# 3.2 Orthogonal Coordinate Systems

A three-dimensional coordinate system allows us to uniquely specify locations of points in space and the magnitudes and directions of vectors. Coordinate systems may be orthogonal or nonorthogonal.

Nonorthogonal systems are very specialized and seldom used in solving practical problems. Many orthogonal coordinate systems have been devised, but the most commonly used are

- the Cartesian (also called rectangular),

- the cylindrical, and

- the spherical coordinate system.

Why do we need more than one coordinate system? Whereas a point in space has the same location and an object has the same shape regardless of which coordinate system is used to describe them, the solution of a practical problem can be greatly facilitated by the choice of a coordinate system that best fits the geometry under consideration. The following subsections examine the properties of each of the aforementioned orthogonal systems, and Section 3-3 describes how a point or vector may be transformed from one system to another.

# 3.2.1 Cartesian Coordinates

The Cartesian coordinate system was introduced in Section 3-1 to illustrate the laws of vector algebra. Instead of repeating these laws for the Cartesian system, we summarize them in Table 3-1. Differential calculus involves the use of differential lengths, areas, and volumes. In Cartesian coordinates, a differential length vector (Fig. 3-8) is expressed as

d l=x ˆ d l x+y ˆ d l y+z ˆ d l z=x ˆ d x+y ˆ d y+z ˆ d z, (3.34)

where d l x=d x is a differential length along x ˆ, and similar interpretations apply to d l y=d y and d l z=d z.

# Table 3-1 Summary of vector relations.

Table 3 1 Full Alternative Text

# Figure 3-8 Differential length, area, and volume in Cartesian coordinates.

Figure 3 8 Full Alternative Text

A differential area vector d s is a vector with magnitude d s equal to the product of two differential lengths (such as d l y and d l z) and direction specified by a unit vector along the third direction (such as x ˆ). Thus, for a differential area vector in the y–z plane,

d s x=x ˆ d l y d l z=x ˆ d y d z(y–z plane), (3.35a)

with the subscript on d s denoting its direction. Similarly,

d s y=y ˆ d x d z(x–z plane), (3.35b)
d s z=z ˆ d x d y(x–y plane). (3.35c)

A differential volume equals the product of all three differential lengths:

d υ=d x d y d z. (3.36)

# 3.2.2 Cylindrical Coordinates

The cylindrical coordinate system is useful for solving problems involving structures with cylindrical symmetry, such as calculating the capacitance per unit length of a coaxial transmission line. In the cylindrical coordinate system, the location of a point in space is defined by three variables: r, ϕ, and z (Fig. 3-9). The coordinate r is the radial distance in the x–y the azimuth angle measured from the positive x axis, and z is as previously defined in the Cartesian coordinate system. Their ranges are 0≤r<∞, 0≤ϕ<2 π, and −∞<z<∞. Point P(r 1,ϕ 1,z 1) in Fig. 3-9 is located at the intersection of three surfaces. These are the cylindrical surface defined by r=r 1, the vertical half-plane defined by ϕ=ϕ 1 (which extends outwardly from the z axis), and the horizontal plane defined by z=z 1.

# Figure 3-9 Point P(r 1,ϕ 1,z 1) in cylindrical coordinates; r 1 is the radial distance from the origin in the x–y plane, ϕ 1 is the angle in the x–y plane measured from the x axis toward the y axis, and z 1 is the vertical distance from the x–y plane.

Figure 3 9 Full Alternative Text

The base unit vectors obey the following right-hand cyclic relations:

and like all unit vectors, r ˆ⋅r ˆ=ϕ ˆ⋅ϕ ˆ=z ˆ⋅z ˆ=1, and r ˆ×r ˆ=ϕ ˆ×ϕ ˆ=z ˆ×z ˆ=0.

In cylindrical coordinates, a vector is expressed as

A=a ˆ|A|=r ˆ A r+ϕ ˆ A ϕ+z ˆ A z, (3.38)

where A r, A ϕ, and A z are the components of A along the r ˆ, ϕ ˆ, and z ˆ directions. The magnitude of A is obtained by applying Eq. (3.17), which gives

|A|=A⋅A+=A r 2+A ϕ 2+A z 2+. (3.39)

The position vector O P→ shown in Fig. 3-9 has components along r and z only. Thus,

R 1=O P→=r ˆ r 1+z ˆ z 1. (3.40)

The dependence of R 1 on ϕ 1 is implicit through the dependence of r ˆ on ϕ 1. Hence, when using Eq. (3.40) to denote the position vector of point P(r 1,ϕ 1,z 1), it is necessary to specify that r ˆ is at ϕ 1.

Figure 3-10 shows a differential volume element in cylindrical coordinates. The differential lengths along r ˆ, ϕ ˆ, and z ˆ are

d l r=d r,d l ϕ=r d ϕ,d l z=d z. (3.41)

# Figure 3-10 Differential areas and volume in cylindrical coordinates.

Figure 3 10 Full Alternative Text

Note that the differential length along ϕ ˆ is r d ϕ, not just d ϕ. The differential length d l in cylindrical coordinates is given by

d l=r ˆ d l r+ϕ ˆ d l ϕ+z ˆ d l z=r ˆ d r+ϕ ˆ r d ϕ+z ˆ d z. (3.42)

As was stated previously for the Cartesian coordinate system, the product of any pair of differential lengths is equal to the magnitude of a vector differential surface area with a surface normal pointing along the direction of the third coordinate. Thus,

d s r=r ˆ d l ϕ d l z=r ˆ r d ϕ d z(ϕ–z cylindrical surface), (3.43a)
d s ϕ=ϕ ˆ d l r d l z=ϕ ˆ d r d z(r–z plane), (3.43b)
d s z=z ˆ d l r d l ϕ=z ˆ r d r d ϕ(r–ϕ plane). (3.43c)

The differential volume is the product of the three differential lengths,

d υ=d l r d l ϕ d l z=r d r d ϕ d z. (3.44)

These properties of the cylindrical coordinate system are summarized in Table 3-1.

# Example 3-4: Distance Vector in Cylindrical Coordinates

Find an expression for the unit vector of vector A shown in Fig. 3-11 in cylindrical coordinates.

# Figure 3-11 Geometry of Example 3-4.

Figure 3 11 Full Alternative Text

Solution: In triangle O P 1 P 2,

O P 2→=O P 1→+A.

Hence,

A=O P 2→−O P 1→=r^r 0−z^h,

and

a ˆ=A|A|=r ˆ r 0−z ˆ h r 0 2+h 2.

We note that the expression for A is independent of ϕ 0. This implies that all vectors from point P 1 to any point on the circle defined by r=r 0 in the x–y plane are equal in the cylindrical coordinate system, which is not true. The ambiguity can be resolved by specifying that A passes through a point whose ϕ=ϕ 0.

# Example 3-5: Cylindrical Area

Find the area of a cylindrical surface described by r=5, 30°≤ϕ≤60°, and 0≤z≤3 (Fig. 3-12).

# Figure 3-12 Cylindrical surface of Example 3-5.

Solution: The prescribed surface is shown in Fig. 3-12. Use of Eq. (3.43a) for a surface element with constant r gives

S=r∫ϕ=30°60°d ϕ∫z=0 3 d z=5 ϕ|π/6 π/3 z|0 3=5 π 2.

Note that ϕ had to be converted to radians before evaluating the integration limits.

# 3.2.3 Spherical Coordinates

In the spherical coordinate system, the location of a point in space is uniquely specified by the variables R, θ, and ϕ (Fig. 3-13). The range coordinate R, which measures the distance from the origin to the point, describes a sphere of radius R centered at the origin. The zenith angle θ is measured from the positive z axis and it describes a conical surface with its apex at the origin, and the azimuth angle ϕ is the same as in cylindrical coordinates. The ranges of R, θ, and ϕ are 0≤R<∞, 0≤θ≤π, and 0≤ϕ<2 π. The base vectors R ˆ, θ ˆ, and ϕ ˆ obey the right-hand cyclic relations:

# Figure 3-13 Point P(R 1,θ 1,ϕ 1) in spherical coordinates.

Figure 3 13 Full Alternative Text

A vector with components A R, A θ, and A ϕ is written as

A=a ˆ|A|=R ˆ A R+θ ˆ A θ+ϕ ˆ A ϕ, (3.46)

and its magnitude is

|A|=A⋅A+=A R 2+A θ 2+A ϕ 2+. (3.47)

The position vector of point P(R 1,θ 1,ϕ 1) is simply

R 1=O P→=R ˆ R 1, (3.48)

while keeping in mind that R ˆ is implicitly dependent on θ 1 and ϕ 1.

As shown in Fig. 3-14, the differential lengths along R ˆ, θ ˆ, and ϕ ˆ are

d l R=d R,d l θ=R d θ,d l ϕ=R sin θ d ϕ. (3.49)

# Figure 3-14 Differential volume in spherical coordinates.

Figure 3 14 Full Alternative Text

Hence, the expressions for the vector differential length d l, the vector differential surface d s, and the differential volume d υ are

d l=R ˆ d l R+θ ˆ d l θ+ϕ ˆ d l ϕ=R ˆ d R+θ ˆ R d θ+ϕ ˆ R sin θ d ϕ, (3.50a)

d s R=R ˆ d l θ d l ϕ=R ˆ R 2 sin θ d θ d ϕ(θ–ϕ spherical surface), (3.50b)
d s θ=θ ˆ d l R d l ϕ=θ ˆ R sin θ d R d ϕ(R–ϕ conical surface), (3.50c)
d s ϕ=ϕ ˆ d l R d l θ=ϕ ˆ R d R d θ(R–θ plane), (3.50d)
d υ=d l R d l θ d l ϕ=R 2 sin θ d R d θ d ϕ. (3.50e)

These relations are summarized in Table 3-1.

# Example 3-6: Surface Area in Spherical Coordinates

The spherical strip shown in Fig. 3-15 is a section of a sphere of radius 3 cm. Find the area of the strip.

# Figure 3-15 Spherical strip of Example 3-6.

Solution: Use of Eq. (3.50b) for the area of an elemental spherical area with constant radius R gives

S=R 2∫θ=30°60°sin θ d θ∫ϕ=0 2 π d ϕ=9(−cos θ)|30°60°ϕ|0 2 π(cm 2)=18 π(cos 30°−cos 60°)=20.7 cm 2.

# Example 3-7: Charge in a Sphere

A sphere of radius 2 cm contains a volume charge density ρ v given by

ρ v=4 cos 2 θ(C/m 3).

Find the total charge Q contained in the sphere.

Solution:

Q=∫υ ρ v d υ=∫ϕ=0 2 π∫θ=0 π∫R=0 2×10−2(4 cos 2 θ)R 2 sin θ d R d θ d ϕ=4∫0 2 π∫0 π R 3 3 0 2×10−2 sin θ cos 2 θ d θ d ϕ=32 3×10−6∫0 2 π−cos 3 θ 3 0 π d ϕ=64 9×10−6∫0 2 π d ϕ=128 π 9×10−6=44.68(μ C).

Note that the limits on R were converted to meters prior to evaluating the integral on R.