1-23

Crash Course on Tensor Analysis

Department of Mathematics, IIT Madras

Avudainayagam ASanyasiraju Y V S S

SatyajitroyUsha R

Coordinator: Prof. Usha R

January 17 - 19, 2014

0Contents

1. Introduction to Vector Algebra 1

2. Orthogonal Curvilinear Coordinates 7

3. Surface Geometry 24

4. General Curvilinear Coordinates 32

5. Tensor Calculus 45

6. Symmetric Tensors 65

7. Tensor Derivatives 81

8. Tensor form of Gradient 106

9. Cartesian Tensors 119

10. Calculus of Cartesian Tensors 147

11. Integral Theorems in Tensors 157

Chapter 1

Introduction to Vector Algebra

1.1 Introduction

1. A vector has two characteristics

(a) magnitude

(b) direction

Force and velocity are two typical examples of a vector.

2. Geometrically, a vector is represented by a directed line segment with the length of thesegment representing the magnitude and the direction of the segment indicating thedirection of the vector.

3. Magnitude of a vector is a non-negative real number.

4. Vectors are in-general denoted by bold letters (like a, v ), however, in writing, anover-bar may be used to represent them, for example, a, v etc.

5. The magnitude of a vector a is denoted by |a|.6. The magnitude of the zero vector is zero

7. Two vectors are equal if they have the same magnitude and the same direction.

8. a, where is a scalar, is called a scalar multiple of the vector a.

9. The direction of a is that of a if > 0 and is that of a if < 010. Every vector can be written as a = |a|a, where a is the unit vector in the direction of

a.

11. Two vectors a and b are collinear, if there exists a scalar such that a = b.

12. In a right handed rectangular system of Cartesian axes with fixed origin O, x1, x2and x3 denote the axes and e1, e2 and e3 denote the unit vectors(base vectors) alongthe corresponding coordinate directions.

1

213. Every vector can be expressed as a linear combination of the base vectors as

a = a1 e1 + a2 e2 + a3 e3 (1.1)

where a1, a2 and a3 (real numbers referred to as the components of a) represent the pro-jection of a on the coordinates axes. A short notation, using summation conventionfor any repeated index, for (1.1) is

a = ai ei (1.2)

14. A typical ith component of a vector a is denoted by [a]i, that is

[a]i = ai (1.3)

15. In component form, the scalar multiple can be written as

a = (a1) e1 + (a2) e2 + (a3) e3 = ( ai) ei (1.4)

Therefore[a]i = ai = [a]i (1.5)

1.2 Parallelogram law

If x and y are any two vectors, then x+ y is defined by the parallelogram law shown in theFig. 1.1.

Figure 1.1: Vector parallelogram law

31.3 Scalar product

The scalar product of any two vectors a and b is given by

a b = |a| |b| cos (1.6)where is the angle between the direction of the vectors a and b.

Remark : We have a b = a1b1 + a2b2 + a3b3 = aibi,a ei = ai = [a]i

ai = |a| cos iwhere i is the angle made by the vector a with the xi-axis.

Further, cos i for i = 1, 2, 3 are the direction cosines of a and ai for i = 1, 2, 3 are thedirection ratios of a. For unit vectors, the components are the direction cosines.

1.4 Vector product

The cross product or vector product of any two vectors a and b is given by

a b = |a| |b| sin n (1.7)where is the angle between the directions of the vectors a and b, n is the unit vectorperpendicular to both a and b. The direction of n is that direction obtained by right handedscrew rotation from a to b. Some properties of the cross product to be noted are:

1. a b = 0 means a and b are collinear2. |a b| is the area of the parallelogram whose adjacent sides are a and b.3. a b = (a2b3 a3b2) e1 + (a3b1 a1b2) e2 + (a1b2 a2b1) e3 in component form.

4. a b =e1 e2 e3a1 a2 a3b1 b2 b3

in determinant form.5. a a = 0 and a b = b a6. Since, for the base vectors, we have

ei ej = ijk ek (1.8)where is defined by

ijk =1 if i, j and k are in cyclic order

1 if i, j and k are in acyclic order0 if any two of i, j and k are equal,

(1.9)

4therefore, the vector product also can be written as

a b = (aiei) (bj ej) = aibj(ei ej) = aibj(ijk ek) (1.10)= ijk ai bj ek = jki aj bk ei = ijk aj bk ei (1.11)

ijk aj bk is the ith component of a b

7. Since a a = 0, we have ijk aj ak = 0

1.5 Scalar triple product

The scalar triple product of three vectors a, b and c is given by

a (b c) =a1 a2 a3b1 b2 b3c1 c2 c3

= (b2c3 b3c2) a1 + (b3c1 c1c2) a2 + (b1c2 b2c1) a3 (1.12)Some properties of scalar triple product are

1. a (b c) = 0 means a, b and c are coplanar, that is, there exists two scalars and such that a = b+ c.

2. Since a b = ijk aj bk, (a b) c = ijk aj bk ci = ijk ai bi ck

1.6 Vector triple product

The vector triple product of three vectors a, b and c satisfies

a (b c) = (a c)b (a b)c (1.13)(a b) c = (a c)b (b c)a (1.14)

1.7 Matrix product in index notation

let A = [aij], B = [bij] and C = [cij] be three square matrices of order 3 such that C = A Bthen, we have

cij =3

i=1

aik bkj [aij] [bij] = aik bkj (1.15)

Some properties of matrix product

1. [aij] [aij] = aik akj

2. [aij] [bij]T = aik bjk

3. [aij]T [bij] = aki bkj

4. ([aij] [bij])T = [bij]

T [aij]T = ajk bki

51.8 Some problems involving the index notation

1.8.1 Problem

Given (with usual notation for aij, bij and ij)

aij = ij bkk + bij

then, express bij in terms of aij.

Solution :

From the given relation, for i = j, we have aij = bij.That is, when i = j, we have bij = 1aij.

For i = j, we have aii = 3(b11 + b22 + b33) + bii = (3 + )bpp. Therefore, wheni = j, we have bpp =

1

3+app (Note that ii = 3)

Putting the above two statements together, we have

bij =1

(aij ijbpp) = 1

(aij

3 + ijapp

)

1.8.2 Problem

aij = aji if and only if ijk ajk = 0.

Solution :

For i = 1, we have

ijk ajk = 1jk ajk =3

j=1

3k=1

1jk ajk

=3

j=1

(1j1 aj1 + 1j2 aj2 + 1j3 aj3)

=3

j=1

(1j2 aj2 + 1j3 aj3) , since 1j1 = 0

= 132 a32 + 123 a23, since all the other terms are zero

= a23 a32

Similarly, we have 2jk ajk = a31 a13 and 3jk ajk = a12 a21. Using these three relations,it is easy to prove the given statement.

61.8.3 Problem

Show that

pqr aip ajq akr =

ai1 ai2 ai3aj1 aj2 aj3ak1 ak2 ak3

1.8.4 Problem

Show thatdet(aij) = pqr a1p a2q a3r = pqr ap1 aq2 ar3

1.9 Some results which involve ijk

1.

ijk pqr det(aij) =

aip aiq airajp ajq ajrakp akq akr

2.

ijk pqr =

ip iq irjp jq jrkp kq kr

3.

ijk pqr = ip jq iq jp4.

ijk pjk = 2 ip

5.ijk ijk = 6

6.

det(aij) =1

6ijk pqraip + ajq + akr

7.

det(aij) =1

6ijk pqrapi + aqj + ark

8.

det(aij) =1

6(aiiajjakk + 2aijajkaki 3aijajiakk)

9.ijk det(aij) = pqraipajqakr = pqrapiaqjark

Chapter 2

Orthogonal Curvilinear Coordinates

2.1 Introduction

For any physical space, a coordinate system, for example as shown in Fig 2.1, is necessary fordefining the problem, locating the objects, evaluating the properties, providing a referencefor discussion etc. Generally three coordinates are essential to define the location of a particlein any 3D space. Similarly, two coordinates are sufficient to locate a particle either movingor stationary in a plane.

Figure 2.1: Coordinate system

2.2 Curvilinear Coordinates

Let (x1, x2, x3) be the rectangular coordinates of any point, say, P in a rectangular coordinatesystem (xi), i = 1, 2, 3.

7

8Expressing x1, x2, x3 in terms of some curvilinear coordinates u1, u2, u3 gives

x1 = x1(u1, u2, u3)

x2 = x2(u1, u2, u3) (2.1)

x3 = x3(u1, u2, u3)

or conversely expressing the curvilinear coordinates u1, u2, u3 in terms of the rectangularcoordinates x1, x2, x3 gives

u1 = u1(x1, x2, x3)

u2 = u2(x1, x2, x3) (2.2)

u3 = u3(x1, x2, x3)

Functions (2.1) and (2.2) are assumed to be single valued and have continuous partial deriva-tives so that the relation between x1, x2, x3 and u1, u2, u3 is unique. Thus, given a pointP (x1, x2, x3), the corresponding unique set of coordinates (u1, u2, u3) can be obtained using(2.2) or the other way using (2.1). The coordinates (u1, u2, u3) are called the curvilinearcoordinates of P .

2.2.1 Orthogonal Curvilinear Coordinates

Enormous simplifications can be achieved by using different coordinate systems in manymathematical problems as in a Boundary value problems for Partial Differential Equationswith boundary conditions specified at the coordinate surfaces. One of them is the orthogonalsystem, wherein, at any point of the space, the vectors aligned with the three coordinatedirections are mutually perpendicular. The variation of a single coordinate, in general, willgenerate a curve in space, rather than a straight line; hence the term curvilinear.

The surfaces, shown in Fig 2.2,

u1 = c1, u2 = c2, u3 = c3 (2.3)

where c1, c2 and c3 are constants, are called coordinate surfaces. These are the surfacesgenerated by holding one coordinate constant and varying the other two. The intersectionof any two of these surfaces generates a coordinate curve or line. If they intersect at rightangles, then such a coordinate system is called orthogonal coordinate system.

In what follows, we present some important results from the orthogonal curvilinear coor-dinate systems.

2.3 Unit Vectors

Let

r = x1i+ x2j + x3k (2.4)

9

Figure 2.2: Coordinate surfaces

be the position vector of the point P .

Using (2.1) in (2.4) gives

r = r(u1, u2, u3) (2.5)

and ru1

is a tangent vector to the u1 curve at the point P . Then the corresponding unittangent vector e1 in the direction of u1 is

e1 =1

| ru1|r

u1(2.6)

or

r

u1= h1 e1, where h1 =

ru1 (2.7)

Similarly, we have

r


ru2

r


ru3

The unit vectors e1, e2 and e3 are in the directions of increasing u1, u2 and u3, respectively.The quantities h1, h2 and h3 are called the scale factors.

10

Figure 2.3: Tangent and normal unit vectors at a point P

ui is a vector normal to the coordinate surface ui = ci for i = 1, 2, 3 at P and we de-note

Ei =ui|ui| , i = 1, 2, 3 (2.8)

as the unit normal vectors to the surfaces ui = ci, i = 1, 2, 3 at P . Therefore, at every pointP of a curvilinear system, there exists two sets of unit vectors (e1, e2, e3) which are tangentto the coordinate curves and (E1, E2, E3) which are normal to the coordinate surfaces asshown in Fig. 2.3.

Note : In any orthogonal coordinate system (e1, e2, e3) and (E1, E2, E3) are identical.

These two sets of vectors are analogous to the i, j, k of the rectangular coordinate systembut they may change from point to point unlike the latter.

2.3.1 Example :

Show that ( ru1

, ru2

, ru3

) and (u1,u2,u3) constitute a reciprocal system of vectors, thatis

r

uiuj =

{1, i = j0, i = j (2.9)

where i and j take any values of 1, 2, and 3.

Proof : We have, from (2.5)

dr =r

u1du1 +

r

u2du2 +

r

u3du3 (2.10)

11

Therefore

u1 . dr = du1 =(u1. r

u1

)du1 +

(u1. r

u2

)du2 +

(u1. r

u3

)du3

u1. ru1

= 1, u1. ru2

= 0, u1. ru3

= 0

Similarly, the other relations can be obtained.

2.4 Representation

Any vector A can be represented in terms of the unit base vectors (e1, e2, e3) or (E1, E2, E3)as

A = a1e1 + a2e2 + a3e3 (2.11)

= A1E1 + A2E2 + A3E3 (2.12)

where a1, a2, a3, A1, A2, A3 are the respective components of A in each system. Also, in termsof the base vectors,

A =A1h1

r

u1+

A2h2

r

u2+

A3h3

r

u3

= C1r

u1+ C2

r

u2+ C3

r

u3

=3

i=1

Cii, Ci =Aihi, i =

r

ui, hi =

rui (2.13)

or

A =a1|u1|u1 +

a2|u2|u2 +

a3|u3|u3

=3

i=1

cii, ci =ai|ui| , i = ui (2.14)

C1, C2 and C3 are called the contravariant components of A and c1, c2 and c3 are the covariantcomponents of A.

2.5 Arc length

The key to deriving expressions for curvilinear coordinates is to consider the arc length alonga curve. From (2.5), we have

dr =r

u1du1 +

r

u2du2 +

r

u3du3

= h1 du1 e1 + h2 du2 e2 + h3 du3 e3 (2.15)

12

Figure 2.4: Volume in curvilinear coordinates

Then the differential of arc length ds is determined from ds2 = dr.dr. For orthogonal systemsds2 = h21 du1 + h

22 du2 + h

23 du3 since

ei.ej =

{1, i = j0, i = j (2.16)

Therefore, we have

ds =

(3

i=1

h2i (dui)2

)1/2(2.17)

The fact that a space curve has an independent geometric significance indicates that thequantity in brackets, in (2.17), must be invariant to the choice of the coordinate system. Wewill consider non-orthogonal systems or arc length in more general curvilinear systems later.

2.6 Volume element

Since u2 and u3 are constant along the u1 = const curve, we have

dr = h1du1e1, along u1 curve (2.18)

and similarly

dr = h2du2e2, along u2 curve

dr = h3du3e3, along u3 curve

13

For a given orthogonal curvilinear coordinate system, the differential volume element cor-responds to the volume of a parallelepiped with adjacent sides h1du1, h2du2 and h3du3 isshown in the Fig 2.4. The volume is given by the scalar triple product, therefore

dv = |(h1du1e1) . (h2du2e2) (h3du3e3)|= h1h2h3 du1du2du3 (2.19)

since we have |e1.e2 e3| = 1.

2.7 Surface element

Expressions for differential surface elements are obtained by using the geometric represen-tation of the cross product. If dsi refers to a surface on which the coordinate ui is heldconstant, we obtain

ds1 = h2 du2 e2 h3 du3 e3 = h2h3du2du3 (2.20)ds2 = h3 du3 e3 h1 du1 e1 = h1h3du1du3 (2.21)ds3 = h1 du1 e1 h2 du2 e2 = h1h2du1du2 (2.22)

2.8 The gradient

An expression for the gradient in orthogonal curvilinear coordinates is obtained by examiningthe differential change in a scalar function associated with a differential change in position.If

d = f1 e1 + f2 e2 + f3 e3 (2.23)

then, our aim is to compute f1, f2 and f3. Letting = (u1, u2, u3), we have

d =i

uidui (2.24)

Also

d = dr. =(

i

hieidui

).

(j

fj ej

)=i

hifidui (2.25)

Comparing (2.24) and (2.25), we see that

fi =1

h1

ui(2.26)

Therefore

=i

eihi

ui, hi =

ui (2.27)

This is the general expression for the gradient operator, valid for any orthogonal curvilinearcoordinate system.

14

2.9 Divergence of a vector

Expression for the divergence of a vector A = (A1, A2, A3) in orthogonal curvilinear coordi-nates:From the derivation of the gradient operator, we have

u1 = e1h1

u1u1

+e2h2

u2u1

+e3h3

u3u1

=e1h1

u1u1

=e1h1

(2.28)

e1 = h1 u1 or h11 = |u1| (2.29)Similarly, we have

e2 = h2 u2 or h12 = |u2|e3 = h3 u3 or h13 = |u3| (2.30)

Using (2.30), we have

u2 u3 = e2 e3h2 h3

=e1

h2 h3(2.31)

So that

e1 = h2h3 (u2 u3) ;e2 = h3h1 (u3 u1) ; (2.32)e2 = h1h2 (u1 u2) ;A1 e1 = A1h2h3 (u2 u3) (2.33)

and

. (A1 e1) = . (A1h2h3 (u2 u3))= (A1h2h3) . (u2 u3) + A1h2h3 ( . u2 u3)= (A1h2h3) . e2

h2 e3

h3

after using (2.29), (2.30) and the fact that the div curl of a vector is zero

. (A1 e1) =(e1h1

u1+

e2h2

u2+

e3h3

u3

)(A1h2h3) .

e2h2

e3h3

=1

h1h2h3

u1(A1h2h3)

Similarly, we also have

. (A2 e2) = 1h1h2h3

u2(A2h3h1)

. (A3 e3) = 1h1h2h3

u3(A3h1h2)

Putting all the three terms together gives

. A = 1h1h2h3

(

u1(A1h2h3) +

u2(A2h3h1) +

u3(A3h1h2)

)(2.34)

which is the general formula for the divergence in the curvilinear coordinates.

15

2.10 Curl of a vector

The curl of a vector in curvilinear coordinates is expressed as follows:We have, from (2.29)

A1 e1 = A1 h1 u1= ((A1 h1)) u1 + a1h1 u1= ((A1 h1)) e1

h1

=

(e1h1

u1+

e2h2

u2+

e3h3

u3

)(A1h1) e1

h1

A1 e1 =(

e2h3h1

)

u3(A1h1)

(e3

h2h1

)

u2(A1h1)

Similarly

A2 e2 =(

e3h1h2

)

u1(A2h2)

(e1

h3h2

)

u3(A2h2) (2.35)

A3 e3 =(

e1h2h3

)

u2(A3h3)

(e2

h1h3

)

u1(A3h3)

When all the components are put together, the curl is written as a determinant form givenby

A = 1h1h2h3

h1e1 h2e2 h3e3

u1

u2

u3

A1h1 A2h2 A3h3

(2.36)

2.11 Laplacian in orthogonal curvilinear coordinates

We have, from the gradient

=(e1h1

u1+

e2h2

u2+

e3h3

u3

)

By setting A = in (2.34), we find that

= . = 1h1h2h3

(

u1

(h2h3h1

u1

)+

u2

(h3h1h2

u2

)+

u3

(h1h1h3

u3

))(2.37)

This completes the results for orthogonal curvilinear coordinates. The remainder of thissection is devoted to the useful special cases.

16

( , !, z)

Figure 2.5: Cylindrical coordinate system

2.12 Some orthogonal curvilinear coordinate systems

2.12.1 Cylindrical coordinate system

Circular cylindrical coordinates, denoted as (, , z), are shown in the Fig 2.5. The cylindricalcoordinates are related to the rectangular coordinates (using (x, y, z) instead of (x1, x2, x3))as

x = cos, y = sin, z = z (2.38)

where 0, 0 2 and < z

17

The position vector of any point P is expressed as

r = xi+ yj + zk

= cosi+ sinj + zk (2.46)


r

= cosi+ sinj (2.47)

r

= sini+ cosj (2.48)

r

z= 1 (2.49)

Therefore, the scale factors are

h1 =

r =cos2 + sin2 = 1 (2.50)

h2 =

r =2 cos2 + 2 sin2 = (2.51)

h2 =

rz = 1 (2.52)

Since, we have ei =1

h1rui

for i = 1, 2, 3, the base vectors for the cylindrical coordinates are

e1 = cosi+ sinj, e2 = sini+ cosj, e3 = k (2.53)Inverting the (2.53), the base vectors in the rectangular coordinate system, in terms of thebase vectors of the cylindrical coordinate system are

i = cos e1 sin e2, j = sin e1 + cos e2, k = e3 (2.54)The gradient, divergence, curl and Laplacian operators in cylindrical coordinate system areas follows:

Gradient of a scalar function f

f =i

eihi

f

ui, hi =

ui = e1f + e2 f + e3fz

Divergence of a vector function A = (A1, A2, A3)

. A = 1h1h2h3

(

u1(A1h2h3) +

u2(A2h3h1) +

(A3h1h2)

)

=1

((A1)

u1+

A2

+(A3)

z

)=

1

(A1)

u1+

A2

+A3z

(2.55)

18

Figure 2.6: Volume and surface elements in cylindrical coordinate system

Curl of a vector function A = (A1, A2, A3)

A = 1h1h2h3

h1e1 h2e2 h3e3

u1

u2

u3

A1h1 A2h2 A3h3

=1

e1 e2 e3

z

A1 A2 A3

(2.56)

Laplacian of a scalar function

= . = 1h1h2h3

(

u1

(h2h3h1

u1

)+

u2

(h3h1h2

u2

)+

u3

(h1h1h3

u3

))

=1

(

)+

1

22

2+

2

z2(2.57)

The differential volume and surface elements are evaluated as

dv = d d dz (2.58)

ds = d dz; ds = d dz; dsz = d d (2.59)

2.12.2 Some problems

Example :

Prove that the cylindrical coordinate system is orthogonal.

19

Example :

Represent the vector A = zi 2xj + yk in cylindrical polar coordinates.

Example :

Express the velocity and acceleration cylindrical polar coordinates.

Example :

Find the arc length in cylindrical polar coordinates.

2.12.3 Spherical coordinate system

Spherical polar coordinates, denoted as (r, , ), are shown in the Fig 2.7. These coordinatesare related to the rectangular coordinates (using (x, y, z) instead of (x1, x2, x3)) as

x = r sin cos, y = r sin sin, z = r cos (2.60)

where 0 0, 0 2 and 0 . This gives

Figure 2.7: Spherical polar coordinate system

r =x2 + y2 + z2, = tan1

y

x, = tan1

x2 + y2

z2(2.61)

The coordinate surfaces are

r = C1; spheres having the center at the origin (2.62)

= C2; planes passing through the zaxis (2.63) = C3; cones having the vertex at the origin, lines if C3 = 0 or , (2.64)

xy plane if C3 = /2

20

The coordinate curves are

intersection of = C1 and = C2; is a straight line (z curve) (2.65)

intersection of = C1 and z = C3; is a circle ( curve) (2.66)

intersection of = C2 and z = C3; is a straight line ( curve) (2.67)

The position vector of any point P is expressed as

r = xi+ yj + zk

= r sin cosi+ r sin sinj + r cos k (2.68)


r

r= sin cosi+ sin sinj + cos k (2.69)

r

= r sin sini+ r sin cosj (2.70)

r

= r cos cosi+ r cos sinj r sin k (2.71)

Therefore, the scale factors are

h1 = hr = 1 ; h2 = h = r; h = r sin (2.72)

Since, we have ei =1

h1rui

for i = 1, 2, 3, the base vectors for the spherical coordinates are

er = sin cosi+ sin sinj + cos k (2.73)

e = cos cosi+ cos sinj sin k (2.74)e = sini+ cosk (2.75)

Inverting the (2.73), the base vectors in the rectangular coordinate system, in terms of thebase vectors of the spherical coordinate system are

i = sin coser + cos cose sine (2.76)j = sin siner + cos sine + cose (2.77)

k = cos er sin e (2.78)

The line elements are

dx2 = sin cosdr + r cos cosd r sin sind (2.79)dy2 = sin sindr + r cos sind + r sin cosd (2.80)

dz2 = cos dr r sin d (2.81)ds2 = dx2 + dy2 + dz2 = dr2 + r2d2 + r2 sin2 d2 (2.82)

The gradient, divergence, curl and Laplacian operators in spherical coordinate system areas follows:

21

Gradient of a scalar function f

f =i

eihi

f

ui, hi =

ui = er fr + er f + er sin f (2.83)

Divergence of a vector function A = (A1, A2, A3)

. A = 1h1h2h3

(

u1(A1h2h3) +

u2(A2h3h1) +

(A3h1h2)

)

=1

r2(A1r

2)

r+

1

r sin

A2 sin

+

1

r sin

A3

(2.84)

Curl of a vector function A = (A1, A2, A3)

A = 1h1h2h3

h1e1 h2e2 h3e3

u1

u2

u3

A1h1 A2h2 A3h3

=1

r2 sin

er re r sin er

A1 rA2 r sin A3

(2.85)Laplacian of a scalar function

= . = 1h1h2h3

(

u1

(h2h3h1

u1

)+

u2

(h3h1h2

u2

)+

u3

(h1h1h3

u3

))

=1

r2

r

(r2

r

)+

1

r2 sin

(sin

)+

1

r2 sin2

2

2(2.86)

The differential volume and surface elements are evaluated as

dv = r2 sin dr d d (2.87)

dsr = r2 sin d d; ds = r sin dr d; ds = r dr d (2.88)

2.12.4 Parabolic cylindrical coordinates

The parabolic cylindrical coordinates are denoted by (u, v, z). These are related to therectangular Cartesian coordinates as

x =1

2(u2 v2), y = uv, z = z (2.89)

where < u 0 and < z

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Figure 2.8: Volume and surface elements in spherical coordinate system

0.5 0 0.51

0.50

0.51

z

x = (u uv v)/2, y = u v, z = 0

x

y

Figure 2.9: Parabolic cylindrical coordinate system

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2.12.5 Parabolic coordinates

The parabolic coordinates are denoted by (u, v, ). These are related to the rectangularCartesian coordinates as

x = uv cos, y = uv sin, z =1

2(u2 v2) (2.91)

where u 0, v 0 and 0 < 2. Therefore, the scale factors are

hu = hv =u2 + v2, h = uv (2.92)

2.12.6 Elliptic cylindric coordinates

The elliptic cylindric coordinates are denoted by (u, v, z). These are related to the rectangularCartesian coordinates as

x = a cosh u cos v, y = a sinh u sin v, z = z (2.93)

where u 0, 0 v < 2 and < z

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