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(We have used s to represent the squared distance from the origin to avoid con-
fusion with the radial distance r in cylindrical coordinates.) The quantity
2
2
s
z
is equal to the squared distance from the z axis, which in cylindrical coordinates
is simply
r . The differential volume dV in cylindrical coordinates is given by
dV
=
rdrdθdz
,
(14.61)
so Equation (14.60) becomes
h π R
22

3
=
ρ
rdrdθdz
33
h
20 0
4
(14.62)
=
1
2
πρhR
.
2
The volume of the cylinder is given by
V πhR
=
, so we can write the moment of
inertia as
2
2
=
1
ρVR
=
1
mR
,
(14.63)
33
2
2
where m ρV
is the mass of the cylinder. Since a cylinder is symmetric about the
z axis, we must have 11
=
  . We can calculate the moment of inertia about the
x axis by evaluating the integral
=
22
(
)
2
2
=
sxρdV
.
(14.64)
11
V
2
2
2
2
2
2
Making the substitutions
s
=+
rz
and
x
=
r
cos
θ
, we have
h π R
22
(
)

2
2
2
2
=
ρ
rzr θ rdrdθdz
+
cos
11
h
h π R
20 0
22
h π R
22

3
2

2
=
ρ
r θdr dθdz ρ
sin
+
zrdrdθdz
.
(14.65)
h
20 0
h
20 0
Evaluating the integrals for the variables r and z in the first term, and evaluating
all three integrals in the second term gives us
2
π
=
ρhR
4
sin
2
θdθ hR
+
3
2
.
(14.66)
1
1
11
4
12
0
Using the trigonometric identity
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