714 Iterated and multiple integrals
sin 0 dO and, except for algebraic sign, the integrand becomes that in
(14) f (r) =fr D +
u
_2 + 2Du + r2)3,(D d.-
When 0 = 0, we have u = r, and when 0 = ir, we have u = -r, and we see
that (14) is correct when we see that - f
rr
=f rr We can discover that
trading (14) for (13) was good business if we know or discover or are told that
(14) can be demolished by the substitution
z - z
v=D22+2Du+r2, u=v 2D
r
D+uv+2 z - r2
2D du- 2Ddv.
Since v = (D - r)22 when it = -r and v = (D + r)2 when it = r, this substitution
gives
(^1) (D+T)2 V + D2 - r2
(15) f(r) =4D2f(D-,)= v3, dv
1! (D+r)'-
4D2 (D-r)2 lv 3' +
(D2 - rz)v-9,]
dv.
Since 0 < r < D, this gives f(r) = 2r/D' Substitution in (12) then gives the
first equality in
(16) F =
k
fo
a
4ar'--82(r) dr =GmMk.
As was shown in the first problem in this list, the integral in (16) is the total
mass M of the sphere, and hence the second equality holds. The result embodied
in (16) is the following famous theorem. If S is a radically homogeneous spherical
ball, then the gravitational force which S exerts upon a particle outside S is equal to
the force resulting from the assumption that the total mass of S is concentrated at the
center of S. To help us understand the significance of this result, we should know
some history. It is said that Newton mistrusted his whole theory of gravitational
attraction (and therefore delayed publication of his theory for 20 years) until he
was able to prove the theorem.
11 If S is a radially homogeneous spherical shell, then the gravitational force
F which S exerts upon a particle inside S is 0. All scientists should know this
fact and some should, when a suitable occasion comes, earn the satisfaction of
discovering the modifications that must be made in the work of Problem 10 to
prove the fact. '
12 With the aid of results of Problems 10 and 11, suppose that the earth is a
homogeneous spherical ball and discuss the gravitational force upon a particle at
the bottom of a very deep well.