Cylindrical Coordinates

Cartesian coordinates (Section 4.2) are not convenient in certain cases. One of these is when the problem has cylindrical symmetry. For example, in the Cartesian coordinate system, the cross-section of a cylinder concentric with the

-axis requires two coordinates to describe:

and

. However, this cross section can be described using a single parameter – namely the radius – which is

in the cylindrical coordinate system. This results in a dramatic simplification of the mathematics in some applications.

The cylindrical system is defined with respect to the Cartesian system in Figure 4.3.1. In lieu of

and

, the cylindrical system uses

, the distance measured from the closest point on the

axis¹, and

, the angle measured in a plane of constant

, beginning at the

axis (

) with

increasing toward the

direction.

Figure 4.3.1: Cylindrical coordinate system and associated basis vectors. Image used with permission (CC BY SA 4.0; K. Kikkeri).

The basis vectors in the cylindrical system are

, and

. As in the Cartesian system, the dot product of like basis vectors is equal to one, and the dot product of unlike basis vectors is equal to zero. The cross products of basis vectors are as follows:

A useful diagram that summarizes these relationships is shown in Figure 4.3.2.

Figure 4.3.2: Cross products among basis vectors in the cylindrical system. (See Figure 4.1.10 for instructions on the use of this diagram.)Image used with permission (CC BY SA 4.0; K. Kikkeri).

The cylindrical system is usually less useful than the Cartesian system for identifying absolute and relative positions. This is because the basis directions depend on position. For example,

is directed radially outward from the

axis, so

for locations along the

-axis but

for locations along the

axis. Similarly, the direction

varies as a function of position. To overcome this awkwardness, it is common to set up a problem in cylindrical coordinates in order to exploit cylindrical symmetry, but at some point to convert to Cartesian coordinates. Here are the conversions:

and

is identical in both systems. The conversion from Cartesian to cylindrical is as follows:

where

is the four-quadrant inverse tangent function; i.e.,

in the first quadrant (

), but possibly requiring an adjustment for the other quadrants because the signs of both

and

are individually significant.²

Similarly, it is often necessary to represent basis vectors of the cylindrical system in terms of Cartesian basis vectors and vice-versa. Conversion of basis vectors is straightforward using dot products to determine the components of the basis vectors in the new system. For example,

in terms of the basis vectors of the cylindrical system is

The last term is of course zero since

. Calculation of the remaining terms requires dot products between basis vectors in the two systems, which are summarized in Table 4.3.1. Using this table, we find

and of course

requires no conversion. Going from Cartesian to cylindrical, we find

$.$	$\hat{\bf \rho}$	$\hat{\bf \phi}$	$\hat{\bf z}$
$\hat{\bf x}$	$\cos\phi$	$-\sin\phi$
$\hat{\bf y}$	$\sin\phi$
$\hat{\bf z}$	$0$		$1$

Table 4.3.1: Dot products between basis vectors in the cylindrical and Cartesian coordinate systems.

Integration Over Length

A differential-length segment of a curve in the cylindrical system is described in general as

Note that the contribution of the

coordinate to differential length is

, not simply

. This is because

is an angle, not a distance. To see why the associated distance is

, consider the following. The circumference of a circle of radius

. If only a fraction of the circumference is traversed, the associated arclength is the circumference scaled by

, where

is the angle formed by the traversed circumference. Therefore, the distance is

, and the differential distance is

As always, the integral of a vector field

over a curve

To demonstrate the cylindrical system, let us calculate the integral of

when

is a circle of radius

in the

plane, as shown in Figure 4.3.3. In this example,

since

and

are both constant along

. Subsequently,

and the above integral is

i.e., this is a calculation of circumference.

Figure 4.3.3: Example in cylindrical coordinates: The circumference of a circle. Image used with permission (CC BY SA 4.0; K. Kikkeri).

Note that the cylindrical system is an appropriate choice for the preceding example because the problem can be expressed with the minimum number of varying coordinates in the cylindrical system. If we had attempted this problem in the Cartesian system, we would find that both xx

and

vary over

, and in a relatively complex way.³

Integration Over Area

Now we ask the question, what is the integral of some vector field

over a circular surface

in the

plane having radius

? This is shown in Figure 4.3.4. The differential surface vector in this case is

Figure 4.3.4: Example in cylindrical coordinates: The area of a circle. Image used with permission (CC BY SA 4.0; K. Kikkeri).

The quantities in parentheses of Equation 4.3.16 are the radial and angular dimensions, respectively. The direction of

indicates the direction of positive flux – see the discussion in Section 4.2 for an explanation. In general, the integral over a surface is

To demonstrate, let’s consider

; in this case

and the integral becomes

which we recognize as the area of the circle, as expected. The corresponding calculation in the Cartesian system is quite difficult in comparison.

Whereas the previous example considered a planar surface, we might consider instead a curved surface. Here we go. What is the integral of a vector field

over a cylindrical surface

concentric with the

axis having radius

and extending from

? This is shown in Figure 4.3.5.

Figure 4.3.5: Example in cylindrical coordinates: The area of the curved surface of a cylinder. Image used with permission (CC BY SA 4.0; K. Kikkeri).

The differential surface vector in this case is

The integral is

which is the area of

, as expected. Once again, the corresponding calculation in the Cartesian system is quite difficult in comparison.

Integration Over Volume

The differential volume element in the cylindrical system is

For example, if

and the volume

is a cylinder bounded by

and

, then

i.e., area times length, which is volume.

Once again, the procedure above is clearly more complicated than is necessary if we are interested only in computing volume. However, if the integrand is not constant-valued then we are no longer simply computing volume. In this case, the formalism is appropriate and possibly necessary.

Footnotes

1
Note that some textbooks use “
$r$
” in lieu of
$\rho$
for this coordinate.
2
Note that this function is available in MATLAB and Octave as
$\text{atan}2\left ( \text{y,x} \right )$
.
3
Nothing will drive this point home more firmly than trying it. It can be done, but it’s a lot more work...