Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 9
To be solved during the tutorial session Thursday/Friday, 24/25 March 2016
You may use Mathematica to sketch the integration regions and solids, and to check the
results of integration.
Moment of inertia: The tendency of a solid to resist a change in rotational motion about
an axis is measured by its moment of inertia about that axis. If the solid occupies a region G
in an xyz-coordinate system, and if its density function δ(x, y, z) is continuous on G, then the
moments of inertia about the x-axis, the y-axis, and the z-axis are denoted by Ix , Iy , and Iz ,
respectively, and are defined by
ZZZ
(y 2 + z 2 )δ(x, y, z)dV ,
Ix =
Z Z ZG
(x2 + z 2 )δ(x, y, z)dV ,
Iy =
(1)
G
ZZZ
Iz =
(x2 + y 2 )δ(x, y, z)dV .
G
Newton’s law of gravitation: Let a solid occupy a region G in an xyz-coordinate system,
and let its density function δ(x, y, z) be continuous on G. Then the gravitational force F =
Fx i + Fy j + Fz k exerted by the solid on a point particle of mass m located at (ξ, η, ζ) is given
by
ZZZ
x−ξ
Fx (ξ, η, ζ) = Gm
δ(x, y, z)dV ,
r3
Z Z ZG
y−η
δ(x, y, z)dV ,
Fy (ξ, η, ζ) = Gm
r3
(2)
Z Z ZG
z−ζ
Fz (ξ, η, ζ) = Gm
δ(x, y, z)dV ,
3
G r
p
r = (x − ξ)2 + (y − η)2 + (z − ζ)2 ,
where G is the gravitational constant.
The force can be obtained from the gravitational potential field U (ξ, η, ζ) as follows
ZZZ
1
U (ξ, η, ζ) = −G
δ(x, y, z)dV ,
G r
∂U (ξ, η, ζ)
∂U (ξ, η, ζ)
∂U (ξ, η, ζ)
Fx (ξ, η, ζ) = −m
, Fy (ξ, η, ζ) = −m
, Fz (ξ, η, ζ) = −m
.
∂ξ
∂η
∂ζ
(3)
In what follows we set m = 1, G = 1, and consider homogeneous solids with δ(x, y, z) = 1.
1. Consider the solid G bounded by the surfaces z = 0, z = h and and x2 + y 2 = a2 .
(a) What are the surfaces z = 0 and z = h?
1
(b) What is the surface x2 + y 2 = a2 ?
(c) Sketch the solid G.
(d) Sketch the projection of the solid G onto the xy-plane.
(e) Find the volume V of the solid G.
(f) Find the centroid of the solid G.
(g) Find the moments of inertia of the solid G.
(h) Find the gravitational force exerted on a point particle by the solid G if the point
particle is located at the origin (0, 0, 0). Find its limit as a → ∞ with h kept fixed,
and its limit as h → ∞ with a kept fixed.
Show the details of your work.
Solution:
(a) They are horizontal planes through (0, 0, 0) and (0, 0, h).
(b) It is a circular cylinder
(c),(d) The solid and its projection R onto the xy-plane are shown below
1.0
0.5
0.0
-0.5
-1.0
-1.0
0.0
-0.5
0.5
1.0
(e) The volume V of G is
ZZZ
ZZ Z
V =
dV =
G
h
Z
2π
Z
dzdA =
R
0
a
hrdrdθ = πa2 h .
(4)
0
0
(f) The x and y coordinates of the centroid are 0 by the symmetry, and the z coordinate
is
ZZZ
Z 2π Z a
1
1
π h2 a2
1
(5)
zc =
z dV =
h2 rdrdθ =
= h.
V
2V 0
V 2
2
G
0
(g) We first find Iz
ZZZ
2
Iz =
2
2π
Z
Z
a
Z
(x + y ) dV =
G
h
2
Z
r dzrdrdθ = 2π
0
ha4
1
1
= 2π
= V a2 = M a2 ,
4
2
2
2
0
0
0
a
r3 hdr
(6)
where V is equal to the mass of the cone because we set δ(x, y, z) = 1. The Ix = Iy =
1
(I + Iy ) by the rotational symmetry, so
2 x
Z 2π Z a Z h
1
1 2
2
2
z 2 dzrdrdθ
Ix =
( (x + y ) + z ) dV = Iz +
2
2
0
0
0
G
Z a
1 3
1
1
2π a2 h3
a2 h2
= Iz + 2π
h rdr = Iz +
= M( + ) .
2
2
3 2
4
3
0 3
ZZZ
(h) Obviously the only nonvanishing component is Fz
ZZZ
Z 2π Z a Z h
z
z
Fz (0, 0, 0) =
dV
=
dzrdrdθ
2
2
2 3/2
2
2 3/2
0
0 (r + z )
G (x + y + z )
0
Z a
Z a
r
1
1
1− √
−√
rdr = 2π
dr
= 2π
r
r2 + h2
r2 + h2
0
0
√
√
= 2πa − 2π
a2 + h2 − h = 2π(a + h − a2 + h2 ) .
(7)
(8)
It is interesting that the result is symmetric under a ↔ h. The limits are easily found by
using the Taylor expansion
r
h2
h2
Fz (0, 0, 0) = 2π(a + h − a 1 + 2 ) ≈ 2π(a + h − a(1 + 2 )) → 2πh as a → ∞ . (9)
a
2a
Similarly, Fz (0, 0, 0) → 2πa as h → ∞. The force remains finite even though the solid
becomes infinitely heavy.
2. Consider the solid G bounded above by the surface
15 p 2
1
− x + y 2 and below by the surface z = (x2 + y 2 ) .
2
2
p
(a) What is the surface z = 15
− x2 + y 2 ?
2
z=
(b) What is the surface z = 21 (x2 + y 2 )?
(c) Sketch the solid G.
(d) Sketch the projection of the solid G onto the xy-plane.
(e) Find the volume V of the solid G.
(f) Find the mass M of the solid G if its density is
√
2
2
e x +y − 1
p
δ(x, y, z) =
.
x2 + y 2 + 5 x2 + y 2
(g) What is the density of the solid G at the origin: δ(0, 0, 0) =?
Show the details of your work.
Solution:
(a) It is a cone
3
(b) It is a paraboloid
(c),(d) The solid and its projection R onto the xy-plane are shown below
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
R is a circle of radius 3 as one can find by solving the equation
3
15
2
− r − 12 r2 = 0.
(e) The volume V of G is
Z 2π Z 3
ZZZ
15
1
135
81
117
( − r − r2 )rdrdθ = 2π(
dV =
−9− )=
π.
V =
2
2
4
8
4
0
0
G
(f) The mass M of G is
Z 2π Z 3 r
ZZZ
e − 1 15
1
δ(x, y, z)dV =
( − r − r2 )rdrdθ
M=
2
r + 5r 2
2
0
0
Z 3G
17
(er − 1)(3 − r)dr = e3 −
=π
π ≈ 36.397 .
2
0
(g) δ(0, 0, 0) = 1/5
4
(10)
(11)