Integration and Differentiation 133THE AREA INTEGRAL. We now take the set G in (3.1.13) to be a subset
of the plane K^2. The set is measurable, and has area A, if the boundary of G
is a continuous, piece-wise smooth simple closed curve. In area integrals, we
use the differential form of the area element dA. In cartesian coordinates,
the area element is dA — dxdy. In the generalized polar coordinates
x = arcos6, y = brsin.6, the area element is dA = ab rdrd6; the usual
polar coordinates correspond to a = b = 1. The area integral is evaluated
by reduction to a suitable iterated integral. The area integral is also known
as the double integral and is often written as JfG f dA. The integral is
defined if / is a continuous scalar field in G. If / is the area density of G,
then fjG f dA is the mass of G. Computation of double integrals is one of
the key skills that must be acquired in the study of multi-variable calculus.
EXERCISE 3.1.14. (a)B Let G be the parallelogram with vertices
(0,0), (3,0), (4,1), (1,1), and f, a continuous function. Write the limits
in the iterated integrals below:fffdA= J ffdxdy = f ffdydx = f ffrdr dO = f f fdO rdr.
GHint: some integrals must be split into several. (b)c Verify that the area of
the ellipse (x^2 /a^2 ) + {y^2 /b^2 ) < 1 with semi-axis a and b is irab. Hint: use
generalized polar coordinates.
THE SURFACE INTEGRAL. We now take the set G in (3.1.13) to be a
surface S defined by the vector-valued function r = r(u, v) of two variables,
where (u,v) belong to some measurable planar set G. We say that the
surface is piece-wise smooth if the function r is continuous on G\ and
the set Gi can be split into finitely many non-overlapping pieces so that,
on each piece, the vector field ru x r„, representing the normal vectors to
the surface, is continuous and is not equal to zero. We say that the surface
is orientable if it is piece-wise smooth and has two sides (the intuitive
idea of a side is enough for our discussion). A piece-wise smooth surface S*
is measurable and its area m(S) is defined as
m(S) = If \\ru x rv\\dA(u,v), (3.1.19)
Giwhere dA(u, v) is the area element of G\. We callda = \\ru x rv\\ dA(u,v)