aligned with the y-axis contribute a net flux of ∂∂uyyδV , while those aligned with the z-axis give a net flux of∂∂uzzδV . In total, the flux throughδV is
I
u·dA= (∂ux
∂x +
∂uy
∂y +∂uz
∂z)δV which is equivalent to
I
u·dA= (∇·u)δV
This result holds for any small volumeδV , and not just a rectangular volume, and provides a physical interpretation of the divergence: the divergence∇·u at a point(x,y,z)is a measure of the net flux of u out of a small volume element located at(x,y,z), as illustrated in Figure 9.
δV δV δV
∇•
u = 0
u u
u
u u u
u
∇•
u < 0
∇•
u> 0
Figure 9: The divergence measures the net outward flow of a vector field u through a small volumeδV . Left: there is a net outward flow of u through the volumeδV , yielding a positive divergence: ∇·u>0. Center: there is a net inward flow (i.e. a negative net outward flow) of u through the volumeδV , yielding a negative divergence:
∇·u<0. Right: there is no net flow out of the volumeδV , what flows intoδV also flows out ofδV .
The velocity field associated with a fluid flow where the fluid has constant densityρhas zero divergence since the conservation of mass demands that what flows into a volume must also flow out of it. That is
I ρu·dA=0
and soρu, and hence u, has zero divergence: ∇·u=0. This is why, as mentioned previously, a field with zero divergence is said to be incompressible.
In your course on electricity and magnetism, you will eventually encounter the equation
∇·E(r) =ρ(r) ε0
, (80)
where E(r)is the electric field at a point r andρ(r)is the charge density at r. In other words, if there is no charge at a point, then the divergence of the field will be zero there. Charge is responsible for net inflow/outflow (depending upon the sign convention) of electric field.
Recall that the electric field at a point r due to the presence of a point charge Q at the origin is E= Q
4πε0r3r. We showed above that∇·E=0 at all points r6= (0,0,0). At(0,0,0), however, the divergence “blows up,” i.e., is infinite. This is because the charge densityρ at(0,0,0)is infinite – we are assuming that a charge Q exists at a point with no volume. Of course, a point charge is a mathematical abstraction that does not correspond to physical reality. We’ll come back to this point later.
4.2 The curl of a vector field
Let us now take the vector product of the del operator with a vector field F:
∇×F = ∂
∂xi+
∂
∂yi+
∂
∂zk
×(F1i+F2j+F3k) (81)
=
i j k
∂x∂ ∂
∂y ∂
∂z
F1 F2 F3
= ∂F3
∂y −∂F2
∂z
i+ ∂F1
∂z −∂F3
∂x
j+ ∂F2
∂x −∂F1
∂y
k.
This is known as the curl of the vector field F. In some books, it is also written as “curl F”. It is a vector quantity, as should be the case when the vector product of two vectors is taken. Note that we can also define the curl of vector fields inR2by setting F3=0.
Examples:
1. ∇×C=0 for any constant vector C(x,y,z) =C1i+C2j+C3k.
2. If F=r=xi+yj+zk, then F1=x, F2=y and F3=z so that
∇×r =
i j k
∂∂x ∂
∂y ∂
∂z
x y z
= ∂z
∂y−∂y
∂z
i+ ∂x
∂z−∂z
∂x
j+ ∂y
∂x−∂x
∂y
k=~0.
3. Now consider F=−yi+xj+0k. Then
∇×F =
i j k
∂∂x ∂
∂y ∂
∂z
−y x 0
= ∂0
∂y−∂x
∂z
i+
∂(−y)
∂z −∂0
∂x
j+ ∂x
∂x−∂(−y)
∂y
k=2k.
We examined this vector field in the previous lecture in the plane, z=0. Since there is no k component, we can translate this field in the positive and negative z directions to obtain the vector field inR3. A view from the positive z-axis looking down onto the xy-plane gives the same picture as shown earlier: In the xy-plane, this vector field can represent the velocity field of a rotating disk (see next section). In this three-dimensional case, it could represent the velocity field of a rotating cylinder.
4.2.1 Physical interpretation of the curl
Consider a two-dimensional circular disk of radius R, lying parallel to the xy-plane, centered on(x,y) = (0,0), rotating about the z-axis with angular velocityΩk, as illustrated in Figure 10. Each point(x,y)on the disk has the velocity u(x,y) =Ω[−yi+xj](by considering the time derivative of the parameterized curve r(t) =cos(Ωt)i+ sin(Ωt)j, it can be verified that this vector field yields the angular velocity of the disk asΩk). We shall next
1. calculate I
C
u·dr for the curve C :|r|=R,
y
x
A top view of the vector field F=−yi+xj+0k
2. calculate∇×u
3. show that(∇×u)z(πR2) = I
C
u·dr=2Ω, where(∇×u)zis the k component of∇×u.
Ωk
R
x y
Figure 10: A disk of radius R, rotating with angular velocityΩk.
1. The curve C, being the locus of points a distance R away from the origin, can be parameterized r(t) = (x(t),y(t)) = (R cos(t),R sin(t)), 0≤t≤2π. This yields dr= (−R sin(t)dt,R cos(t)dt)and
u·dr= (−ΩR sin(t),ΩR cos(t))·(−R sin(t)dt,R cos(t)dt) =ΩR2 This yields
I C
u·dr=2πΩR2. 2. ∇×u=
∂(Ωx)
∂x −
∂(−Ωy)
∂y
k=2Ωk
3. From the above two calculations, we see that(∇×u)z(πR2) = I
C
u·dr=2Ω
We can generalize the above calculations for a two-dimensional vector field u. Consider a small areaδA of the domain of the vector field, with unit normal ˆn, encircled by a curve C. IfδA is small enough that∇×u is constant over the area, then it can be shown that
(∇×u)·ˆnδA= I
C
u·dr
