Integral forms for grad div and curl

LINE, SURFACE AND VOLUME INTEGRALS
11.7 Integral forms for grad, div and curl
In the previous chapter we defined the vector operators grad, div and curl in purely
mathematical terms, which depended on the coordinate system in which they were
expressed. An interesting application of line, surface and volume integrals is the
expression of grad, div and curl in coordinate-free, geometrical terms. If φ is a
scalar field and a is a vector field then it may be shown that at any point P
0
1
φ dS
V →0 V S
0
1
a · dS
∇ · a = lim
V →0 V S
0
1
dS × a
∇ × a = lim
V →0 V S
∇φ = lim
(11.14)
(11.15)
(11.16)
where V is a small volume enclosing P and S is its bounding surface. Indeed,
we may consider these equations as the (geometrical) definitions of grad, div and
curl. An alternative, but equivalent, geometrical definition of ∇ × a at a point P ,
which is often easier to use than (11.16), is given by
(∇ × a) · n̂ = lim
A→0
1
A
0
a · dr ,
(11.17)
C
where C is a plane contour of area A enclosing the point P and n̂ is the unit
normal to the enclosed planar area.
It may be shown, in any coordinate system, that all the above equations are
consistent with our definitions in the previous chapter, although the difficulty of
proof depends on the chosen coordinate system. The most general coordinate
system encountered in that chapter was one with orthogonal curvilinear coordinates u1 , u2 , u3 , of which Cartesians, cylindrical polars and spherical polars are all
special cases. Although it may be shown that (11.14) leads to the usual expression
for grad in curvilinear coordinates, the proof requires complicated manipulations
of the derivatives of the basis vectors with respect to the coordinates and is not
presented here. In Cartesian coordinates, however, the proof is quite simple.
Show that the geometrical definition of grad leads to the usual expression for ∇φ in
Cartesian coordinates.
Consider the surface S of a small rectangular volume element ∆V = ∆x ∆y ∆z that has its
faces parallel to the x, y, and z coordinate surfaces; the point P (see above) is at one corner.
We must calculate the surface integral (11.14) over each of its six faces. Remembering that
the normal to the surface points outwards from the volume on each face, the two faces
with x = constant have areas ∆S = −i ∆y ∆z and ∆S = i ∆y ∆z respectively. Furthermore,
over each small surface element, we may take φ to be constant, so that the net contribution
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11.7 INTEGRAL FORMS FOR grad, div AND curl
to the surface integral from these two faces is then
∂φ
∆x − φ ∆y ∆z i
[(φ + ∆φ) − φ] ∆y ∆z i = φ +
∂x
∂φ
=
∆x ∆y ∆z i.
∂x
The surface integral over the pairs of faces with y = constant and z = constant respectively
may be found in a similar way, and we obtain
0
∂φ
∂φ
∂φ
i+
j+
k ∆x ∆y ∆z.
φ dS =
∂x
∂y
∂z
S
Therefore ∇φ at the point P is given by
∂φ
∂φ
∂φ
1
∇φ = lim
i+
j+
k ∆x ∆y ∆z
∆x,∆y,∆z→0 ∆x ∆y ∆z
∂x
∂y
∂z
∂φ
∂φ
∂φ
i+
j+
k. =
∂x
∂y
∂z
We now turn to (11.15) and (11.17). These geometrical definitions may be
shown straightforwardly to lead to the usual expressions for div and curl in
orthogonal curvilinear coordinates.
By considering the infinitesimal volume element dV = h1 h2 h3 ∆u1 ∆u2 ∆u3 shown in figure 11.10, show that (11.15) leads to the usual expression for ∇·a in orthogonal curvilinear
coordinates.
Let us write the vector field in terms of its components with respect to the basis vectors
of the curvilinear coordinate system as a = a1 ê1 + a2 ê2 + a3 ê3 . We consider first the
contribution to the RHS of (11.15) from the two faces with u1 = constant, i.e. P QRS
and the face opposite it (see figure 11.10). Now, the volume element is formed from the
orthogonal vectors h1 ∆u1 ê1 , h2 ∆u2 ê2 and h3 ∆u3 ê3 at the point P and so for P QRS we
have
∆S = h2 h3 ∆u2 ∆u3 ê3 × ê2 = −h2 h3 ∆u2 ∆u3 ê1 .
Reasoning along the same lines as in the previous example, we conclude that the contribution to the surface integral of a · dS over P QRS and its opposite face taken together is
given by
∂
∂
(a · ∆S) ∆u1 =
(a1 h2 h3 ) ∆u1 ∆u2 ∆u3 .
∂u1
∂u1
The surface integrals over the pairs of faces with u2 = constant and u3 = constant
respectively may be found in a similar way, and we obtain
0
∂
∂
∂
a · dS =
(a1 h2 h3 ) +
(a2 h3 h1 ) +
(a3 h1 h2 ) ∆u1 ∆u2 ∆u3 .
∂u1
∂u2
∂u3
S
Therefore ∇ · a at the point P is given by
0
1
∇·a=
lim
a · dS
∆u1 ,∆u2 ,∆u3 →0 h1 h2 h3 ∆u1 ∆u2 ∆u3 S
∂
∂
1
∂
(a1 h2 h3 ) +
(a2 h3 h1 ) +
(a3 h1 h2 ) . =
h1 h2 h3 ∂u1
∂u2
∂u3
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LINE, SURFACE AND VOLUME INTEGRALS
z
h1 ∆u1 ê1
R
T
S
h2 ∆u2 ê2
Q
P
h3 ∆u3 ê3
y
x
Figure 11.10 A general volume ∆V in orthogonal curvilinear coordinates
u1 , u2 , u3 . P T gives the vector h1 ∆u1 ê1 , P S gives h2 ∆u2 ê2 and P Q gives
h3 ∆u3 ê3 .
By considering the infinitesimal planar surface element P QRS in figure 11.10, show that
(11.17) leads to the usual expression for ∇ × a in orthogonal curvilinear coordinates.
The planar surface P QRS is defined by the orthogonal vectors h2 ∆u2 ê2 and h3 ∆u3 ê3
at the point P . If we traverse the loop in the direction P SRQ then, by the right-hand
convention, the unit normal to the plane is ê1 . Writing a = a1 ê1 + a2 ê2 + a3 ê3 , the line
integral around the loop in this direction is given by
0
∂
a · dr = a2 h2 ∆u2 + a3 h3 +
(a3 h3 ) ∆u2 ∆u3
∂u2
P S RQ
∂
− a2 h2 +
(a2 h2 ) ∆u3 ∆u2 − a3 h3 ∆u3
∂u3
∂
∂
=
(a3 h3 ) −
(a2 h2 ) ∆u2 ∆u3 .
∂u2
∂u3
Therefore from (11.17) the component of ∇ × a in the direction ê1 at P is given by
0
1
a · dr
(∇ × a)1 = lim
∆u2 ,∆u3 →0 h2 h3 ∆u2 ∆u3 P S RQ
∂
1
∂
(h3 a3 ) −
(h2 a2 ) .
=
h2 h3 ∂u2
∂u3
The other two components are found by cyclically permuting the subscripts 1, 2, 3. Finally, we note that we can also write the ∇2 operator as a surface integral by
setting a = ∇φ in (11.15), to obtain
0
1
∇φ · dS .
∇2 φ = ∇ · ∇φ = lim
V →0 V S
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