 # Surface integrals

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Surface integrals
```11.5 SURFACE INTEGRALS
independent of the path taken. Since a is conservative, we can write a = ∇φ. Therefore, φ
must satisfy
∂φ
= xy 2 + z,
∂x
which implies that φ = 12 x2 y 2 + zx + f(y, z) for some function f. Secondly, we require
∂φ
∂f
= x2 y +
= x2 y + 2,
∂y
∂y
which implies f = 2y + g(z). Finally, since
∂g
∂φ
=x+
= x,
∂z
∂z
we have g = constant = k. It can be seen that we have explicitly constructed the function
φ = 12 x2 y 2 + zx + 2y + k. The quantity φ that ﬁgures so prominently in this section is called the scalar
potential function of the conservative vector ﬁeld a (which satisﬁes ∇ × a = 0), and
is unique up to an arbitrary additive constant. Scalar potentials that are multivalued functions of position (but in simple ways) are also of value in describing
some physical situations, the most obvious example being the scalar magnetic
potential associated with a current-carrying wire. When the integral of a ﬁeld
quantity around a closed loop is considered, provided the loop does not enclose
a net current, the potential is single-valued and all the above results still hold. If
the loop does enclose a net current, however, our analysis is no longer valid and
extra care must be taken.
If, instead of being conservative, a vector ﬁeld b satisﬁes ∇ · b = 0 (i.e. b
is solenoidal) then it is both possible and useful, for example in the theory of
electromagnetism, to deﬁne a vector ﬁeld a such that b = ∇ × a. It may be shown
that such a vector ﬁeld a always exists. Further, if a is one such vector ﬁeld then
a = a + ∇ψ + c, where ψ is any scalar function and c is any constant vector, also
satisﬁes the above relationship, i.e. b = ∇ × a . This was discussed more fully in
subsection 10.8.2.
11.5 Surface integrals
As with line integrals, integrals over surfaces can involve vector and scalar ﬁelds
and, equally, can result in either a vector or a scalar. The simplest case involves
entirely scalars and is of the form
φ dS.
(11.8)
S
As analogues of the line integrals listed in (11.1), we may also encounter surface
integrals involving vectors, namely
φ dS,
a · dS,
a × dS.
(11.9)
S
S
S
389
LINE, SURFACE AND VOLUME INTEGRALS
S
dS
dS
S
V
C
(a)
(b)
Figure 11.5 (a) A closed surface and (b) an open surface. In each case a
normal to the surface is shown: dS = n̂ dS.
All the above integrals are taken over some surface S, which may be either
open or closed, and are therefore, in general, double integrals. Following
the
is
replaced
notation
for
line
integrals,
for
surface
integrals
over
a
closed
surface
S
/
by S .
The vector diﬀerential dS in (11.9) represents a vector area element of the
surface S. It may also be written dS = n̂ dS, where n̂ is a unit normal to the
surface at the position of the element and dS is the scalar area of the element used
in (11.8). The convention for the direction of the normal n̂ to a surface depends
on whether the surface is open or closed. A closed surface, see ﬁgure 11.5(a),
does not have to be simply connected (for example, the surface of a torus is not),
but it does have to enclose a volume V , which may be of inﬁnite extent. The
direction of n̂ is taken to point outwards from the enclosed volume as shown.
An open surface, see ﬁgure 11.5(b), spans some perimeter curve C. The direction
of n̂ is then given by the right-hand sense with respect to the direction in which
the perimeter is traversed, i.e. follows the right-hand screw rule discussed in
subsection 7.6.2. An open surface does not have to be simply connected but for
our purposes it must be two-sided (a Möbius strip is an example of a one-sided
surface).
The formal deﬁnition of a surface integral is very similar to that of a line
integral. We divide the surface S into N elements of area ∆Sp , p = 1, 2, . . . , N,
each with a unit normal n̂p . If (xp , yp , zp ) is any point in ∆Sp then the second type
of surface integral in (11.9), for example, is deﬁned as
a · dS = lim
S
N→∞
N
a(xp , yp , zp ) · n̂p ∆Sp ,
p=1
where it is required that all ∆Sp → 0 as N → ∞.
390
11.5 SURFACE INTEGRALS
z
k
dS
α
S
y
R
dA
x
Figure 11.6 A surface S (or part thereof) projected onto a region R in the
xy-plane; dS is a surface element.
11.5.1 Evaluating surface integrals
We now consider how to evaluate surface integrals over some general surface. This
involves writing the scalar area element dS in terms of the coordinate diﬀerentials
of our chosen coordinate system. In some particularly simple cases this is very
straightforward. For example, if S is the surface of a sphere of radius a (or some
part thereof) then using spherical polar coordinates θ, φ on the sphere we have
dS = a2 sin θ dθ dφ. For a general surface, however, it is not usually possible to
represent the surface in a simple way in any particular coordinate system. In such
cases, it is usual to work in Cartesian coordinates and consider the projections of
the surface onto the coordinate planes.
Consider a surface (or part of a surface) S as in ﬁgure 11.6. The surface S is
projected onto a region R of the xy-plane, so that an element of surface area dS
projects onto the area element dA. From the ﬁgure, we see that dA = | cos α| dS,
where α is the angle between the unit vector k in the z-direction and the unit
normal n̂ to the surface at P . So, at any given point of S, we have simply
dS =
dA
dA
=
.
| cos α|
|n̂ · k|
Now, if the surface S is given by the equation f(x, y, z) = 0 then, as shown in subsection 10.7.1, the unit normal at any point of the surface is given by n̂ = ∇f/|∇f|
evaluated at that point, cf. (10.32). The scalar element of surface area then becomes
dS =
|∇f| dA
|∇f| dA
dA
=
=
,
|n̂ · k|
∇f · k
∂f/∂z
391
(11.10)
LINE, SURFACE AND VOLUME INTEGRALS
where |∇f| and ∂f/∂z are evaluated on the surface S. We can therefore express
any surface integral over S as a double integral over the region R in the xy-plane.
Evaluate the surface integral I = S a · dS, where a = xi and S is the surface of the
hemisphere x2 + y 2 + z 2 = a2 with z ≥ 0.
The surface of the hemisphere is shown in ﬁgure 11.7. In this case dS may be easily
expressed in spherical polar coordinates as dS = a2 sin θ dθ dφ, and the unit normal to the
surface at any point is simply r̂. On the surface of the hemisphere we have x = a sin θ cos φ
and so
a · dS = x (i · r̂) dS = (a sin θ cos φ)(sin θ cos φ)(a2 sin θ dθ dφ).
Therefore, inserting the correct limits on θ and φ, we have
π/2
2π
2πa3
dθ sin3 θ
dφ cos2 φ =
.
I = a · dS = a3
3
0
0
S
We could, however, follow the general prescription above and project the hemisphere S
onto the region R in the xy-plane that is a circle of radius a centred at the origin. Writing
the equation of the surface of the hemisphere as f(x, y) = x2 + y 2 + z 2 − a2 = 0 and using
(11.10), we have
|∇f| dA
x (i · r̂)
I = a · dS = x (i · r̂) dS =
.
∂f/∂z
S
S
R
Now ∇f = 2xi + 2yj + 2zk
= 2r, so on the surface S we have |∇f| = 2|r| = 2a. On S we
also have ∂f/∂z = 2z = 2 a2 − x2 − y 2 and i · r̂ = x/a. Therefore, the integral becomes
x2
I=
dx dy.
2
a − x2 − y 2
R
Although this integral may be evaluated directly, it is quicker to transform to plane polar
coordinates:
ρ2 cos2 φ
I=
ρ dρ dφ
a2 − ρ 2
R
a
2π
ρ3 dρ
cos2 φ dφ
.
=
a2 − ρ 2
0
0
Making the substitution ρ = a sin u, we ﬁnally obtain
π/2
2π
2πa3
cos2 φ dφ
a3 sin3 u du =
I=
.
3
0
0
In the above discussion we assumed that any line parallel to the z-axis intersects
S only once. If this is not the case, we must split up the surface into smaller
surfaces S1 , S2 etc. that are of this type. The surface integral over S is then the
sum of the surface integrals over S1 , S2 and so on. This is always necessary for
closed surfaces.
Sometimes we may need to project a surface S (or some part of it) onto the
zx- or yz-plane, rather than the xy-plane; for such cases, the above analysis is
easily modiﬁed.
392
11.5 SURFACE INTEGRALS
z
dS
a
S
a
a
y
dA = dx dy
C
x
Figure 11.7 The surface of the hemisphere x2 + y 2 + z 2 = a2 , z ≥ 0.
11.5.2 Vector areas of surfaces
The vector area of a surface S is deﬁned as
S = dS,
S
where the surface integral may be evaluated as above.
Find the vector area of the surface of the hemisphere x2 + y 2 + z 2 = a2 with z ≥ 0.
As in the previous example, dS = a2 sin θ dθ dφ r̂ in spherical polar coordinates. Therefore
the vector area is given by
a2 sin θ r̂ dθ dφ.
S=
S
Now, since r̂ varies over the surface S, it also must be integrated. This is most easily
achieved by writing r̂ in terms of the constant Cartesian basis vectors. On S we have
r̂ = sin θ cos φ i + sin θ sin φ j + cos θ k,
so the expression for the vector area becomes
π/2
2π
2
2
cos φ dφ
sin θ dθ + j a2
S=i a
0
0
2π
+ k a2
π/2
dφ
0
2π
2
sin φ dφ
0
π/2
sin θ dθ
0
sin θ cos θ dθ
0
= 0 + 0 + πa2 k = πa2 k.
Note that the magnitude of S is the projected area, of the hemisphere onto the xy-plane,
and not the surface area of the hemisphere. 393
LINE, SURFACE AND VOLUME INTEGRALS
C
dr
r
O
Figure 11.8 The conical surface spanning the perimeter C and having its
vertex at the origin.
The hemispherical shell discussed above is an example of an open surface. For
a closed surface, however, the vector area is always zero. This may be seen by
projecting the surface down onto each Cartesian coordinate plane in turn. For
each projection, every positive element of area on the upper surface is cancelled
by the corresponding
/ negative element on the lower surface. Therefore, each
component of S = S dS vanishes.
An important corollary of this result is that the vector area of an open surface
depends only on its perimeter, or boundary curve, C. This may be proved as
follows. If surfaces S1 and S2 have the same perimeter then S1 − S2 is a closed
surface, for which
0
dS −
dS =
S1
dS = 0.
S2
Hence S1 = S2 . Moreover, we may derive an expression for the vector area of
an open surface S solely in terms of a line integral around its perimeter C.
Since we may choose any surface with perimeter C, we will consider a cone
with its vertex at the origin (see ﬁgure 11.8). The vector area of the elementary
triangular region shown in the ﬁgure is dS = 12 r × dr. Therefore, the vector area
of the cone, and hence of any open surface with perimeter C, is given by the line
integral
S=
1
2
0
r × dr.
C
For a surface conﬁned to the xy-plane, r = xi + yj and dr = dx i + dy j,
and we/ obtain for this special case that the area of the surface is given by
A = 12 C (x dy − y dx), as we found in section 11.3.
394
11.5 SURFACE INTEGRALS
Find the vector area of the surface
of the hemisphere x2 + y 2 + z 2 = a2 , z ≥ 0, by
/
evaluating the line integral S = 12 C r × dr around its perimeter.
The perimeter C of the hemisphere is the circle x2 + y 2 = a2 , on which we have
dr = −a sin φ dφ i + a cos φ dφ j.
r = a cos φ i + a sin φ j,
Therefore the cross product r × dr is given
i
j
a cos φ
a sin φ
r × dr = −a sin φ dφ a cos φ dφ
and the vector area becomes
S = 12 a2 k
by
k
0
0
2π
= a2 (cos2 φ + sin2 φ) dφ k = a2 dφ k,
dφ = πa2 k. 0
11.5.3 Physical examples of surface integrals
There are many examples of surface integrals in the physical sciences. Surface
integrals of the form (11.8) occur
in computing the total electric charge on a
surface or the mass of a shell, S ρ(r) dS, given the charge or mass density ρ(r).
For surface integrals involving vectors, the second form
in (11.9) is the most
common. For a vector ﬁeld a, the surface integral S a · dS is called the ﬂux
of a through S. Examples of physically important ﬂux integrals are numerous.
For example, let us consider a surface S in a ﬂuid with density ρ(r) that has a
velocity ﬁeld v(r). The mass of ﬂuid crossing an element of surface area dS in
time dt is dM = ρv · dS dt. Therefore the net total mass ﬂux of ﬂuid crossing S
ﬂux of energy
is M = S ρ(r)v(r) · dS. As a another example, the electromagnetic
/
out of a given volume V bounded by a surface S is S (E × H) · dS.
The solid angle, to be deﬁned below, subtended at a point O by a surface (closed
or otherwise) can also be represented by an integral of this form, although it is
not strictly a ﬂux integral (unless we imagine isotropic rays radiating from O).
The integral
r̂ · dS
r · dS
=
,
(11.11)
Ω=
3
r
r2
S
S
gives the solid angle Ω subtended at O by a surface S if r is the position vector
measured from O of an element of the surface. A little thought will show that
(11.11) takes account of all three relevant factors: the size of the element of
surface, its inclination to the line joining the element to O and the distance from
O. Such a general expression is often useful for computing solid angles when the
three-dimensional geometry is complicated. Note that (11.11) remains valid when
the surface S is not convex and when a single ray from O in certain directions
would cut S in more than one place (but we exclude multiply connected regions).
395
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