Calculus: Analytic Geometry and Calculus, with Vectors

688 Iterated and multiple integrals

ferent ways of proving that, when p > 0,

(13.532) lim As = 1.

AP-+o, m-+o PD P A¢

It is, in fact, easy to work out an exact formula for AS, because AS is

the difference of the areas of two sectors having central angle L. The

larger sector has radius p + tp and the smaller sector has radius p, so

AS = 2(p + Ap)2 0o -jp2'o = 1(2p iP + (Op)') A0

and hence

(13.533) As =(p+ 2P 1 LO AP.

This can be used to prove (13.532), and it can also give us another idea.

If we let p* = p + pp/2, then we obtain the exact formula

(13.534) AS = P* 0o AP.

One who wishes to do so may insist that the formula

(13.535) AS = p A0 Ap

is an exact formula obtained by setting Ep, = p,+i - p E¢k = 40k+l - ok,

P* = p, + '(P,+i - p,), and discarding all subscripts and stars. When we

put f(P) in the form f(p,¢), where p and q5 are polar coordinates, we are

therefore able to put (13.51) in the form

(13.54) Ifs f(p,o)p do dp = lim jf(P,db)P A0 A p.

Assuming that f is bounded over S and is sufficiently continuous to make

all of the integrals exist, we can use Theorem 13.38 to express this double

integral as an iterated integral. For example, when S is the set featured

in Figure 13.52, we observe that the point having polar coordinates

(p,o) lies in S when a --<_ o _<_ P and, for each such 0, gl(¢) _< p 5 g2(o),

so

(13.55)

ffs f(p,o)p do dP = f " do f f(P,4)P dp.

Except in cases where S is a circular sector (which may be a whole circular

disk) or the difference of two circular sectors (which may be a whole

circular ring), it is usually not convenient to use formulas of the form

(13.551) Ifsf(p,o)p do dp =

fab

dp fh

P(a)

f(p,o)p do

in which the first integration is with respect to 0. The following three

examples serve to show how double and iterated integrals in polar coordi-

nates can be set up. It is not recommended that the formulas be

remembered; we reconstruct them whenever we want to use them.