Calculus: Analytic Geometry and Calculus, with Vectors

690 Iterated and multiple integrals

It is particularly easy to evaluate this integral for the special case in
which there is a constant R for which R when a < ¢ <_ R. In
this case, S is a circular sector and
3
(13.565) JBI =

2

3 [cos a - cos S].

In case a = 0 and # = n, the solid body B is a complete spherical ball of

radius R and the right side reduces to the correct volume jrR3.

Example 2 Let a < 0 z-= a + 21r, let f(O) > 0 when a <_ 0 < 0, let f

be Riemann integrable over the interval a S 0 <_ i3, let S be the plane

set of points having polar coordinates (p,o) for which a S 4) S i and

0 < p < f (o), and let a lamina (or flat plate) cover S and have areal

density (mass per unit area) 5(p,¢) at the point (p,¢) of S. The problem

is to find the pth moment of the lamina about the y axis shown in Figure

0=0

Figure 13.57

13.57. As before, we use p £p L¢ as an approximation to the area of a

subset of S. The next step is to use

b(P,o)p Ep AO

as an approximation to the mass of a part of the lamina. Multiplying

this by xp or (p cos ¢)p, the pth power of the x coordinate of a point in

the part, gives an approximation to the pth moment about the line x = 0

of the part of the lamina. This leads to the formula

Mao = lim T"(p cos o)p&(p,¢)p AP z4

for the required pth moment. The right member is a double integral,

and expressing it as an iterated integral gives

cosp ¢ do

AO)

fo S(P,o)Pp+1 dp.

Example 3 We now require an integral for the polar moment of

inertia M`'of the lamina of Example 2 about the line L through the

origin perpendicular to the plane of the lamina. We use p LIp L4 as

an approximation to the area of a subset of the lamina and then use

S(p,c)p Lip 0o as an approximation to the mass of the subset. Multi-

plying this by p2, the square of the distance from the line L to the subset,