13.5 Integrals in polar coordinates 687and putting x = ut gives the second equality Therefore(6) p!4! = (p + q + 1)! 101 tP(1- t)4 dt
and (1) follows.13.5 Integrals in polar coordinates In some cases a plane set S anda
function f defined over S are such that the double integraldefined as in
(13.37) by(13.51) f f S f(P) dS= limI f(P)AS
can be advantageously expressed in terms of polar coordinatesp and c_
For example, suppose that, as in Figure 13.52, S is the set of points0=RA& (P,+110k+1
P=g2(0)
(P,,Ok+,) (p,o + 6¢)Figure 13.52 Figure 13.53APFigure 13.531having polar coordinates p, 0 for which gl(4,) <- p S g2(4,) anda < 0 <
0, where gi and g2 are continuous functions for which 0 < gl(4,) S g2(¢)
when a S 0 < /3. Our first step is to partition S into subsets Si, S2,
, S by radial lines having the polar equations ¢ = 4,o, 0 = 01,
, 0 = yam, where a = ¢o < .01 < < 4,,,, = 9, and by circles
or circular arcs having the equations p = po, p = pi, , p = as
in Figure 13.52. A typical one of the subsets is a part of a sector having
corners at the points whose polar coordinates are shown in Figure 13.53.
We set Op, = p,+1 - p, and D4ik = Ok+l- 4,k and then simplify the things
we write by discarding all subscripts to obtain the polar coordinates
shown in Figure 13.531. The symbol AS then represents the area of the
shaded set in Figure 13.531. This shaded set is not a rectangle, because
the straight sides are not parallel, and the inner and outer sides are arcs of
circles which are not parallel line segments. It is, however, thoroughly rea-
sonable to have the opinion that, when p > 0 and Op and 0¢ are small,
the shaded set is "nearly" rectangular and that AS should be closely
approximated by the product of /p (the length of one of the straight
sides) and p 0¢ (the length of the inner curved side). This suggests that
we should be able to use p A4, Op as an approximation to AS. Much
more can be said about this approximation business, and there are dif-