Calculus: Analytic Geometry and Calculus, with Vectors

13.5 Integrals in polar coordinates 693

7r and p > a1/cos 4> if 101 < it/2. It follows that if P lies outside our solid,
then Fz < GmM/a These results show that if we transfer material from the
inside to the outside of our solid, we decrease the force which pulls m in the
direction of the positive x axis. This fact and the answer to Problem 6 give the
following interesting conclusion. Of all solids having volume 4aa3/15 and uni-
form density S, our solid S is the one and only solid which exerts, upon a particle
at the origin, the greatest force in the direction of the positive x axis. The force
Fs exerted upon m by a spherical ball B which has the same volume and density
as S, and which has its center on the positive x axis and has the origin on its
surface, is the first of the vectors

FB = 25 -I,iraGm8i, Fs = 3aGmSi,

and the force exerted by our solid S is the second one. Thus we have a new

proof of the inequality 3 25 < 3. Tables say that

325=2.924018.

Spherical balls are not the best, but the best is only about 2.5 per cent better.

8 Partly because the result is thoroughly important in probability and

statistics and elsewhere, and partly because understandings of multiple and

iterated integrals should be developed, this problem requires learning a standard

method by which the formulas

(1) f o,a -' dx = 2 , 1 f dx = 1

and

1 r

(2) J

• e°> dx = 1, f xae z'12d' dx = Qs

are derived. Supposing that h > 0, we define F(h) by the first of the two equiva-

lent formulas

(3) F(h) = foh e -z' dx' F(h) =foh

e-" dy.

Then

r r

(4) [F(h)1Z =[ f h e-,2 dxI [ foh e dyJ =J0 dx Oh dy

1

ffQ(h) e(z'+u') dx dy

where we have, in order, the product of two integrals, an iteratedintegral and,

finally, a double integral over the square region Q(h) of Figure 13.591. This

Figure 13.591