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Unformatted text preview: 3 Vector Integral Calculus 31 Vector integrals; the line integral of VW‘ We found in Chapter 2 that there were various ways of taking derivatives of
ﬁelds. Some gave vector ﬁelds; some gave scalar ﬁelds. Although we developed
many different formulas, everything in Chapter 2 could be summarized in one rule:
the operators a/ax, a/ay, and 6/62 are the three components of a vector operator
V. We would now like to get some understanding of the signiﬁcance of the deriva
tives of ﬁelds. We will then have a better feeling for what a vector ﬁeld equation
means. We have already discussed the meaning of the gradient operation (V on a
scalar). Now we turn to the meanings of the divergence and curl operations.
The interpretation of these quantities is best done in terms of certain vector
integrals and equations relating such integrals. These equations cannot, unfor
tunately, be obtained from vector algebra by some easy substitution, so you will
just have to learn them as something new. Of these integral formulas, one is
practically trivial, but the other two are not. We will derive them and explain their
implications. The equations we shall study are really mathematical theorems.
They will be useful not only for interpreting the meaning and the content of the
divergence and the curl, but also in working out general physical theories. These
mathematical theorems are, for the theory of ﬁelds, what the theorem of the con
servation of energy is to the mechanics of particles. General theorems like these
are important for a deeper understanding of physics. You will ﬁnd, though, that
they are not very useful for solving problems—except in the simplest cases. It is
delightful, however, that in the beginning of our subject there will be many simple
problems which can be solved with the three integral formulas we are going to
treat. We will see, however, as the problems get harder, that we can no longer use
these simple methods. We take up ﬁrst an integral formula involving the gradient. The relation
contains a very simple idea: Since the gradient represents the rate of change of a
ﬁeld quantity, if we integrate that rate of change, we should get the total change.
Suppose we have the scalar ﬁeld “x, y, 2). At any two points (1) and (2), the
function it will have the values M1) and M2), respectively. [We use a convenient
notation, in which (2) represents the point (x2, y2, 22) and M2) means the same
thing as :la(x2, y2, 22).] If I‘(gamma) is any curve joining (1) and (2), as in Fig. 3—1,
the following relation is true: (2) iI/(Z)  WU) = [m (Vid'ds (3.1)
alongl‘ THEOREM 1. The integral is a line integral, from (I) to (2) along the curve I‘, of the dot product
of Wx—a vector—with ds—another vector which is an inﬁnitesimal line element
of the curve 1‘ (directed away from (1) and toward (2)). First, we should review what we mean by a line integral. Consider a scalar
function f(x, y, z), and the curve I‘ joining two points (1) and (2). We mark off
the curve at a number of points and join these points by straightline segments, as
shown in Fig. 3—2. Each segment has the length As“ where iis an index that runs 1,2, 3,... By the line integral
(2)
f w
(1) along I‘
3] 3—1 Vector integrals; the line
integral of V‘P‘ 3—2 The ﬂux of a vector ﬁeld 3—3 The ﬂux from a cube; Gauss’
theorem 3—4 Heat conduction; the diffusion
equation 3—5 The circulation of a vector ﬁeld 3—6 The circulation around a square;
Stokes’ theorem 3—7 Curlfree and divergencefree
ﬁelds 3—8 Summary (2) Curve I‘
ds
(I) Fig. 3—1. The terms used in Eq. (3.1).
The vector Vil/ is evaluated at the line
element d3. (2) (I) e b
A52 Fig. 32. The line integral is the
limit of a sum. Closed
Surface 5 Fig. 33. The closed surface S
deﬁnes the volume V. The unit vector n
is the outward normal to the surface
element do, and II is the heatﬂow vector
at the surface element. we mean the limit of the sum Zﬁ A35, where f, is the value of the function at the ith segment. The limiting value is what
the sum approaches as we add more and more segments (in a sensible way, so that
the largest As, —> 0). The integral in our theorem, Eq. (3.1), means the same thing, although it
looks a little different. Instead off, we have another scalar—the component of
V1; in the direction of As. If we write (W), for this tangential component, it is clear that
(VIII)1AS = (Vip) As. (3.2) The integral in Eq. (3.1) means the sum of such terms. Now let’s see why Eq. (3.1) is true. In Chapter 1, we showed that the com
ponent of Vzp along a small displacement AR was the rate of change of w in the
direction of AR. Consider the line segment As from (1) to point a in Fig. 3—2.
According to our deﬁnition, 4% = M0) — M1) = (V¢)1'A51 (33)
Mb)  “61) = (V¢)2 ‘ A52, 0“) where, of course, (Vinl means the gradient evaluated at the segment Asl, and
(V102, the gradient evaluated at A12. If we add Eqs. (3.3) and (3.4), we get Also, we have Mb) " \l/(l) = (Vi/)1 'Asi + (V¢)2'AS2 (35)
You can see that if we keep adding such terms, we get the result
1P0)  M1) = 2(V¢)i'Asi. (35) The lefthand side doesn’t depend on how we choose our intervals—if (1) and (2)
are kept always the same—so we can take the limit of the righthand side. We have
therefore proved Eq. (3.1). You can see from our proof that just as the equality doesn’t depend on how the points a, b, c, . . . . are chosen, similarly it doesn’t depend on what we choose
for the curve I‘ to 10m (1) and (2). Our theorem is correct for any curve from (1)
to (2). One remark on notation: You will see that there is no confusion if we write, for convenience,
(vi/z)  ds = w  ds. (3.7) With this notation, our theorem is (2) M2)  30(1) = fa) WI  ds. (38) any curve from
(1) to (2) THEOREM 1. 3—2 The ﬂux of a vector ﬁeld Before we consider our next integral theorem—a theorem about the divergence
—we would like to study a certain idea which has an easily understood physical
signiﬁcance in the case of heat ﬂow. We have deﬁned the vector [1, which represents
the heat that ﬂows through a unit area in a unit time. Suppose that inside a block
of material we have some closed surface S which encloses the volume V (Fig. 3—3).
We would like to ﬁnd out how much heat is ﬂowing out of this volume. We can,
of course, ﬁnd it by calculating the total heat ﬂow out of the surface S. We write do for the area of an element of the surface. The symbol stands for
a twodimensional differential. If, for instance, the area. happened to be in the
xyplane we would have da = dx dy.
3—2 Later we shall have integrals Over volume and for these it is convenient to con
sider a differential volume that is a little cube. So when we write dV we mean dV = dxdy dz. Some people like to write dza instead of dc to remind themselves that it is
kind of a secondorder quantity. They would also write d3V instead of dV. We
will use the simpler notation, and assume that you can remember that an area
has two dimensions and a volume has three. The heat ﬂow out through the surface element da is the area times the com
ponent of h perpendicular to da. We have already deﬁned it as a unit vector pointing outward at right angles to the surface (Fig. 3—3). The component of I: that we
want is
hn = h  n. (3.9) The heat ﬂow out through da is then hnda. (3.10) To get the total heat ﬂow through any surface we sum the contributions from all the elements of the surface. In other words, we integrate (3.10) over the whole
surface: Total heat ﬂow outward through S = f h  n da. (3.11)
s We are also going to call this surface integral “the ﬂux of I: through the sur
face.” Originally the word ﬂux meant ﬂow, so that the surface integral just means
the ﬂow of I: through the surface. We may think: It is the “current density” of
heat ﬂow and the surface integral of it is the total heat current directed out of the
surface; that is, the thermal energy per unit time (joules per second). We would like to generalize this idea to the case where the vector does not
represent the ﬂow of anything; for instance, it might be the electric ﬁeld. We can
certainly still integrate the normal component of the electric ﬁeld over an area if
we wish. Although it is not the ﬂow of anything, we still call it the “ﬂux.” We say Flux of E through the surface S = [s E  n da. (3.12) We generalize the word “ﬂux” to mean the “surface integral of the normal com
ponent” of a vector. We will also use the same deﬁnition even when the surface
considered is not a closed one, as it is here. Returning to the special case of heat ﬂow, let us take a situation in which
heat is conserved. For example, imagine some material in which after an initial
heating no further heat energy is generated or absorbed. Then, if there is a net
heat ﬂow out of a closed surface, the heat content of the volume inside must
decrease. So, in circumstances in which heat would be conserved, we say that . _ _9'_Q.
[Sh nda _ dt, (3.13) where Q is the heat inside the surface. The heat ﬂux out of S is equal to minus the
rate of change with respect to time of the total heat Q inside of S. This interpreta
tion is possible because we are speaking of heat flow and also because we supposed
that the heat was conserved. We could not, of course, speak of the total heat
inside the volume if heat were being generated there. Now we shall point out an interesting fact about the ﬂux of any vector. You
may think of the heat ﬂow vector if you wish, but what we say will be true for any
vector ﬁeld C. Imagine that we have a closed surface S that encloses the volume V.
We now separate the volume into two parts by some kind of a “cut,” as in Fig.
3—4. Now we have two closed surfaces and volumes. The volume V1 is enclosed
in the surface S1, which is made up of part of the original surface 5,, and of the
surface of the cut, Sub. The volume V2 is enclosed by 52, which is made up of
the rest of the original surface 5,, and closed off by the cut Sub. Now consider the 3—3 Fig. 34. A volume V contained inside the surface
S is divided into two pieces by a ”cut" at the surface
Sub. We now have the volume V] enclosed in the
surface 51 = 5;. + Sub and the volume V2 enclosed
in the surface 52 = $1, + Sub. (x, y. m) s Fig. 3—5. Computation of the flux of
C out of a small cube. cut following question: Suppose we calculate the ﬂux out through surface S1 and
add to it the ﬂux through surface S2. Does the sum equal the ﬂux through the
whole surface that we started with? The answer is yes. The ﬂux through the part
of the surfaces Sal, common to both 51 and S 2 just exactly cancels out. For the
ﬂux of the vector C out of V1, we can write Flux through SI = f C  n da + C  n1 da, (3.14)
Si: Sub
and for the ﬂux out of V2,
mem$$=mew+sme (M3
Sb ab Note that in the second integral we have written 111 for the outward normal for
5..., when it belongs to 51, and n2 when it belongs to S2, as shown in Fig. 3—4.
Clearly, n1 = —n2, so that C'nl do = — Cngda. (3J6)
Sub Sub
If we now add Eqs. (3.14) and (3.15), we see that the sum of the ﬂuxes through
S1 and $2 is just the sum of two integrals which, taken together, give the ﬂux
through the original surface S = SCl + Sb. We see that the ﬂux through the complete outer surface S can be considered
as the sum of the ﬂuxes from the two pieces into which the volume was broken.
We can similarly subdivide again—say by cutting V1 into two pieces. You see
that the same arguments apply. So for any way of dividing the original volume, it
must be generally true that the ﬂux through the outer surface, which is the original
integral, is equal to a sum of the ﬂuxes out of all the little interior pieces. 3—3 The ﬂux from a cube; Gauss’ theorem We now take the special case of a small cube* and ﬁnd an interesting formula
for the ﬂux out of it. Consider a cube whose edges are lined up with the axes as in
Fig. 3—5. Let us suppose that the coordinates of the corner nearest the origin
are x, y, 2. Let Ax be the length of the cube in the xdirection, Ay be the length
in the ydirection, and A2 be the length in the zdirection. We wish to ﬁnd the
ﬂux of a vector ﬁeld C through the surface of the cube. We shall do this by making
a sum of the ﬂuxes through each of the six faces. First, consider the face marked
1 in the figure. The ﬂux outward on this face is the negative of the xcomponent
of C, integrated over the area of the face. This ﬂux is —jqwa
Since we are considering a small cube, we can approximate this integral by the * The followmg development applies equally well to any rectangular parallelepiped.
3—4 value of CI at the center of the face—which we call the point (l)——multiplied by
the area of the face, Ay Az: Flux out of 1 = —CI(1)Ay Az.
Similarly, for the ﬂux out of face 2, we write Flux out of 2 = Cz(2) Ay Az. Now C;(1) and Cx(2) are, in general, slightly different. If Ax is small enough, we
can write 6C;
Cr(2) — Cx(l) + 6x Ax.
There are, of course, more terms, but they will involve (A,,)2 and higher powers,
and so will be negligible if we consider only the limit of small Ax. So the ﬂux through face 2 is 8C3
6x Flux out of 2 = [Cz(1) + Ax] Ay Az. Summing the ﬂuxes for faces 1 and 2, we get 6C, Flux out of 1 and 2 =
6x Ax Ay A2. The derivative should really be evaluated at the center of face 1; that is, at
[x, y + (Ay/2), z + (AZ/2)]. But in the limit of an inﬁnitesimal cube, we make
a negligible error if we evaluate it at the corner (x, y, 2). Applying the same reasoning to each of the other pairs of faces, we have Flux out of3 and 4 = 9T? Ax Ay A2
and C
Flux out of 5 and 6 = 952—: Ax Ay Az. The total ﬂux through all the faces is the sum of these terms. We ﬁnd that . _ 6Cz 6—Cy 6C;
[C Mia—(ax 6y + aZ>AxAyAz, cube and the sum of the derivatives is just V  C. Also, Ax Ay A2 2 AV, the volume of
the cube. So we can say that for an inﬁnitesimal cube / Cnda = (VC)AV. (3.17) surface We have shown that the outward ﬂux from the surface of an inﬁnitesimal cube is
equal to the divergence of the vector multiplied by the volume of the cube. We
now see the “meaning” of the divergence of a vector. The divergence of a vector
at the point P is the ﬂux—the outgoing “ﬂow” of C—per unit volume, in the neigh
borhood of P. We have connected the divergence of C to the ﬂux of C out of each inﬁnitesimal
volume. For any ﬁnite volume we can use the fact we proved above—that the
total ﬂux from a volume is the sum of the ﬂuxes out of each part. We can, that is,
integrate the divergence over the entire volume. This gives us the theorem that the
integral of the normal component of any vector over any closed surface can also be
written as the integral of the divergence of the vector over the volume enclosed
by the surface. This theorem is named after Gauss. GAuss’ THEOREM.
[cnda =/ vCdV, (3.18)
S V where S is any closed surface and V is the volume inside it.
3—5 Fig. 3—6. In the region near a point
source of heat, the heat ﬂow is radially
outward. 34 Heat conduction; the diﬂ'usion equation Let’s consider an example of the use of this theorem, just to get familiar
with it. Suppose we take again the case of heat ﬂow in, say, a metal. Suppose we
have a simple situation in which all the heat has been previously put in and the
body is just cooling off. There are no sources of heat, so that heat is conserved.
Then how much heat is there inside some chosen volume at any time? It must be
decreasing by just the amount that ﬂows out of the surface of the volume. If our
volume is a little cube, we would write, following Eq. (3.17), Heat out = fh'nda = VltAV. (3.19) cube But this must equal the rate of loss of the heat inside the cube. If q is the heat per
unit volume, the heat in the cube is q AV, and the rate of loss is d _ dq
— 3t (q AV) — — 7t AV. (3.20) Comparing (3.19) and (3.20), we see that dq _ _
 3? — v h. (3.21) Take careful note of the form of this equation; the form appears often in phys
ics. It expresses a conservation law—here the conservation of heat. We have
expressed the same physical fact in another way in Eq. (3.13). Here we have the
dzﬁbrential form of a conservation equatiofi, while Eq. (3.13) is the integral form. We have obtained Eq. (3.21) by applying Eq. (3.13) to an inﬁnitesimal cube.
We can also go the other way. For a big volume V bounded by S, Gauss’ law
says that h'nda = V th. (3.22)
f. Using (3.21), the integral on the righthand side is found to be just —dQ/dt,
and again we have Eq. (3.13). Now let’s consider a different case. Imagine that we have a block of material
and that inside it there is a very tiny hole in which some chemical reaction is
taking place and generating heat. Or we could imagine that there are some wires
running into a tiny res13tor that is being heated by an electric current. We shall
suppose that the heat is generated practically at a point, and let W represent the
energy liberated per second at that point. We shall suppose that in the rest of the
volume heat is conserved, and that the heat generation has been going on for a
long time—so that now the temperature is no longer changing anywhere. The
problem is: What does the heat vector h look like at various places in the metal?
How much heat ﬂow is there at each point? We know that if we integrate the normal component of h over a closed surface
that encloses the source, we will always get W. All the heat that is being generated
at the point source must ﬂow out through the surface, since we have supposed
that the ﬂow is steady. We have the difficult problem of ﬁnding a vector ﬁeld
which, when integrated over any surface, always gives W. We can, however, ﬁnd
the ﬁeld rather easily by taking a somewhat special surface. We take a Sphere of
radius R, centered at the source, and assume that the heat flow is radial (Fig. 3—6).
Our intuition tells us that h should be radial if the block of material is large and
we don’t get too close to the edges, and it should also have the same magnitude
at all points on the sphere. You see that we are adding a certain amount of guess
work—usually called “physical intuition”~—to our mathematics in order to ﬁnd
the answer. When h is radial and spherically symmetric, the integral of the normal com
ponent of ’1 over the area is very simple, because the normal component is just 3—6 the magnitude of It and is constant. The area over which we integrate is 41rR2.
We have then that [ShMa = 11.471? (3.23) (where h is the magnitude of h). This integral should equal W, the rate at which
heat is produced at the source. We get W
h _ 47rR2 ’
or
W
h = 4sz e,, (3.24) where, as usual, er represents a unit vector in the radial direction. Our result
says that h is proportional to W and varies inversely as the square of the distance
from the source. The result we have just obtained applies to the heat ﬂow in the vicinity of a
point source of heat. Let’s now try to ﬁnd the equations that hold in the most
general kind of heat flow, keeping only the condition that heat is conserved.
We will be dealing only with what happens at places outside of any sources or
absorbers of heat. The differential equation for the conduction of heat was derived in Chapter 2.
According to Eq. (2.44), h = ——x VT. (3.25) (Remember that this relationship is an approximate one, but fairly good for some
materials like metals.) It is applicable, of course, only in regions of the material
where there is no generation or absorption of heat. We derived above another
relation, Eq. (3.21), that holds when heat is conserved. If we combine that equation
with (3.25), we get dq —7t= vh= — v(er),
01'
g = KV'VT= xva, (3.26) if x is a constant. You remember that q is the amount of heat in a unit volume
and V  V = V2 is the Laplacian operator If we now make one more assumption we can obtain a very interesting equa
tion. We assume that the temperature of the material is proportional to the heat
content per unit volume—that is, that the material has a deﬁnite speciﬁc heat.
When this assumption is valid (as it often is), we can write Aq = c, AT
or
d4 _ LT.
E —...
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 Spring '09
 LeeKinohara
 Physics

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