Calculus: Analytic Geometry and Calculus, with Vectors

(lu) #1

268 Integrals


In case 0 < b < a, this gives


(11) U-2b2C(b l a)-(a-b)-


bb
+a2 b-aJ=O

and (5) shows that F = 0 as we wish to prove. In case 0 < a < b, (10) gives

_ 1 b2 - a2 b2 - a2 2a
(12) LT-2b2C(b+ a)-(b-a) b+a T

b-a]b2

Putting this in (5) gives
F -Gm(4wra21)i-GmM.
b2 b2^11

where M = 4ira2S, the total mass of the spherical shell. This is the desired result
(1) and the fundamental facts about attractions of spherical shells are now
established.
17 Use the method of Problem 16 but modify the details in appropriate
places to obtain the force FH exerted upon m* by the hemispherical shell that
remains after removal of the part of the spherical shell whose points have nega-
tive x coordinates.
18 A spherical ball (or solid sphere) is said to be radially homogeneous if
there is a function S such that the ball has density (mass per unit volume) S(r)
at each point having distance r from the center of the sphere. Supposing that
0 < a < b, find the gravitational force exerted upon a particle m* of mass m
located at the point (-b,0,0) in E3 by a radially homogeneous spherical ball
(like an idealized earth or golf ball) B which has radius a, which has its center at
the origin, and which has a density function 6
which is not necessarily constant but is inte-
grable. Solution: As suggested by Figure
4.795, we make a partition P of the interval
-b -a^0 xk a x 0 5 x 5 a. When xk_1 < xk 5 xk as usual,
Figure 4.795 the points of B having distance xk* from 0 such
that xk_I < xk 5 xk form a spherical shell
whose volume is approximately the product of the area 47rxk* 2 and the thickness
Ark. The mass Mk of this shell is therefore approximately

(1) 1Vfk = 4,rxk 26(xk) AXk.

Considering the shell to be concentrated upon the sphere having center at 0
and having radius xk* enables us to use a result of Problem 16 to show that the
force AFk which the shell exerts upon m* is approximately

(2) AFk=Gm[47rxk 2S(xk) Axk)
i.
b2

The limit of the sum of these things should be F. But the sum is a Riemann
sum and its limit is the integral in the formula

(3) Gmi
F = 72
o

4ax2S(x) dz.
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