CYLINDRICAL COORDINATES: SYSTEM, CHANGE AND EXERCISES - MATHS - 2024

The cylindrical coordinates are used to locate points in three dimensional space and consist of a radial coordinate ρ, φ azimuthal coordinate and z coordinate of height.

A point P located in space is projected orthogonally on the XY plane giving rise to the point P 'in that plane. The distance from the origin to the point P 'defines the coordinate ρ, while the angle between the X axis and the ray OP' defines the coordinate φ. Finally, the z coordinate is the orthogonal projection of point P on the Z axis. (see figure 1).

Figure 1. Point P of cylindrical coordinates (ρ, φ, z). (Own elaboration)

The radial coordinate ρ is always positive, the azimuthal coordinate φ varies from zero radians to two pi radians, while the z coordinate can take any real value:

0 ≤ ρ <∞

0 ≤ φ <2π

- ∞ <z <+ ∞

Change of coordinates

It is relatively easy to obtain the Cartesian coordinates (x, y, z) of a point P from its cylindrical coordinates (ρ, φ, z):

x = ρ cos (φ)

y = ρ sin (φ)

z = z

But it is also possible to obtain the polar coordinates (ρ, φ, z) starting from the knowledge of the Cartesian coordinates (x, y, z) of a point P:

ρ = √ (x ² + y ²)

φ = arctan (y / x)

z = z

Vector base in cylindrical coordinates

The base of cylindrical unit vectors Uρ, Uφ, Uz is defined.

The vector Uρ is tangent to the line φ = ctte and z = ctte (pointing radially outward), the vector Uφ is tangent to the line ρ = ctte and z = ctte and finally Uz has the same direction of the Z axis.

Figure 2. Cylindrical coordinate base. (wikimedia commons)

In the cylindrical unit base, the position vector r of a point P is written vectorially like this:

r = ρ Uρ + 0 Uφ + z Uz

On the other hand, an infinitesimal displacement d r from point P is expressed as follows:

d r = dρ Uρ + ρ dφ Uφ + dz Uz

Similarly, an infinitesimal element of volume dV in cylindrical coordinates is:

dV = ρ dρ dφ dz

Examples

There are countless examples of the use and application of cylindrical coordinates. In cartography, for example, the cylindrical projection is used, based precisely on these coordinates. There are more examples:

Example 1

Cylindrical coordinates have applications in technology. As an example we have the CHS (Cylinder-Head-Sector) system of data location on a hard disk, which actually consists of several disks:

- The cylinder or track corresponds to the coordinate ρ.

- The sector corresponds to the position φ of the disk that rotates at high angular speed.

- The head corresponds to the z-position of the reading head on the corresponding disk.

Each byte of information has a precise address in cylindrical coordinates (C, S, H).

Figure 2. Location of information in cylindrical coordinates on a hard disk system. (wikimedia commons)

Example 2

Construction cranes fix the position of the load in cylindrical coordinates. The horizontal position is defined by the distance to the axis or arrow of the crane ρ and by its angular position φ with respect to some reference axis. The vertical position of the load is determined by the z coordinate of the height.

Figure 3. The position of the load on a construction crane can easily be expressed in cylindrical coordinates. (image pixabay - annotations R. Pérez)

Solved exercises

Exercise 1

There are points P1 with cylindrical coordinates (3, 120º, -4) and point P2 with cylindrical coordinates (2, 90º, 5). Find the Euclidean distance between these two points.

Solution: First, we proceed to find the Cartesian coordinates of each point following the formula that was given above.

P1 = (3 * cos 120º, 3 * sin 120º, -4) = (-1.5, 2.60, -4)

P2 = (2 * cos 90º, 2 * sin 90º, 5) = (0, 2, 5)

The Euclidean distance between P1 and P2 is:

d (P1, P2) = √ ((0 - (-1.5)) ² + (2 - 2.60) ² + (5 - (- 4)) ²) =…

… √ (2.25 + 0.36 + 81) = 9.14

Exercise 2

Point P has Cartesian coordinates (-3, 4, 2). Find the corresponding cylindrical coordinates.

Solution: We proceed to find the cylindrical coordinates using the relationships given above:

ρ = √ (x ² + y ²) = √ ((- 3) ² + 4 ²) = √ (9 + 16) = √ (25) = 5

φ = arctan (y / x) = arctan (4 / (- 3)) = -53.13º + 180º = 126.87º

z = 2

It should be remembered that the arctangent function is multivalued with 180º periodicity. Also, angle φ must belong to the second quadrant, since the x and y coordinates of point P are in that quadrant. This is the reason why 180º has been added to the result φ.

Exercise 3

Express in cylindrical coordinates and in Cartesian coordinates the surface of a cylinder with radius 2 and whose axis coincides with the Z axis.

Solution: It is understood that the cylinder has an infinite extension in the z direction, so the equation of said surface in cylindrical coordinates is:

ρ = 2

To obtain the Cartesian equation of the cylindrical surface, the square of both members of the previous equation is taken:

ρ ² = 4

We multiply both members of the previous equality by 1 and apply the fundamental trigonometric identity (sin ² (φ) + cos ² (φ) = 1):

1 * ρ ² = 1 * 4

(sin ² (φ) + cos ² (φ)) * ρ ² = 1 * 4

The parenthesis is developed to obtain:

(ρ sin (φ)) ² + (ρ cos (φ)) ² = 4

We remember that the first parentheses (ρ sin (φ)) is the y coordinate of a point in polar coordinates, while the parentheses (ρ cos (φ)) represents the x coordinate, so that we have the equation of the cylinder in coordinates Cartesian:

y ² + x ² = 2 ²

The above equation should not be confused with that of a circumference in the XY plane, since in this case it would look like this: {y ² + x ² = 2 ²; z = 0}.

Exercise 4

A cylinder of radius R = 1 m and height H = 1m has its mass distributed radially according to the following equation D (ρ) = C (1 - ρ / R) where C is a constant of value C = 1 kg / m ³. Find the total mass of the cylinder in kilograms.

Solution: The first thing is to realize that the function D (ρ) represents the volumetric mass density, and that the mass density is distributed in cylindrical shells of decreasing density from the center to the periphery. An infinitesimal element of volume according to the symmetry of the problem is:

dV = ρ dρ 2π H

Hence, the infinitesimal mass of a cylindrical shell will be:

dM = D (ρ) dV

Therefore, the total mass of the cylinder will be expressed by the following definite integral:

M = ∫ _or ^R D (ρ) dV = ∫ _or ^R C (1 - ρ / R) ρ dρ 2π H = 2π HC ∫ _or ^R (1 - ρ / R) ρ dρ

The solution of the indicated integral is not difficult to obtain, its result being:

∫ _or ^R (1 - ρ / R) ρ dρ = (⅙) R ²

Incorporating this result in the expression of the mass of the cylinder, we obtain:

M = 2π HC (⅙) R ² = ⅓ π HCR ² =

⅓ π 1m * 1kg / m ³ * 1m ² = π / 3 kg ≈ 1.05 kg

References

1. Arfken G and Weber H. (2012). Mathematical methods for physicists. A comprehensive guide. 7th edition. Academic Press. ISBN 978-0-12-384654-9
2. Calculation cc. Solved problems of cylindrical and spherical coordinates. Recovered from: calculo.cc
3. Weisstein, Eric W. "Cylindrical Coordinates." From MathWorld – A Wolfram Web. Recovered from: mathworld.wolfram.com
4. wikipedia. Cylindrical coordinate system. Recovered from: en.wikipedia.com
5. wikipedia. Vector fields in cylindrical and spherical coordinates. Recovered from: en.wikipedia.com