Exercises

6.5 EXERCISES
and similarly for Jyz and Jxz . On taking the determinant of (6.14), we therefore
obtain
Jxz = Jxy Jyz
or, in the usual notation,
∂(x1 , . . . , xn )
∂(x1 , . . . , xn ) ∂(y1 , . . . , yn )
=
.
∂(z1 , . . . , zn )
∂(y1 , . . . , yn ) ∂(z1 , . . . , zn )
(6.16)
As a special case, if the set zi is taken to be identical to the set xi , and the
obvious result Jxx = 1 is used, we obtain
Jxy Jyx = 1
or, in the usual notation,
−1
∂(y1 , . . . , yn )
∂(x1 , . . . , xn )
=
.
∂(y1 , . . . , yn )
∂(x1 , . . . , xn )
(6.17)
The similarity between the properties of Jacobians and those of derivatives is
apparent, and to some extent is suggested by the notation. We further note from
(6.15) that since |A| = |AT |, where AT is the transpose of A, we can interchange the
rows and columns in the determinantal form of the Jacobian without changing
its value.
6.5 Exercises
6.1
6.2
6.3
6.4
6.5
Identify the curved wedge bounded by the surfaces y 2 = 4ax, x + z = a and
z = 0, and hence calculate its volume V .
Evaluate the volume integral of x2 + y 2 + z 2 over the rectangular parallelepiped
bounded by the six surfaces x = ±a, y = ±b and z = ±c.
Find the volume integral of x2 y over the tetrahedral volume bounded by the
planes x = 0, y = 0, z = 0, and x + y + z = 1.
Evaluate the surface integral of f(x, y) over the rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b
for the functions
x
(a) f(x, y) = 2
,
(b) f(x, y) = (b − y + x)−3/2 .
x + y2
Calculate the volume of an ellipsoid as follows:
(a) Prove that the area of the ellipse
y2
x2
+ 2 =1
a2
b
is πab.
(b) Use this result to obtain an expression for the volume of a slice of thickness
dz of the ellipsoid
y2
z2
x2
+ 2 + 2 = 1.
a2
b
c
Hence show that the volume of the ellipsoid is 4πabc/3.
207
MULTIPLE INTEGRALS
6.6
6.7
The function
Zr
e−Z r/2a
Ψ(r) = A 2 −
a
gives the form of the quantum-mechanical wavefunction representing the electron
in a hydrogen-like atom of atomic number Z, when the electron is in its first
allowed spherically symmetric excited state. Here r is the usual spherical polar
coordinate, but, because of the spherical symmetry, the coordinates θ and φ do
not appear explicitly in Ψ. Determine the value that A (assumed real) must have
if the wavefunction is to be correctly normalised, i.e. if the volume integral of
|Ψ|2 over all space is to be equal to unity.
In quantum mechanics the electron in a hydrogen atom in some particular state
is described by a wavefunction Ψ, which is such that |Ψ|2 dV is the probability of
finding the electron in the infinitesimal volume dV . In spherical polar coordinates
Ψ = Ψ(r, θ, φ) and dV = r2 sin θ dr dθ dφ. Two such states are described by
1/2 3/2
1
1
2e−r/a0 ,
Ψ1 =
4π
a0
Ψ2 = −
3
8π
1/2
sin θ eiφ
1
2a0
3/2
re−r/2a0
√ .
a0 3
(a) Show that each Ψi is normalised, i.e. the integral over all space |Ψ|2 dV is
equal to unity – physically, this means that the electron must be somewhere.
(b) The (so-called) dipole matrix element between the states 1 and 2 is given by
the integral
px = Ψ∗1 qr sin θ cos φ Ψ2 dV ,
where q is the charge on the electron. Prove that px has the value −27 qa0 /35 .
6.8
6.9
A planar figure is formed from uniform wire and consists of two equal semicircular arcs, each with its own closing diameter, joined so as to form a letter ‘B’. The
figure is freely suspended from its top left-hand corner. Show that the straight
edge of the figure makes an angle θ with the vertical given by tan θ = (2 + π)−1 .
A certain torus has a circular vertical cross-section of radius a centred on a
horizontal circle of radius c (> a).
(a) Find the volume V and surface area A of the torus, and show that they can
be written as
V =
π2 2
(r − ri2 )(ro − ri ),
4 o
A = π 2 (ro2 − ri2 ),
where ro and ri are, respectively, the outer and inner radii of the torus.
(b) Show that a vertical circular cylinder of radius c, coaxial with the torus,
divides A in the ratio
πc + 2a : πc − 2a.
6.10
A thin uniform circular disc has mass M and radius a.
(a) Prove that its moment of inertia about an axis perpendicular to its plane
and passing through its centre is 12 Ma2 .
(b) Prove that the moment of inertia of the same disc about a diameter is 14 Ma2 .
208
6.5 EXERCISES
This is an example of the general result for planar bodies that the moment of
inertia of the body about an axis perpendicular to the plane is equal to the sum
of the moments of inertia about two perpendicular axes lying in the plane; in an
obvious notation
Iz = r2 dm = (x2 + y 2 ) dm = x2 dm + y 2 dm = Iy + Ix .
6.11
In some applications in mechanics the moment of inertia of a body about a
single point (as opposed to about an axis) is needed. The moment of inertia, I,
about the origin of a uniform solid body of density ρ is given by the volume
integral
I = (x2 + y 2 + z 2 )ρ dV .
V
Show that the moment of inertia of a right circular cylinder of radius a, length
2b and mass M about its centre is
2
b2
a
M
.
+
2
3
6.12
The shape of an axially symmetric hard-boiled egg, of uniform density ρ0 , is
given in spherical polar coordinates by r = a(2 − cos θ), where θ is measured
from the axis of symmetry.
(a) Prove that the mass M of the egg is M = 40
πρ0 a3 .
3
(b) Prove that the egg’s moment of inertia about its axis of symmetry is
6.13
6.14
6.15
6.16
In spherical polar coordinates r, θ, φ the element of volume for a body that
is symmetrical about the polar axis is dV = 2πr 2 sin θ dr dθ, whilst its element
of surface area is 2πr sin θ[(dr)2 + r2 (dθ)2 ]1/2 . A particular surface is defined by
r = 2a cos θ, where a is a constant and 0 ≤ θ ≤ π/2. Find its total surface area
and the volume it encloses, and hence identify the surface.
By expressing both the integrand and the surface element in spherical polar
coordinates, show that the surface integral
x2
dS
x2 + y 2
√
over the surface x2 + y 2 = z 2 , 0 ≤ z ≤ 1, has the value π/ 2.
By transforming to cylindrical polar coordinates, evaluate the integral
I=
ln(x2 + y 2 ) dx dy dz
over the interior of the conical region x2 + y 2 ≤ z 2 , 0 ≤ z ≤ 1.
Sketch the two families of curves
y 2 = 4u(u − x),
6.17
342
Ma2 .
175
y 2 = 4v(v + x),
where u and v are parameters.
By transforming to the uv-plane, evaluate the integral of y/(x2 + y 2 )1/2 over
the part of the quadrant x > 0, y > 0 that is bounded by the lines x = 0, y = 0
and the curve y 2 = 4a(a − x).
By making two successive simple changes of variables, evaluate
I=
x2 dx dy dz
209
MULTIPLE INTEGRALS
over the ellipsoidal region
y2
z2
x2
+ 2 + 2 ≤ 1.
2
a
b
c
6.18
6.19
6.20
6.21
Sketch the domain of integration for the integral
1 1/y 3
y
I=
exp[y 2 (x2 + x−2 )] dx dy
0
x=y x
and characterise its boundaries in terms of new variables u = xy and v = y/x.
Show that the Jacobian for the change from (x, y) to (u, v) is equal to (2v)−1 , and
hence evaluate I.
Sketch the part of the region 0 ≤ x, 0 ≤ y ≤ π/2 that is bounded by the curves
x = 0, y = 0, sinh x cos y = 1 and cosh x sin y = 1. By making a suitable change
of variables, evaluate the integral
I=
(sinh2 x + cos2 y) sinh 2x sin 2y dx dy
over the bounded subregion.
Define a coordinate system u, v whose origin coincides with that of the usual
x, y system and whose u-axis coincides with the x-axis,
whilst the v-axis makes
an angle α with it. By considering the integral I = exp(−r 2 ) dA, where r is the
radial distance from the origin, over the area defined by 0 ≤ u < ∞, 0 ≤ v < ∞,
prove that
∞ ∞
α
exp(−u2 − v 2 − 2uv cos α) du dv =
.
2 sin α
0
0
As stated in section 5.11, the first law of thermodynamics can be expressed as
dU = T dS − P dV .
2
By calculating and equating ∂ U/∂Y ∂X and ∂2 U/∂X∂Y , where X and Y are an
unspecified pair of variables (drawn from P , V , T and S), prove that
∂(S, T )
∂(V , P )
=
.
∂(X, Y )
∂(X, Y )
Using the properties of Jacobians, deduce that
∂(S, T )
= 1.
∂(V , P )
6.22
The distances of the variable point P , which has coordinates x, y, z, from the fixed
points (0, 0, 1) and (0, 0, −1) are denoted by u and v respectively. New variables
ξ, η, φ are defined by
ξ = 12 (u + v),
η = 12 (u − v),
and φ is the angle between the plane y = 0 and the plane containing the three
points. Prove that the Jacobian ∂(ξ, η, φ)/∂(x, y, z) has the value (ξ 2 − η 2 )−1 and
that
(u − v)2
16π
u+v
dx dy dz =
exp −
.
uv
2
3e
all space
6.23
This is a more difficult question about ‘volumes’ in an increasing number of
dimensions.
210