Calculus: Analytic Geometry and Calculus, with Vectors

4.7 Mass, linear density, and moments 267

be describedroughly as a circular ribbon having radius a sin Sk, width ai Ok,
length 27r a sinSk, area 27r a2 sin Ok ABA;, and mass Mk, where

(3) Mk = 21rba2 sin Ok* MBk.

Considering this ribbon to be a circular ring of mass Mk and radius a sin Ok* which
has its center on the xaxis and which lies in the plane having the equation x =

a cosSk, we use formula (1) of Problem 12 with c = -b to obtain the formula

(b + a cos BL*) sin 84

(4) AFk = 27rGm3a2i AOk

[(b + a cosOk)2+(a sin9k)213

for an approximation to the force upon m* produced by one element of the

spherical shell. The limit of the sum of these things should be the force F that

tie are seeking. But the sum is a Riemann sum and its limit is a Riemann

integral. This leads us to the formula

(5) F = 27rGmbaiU,

where U is the unruly integral defined by

f ,r (b + a cos 8)a sin 8

(6) U- JfTheo [b2 + 2ab cos 0 + a2]3dB.

hypothesis that b > 0 and b 5-6 a implies that the denominators in (4) and

(6) are never zero and hence that the integrand in (6) is continuous. Before

making a serious attack on the integral, we can observe that it is certainly positive

when b > a and that it is 0 when b = 0. To simplify the integral, we make the

substitution (or change of variable)

(5) a cos 0 = x,

so that -a sin O d8 = dx. Since x = a when 8 = 0 and x = -a when 8 =7r,

rules which have not yet been adequately treated imply that

ra b+x

(6)

U=J-a[b2+

To simplify the integral some more when b 3-6 0, we make the substitution

b z - a2

b2+2bx+a2=t, x=t 2b a

Since dx = (1/2b) dt, t = (b - a)2 when x = -a, and t = (b + a)2 when x = a.

substitution in (6) gives

b2-a2

1 ,b+

(7) U = 2b

r

J(bba)2

t 2b dt.

Thus

(8)

(9)

(b+a)s

U = zr [t-3, + (b2 - a2)t] dt,

4b J (b-a)2

r bIa)_

U = 1 It3h - (b2 - a2)t-3h]

2b2

1 b2-a2 b^2 -al

(10) -1b-aIJ