Mathematics of Physics and Engineering

134 Functions of Several Variables

the surface area element of S.

EXERCISE 3.1.15. B Justify (3.1.19). Hint. Verify that ||r„ x rv\AuAv is
the area of the parallelogram in the tangent plane to S at the point r(u,v), with
sides Au, Av parallel to the vectors ru, rv, respectively. For small Au, Av, this
area closely approximates a curved parallelogram-like surface element of S with
vertices atr(u,v), r(u + Au,v), r(u,v + Av), r(u + Au,v + Av). Summing up
over all such surface elements and passing to the limit as Au —• 0, At; —» 0 gives
(3.1.19). For a slightly different derivation of (3.1.19), see Exercise 3.1.35(c) on
page 149 below.

The surface integral of the first kind Jf f da is defined for a
s
continuous scalar field / by the equality

J Jfda = J J f(r(u,v)) \\ru x r„||dA(u,i;), (3.1.20)

S Gi

where dA(u, v) is the area element. If f(P) represents the areal density of
the surface at point P, then JJ f da is the mass of the surface. The surface
s
integral of the second kind JJ F-dcr is defined for a continuous vector
s
field F with values in E^3 and an orientable surface 5" by the equality

ffFd(T= fJ F(r(u,v)) • (ru x rv)dA(u,v); (3.1.21)

S Gi

the integral is also called the flux of F across S. A surface integral of the

second kind can be reduced to a surface integral of the first kind:

IIFd(T= ffF-nsda, (3.1.22)

where fis is the unit normal vector to the surface. The direction of ns
defines the orientation of S; the integral reverses the sign if the orientation
of S is reversed.
If F = pv, where v is the velocity of a fluid flow, and p is the density of
the fluid, then J J F • da is the mass of the fluid passing across the surface
s
S per unit time; verify that

F • fis da = pv • fis da