Module MA1132 (Frolov), Advanced Calculus
Homework Sheet 9
Due: at the beginning of 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
δ(x, y, z)dV ,
Fx (ξ, η, ζ) = Gm
r3
G
ZZZ
y−η
Fy (ξ, η, ζ) = Gm
δ(x, y, z)dV ,
r3
(2)
G
ZZZ
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. Considerpthe solid G bounded above by the surface z = h and below by the surface
az 2 = h2 x2 + y 2 .
1
(a) What is the surface z = h?
(b) Describe the surface az 2 = h2
p
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 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) It is a horizontal plane through (0, 0, h)
(b) It is a surface of revolution obtained by rotating the parabola az 2 = h2 x in the
xz-plane about the
√ z-axes. In polar coordinates which we’ll be using it is given by the
equation z = √ha r.
(c),(d) The solid and its projection R onto the xy-plane are shown below. R is a circle of
radius a.
(e) The volume V of G is
ZZZ
ZZ Z
V =
dV =
G
R
h
h √
√
r
a
Z
2π
Z
dzdA =
0
0
a
h √
ha2 2ha2
1
(h − √ r)rdrdθ = 2π(
−
) = πa2 h .
2
5
5
a
(4)
(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
h2
π h2 a2 h2 a2
zc =
z dV =
(h2 − r)rdrdθ = (
−
)
V
2V 0
a
V 2
3
G
0
(5)
πa2 h2
5
=
= h.
6V
6
2
(g) We first find Iz
ZZZ
2
Iz =
2π
Z
2
Z
a
Z
(x + y ) dV =
G
0
2
h √
√
r
a
0
a
Z
r dzrdrdθ = 2π
0
h √
r2 (h − √ r)rdr
a
(6)
4
4
= 2π(
h
2ha
π
5
5
ha
−
) = ha4 = V a2 = M a2 ,
4
9
18
18
18
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
Ix =
( (x + y ) + z ) dV = Iz +
z 2 dzrdrdθ
√
h
2
√
0
0
G 2
r
a
Z a
1
1 3
h3
1
2π a2 h3 2a2 h3
1
π
= Iz + 2π
(h − 3/2 r3/2 )rdr = Iz +
(
−
) = Iz + a2 h3 (7)
2
a
2
3
2
7
2
7
0 3
2
2
h
a
= 5M ( + ) .
36
7
ZZZ
(h) Obviously the only nonvanishing component is Fz
Z 2π Z a Z h
z
z
Fz (0, 0, 0) =
dV
=
dzrdrdθ
2
2
2 3/2
h √ (r 2 + z 2 )3/2
√
G (x + y + z )
0
0
r
a


Z a
Z a
Z as
1
r
1
r
 rdr = 2π
q
√
−√
dr
−
2π
= 2π
dr .
2
h
2 + h2
2 + h2
h2
r
+
r
r
2
0
0
0
a
r + r
ZZZ
a
(8)
The first integral is computed by first making the substitution
s
a
r
√
=
t
,
r
=
0
→
t
=
0
,
r
=
a
→
t
=
0
a
2
r + ha
a2 + h2
h2 t2
,
r=
a 1 − t2
(9)
h2
2t
h2
1
dr =
dt
=
,
d
2
2
a (1 − t )
a 1 − t2
so that
Z
2π
0
a
s
r
h2
dr
=
2π
2
a
r + ha
Z
0
ta
2t
h2
t
dt
=
2π
(1 − t2 )2
a
Z
ta
td
0
1
.
1 − t2
Then one uses the integration by parts and gets
Z
Z
h2 ta
1
h2
1 ta
h2 ta 1
td
= 2π t
dt
2π
− 2π
a 0
1 − t2
a 1 − t2 0
a 0 1 − t2
Z
√
h2 ta 1
2
2
= 2π a + h − 2π
dt .
a 0 1 − t2
3
(10)
(11)
Finally one uses the partial fraction decomposition and gets
Z
Z 1
1
h2 1 + t ta
h2 ta 1
h2 ta
+
dt
=
π
ln
2π
dt
=
π
a 0 1 − t2
a 0
1+t 1−t
a
1−t 0
√
√
h2
a2 + h2 + a
h2
a2 + h2 + a
=π ln √
= 2π ln
a
a
h
a2 + h2 − a
Thus, the first integral is equal to
√
Z as
√
r
a2 + h2 + a
h2
2
2
ln
.
2π
a
+
h
−
2π
dr
=
2π
2
a
h
r + ha
0
The second integral in (8) gives
Z a
√
r
√
− 2π
dr = −2π
a2 + h2 − h .
r2 + h2
0
(12)
(13)
(14)
Summing up the two contribution one finds
h2
Fz (0, 0, 0) = 2πh − 2π ln
a
√
a2 + h2 + a
.
h
(15)
The limits are easily found by using the Taylor expansion
r
h2
a2 a
h2
a2
a
Fz (0, 0, 0) = 2πh − 2π ln( 1 + 2 + ) ≈ 2πh − 2π ln(1 + 2 + )
a
h
h
a
2h
h
(16)
2
2
2
3
3
2
h a
a
a
a
a
a
≈ 2πh − 2π ( + 2 − 2 − 3 + 3 ) = π
→ 0 as h → ∞ .
a h 2h
2h
2h
3h
3h
Interestingly, the force vanishes. Similarly, Fz (0, 0, 0) → 2πh as a → ∞. The force
remains finite even though the solid becomes infinitely heavy.
2. Consider the solid G bounded by the surfaces z = 0, z = h and x2 + y 2 = a2 .
(a) What is the solid?
(b) Find the gravitational force exerted on a point particle by the solid G if the point
particle is located on the z-axis at (0, 0, ζ) (ζ can be any real number)
(c) Plot its graph and the graph of ∂Fz /∂ζ for a = 1, h = 2 (use Mathematica).
(d) Find the limit of Fz as a → ∞ with h and ζ kept fixed.
(e) Find the limit of Fz as h → ∞ with a and ζ kept fixed.
Solution:
(a) It is a circular solid cylinder of radius a and height h with the centre located at the
origin and the axis of symmetry coinciding with the z-axis.
4
(b) Obviously the only nonvanishing component is Fz
ZZZ
Z 2π Z a Z h
z−ζ
z−ζ
Fz (0, 0, ζ) =
dV =
dzrdrdθ
2
2
2
3/2
2
2 3/2
G (x + y + (z − ζ) )
0
0
0 (r + (z − ζ) )
!
Z a
1
1
p
−p
= 2π
rdr
r2 + (h − ζ)2
r2 + ζ 2
0
p
p
a2 + ζ 2 − |ζ| − a2 + (h − ζ)2 + |h − ζ| .
= 2π
(17)
(c) The plots of Fz and ∂Fz /∂ζ are shown below (a = 1, h = 2)
(d) The limit a → ∞ is easily found by using the Taylor expansion

2πh
if
ζ≤0

2πh − 4πζ if 0 ≤ ζ ≤ h
Fz (0, 0, ζ) → 2π (−|ζ| + |h − ζ|) =

−2πh
if
ζ≥h
(18)
Thus in this limit the gravitational force is constant outside the solid, and changes linearly
with the distance to the boundary when the particle is inside it.
(e) The limit h → ∞ gives
p
2
2
a + ζ − |ζ| > 0 .
Fz (0, 0, ζ) → 2π
(19)
It is interesting that it is symmetric under ζ → −ζ. Can you give a simple explanation
to this result?
p
3. Consider the solid G bounded below by the surface z = rα , α > 0, r = x2 + y 2 and
above by the plane z = 1. Note that the surface z = r is a cone, and z = r2 is a
paraboloid.
(a) Sketch the surface z = rα for α = 1/2, α = 1, α = 2, and the projection of the solid
G onto the xy-plane.
(b) Find the volume V of the solid G, and its limit as α → ∞.
(c) Find the centroid of the solid G, and its limit as α → ∞.
(d) Find the moments of inertia of the solid G, and its limit as α → ∞.
5
(e) Use Mathematica to 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).
(f) Use Mathematica to plot Fz as a function of α, and find its limit as α → ∞.
(g) Explain the limiting values obtained in (b), (c), (d), (f).
Show the details of your work.
Solution:
(a) The surface z = rα and the projection R of the solid G onto the xy-plane are shown
below. R is a circle of radius 1.
(b) The volume V of G is
ZZ Z
ZZZ
dV =
V =
R
G
1
Z
2π
Z
Z
α
(1 − r )rdrdθ = 2π
dzdA =
rα
1
0
1
(r + rα+1 )dr
0
0
1
1
α
= 2π( −
)=π
,
2 α+2
α+2
lim V = π .
(20)
α→∞
(c) The x and y coordinates of the centroid are 0 by the symmetry, and the z coordinate
is
ZZZ
Z
Z
π 1
1
π 1
2α
zc =
z dV =
(1 − r )rdr =
(r − r2α+1 )dr
V
V
V
0
0
G
π 1
1
π α
2+α
(21)
= ( −
)=
=
,
V 2 2 + 2α
2V 1 + α
2 + 2α
1
lim zc = .
α→∞
2
(d) We first find Iz
ZZZ
Z 1
Z 1
2
2
α 3
Iz =
(x + y ) dV = 2π
(1 − r )r dr = 2π
(r3 − rα+3 )dr
G
0
1
1
α
= 2π( −
)=π
,
4 4+α
8 + 2α
π
lim Iz = .
α→∞
2
6
0
(22)
The Ix = Iy = 21 (Ix + Iy ) by the rotational symmetry, so
Z 1Z 1
1 2
1
2
2
z 2 dzrdr
( (x + y ) + z ) dV = Iz + 2π
Ix =
2
2
α
G
Z 1
Z0 1 r
1
1
1
2π
= Iz + 2π
(1 − r3α )rdr = Iz +
(r − r3α+1 )dr
2
2
3 0
0 3
1
2π 1
1
α
α
α(18 + 7α)
= Iz +
( −
)=π
+π
=π
,
2
3 2 2 + 3α
4(4 + α)
2 + 3α
4(4 + α)(2 + 3α)
7π
.
lim Ix =
α→∞
12
ZZZ
(23)
(e) Obviously the only nonvanishing component is Fz
ZZZ
Z 1Z 1
z
z
Fz (0, 0, 0) =
dV = 2π
dzrdr
2
2
2
3/2
2
2 3/2
G (x + y + z )
0
rα (r + z )
Z 1
Z 1
1
1
1
r
√
√
= 2π
−√
rdr = 2π
−√
dr
r2 + r2α
r2 + 1
r2α−2 + 1
r2 + 1
0
0

α−2
√
; 4−3α
;−1)
 2π − 2 F1 ( 21 , 2(α−1)
2−2α
− 2+1
if 0 ≤ α ≤ 1
α−2
=
,
√

1
1
1
2π 2 F1 2 , 2α−2 ; 1 + 2α−2 ; −1 − 2 + 1 if
α≥1
(24)
where 2 F1 is a hypergeometric function.
(f) The plot of Fz is shown below
The limit is
√ Fz (0, 0, 0) → 2π 2 − 2 ≈ 3.6806 as α → ∞ .
(25)
(g) In the limit α → ∞ the solid becomes a cylinder of radius 1 and of height 1. The
result in (f) agrees with (17) where one sets a = h = 1 and ζ = 0.
7