LINE, SURFACE AND VOLUME INTEGRALS
drrO
C
Figure 11.8 The conical surface spanning the perimeterCand having its
vertex at the origin.The hemispherical shell discussed above is an example of an open surface. Fora 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 ofS=
∮
SdSvanishes.
An important corollary of this result is that the vector area of an open surfacedepends only on its perimeter, or boundary curve,C. This may be proved as
follows. If surfacesS 1 andS 2 have the same perimeter thenS 1 −S 2 is a closed
surface, for which
∮
dS=∫S 1dS−∫S 2dS= 0.HenceS 1 =S 2. Moreover, we may derive an expression for the vector area of
an open surfaceS solely in terms of a line integral around its perimeterC.
Since we may choose any surface with perimeterC, 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 isdS=^12 r×dr. Therefore, the vector area
of the cone, and hence ofanyopen surface with perimeterC, is given by the line
integral
S=1
2∮Cr×dr.For a surface confined to thexy-plane, r=xi+yj anddr=dxi+dyj,
and we obtain for this special case that the area of the surface is given by
A=^12
∮
C(xdy−ydx), as we found in section 11.3.