The surface S, bounded by a simple closed curve, is an open surface; and the
normal to an open surface can point in two opposite directions. We adopt the
usual convention, namely the right hand rule: when the ®ngers of the right hand
follow the direction of dl, the thumb points in the da direction, as shown in Fig.
1.21.
Note that Eq. (1.80) does not specify the shape of the surface S other than that
it be bounded by ÿ; thus there are many possibilities in choosing the surface. But
Stokes' theorem enables us to reduce the evaluation of surface integrals which
depend upon the shape of the surface to the calculation of a line integral which
depends only on the values of A along the common perimeter.
To prove the theorem, we ®rst expand the left hand side of Eq. (1.80); with the
aid of Eq. (1.50), it becomes
Z
S
rA da
Z
S
@A
1
@x
3
da
2
ÿ
@A
1
@x
2
da
3
Z
S
@A
2
@x
1
da
3
ÿ
@A
2
@x
3
da
1
Z
S
@A
3
@x
2
da
1
ÿ
@A
3
@x
1
da
2
; 1:81
where we have grouped the terms by components of A. We next subdivide the
surface S into a large number of small strips, and integrate the ®rst integral on the
right hand side of Eq. (1.81), denoted by I
1
, over one such a strip of width dx
1
,
which is parallel to the x
2
x
3
plane and a distance x
1
from it, as shown in Fig. 1.21.
Then, by integrating over x
1
, we sum up the contributions from all of the strips.
Fig. 1.21 also shows the projections of the strip on the x
1
x
3
and x
1
x
2
planes that
will help us to visualize the orientation of the surface. The element area d a is
shown at an intermediate stage of the integration, when the direction angles have
values such that and ÿ are less than 908 and þ is greater than 908. Thus,
da
2
ÿdx
1
dx
3
and da
3
dx
1
dx
2
and we can write
I
1
ÿ
Z
strips
dx
1
Z
Q
P
@A
1
@x
2
dx
2
@A
1
@x
3
dx
3
: 1:82
Note that dx
2
and dx
3
in the parentheses are not independent because x
2
and x
3
are related by the equation for the surface S and the value of x
1
involved. Since
the second integral in Eq. (1.82) is being evaluated on the strip from P to Q for
which x
1
const., dx
1
0 and we can add @A
1
=@x
1
dx
1
0 to the integrand to
make it dA
1
:
@A
1
@x
1
dx
1
@A
1
@x
2
dx
2
@A
1
@x
3
dx
3
dA
1
:
And Eq. (1.82) becomes
I
1
ÿ
Z
strips
dx
1
Z
Q
P
dA
1
Z
strips
A
1
PÿA
1
Qdx
1
:
41
VECTOR INTEGRATION AND INTEGRAL THEOREMS