Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 10
To be solved during the tutorial session Thursday/Friday, 31/32 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 surface x2 + y 2 + z 2 = a2 .
(a) What is the surface x2 + y 2 + z 2 = a2 ?
1
(b) Find the volume V of the solid G.
(c) Find the moments of inertia of the solid G.
(d) Find the gravitational force F(ξ, η, ζ) exerted on a point particle located at (ξ, η, ζ)
by the solid G.
(e) Find the gravitational potential field U (ξ, η, ζ) of the solid G.
(f) Plot Fz (0, 0, ζ) and U (0, 0, ζ) for a = 1.
Solution:
(a) It is a sphere of radius a centred at the origin, and the solid is a ball of radius a.
(b) We use spherical coordinates x = r cos θ sin φ, y = r sin θ sin φ, z = r cos φ.
The volume V of G is
ZZZ
Z
V =
2π
π
Z
Z
dV =
G
0
0
0
a
r2 sin φdrdφdθ =
4πa3
.
3
(4)
(c) Due to the rotational symmetry Ix = Iy = Iz = 13 (Ix + Iy + Iz ), so
ZZZ
Z Z Z
2 2π π a 4
8πa5
2
2
2
2
2
2
(x + y + z ) dV =
r sin φdrdφdθ =
= V a2 = M a2 ,
Ix =
3
3 0
15
5
5
G
0
0
(5)
where V is equal to the mass of the ball because we set δ(x, y, z) = 1.
(d) Due to the rotational symmetry the force F points towards the origin, and therefore
has the form
p
ρ~
(6)
ρ| = ξ 2 + η 2 + ζ 2 .
F(ξ, η, ζ) = −f (ρ) , ρ~ = ξ i + η j + ζ k , ρ = |~
ρ
Thus to compute f (ρ) it is sufficient to choose (ξ, η, ζ) = (0, 0, ρ), ρ ≥ 0, and compute
Fz (0, 0, ρ)
ZZZ
z−ρ
dV
f (ρ) = −Fz (0, 0, ρ) = −
2
2
2 3/2
G (x + y + (z − ρ) )
Z 2π Z a Z π
r cos φ − ρ
=−
r2 sin φdφdrdθ
2 − 2ρr cos φ + ρ2 )3/2
(r
0
0
0
Z aZ 1
−rt − ρ
= −2π
r2 dtdr
2
2
3/2
0
−1 (r + 2ρrt + ρ )
Z aZ 1
Z Z
π a 1
π
r2
ρ2 − r 2
p
=
dtdr +
r2 dtdr
ρ 0 −1 r2 + 2ρrt + ρ2
ρ 0 −1 (r2 + 2ρrt + ρ2 )3/2
Z
Z
(7)
π a p 2
π a
ρ2 − r 2
1
2
p
= 2
r r + 2ρrt + ρ |−1 dr − 2
|1−1 rdr
ρ 0
ρ 0
r2 + 2ρrt + ρ2
Z a
Z a
π
π
ρ2 − r 2
= 2
(r(r + ρ) − r|r − ρ|)dr − 2
(ρ − r −
)rdr
ρ 0
ρ 0
|r − ρ|
Z Z
π a
2r2
2π a
2
= 2
2r −
|r − ρ| dr = 2
(1 − sign(r − ρ)) r2 dr
ρ 0
r−ρ
ρ 0
(
4πa3
=M
if ρ ≥ a
3ρ2
ρ2
=
,
Mρ
4πρ
= ρ2 if ρ ≤ a
3
2
3
where Mρ = 4πρ
is the mass of a ball of radius ρ and density 1. Thus if the particle is
3
outside of the ball then the gravitational force due to the sphere coincides with the force
of a point particle of the same mass, while if it is inside the ball only the part of the ball
of radius ρ acts on it. The outer shell makes no contribution.
(e) Due to the rotational symmetry the gravitational potential field depends only on the
distance from the particle to the centre of the ball
U (ξ, η, ζ) = U (ρ) .
(8)
Thus to compute U it is sufficient to choose (ξ, η, ζ) = (0, 0, ρ), ρ ≥ 0
ZZZ
1
p
dV
U (ρ) = −
x2 + y 2 + (z − ρ)2
G
Z 2π Z a Z π
1
p
=−
r2 sin φdφdrdθ
2
2
r − 2ρr cos φ + ρ
0
0
0
Z aZ 1
Z
2π a p 2
1
2
p
= −2π
r dtdr = −
r r + 2ρrt + ρ2 |1−1 dr
2
2
ρ
r + 2ρrt + ρ
0
−1
0
Z
2π a
=−
(r(r + ρ) − r|r − ρ|)dr
ρ 0
Ra 2
− 4π
r dr
if ρ ≥ a
R ρ 0
Ra
=
4π ρ 2
− ρ 0 r dr − 4π ρ rdr if ρ ≤ a
(
3
= − Mρ
if ρ ≥ a
− 4πa
3ρ
=
.
2
4πρ2
2πρ
− 3 − 2πa2 + 2πρ2 = 3 − 2πa2 if ρ ≤ a
(9)
Thus if the particle is outside of the ball then the gravitational potential field due to the
ball coincides with the force of a point particle of the same mass, while if it is inside the
ball then the potential is a sum of a term which is independent of a and a term which is
independent of ρ.
(f) The plots are below
1
2
3
4
1
2
3
4
-1
-1
-2
-3
-2
-4
-3
-5
-6
-4
2. Consider the spherical shell, i.e. the solid G: a21 ≤ x2 + y 2 + z 2 ≤ a22 . Use the results
obtained for a sphere, and express your answers in terms of the shell’s mass.
3
(a) Find the volume V of the solid G.
(b) Find the moments of inertia of the solid G.
(c) Find the gravitational force F(ξ, η, ζ) exerted on a point particle located at (ξ, η, ζ)
by the solid G.
(d) Find the gravitational potential field U (ξ, η, ζ) of the solid G.
(e) Plot Fz (0, 0, ζ) and U (0, 0, ζ) for a1 = 1 and a2 = 2.
Solution:
(a) The volume V of G is
4π(a32 − a31 )
.
3
(10)
8π(a52 − a51 )
2 a5 − a51
2 a52 − a51
= V 23
=
M
,
15
5 a2 − a31
5 a32 − a31
(11)
V = V2 − V1 =
(b)
Ix = Iy = Iz =
where V is equal to the mass of the shell because we set δ(x, y, z) = 1.
(c) Due to the rotational symmetry the force F points towards the origin
ρ~
F(ξ, η, ζ) = −f (ρ) ,
ρ
ρ~ = ξ i + η j + ζ k ,
and
f (ρ) = −Fz (0, 0, ρ) =





4π(a32 −a31 )
=M
3ρ2
ρ2
4πa31
4πρ
ρ
− 3ρ2 = M
3
ρ2
4πρ
− 4πρ
=0
3
3
ρ = |~
ρ| =
if
p
ξ 2 + η2 + ζ 2 ,
(12)
ρ ≥ a2
if a1 ≤ ρ ≤ a2 ,
if
ρ ≤ a1
(13)
4π(ρ3 −a3 )
1
is the mass of a shell of radii a1 and ρ and density 1. Thus if the
where Mρ =
3
particle is outside of the shell then the gravitational force due to the shell coincides with
the force of a point particle of the same mass, while if it is inside the shell only the part
of the shell of radius ρ acts on it. Finally if it is inside the cavity the force is 0.
(d) Due to the rotational symmetry the gravitational potential field depends only on the
distance from the particle to the centre of the shell
U (ξ, η, ζ) = U (ρ) .
and

4π(a32 −a31 )

= − Mρ if
ρ ≥ a2
 − 3ρ
3
2
4πa
2πρ
2
1
U (ρ) =
− 2πa2 + 3ρ if a1 ≤ ρ ≤ a2 .

 3
2πa21 − 2πa22
if
ρ ≤ a1
(14)
(15)
Thus if the particle is outside of the shell then the gravitational potential field due to the
shell coincides with the force of a point particle of the same mass, while if it is inside the
cavity then the potential is a constant.
(e) The plots are below
4
2
4
6
8
2
-2
-5
-4
-10
-6
-15
4
6
8
3. Consider the cylinder G bounded by the planes z = 0, z = h and by the surface
x 2/3 y 2/3
+
= 1.
a
b
(a) Sketch the solid G and its projection onto the xy-plane.
(b) G can be expressed parametrically by a generalisation of cylindrical coordinates
x = a r cos3 θ , y = b r sin3 θ , z = u ,
0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2π , 0 ≤ u ≤ h .
(16)
Think about (16) as a change go variable. Find the Jacobian of the change.
(c) Find the volume V of the solid G.
(d) Find the moments of inertia of the solid G (you may use Mathematica).
Show the details of your work.
Solution:
(a) The solid and its projection R onto the xy-plane are shown below (a = b = 1)
1.0
0.5
-1.0
0.5
-0.5
1.0
-0.5
-1.0
(b) The Jacobian is
J=
∂(x, y, z)
= 3 a b r sin2 θ cos2 θ .
∂(r, θ, u)
5
(17)
(c) We use the coordinates r, φ, u. The volume V of G is
ZZZ
Z
2π
Z
h
Z
1
dV =
3 a b r sin2 θ cos2 θdrdudθ
G
0
0
Z 2π 0
Z 2π
3
1 2
3
2
2
sin θ cos θdθ = a b h
sin 2θdθ
= abh
2
2
4
0
0
3π
=
abh.
8
V =
(d) We have
ZZZ
1
1
Ix =
(y 2 + z 2 ) dV =
πabh 21b2 + 64h2 =
21b2 + 64h2 M ,
512
192
Z Z ZG
1
1
Iy =
(x2 + z 2 ) dV =
πabh 21a2 + 64h2 =
21a2 + 64h2 M ,
512
192
Z Z ZG
7 2
21
πabh a2 + b2 =
a + b2 M .
Iz =
(y 2 + x2 ) dV =
512
64
G
where the mass M of G is equal to V because we set δ(x, y, z) = 1.
6
(18)
(19)