Page 233 - Schaum's Outline of Theory and Problems of Advanced Calculus

224 MULTIPLE INTEGRALS [CHAP. 9

The result can be expressed in terms of the mass M of the region, since by Problem 9.17,
6
4 a a 6 2M 2 2
a so that ¼ Ma
M ¼ volume density ¼ I z ¼ ¼ 4
2 3 3 a 3
Note that in setting up the integral for I z we can think of dz d d as being the mass of the
2
cubical volume element, dz d d ,asthe moment of inertia of this mass with respect to the z-axis
ðð ð
2
and dz d d as the total moment of inertia about the z-axis. The limits of integration are
r
determined as in Problem 9.17.
2
2
2
2 2
2
p
(b) The radius of gyration is the value K such that MK ¼ Ma , i.e., K ¼ a or K ¼ a 2=3.
ffiffiffiffiffiffiffiffi
3 3
The physical significance of K is that if all the mass M were concentrated in a thin cylindrical shell
of radius K,then the moment of inertia of this shell about the axis of the cylinder would be I z .
9.19. (a) Find the volume of the region
bounded above by the sphere
2
2
2
x þ y þ z ¼ a 2 and below by the
2
2
2
2
2
cone z sin ¼ðx þ y Þ cos , where
is a constant such that 0 @ @ .
(b)From the result in (a), find the
volume of a sphere of radius a.
In spherical coordinates the equation
of the sphere is r ¼ a and that of the
cone is ¼ . This can be seen directly
or by using the transformation equations
x ¼ r sin cos ; y ¼ r sin sin , z ¼ r cos .
2
2
2
2
2
For example, z sin ¼ðx þ y Þ cos
becomes, on using these equations,
2 2 2
r cos sin ¼
2
2
2
2
2
2
2
ðr sin cos þ r sin sin Þ cos Fig. 9-19
2
2
2
2
2
2
i.e., r cos sin ¼ r sin cos
from which tan ¼ tan and so ¼ or ¼ .Itissufficient to consider one of these, say, ¼ .
Required volume ¼ 4 times volume (shaded) in Fig. 9-19
ðaÞ
=2 a
ð ð ð
2
¼ 4 r sin dr d d
¼0 ¼0 r¼0
ð =2 ð r 3
¼ 4 sin d d
¼0 ¼0 3 r¼0
4a 3 ð =2 ð
sin d d
3 ¼0 ¼0
¼
4a 3 ð =2
cos d
¼
3 ¼0 ¼0
2 a 3
3
¼ ð1 cos Þ
The integration with respect to r (keeping and constant) from r ¼ 0to r ¼ a corresponds to
summing the volumes of all cubical elements (such as indicated by dV)inacolumn extending from
r ¼ 0to r ¼ a. The subsequent integration with respect to (keeping constant) from ¼ 0to ¼ =4
corresponds to summing the volumes of all columns in the wedge-shaped region. Finally, integration
with respect to corresponds to adding volumes of all such wedge-shaped regions.