The Cylindrical Coordinate System

1.5 Systems of Coordinates

1.5.2 The Cylindrical Coordinate System

The need for a cylindrical coordinate system should be apparent fromFigure1.26a, where the cylindrical surface (such as a pipe) is located along thezaxis in the Cartesian system of coordinates. To describe a pointP1on the surface, we must give the three coordinatesP1(x1,y1,z1). A second point on the cylindrical surface,P2(x2,y2,z2) is shown and a vector connects the two points. The vector can be immediately written as

A¼^xðx2x1Þ þ^yðy₂y₁Þ þ^zðz2z1Þ ð1:53Þ On the other hand, we observe that pointsP1andP2are at a constant distance from thezaxis, equal to the radius of the cylinder. To draw this cylinder all we need is to draw a circle at a constant radius and repeat the process for each value ofz. In doing so, a point at radiusris rotated an angle of 2πto describe the circle. This suggests that the process above is easiest to describe in terms of a radial distance (radius of the point), an angle of rotation, and length of the cylinder in thezdirection.

The result is the cylindrical system of coordinates. It is also called a circular-cylindrical system to distinguish it from the polar-cylindrical system, but we will use the short form“cylindrical”throughout this book.

x

y z

P₂ (x₂ , y₂ , z₂) P₁ (x₁ , y₁ , z₁) r

a b

x

y z

φ r

h Figure 1.26 (a) Two points

on a cylindrical surface described in terms of their Cartesian coordinates. (b) The cylindrical coordinate system and its relation to the Cartesian system

1.5 Systems of Coordinates 27

The cylindrical system is shown inFigure1.26b. The axes are orthogonal and theϕaxis is positive in the counterclock- wise direction when viewed from the positivezaxis. Similarly,ris positive in the direction away from thezaxis whereas the zaxis is the same as for the Cartesian coordinate system and serves as the axis of the cylinder. Theraxis extends from zero to +1, thezaxis extends from –1to +1, and theϕaxis varies from 0 to 2π. Theϕangle is also called theazimuthal angleand is given with reference to thexaxis of a superposed Cartesian system. This reference also allows transformation between the two systems of coordinates.

A general vector in the cylindrical coordinate system is given as

A¼^rArðr;ϕ;zÞ þϕ^A_ϕðr;ϕ;zÞ þ^zAzðr;ϕ;zÞ ð1:54Þ All other aspects of vector algebra that we have defined are preserved. The unit vector, the magnitude of the vector, as well as vector and scalar products are evaluated in an identical fashion although the fact that one of the coordinates is an angle must be taken into account, as we shall see shortly. Since the three coordinates are orthogonal to each other, the scalar and vector products of the unit vectors are

^

r

^{^r ¼ϕ^}

^{ϕ^ ¼^z}

^{^z ¼1 ð1:55Þ}

^

r

^{ϕ^ ¼^r}

^{^z ¼ϕ^}

^{^z ¼ϕ^}

^{^r ¼^z}

^{^r ¼^z}

^{ϕ^ ¼0 ð1:56Þ}

^

r ^r ¼ϕ^ ϕ^ ¼^z^z ¼0 ð1:57Þ

^

r ϕ^ ¼^z, ϕ^ ^z ¼^r, ^z ^r ¼ϕ^, ϕ^ ^r ¼ ^z, ^z ϕ^ ¼ ^r, ^r ^z ¼ ϕ^ ð1:58Þ Next, we need to define the differentials of length, surface, and volume in the cylindrical system. This is shown in Figure1.27. The differential lengths in therandzdirections are ^rdr and^zdz, correspondingly. In theϕ direction, the differential of length is an arc of lengthrdϕ, as shown inFigure1.27. Thus, the differential length in cylindrical coordinates is

d1¼^rdrþϕ^rdϕþ^zdz ð1:59Þ

The differentials of area are

dsr¼rdϕdz, ds_ϕ¼drdz, dsz¼rdϕdr ð1:60Þ

The differential volume is therefore

dv¼dr rdϕð Þdz¼r drdϕdz ð1:61Þ

rdφ dφ

φ

ds

ds

r x

y z

ds=(rdφ)dz

z

−z r

−r φ

− φ

r

dr dz

Figure 1.27 Differentials of length, surface, and volume in cylindrical coordinates

The oriented differentials of surface are defined as for the Cartesian coordinate system in terms of unit vectors normal to the surface of a cylindrical object (seeFigure1.27):

dsr¼ ^rrdϕdz, ds_ϕ¼ ϕdrdz,^ dsz¼ ^zrdϕdr ð1:62Þ Of course, the surface vector for a general surface will vary, but it must be normal to the surface and is considered positive if it points out of the volume enclosed by the surface.

The fundamental principle that we followed in defining systems of coordinates is that the fields are independent of the system of coordinates. This also means that we can transform from one system of coordinates to another at will. All we need is to identify the transformation necessary and ensure that this transformation is unique. To find the transformation between the cylindrical and Cartesian systems, we superimpose the two systems on each other so that thezaxes of the systems coincide as inFigure 1.28a. In the Cartesian system, the point Phas coordinates (x,y,z). In the cylindrical system, the coordinates arer,ϕ,andz.

If we assume (r,ϕ,z) are known, we can transform these into the Cartesian coordinates using the relations in Figure1.28a:

x¼rcosϕ, y¼rsinϕ, z¼z ð1:63Þ

Similarly, we can write for the inverse transformation (assumingx,y, andzare known) either directly fromFigure1.28a or fromEq. (1.63):

r¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þy²

p , ϕ¼ tan¹ y x

, z¼z ð1:64Þ

These are the transformations for a single point. We also need to transform the unit vectors and, finally, the vectors from one system to the other. To transform the unit vectors, we useFigure 1.28b. First, we note that the unit vector in the z direction remains unchanged as expected. To calculate the unit vectors in therandϕdirections, we resolve the unit vectors^x and^y onto ther–ϕplane by calculating their projections in therandϕdirections. Since all unit vectors are of unit length, they all fall on the unit circle shown. Thus,

ϕ^ ¼ ^xsinϕþ^ycosϕ, ^r ¼^xcosϕþ^ysinϕ, ^z ¼^z ð1:65Þ The inverse transformation is also obtained fromFigure1.28bby an identical process:

^

x ¼^rcosϕϕ^sinϕ, ^y ¼^rsinϕþϕ^cosϕ, ^z ¼^z ð1:66Þ φ

x y

z

r

y

a b

f r

φ

rcosφ ^x y

φ φ φ

P(r,φ^,z) P(x,y,z)

x y

z

x

φ

r

−xsinφ ycos φ

90º−φ φ

f=−xsinφ+ycosφ r=xcosφ⁺ysinφ y

x

rsinφ

f

Figure 1.28 (a) Relation between unit vectors in the Cartesian and cylindrical systems. The circle has unit radius.

(b) Calculation of unit vectors in cylindrical coordinates as projections of the unit vectors in Cartesian coordinates

1.5 Systems of Coordinates 29

Important Note: Unlike in Cartesian coordinates, the unit vectors^r andϕ^ (in cylindrical coordinates) are not constant;

both depend onϕ(whereas^zis constant). Therefore, whenever they are used, such as in integration, this fact must be taken into account. It will often become necessary to transform the unit vectors into Cartesian coordinates usingEq. (1.66)to avoid this difficulty.

To obtain the transformation necessary for a vector, we use the properties of the scalar product to find the scalar components of the vector in one system of coordinates in the directions of the unit vectors of the other system. For a vector Agiven in the cylindrical system, we can write

Ax¼^x

^{A¼^x}

^{^rArþϕ^A}ϕþ^zAz

¼^ð^x

^{^rÞArþ^x}

^{ϕ^A}ϕþ^ð^x

^{^zÞAz ð1:67Þ}

FromFigure1.28b,^x

^{^r ¼ cosϕ,^x}

^{ϕ^ ¼ sinϕ, and^x}

^{^z ¼}0. Therefore,

Ax¼ArcosϕA_ϕsinϕ ð1:68Þ

Similarly, calculating the productsAy¼^y

^{AandAz¼^z}

^{A, we get}

Ay¼ArsinϕþA_ϕcosϕ and Az¼Az ð1:69Þ To find the inverse transformation, we can write vectorAin the Cartesian coordinate system and repeat the process above by finding its projections on ther,ϕ, andzaxes. Alternatively, we calculate the inverse transformation by first writing Eqs. (1.68)and(1.69)as a system of equations:

Ax

Ay

Az

2 4

3 5¼

cosϕ sinϕ 0 sinϕ cosϕ 0

0 0 1

2 4

3 5 Ar

A_ϕ Az

2 4

3

5 ð1:70Þ

Calculating the inverse of this system, we get Ar

A_ϕ Az

2 4

3

5¼ cosϕ sinϕ 0 sinϕ cosϕ 0

0 0 1

2 4

3 5 Ax

Ay

Az

2 4

3

5 ð1:71Þ

Now, for a general vector given in the cylindrical coordinate system, we can write the same vector in the Cartesian system by using the scalar components fromEq. (1.70)and adding the unit vectors. Similarly, if a vector in the Cartesian system must be transformed into the cylindrical system, we useEq. (1.71)to evaluate its components.

Example 1.16 Two points in cylindrical coordinates are given asP1(r1,ϕ1,z1) andP2(r2,ϕ2,z2). Find the expression of the vector pointing fromP1toP2:

(a) In Cartesian coordinates.

(b) In cylindrical coordinates.

(c) Calculate the length of the vector.

Solution: The cylindrical coordinates are first converted into Cartesian coordinates. After the components of the vector are found, these are transformed back into cylindrical coordinates. Calculation of the length of the vector must be done in Cartesian coordinates in which each coordinate represents the same quantity (for example, length):

(a) The coordinates of pointsP1andP2in Cartesian coordinates are [Eq. (1.63)]

x1¼r1cosϕ1, y₁ ¼r1sinϕ1, z1¼z1

x2¼r2cosϕ2, y₂ ¼r2sinϕ2, z2¼z2

Thus, the vectorAin Cartesian coordinates is A¼^xðx2x1Þ þ^y

y₂y₁ þ^z

z2z1

¼^xðr2cosϕ2r1cosϕ1Þ þ^y

r2sinϕ2r1sinϕ1

þ^z z2z1

(b) To find the components of the vector in cylindrical coordinates, we write fromEq. (1.71) Ar

A_ϕ Az

2 4

3 5¼

cosϕ sinϕ 0 sinϕ cosϕ 0

0 0 1

2 4

3

5 r2cosϕ2r1cosϕ1

r2sinϕ2r1sinϕ1

z2z1

2 4

3 5

Expanding, we get:

Ar¼ðr2cosϕ2r1cosϕ1Þcosϕþðr2sinϕ2r1sinϕ1Þsinϕ A_ϕ¼ ðr2cosϕ2r1cosϕ1Þsinϕþðr2sinϕ2r1sinϕ1Þcosϕ

Az¼z2z1

The vector in cylindrical coordinates is

Aðr;ϕ;zÞ ¼^r½ðr2cosϕ2r1cosϕ1Þcosϕþðr2sinϕ2r1sinϕ1Þsinϕ

þϕ^½ðr2cosϕ2r1cosϕ1Þsinϕþðr2sinϕ2r1sinϕ1Þcosϕ þ^z½z2z1

Note that therandϕcomponents of the vector in cylindrical coordinates are not constant (they depend on the angleϕin addition to the coordinates of pointsP1andP2). In Cartesian coordinates, the components only depend on the two end points. Because of this, it is often necessary to transform from cylindrical to Cartesian coordinates, especially when the magnitudes of vectors need to be evaluated.

(c) To calculate the length of the vector, we use the representation in Cartesian coordinates because it is easier to evaluate.

The length of the vector is

j j ¼A ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A²_xþA²_yþA²_z q

¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2cosϕ2r1cosϕ1

ð Þ²þðr2sinϕ2r1sinϕ1Þ²þðz2z1Þ² q

¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²₂þr²₁2r2r1ðcosϕ2cosϕ1þsinϕ2sinϕ1Þ þðz2z1Þ² q

With cosϕ2cosϕ1+ sinϕ2sinϕ1¼cos(ϕ2–ϕ1), we get

j j ¼A ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²₂þr²₁2r2r1cosðϕ2ϕ1Þ þðz2z1Þ² q

This expression may be used to calculate the magnitude of a vector when two points are given in cylindrical coordinates without first transforming into Cartesian coordinates.

Example 1.17 A vector is given in cylindrical coordinates as:A¼^r2þϕ^3^z1. Describe this vector in Cartesian coordinates.

Solution: The vector in Cartesian coordinates is written directly fromEqs. (1.70), (1.63),and(1.64):

Ax

Ay

Az

2 64

3 75¼

cosϕ sinϕ 0 sinϕ cosϕ 0

0 0 1

2 64

3 75

Ar

A_ϕ Az

2 64

3 75¼

cosϕ sinϕ 0 sinϕ cosϕ 0

0 0 1

2 64

3 75

2 3 1 2 64

3 75

¼

2 cosϕ3sinϕ 2sinϕþ3cosϕ

1 2

64

3 75¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi2x x²þy²

p 3x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þy² p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi2y x²þy²

p þ 3x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þy² p 1 2

66 66 66 64

3 77 77 77 75

1.5 Systems of Coordinates 31

where the following substitutions were made usingEq. (1.63):

cosϕ¼x

r, sinϕ¼y

r, r¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þy² p

Thus, vectorAis

A¼^x 2x ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þy²

p 3y

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þy² p

!

þ^y 2y ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þy²

p þ 3x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þy² p

! ^z1

Exercise 1.9 TransformA¼^x2^y5þ^z3 into cylindrical coordinates at point (x¼–2, y¼3, z¼1).

Answer A¼^r5:27ϕ^1:11þ^z3