Hassani S. Mathematical Methods: For Students of Physics and Related Fields
Подождите немного. Документ загружается.
204 Planar and Spatial Vectors
The reader may verify that, not only A
−1
A = 1, but also AA
−1
= 1.
Equation (6.48) gives the components b
1
and b
2
of a new vector obtained
from an old vector with components x and y when the matrix A acts on the
latter. We want to see what conditions A must satisfy for it to transform
vectors in a basis into vectors of a new basis. Let B = {a
1
, a
2
} be the old
basis. The components of a
1
in B are x =1andy = 0; so by (6.48), a
1
,
the vector obtained from a
1
by the action of A, has components b
1
= a
11
and
b
2
= a
21
. The components of a
2
in B are x =0andy =1;soa
2
, the vector
obtained from a
2
by the action of A, has components c
1
= a
12
and c
2
= a
22
.
The vectors (b
1
,b
2
)and(c
1
,c
2
) form a basis if and only if they are linearly
independent, i.e.,
(b
1
,b
2
)=k(c
1
,c
2
)=(kc
1
,kc
2
) ⇒ b
1
= kc
1
,b
2
= kc
2
,
does not hold for any constant k.Thisisequivalenttosayingthat
b
1
c
1
=
b
2
c
2
or b
1
c
2
− b
2
c
1
=0.
Expressing the b’s and c’s in terms of a
ij
’s, we recognize the last relation as
a condition on the determinant of A. Using Theorem 6.3.1, we thus have
Box 6.3.1. A transformation (or a matrix) transforms a basis into an-
other basis if and only if it is invertible.
Let us now consider three equations in three unknowns:
a
11
x + a
12
y + a
13
z = b
1
,
a
21
x + a
22
y + a
23
z = b
2
, (6.53)
a
31
x + a
32
y + a
33
z = b
3
,
which can also be written in matrix form as
⎛
⎝
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
⎞
⎠
⎛
⎝
x
y
z
⎞
⎠
=
⎛
⎝
b
1
b
2
b
3
⎞
⎠
⇒ Ax = b. (6.54)
We eliminate z from the set of equations by multiplying the first equation
of (6.53) by a
23
and the second by a
13
and subtracting. This will give onefrom three
equations in three
unknowns to two
equations in two
unknowns, and
from the
determinant of a
2 × 2 matrix to
that of a 3 ×3
matrix
equation in x and y. Similarly, multiplying the first equation by a
33
and the
third by a
13
and subtracting gives another equation in x and y.Thesetwo
equations are
(a
11
a
23
− a
21
a
13
)
≡a
11
x +(a
12
a
23
− a
22
a
13
)
≡a
12
y = a
23
b
1
− a
13
b
2
≡b
1
,
(a
11
a
33
− a
31
a
13
)
≡a
21
x +(a
12
a
33
− a
32
a
13
)
≡a
22
y = a
33
b
1
− a
13
b
3
≡b
2
. (6.55)
6.3 Determinant 205
Thus, we have reduced the three equations in three unknowns to two equa-
tions in two unknowns. We know how to find the solution for this set of
equations. These solutions are given in Equations (6.49) and (6.51). In order
for this equation to have a solution, the determinant of the coefficients must
not vanish. Let us calculate this determinant:
a
11
a
22
− a
12
a
21
=(a
11
a
23
− a
21
a
13
)(a
12
a
33
− a
32
a
13
)
− (a
12
a
23
− a
22
a
13
)(a
11
a
33
− a
31
a
13
)
= a
11
a
23
a
12
a
33
− a
11
a
23
a
32
a
13
− a
21
a
13
a
12
a
33
+ a
21
a
13
a
32
a
13
− a
12
a
23
a
11
a
33
+ a
12
a
23
a
31
a
13
+ a
22
a
13
a
11
a
33
− a
22
a
13
a
31
a
13
= a
13
[a
11
(a
22
a
33
− a
23
a
32
) − a
12
(a
21
a
33
− a
31
a
23
)
+ a
13
(a
21
a
32
− a
22
a
31
)]
= a
13
a
11
det
a
22
a
23
a
32
a
33
− a
12
det
a
21
a
23
a
31
a
33
+ a
13
det
a
21
a
22
a
31
a
32
.
If the original set of equations is to have a solution, the expression in the
square brackets must not vanish. We call this expression the determinant
of the 3 × 3matrixA. We can give a cookbook recipe for calculating the
determinant; but first we need the following definition:
Box 6.3.2. The cofactor A
ij
of an element a
ij
of a matrix A is defined
as the product of (− 1)
i+j
(i.e., +1 if i + j is even and −1 if i + j is odd)
and the determinant of the smaller matrix (2 × 2,ifA is a 3 ×3 matrix)
obtained from A when its ith row and jth column are deleted.
The following recipe applies to any (square) matrix, not just to 3 × 3
matrices:
Box 6.3.3. The determinant of A is obtained by multiplying each ele-
ment of a row (or a column) by its cofactor and adding the products.
If det A = 0, then Equation (6.49) gives
x =
a
22
b
1
− a
12
b
2
a
13
det A
.
The numerator is
a
22
b
1
− a
12
b
2
=(a
12
a
33
− a
32
a
13
)(a
23
b
1
− a
13
b
2
)
− (a
12
a
23
− a
22
a
13
)(a
33
b
1
− a
13
b
3
)
= a
13
.
(a
22
a
33
− a
32
a
23
)
≡C
11
b
1
+(a
32
a
13
− a
12
a
33
)
≡C
12
b
2
+(a
12
a
23
− a
22
a
13
)
≡C
13
b
3
/
= a
13
(C
11
b
1
+ C
12
b
2
+ C
13
b
3
).
206 Planar and Spatial Vectors
Therefore,
x =
C
11
b
1
+ C
12
b
2
+ C
13
b
3
det A
. (6.56)
Similarly, using Equation (6.51), we find
y =
a
11
b
2
− a
21
b
1
a
13
det A
with
a
11
b
2
− a
21
b
1
=(a
11
a
23
− a
21
a
13
)(a
33
b
1
− a
13
b
3
)
− (a
11
a
33
− a
31
a
13
)(a
23
b
1
− a
13
b
2
)
= a
13
.
(a
31
a
23
− a
21
a
33
)
≡C
21
b
1
+(a
11
a
33
− a
31
a
13
)
≡C
22
b
2
+(a
21
a
13
− a
11
a
23
)
≡C
23
b
3
/
= a
13
(C
21
b
1
+ C
22
b
2
+ C
23
b
3
),
so that
y =
C
21
b
1
+ C
22
b
2
+ C
23
b
3
det A
. (6.57)
With x and y thus determined, we can substitute them in any of the three
original equations and find z. Let us use the first equation; then
z =
b
1
− a
11
x − a
12
y
a
13
=
b
1
− a
11
C
11
b
1
+ C
12
b
2
+ C
13
b
3
det A
− a
12
C
21
b
1
+ C
22
b
2
+ C
23
b
3
det A
a
13
=
b
1
(det A − a
11
C
11
− a
12
C
21
) − b
2
(a
11
C
12
+ a
12
C
22
) − b
3
(a
11
C
13
+ a
12
C
23
)
a
13
det A
.
The numerator N can be calculated:
N = b
1
[a
11
(a
22
a
33
− a
23
a
32
) − a
12
(a
21
a
33
− a
31
a
23
)+a
13
(a
21
a
32
− a
22
a
31
)
− a
11
(a
22
a
33
− a
32
a
23
) −a
12
(a
31
a
23
− a
21
a
33
)]
− b
2
[a
11
(a
32
a
13
− a
12
a
33
)+a
12
(a
11
a
33
− a
31
a
13
)]
− b
3
[a
11
(a
12
a
23
− a
22
a
13
)+a
12
(a
21
a
13
− a
11
a
23
)]
= a
13
.
(a
21
a
32
− a
22
a
31
)
≡C
31
b
1
+(a
12
a
31
− a
11
a
32
)
≡C
32
b
2
+(a
11
a
22
− a
12
a
21
)
≡C
33
b
3
/
= a
13
(C
31
b
1
+ C
32
b
2
+ C
33
b
3
).
It now follows that
z =
C
31
b
1
+ C
32
b
2
+ C
33
b
3
det A
. (6.58)
We can put Equations (6.56), (6.57), and (6.58) in matrix form:
⎛
⎝
x
y
z
⎞
⎠
=
1
det A
⎛
⎝
C
11
C
12
C
13
C
21
C
22
C
23
C
31
C
32
C
33
⎞
⎠
⎛
⎝
b
1
b
2
b
3
⎞
⎠
⇒ x =
1
det A
Cb. (6.59)
6.4 The Jacobian 207
This is the inverse of Equation (6.54). The reader may verify that multiplying
A on either side of C/ det A yields the identity matrix, so that C/ det A is indeed
the inverse of A. The rule for calculating this inverse is as follows. Construct
a matrix out of the cofactors and denote it by A:
A ≡
⎛
⎝
A
11
A
12
A
13
A
21
A
22
A
23
A
31
A
32
A
33
⎞
⎠
(6.60)
and note that
⎛
⎝
C
11
C
12
C
13
C
21
C
22
C
23
C
31
C
32
C
33
⎞
⎠
=
8
A
so, we obtain the important result
inverse of a 3 × 3
matrix
A
−1
=
1
det A
8
A =
1
det A
⎛
⎝
A
11
A
21
A
31
A
12
A
22
A
32
A
13
A
23
A
33
⎞
⎠
. (6.61)
Equation (6.61), although derived for a 3 ×3 matrix, applies to all matrices,
including a 2 ×2 one whose inverse was given in Theorem 6.3.1, as the reader
is asked to verify.
As in the case of 2 × 2 matrices, a transformation in space that takes a
basis onto another basis is invertible.
6.4 The Jacobian
With the machinery of determinants at our disposal, we can formalize the
geometric construction of area and volume elements in Chapter 2 to a pro-
cedure which can be used for all coordinate transformations. We start with
two dimensions and consider the coordinate transformation
x = f(u, v),y= g(u, v). (6.62)
Our goal is to write the element of area in the (u, v) coordinate system. This
is the area formed by infinitesimal elements in the direction of u and v, i.e.,
elements in the direction of the primary curves of the (u, v)coordinatesystem.
For an arbitrary change du and dv in u and v, the Cartesian coordinates
change as follows:
dx =
∂f
∂u
du +
∂f
∂v
dv,
dy =
∂g
∂u
du +
∂g
∂v
dv.
208 Planar and Spatial Vectors
The element in the direction of the first primary curve is obtained by holding
v constant and letting u vary. This corresponds to setting dv =0intheabove
equations. It follows that the first primary (vector) length element is
d
l
1
=
ˆ
e
x
dx
1
+
ˆ
e
y
dy
1
=
ˆ
e
x
∂f
∂u
du +
ˆ
e
y
∂g
∂u
du. (6.63)
Similarly, the second primary (vector) length element, obtained by fixing u
and letting v vary, is
d
l
2
=
ˆ
e
x
dx
2
+
ˆ
e
y
dy
2
=
ˆ
e
x
∂f
∂v
dv +
ˆ
e
y
∂g
∂v
dv. (6.64)
When we derived the elements of area and volume in the three coordinate
systems in Chapter 2, we used the fact that the set of unit vectors in each
system were mutually perpendicular. Therefore, the area and volume elements
were obtained by mere multiplication of length elements. We are not assuming
that
ˆ
e
u
and
ˆ
e
v
are perpendicular. Thus, we cannot simply multiply the lengths
to get the area. However, we can use the result of Example 1.1.2 which gives
the area of a parallelogram formed by two non-collinear vectors. Writing the
cross product in terms of the determinant, we have
d
l
1
× d
l
2
=det
⎛
⎜
⎜
⎜
⎜
⎝
ˆ
e
x
ˆ
e
y
ˆ
e
z
∂f
∂u
du
∂g
∂u
du 0
∂f
∂v
dv
∂g
∂v
dv 0
⎞
⎟
⎟
⎟
⎟
⎠
=
ˆ
e
z
det
⎛
⎝
∂f
∂u
∂g
∂u
∂f
∂v
∂g
∂v
⎞
⎠
du dv
and the area is simply the absolute value of this cross product:
Jacobian matrix
and Jacobian
da =
det
⎛
⎝
∂f
∂u
∂g
∂u
∂f
∂v
∂g
∂v
⎞
⎠
du dv ≡
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
du dv, (6.65)
where we substituted x and y for f and g and introduced a new notation
for the (absolute value of the) determinant. The matrix whose determinant
multiplies du dv is called the Jacobian matrix, and the absolute value of its
determinant, the Jacobian.
Example 6.4.1.
Let us apply Equation (6.65) to polar coordinates. The trans-
formation is
x = f (r, θ)=r cos θ, y = g(r, θ)=r sin θ.
This gives
∂x
∂r
=
∂f
∂r
=cosθ,
∂x
∂θ
=
∂f
∂θ
= −r sin θ,
∂y
∂r
=
∂g
∂r
=sinθ,
∂y
∂θ
=
∂g
∂θ
= r cos θ,
6.4 The Jacobian 209
and
da =
∂x
∂r
∂y
∂r
∂x
∂θ
∂y
∂θ
dr dθ =
cos θ sin θ
−r sin θrcos θ
dr dθ
=(r cos
2
θ + r sin
2
θ) dr dθ = rdrdθ,
which is the familiar element of area in polar coordinates.
The procedure discussed above for two dimensions can be generalized to
three dimensions using the result of Example 1.1.3 which gives the volume of a
parallelepiped formed by three non-coplanar vectors. Suppose the coordinate
transformations are of the form
x = f(u, v, w),y= g(u, v, w),z= h(u, v, w).
Then
dx =
∂f
∂u
du +
∂f
∂v
dv +
∂f
∂w
dw,
dy =
∂g
∂u
du +
∂g
∂v
dv +
∂g
∂w
dw,
dz =
∂h
∂u
du +
∂h
∂v
dv +
∂h
∂w
dw.
The first primary element of length is obtained by fixing v and w and
allowing u to vary; similarly for the second and third primary elements of
length. We therefore have
d
l
1
=
ˆ
e
x
dx
1
+
ˆ
e
y
dy
1
+
ˆ
e
z
dz
1
=
ˆ
e
x
∂f
∂u
du +
ˆ
e
y
∂g
∂u
du +
ˆ
e
z
∂h
∂u
du,
d
l
2
=
ˆ
e
x
dx
2
+
ˆ
e
y
dy
2
+
ˆ
e
z
dz
2
=
ˆ
e
x
∂f
∂v
dv +
ˆ
e
y
∂g
∂v
dv +
ˆ
e
z
∂h
∂v
dv,
d
l
3
=
ˆ
e
x
dx
3
+
ˆ
e
y
dy
3
+
ˆ
e
z
dz
3
=
ˆ
e
x
∂f
∂w
dw +
ˆ
e
y
∂g
∂w
dw +
ˆ
e
z
∂h
∂w
dw.
Example 1.1.3 now yields
dV =
d
l
1
·(d
l
2
× d
l
3
)
=
det
⎛
⎜
⎜
⎜
⎜
⎜
⎝
∂f
∂u
du
∂g
∂u
du
∂h
∂u
du
∂f
∂v
dv
∂g
∂v
dv
∂h
∂v
dv
∂f
∂w
dw
∂g
∂w
dw
∂h
∂w
dw
⎞
⎟
⎟
⎟
⎟
⎟
⎠
.
We summarize the foregoing argument in
Theorem 6.4.2. For the coordinates u, v,andw, related to the Cartesian
coordinates by x = f(u, v, w), y = g(u, v, w),andz = h(u, v, w),thevolume
element is given by
210 Planar and Spatial Vectors
dV =
∂x
∂u
∂y
∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
∂y
∂w
∂z
∂w
du dv dw. (6.66)
The (absolute value of the) determinant multiplying du dv dw is called the
Jacobian defined
Jacobian of the coordinate transformation.
Historical Notes
Determinants were mathematical objects created in the process of solving a system
of linear equations. As early as 1693 Leibniz used a systematic set of indices for the
coefficients of a system of three equations in two unknowns. By eliminating the two
unknowns from the set of three equations, he obtained an expression involving the
coefficients that “determined” whether a solution existed for the set of equations.
The solution of simultaneous linear equations in two, three, and four unknowns
by the method of determinants was created by Maclaurin around 1729. Though
not as good in notation, his rule is the one we use today and which Cramer used
in connection with his study of the conic sections. In 1764, Bezout systematized
the process of determining the signs of the terms of a determinant for n equations
in n unknowns and showed that the vanishing of the determinant is a necessary
condition for nonzero solutions to exist.
Vandermonde was the first to give a connected and logical exposition of the
theory of determinants detached from any system of linear equations, although he
used his theory mostly as applied to such systems. He also gave a rule for expanding
a determinant by using second-order minors and their complementary minors. In
the sense that he concentrated on determinants, he is aptly considered the founder
of the theory.
One of the consistent workers in determinant theory over a period of over fifty
years was James Joseph Sylvester.
In 1833 he became a student at St. John’s College, Cambridge, and took the
difficult tripos examination in the same year along with two other famous math-
ematicians, Gregory and Green (the creator of the important Green’s functions).
Sylvester came second, Green who was 20 years older than the other two came fourth
with Duncan Gregory fifth. (The first-place winner did little work of importance
after graduating.)
James Joseph
Sylvester
1814–1897
At this time it was necessary for a student to sign a religious oath to the Church
of England before graduating and Sylvester, being Jewish, refused to take the oath,
so could not graduate. For the same reason he was not eligible for a Smith’s prize
nor for a Fellowship.
From 1838 Sylvester started to teach physics at the University of London, one
of the few places which did not bar him because of his religion. Three years later
he was appointed to a chair in the University of Virginia but he resigned after a few
months. A student who had been reading a newspaper in one of Sylvester’s lectures
insulted him and Sylvester struck him with a sword stick. The student collapsed in
shock and Sylvester believed (wrongly) that he had killed him. He fled to New York
boarding the first available ship back to England.
On his return, Sylvester worked as an actuary and lawyer but gave private
mathematics lessons. His pupils included Florence Nightingale. By good fortune
6.5 Problems 211
Cayley was also a lawyer, and both worked at the courts of Lincoln’s Inn in London.
Cayley and Sylvester discussed mathematics as they walked around the courts and,
although very different in temperament, they became life-long friends.
Sylvester tried hard to return to mathematics as a profession, and he applied
unsuccessfully for a lectureship in geometry at Gresham College, London, in 1854.
Another failed application was for the chair in mathematics at the Royal Military
Academy at Woolwich, but, after the successful applicant died within a few months
of being appointed, Sylvester became professor of mathematics at Woolwich. Being
at a military academy, Sylvester had to retire at age 55. At first it looked as though
he might give up mathematics since he had published his only book at this time,
and it was on poetry. Apparently Sylvester was proud of this work, entitled The
Laws of Verse, since after this he sometimes signed himself “J. J. Sylvester, author
of The Laws of Verse.”
In 1877 Sylvester accepted a chair at the Johns Hopkins University and founded
in 1878 the American Journal of Mathematics, the first mathematical journal in the
USA.
In 1883 Sylvester, although 68 years old at this time, was appointed to the
Savilian chair of geometry at Oxford. However he only liked to lecture on his own
research and this was not well liked at Oxford where students wanted only to do well
in examinations. In 1892, at the age of 78, Oxford appointed a deputy professor
in his place and Sylvester, by this time partially blind and suffering from loss of
memory, returned to London where he spent his last years at the Athenaeum Club.
Sylvester did important work on matrix and determinant theory, a topic in which
he became interested during the walks with Cayley while they were at the courts
of Lincoln’s Inn. In particular he used matrix theory to study higher-dimensional
geometry. He also devised an improved method of determining conditions under
which a system of polynomial equations has a solution.
The formula for the derivative of a determinant when the elements are functions
of a variable was first given in 1841 by Jacobi whohadearlierusedtheminthe
change of variables in a multiple integral. In this context the determinant is called
the Jacobian of the transformation (as discussed in the current section of this book).
6.5 Problems
6.1. What vector is obtained when the vector a
2
of a basis {a
1
, a
2
} is actively
transformed with the matrix
01
00
!
.
6.2. Show that the nonzero matrix A =
10
00
!
cannot have an inverse. Hint:
Suppose that B =
ab
cd
!
is the inverse of A.CalculateAB and BA, set them
equal to the unit matrix and show that no solution exists for a, b, c,andd.
6.3. Let A =
a
1
b
1
c
1
d
1
!
and B =
a
2
b
2
c
2
d
2
!
be arbitrary matrices. Find AB, A
t
,
and B
t
and show that (AB)
t
= B
t
A
t
.
6.4. Find the angle between 1 + t and 1 − t when the inner product is inte-
gration over the interval (0, 1).
6.5. Instead of (0, 1), choose (−1, 1) as the interval of integration for P
1
[t].
From the basis {1,t}, construct an orthonormal basis using the Gram–Schmidt
process.
212 Planar and Spatial Vectors
6.6. Take the interval of the integration to be (−1, +1), and find the inner
product matrix for the basis {1,t} of P
1
[t].
6.7. Find the angle between two vectors a and b, whose components in an
orthonormal basis are, respectively, (1, 2) and (2, −3). Use the Gram–Schmidt
process to find the orthonormal vectors obtained from a and b.
6.8. Use the Gram–Schmidt process to find an orthonormal basis in three
dimensions from each of the following:
(a) (−1, 1, 1), (1, −1, 1), (1, 1, −1) (b) (1, 2, 2), (0, 0, 1), (0, 1, 0)
6.9. (a) Find the inner product matrix associated with the basis vectors
a
1
=
ˆ
e
x
+
ˆ
e
y
, a
2
=
ˆ
e
x
+
ˆ
e
z
,anda
3
=
ˆ
e
y
+
ˆ
e
z
.
(b) Calculate the inner product of two vectors a and b, whose components in
the basis above are, respectively, (1, −1, 2) and (0, 2, 3).
(c) Use the Gram–Schmidt process to find three orthonormal vectors out of
the basis of (a).
6.10. Use Gram–Schmidt process to find orthonormal vectors out of the three
vectors (2, −1, 3), (−1, 1, −2), and (3, 1, 2). What do you get as the last
vector? What can you say about the linear independence of the original
vectors?
6.11. What is the angle between the second and fourth vectors in the standard
basis of P
3
[t] when the interval of integration of the inner product is (0, 1)?
Between the first and fourth vectors?
6.12. Calculate the inner product matrix for the standard basis of P
3
[t]when
the interval of integration of the inner product is (−1, +1). Now find the angle
between all vectors in that basis.
6.13. The inner product matrix in a basis {a
1
, a
2
} is given by
G =
2 −1
−13
.
(a) Calculate the cosine of the angle between a
1
and a
2
.
(b) Suppose that a = −a
1
+ a
2
and b =2a
1
− a
2
.Calculate|a|, |b|, a · b,
and the cosine of the angle between a and b.
6.14. Let a
1
=1+t and a
2
=1− t be a basis of P
1
[t]. Define the inner
product as the integral of products of polynomials over the interval (0,a)with
a>0.
(a) Determine a such that a
1
and a
2
are orthogonal.
(b) Given this value of a,calculate|a
1
| and |a
2
|.
(c) Find two orthogonal polynomials {
ˆ
e
1
,
ˆ
e
2
} of unit length that form a basis
for P
1
[t].
(d) Write the polynomial b =3−2t as a linear combination of
ˆ
e
1
and
ˆ
e
2
.
(e) Calculate b ·b using the definition of the inner product.
(f) Calculate b ·b by squaring (and then adding) the components in {
ˆ
e
1
,
ˆ
e
2
}.
6.5 Problems 213
6.15. Show that the matrix C defined in Equations (6.56)–(6.59) is indeed
the transpose of the matrix A of cofactors of A.
6.16. Show directly that the matrix given in Equation (6.61) is indeed the
inverse of the matrix A.
6.17. From the transformation rules (1.8) and (1.9) giving the Cartesian
coordinates as functions of cylindrical and spherical coordinates, and using
the Jacobian (6.66), find the volume elements in cylindrical and spherical
coordinates
6.18. The elliptic coordinates are given by
x = a cosh u cos θ
y = a sinh u sin θ.
Using the Jacobian for two variables (6.65), find the element of area for the
elliptic coordinate system.
6.19. The elliptic cylindrical coordinates are given by
x = a cosh u cos θ
y = a sinh u sin θ
z = z
Using the Jacobian for three variables (6.66), find the element of volume for
the elliptic cylindrical coordinate system.
6.20. The prolate spheroidal coordinates are given by
x = a sinh u sin θ cos ϕ
y = a sinh u sin θ sin ϕ
z = a cosh u cos θ
Using the Jacobian for three variables (6.66), find the element of volume for
the prolate spheroidal coordinate system.
6.21. The toroidal coordinates are given by
x =
a sinh θ
cos ϕ
cosh θ −cos u
y =
a sinh θ sin ϕ
cosh θ −cos u
z =
a sin u
cosh θ −cos u
Using the Jacobian for three variables (6.66), find the element of volume for
the toroidal coordinate system.