Calculus: Analytic Geometry and Calculus, with Vectors

672 Iterated and multiple integrals

provided the integrals exist. With the aid of series and termwise integration
(which can be justified), continue the work to obtain

rr

z

(2) J J s l ---X y

dx dy = 1 kz=

j

Remark: Persons who study more advanced mathematical analysis may encounter

the following elegant theorem. If S is a set in E,,, if S1 is a subset of S having

measure 0, if

(3) f(P) = uI(P) + uz(P) + ua(P) + ...

when P is in S but not in S1, and if uk(P) > 0 for each k i% hen P is in S, then the

formula

(4) ff(P)dS=

fu3(P)dS+ fuz(P)dS+ J ua(P)dS+. ..

is valid provided the integrals exist as Riemann integrals or Riemann-Cauchy

integrals or Lebesgue integrals.

9 Obtain the result (2) of the preceding problem by termwise integration

(which can be justified) of the series in

1 =1+xy+xzyz+xaya+..

1 - xy

10 This long problem is for persons who wish to become good mathematicians

or physicists or engineers and who realize that the best ones start learning about

important things while they are young and then continue to learn more. Make

a large copy of Figure 13.291 showing the set S which is the set of points (x,y)

for which a 5 x < b and f, (x) < y 5 f2 (x) and is also the set of points (x,y)

for which c 5 y < d and gi(y) S x < g2(y). Let C be the curve consisting of the

boundary of S traversed once in the positive direction. Let F and G be continuous

over S and let

(1) I = ff5F(x,y) dS, J =

ffsG(x,y) dS.

Show that

rd g2(y) b r12(x)

(2) I =Jdylog(y) F(x,y) dx, J =

f

dxJf,(x) G(x,y) dy

We enter a gate to an important scientific garden when we suppose that P and Q

are functions which are defined and have continuous first-order partial derivatives

over S and put

(3) F(x,y) =aQ= Q, (x,y), G(x,y) = 8y= PY(x,y)

Show that in this case

(4) I = f.

d

dyJDg(Y) Q,(x,y) dx =Jr

LdQ(g2(v), y) dy - f d Q(gl(y), y) dy