Calculus: Analytic Geometry and Calculus, with Vectors

692 Iterated and multiple integrals

appearing among the problems of Section 8.4, and the fact that (-i)! _ /jr-
we can put the result in the form

IS =

(p-11!

4ap+4 w/2 2 / 7r a9t4 ``^2 )

p -}- 4 Jo sin' q5 dp = p + 4

02/!

which is valid even when p is not an integer.

5 A circular disk has radius a and constant density 6. Find its moment of

inertia about each of the following:

(a) A diameter (line, not number) fins.:a4S

(b) A tangent in its plane Ins :a4S

6 Find the volume of the solid S obtained by rotating, about the x axis,

the region in the first quadrant bounded by the x axis and the graph of the polar

equation p = a cos q6. Outline of solution: The volume ISI of S is approximated

by the sum of "elementary" rings a sample one of which has cross-sectional area

Pk AN Aq'k and length27rpk sin ¢k This leads to the formula

ISI = 2ir Joy /2 sin q5 do Ioa

eos 0

P2 dp.

AIns.: 4iras/15.

7 Supposing that the solid S of Problem 6 has constant density S, find the

gravitational force F which it exerts upon a particle m* of mass m concentrated

at the origin. Outline of solution: The sample elementary ring of the preceding

problem has mass Mk, where

Mk = 27r& sin 4kpk AOk APk

The force AFk which this ring exerts upon m* is the same as the i component of

the force exerted upon m* by a single particle of the same mass Mk concentrated

at the point having polar coordinates (pk,lk). Therefore,

AFk _ iGmMkcos

Pk

OA;.

This leads to the formula

F = 2irGm&i fr12 sin 0 cos 0 r0a cos m dp

The answer is F = --raGmSi. Remark: We embark on a little excursion to see

that the solid S of this problem and the preceding one is a most remarkable solid.

If a particle P of mass M is located at the point in our plane having polar coordi-

nates (p,o), or at a point which is obtained by rotating it part of the way around

the x axis, then the scalar component F. in the direction of thex axis which it

exerts upon m* is

F. =

GmM

Mcos.

If P lies inside our solid S, then p < a _\/cos 0,so p2 < a2 cos ¢, so (cos cb)/p2 >

1/a2, so F. > GmM/a2. If P lies outside our solid S, then F. S 0 if 7r/2 _< jq5I <