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Cylindrical and Spherical Coordinates · Tutorial 278 of 999

Angle Conventions And Equivalent Cylindrical Coordinates

Choose consistent angle representatives for cylindrical coordinates and recognize when different angle values describe the same point.

Intermediate 10 min read

What You'll Learn

  • State the standard cylindrical-coordinate convention and the role of the angle theta.
  • Choose the representative of a direction in either of two common angle intervals.
  • Explain why adding an integer multiple of two pi preserves a direction.
  • Convert angles between the intervals zero to two pi and negative pi to pi.
  • Identify why theta is unrestricted when the radial coordinate is zero.

Why Angle Conventions Matter

Cylindrical coordinates describe a point by its distance from the \(z\)-axis, its direction around that axis, and its height. The direction is specified by an angle, but the same direction can be named by infinitely many angle values. An agreed interval lets us choose one representative in most cases and makes descriptions consistent.

This choice is not merely a matter of notation. If a problem asks for an angle in a particular interval, an equivalent angle outside that interval does not meet the requested convention. There is also an important exception: on the \(z\)-axis, the angle does not identify a direction at all, because every value of the angle gives the same point.

Definition: In cylindrical coordinates, a point is represented by \((r,\theta,z)\), where \(r\ge 0\) is its distance from the \(z\)-axis, \(\theta\) is an angle in the \(xy\)-plane measured from the positive \(x\)-axis, and \(z\) is its height. The corresponding Cartesian coordinates are \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\).

Here the usual convention \(r\ge 0\) is important. The angle specifies the direction of the horizontal displacement from the \(z\)-axis; it does not specify a signed distance. The angle conventions in this tutorial concern how to name that direction.

Equivalent Angles

A complete turn around the origin changes an angle by \(2\pi\) radians but returns to the same direction. More generally, adding or subtracting any whole number of complete turns leaves the direction unchanged.

Definition: Two angles \(\theta_1\) and \(\theta_2\) are equivalent if \(\theta_2=\theta_1+2\pi k\) for some integer \(k\). They point in the same direction in the \(xy\)-plane.

For a fixed positive radius, equivalent angles give the same horizontal coordinates. This follows from the periodicity of sine and cosine: for every integer \(k\), \(\cos(\theta+2\pi k)=\cos\theta\) and \(\sin(\theta+2\pi k)=\sin\theta\). Thus the corresponding cylindrical coordinates give the same Cartesian point when \(r\) and \(z\) are fixed.

$$ r\cos(\theta+2\pi k)=r\cos\theta,\qquad r\sin(\theta+2\pi k)=r\sin\theta. $$

Conversely, for a positive radius, if two angles give the same direction, their difference is an integer multiple of \(2\pi\). Geometrically, each angle identifies a point on the unit circle, and one full revolution is the \(2\pi\)-radian period. Therefore, when \(r>0\), the possible angle values for a fixed direction are exactly \(\theta+2\pi k\), with \(k\in\mathbb Z\).

Key takeaway: When \(r>0\), the direction is determined by an equivalence class of angles, not by one unrestricted real number. A specified interval selects a preferred representative from that class.

Two Common Angle Intervals

Two standard conventions for cylindrical angles are \(0\le\theta<2\pi\) and \(-\pi<\theta\le\pi\). Each interval has length \(2\pi\), so it represents every direction once. The endpoint choices matter: including both ends would count one direction twice, since the endpoints differ by \(2\pi\).

ConventionIntervalWhere the negative \(x\)-axis is represented
Nonnegative angles\(0\le\theta<2\pi\)\(\theta=\pi\)
Symmetric angles\(-\pi<\theta\le\pi\)\(\theta=\pi\), not \(-\pi\)

With the interval \(0\le\theta<2\pi\), angles increase counterclockwise from the positive \(x\)-axis. Directions just below the negative \(x\)-axis have angles just less than \(2\pi\); directions just above it have angles just greater than \(\pi\). With the interval \(-\pi<\theta\le\pi\), directions immediately on either side of the negative \(x\)-axis have angles near \(\pi\) and \(-\pi\). The negative \(x\)-axis itself is represented by \(\pi\), because \(-\pi\) is excluded.

To change an angle between these conventions, add or subtract \(2\pi\) until the result lies in the requested interval. The direction does not change; only its representative does.

Worked Example: Changing an Angle to the Symmetric Interval

Write the cylindrical point \((4,11\pi/6,-2)\) using the convention \(-\pi<\theta\le\pi\), then verify that the point is unchanged.

The given angle \(11\pi/6\) is greater than \(\pi\), so subtract one full turn:

$$ \frac{11\pi}{6}-2\pi = \frac{11\pi}{6}-\frac{12\pi}{6} = -\frac{\pi}{6}. $$

Since \(-\pi<-\pi/6\le\pi\), the requested representation is \((4,-\pi/6,-2)\). To check it, calculate the horizontal coordinates:

$$ x=4\cos\left(-\frac{\pi}{6}\right) =4\cdot\frac{\sqrt{3}}{2} =2\sqrt{3}, \qquad y=4\sin\left(-\frac{\pi}{6}\right) =4\cdot\left(-\frac{1}{2}\right) =-2. $$

The original angle gives the same values because \(11\pi/6=-\pi/6+2\pi\). Thus both triples represent the Cartesian point \((2\sqrt{3},-2,-2)\).

How to Select an Angle from a Point

When a point is given in Cartesian coordinates, its horizontal direction is determined by \((x,y)\). For \(r>0\), \(r=\sqrt{x^2+y^2}\), and an angle must point in the same direction as that pair. In practical terms, locate the point in the \(xy\)-plane, determine its quadrant or axis, and then choose the equivalent angle that lies in the required interval.

The equation \(\tan\theta=y/x\), when \(x\ne0\), can help find a reference angle, but it does not by itself determine the quadrant. Tangent has period \(\pi\), so angles that differ by \(\pi\) have the same tangent even though they point in opposite directions. The signs of \(x\) and \(y\) must be used to select the correct direction.

Worked Example: Choosing Equivalent Angles for a Point

Find cylindrical coordinates for the Cartesian point \((-3,-3\sqrt{3},1)\), using each of the intervals \(0\le\theta<2\pi\) and \(-\pi<\theta\le\pi\).

First calculate the radius:

$$ r=\sqrt{x^2+y^2} =\sqrt{(-3)^2+(-3\sqrt{3})^2} =\sqrt{9+27} =\sqrt{36} =6. $$

Both \(x\) and \(y\) are negative, so the point lies in the third quadrant. The ratio of the absolute values is \(|y/x|=\sqrt{3}\), giving reference angle \(\pi/3\). In the interval \(0\le\theta<2\pi\), the third-quadrant angle is \(\pi+\pi/3=4\pi/3\). In the interval \(-\pi<\theta\le\pi\), the equivalent angle is \(4\pi/3-2\pi=-2\pi/3\).

For the first convention, check the Cartesian coordinates:

$$ x=6\cos\left(\frac{4\pi}{3}\right) =6\left(-\frac{1}{2}\right) =-3, \qquad y=6\sin\left(\frac{4\pi}{3}\right) =6\left(-\frac{\sqrt{3}}{2}\right) =-3\sqrt{3}. $$

For the second convention, the values agree:

$$ x=6\cos\left(-\frac{2\pi}{3}\right)=-3, \qquad y=6\sin\left(-\frac{2\pi}{3}\right)=-3\sqrt{3}. $$

In both cases \(z=1\). The two cylindrical descriptions are \((6,4\pi/3,1)\) and \((6,-2\pi/3,1)\), and they represent the same point.

Angles on the Coordinate Axes

Axis directions make the endpoint conventions especially visible. The positive \(x\)-axis has angle \(0\) in the interval \(0\le\theta<2\pi\), and also angle \(0\) in the interval \(-\pi<\theta\le\pi\). The negative \(x\)-axis has angle \(\pi\) in both intervals. The positive \(y\)-axis has angle \(\pi/2\); the negative \(y\)-axis has angle \(3\pi/2\) in the nonnegative convention and \(-\pi/2\) in the symmetric convention.

Worked Example: The Negative \(y\)-Axis

Write the point \((0,-5,4)\) in cylindrical coordinates under both angle conventions.

The radius is

$$ r=\sqrt{0^2+(-5)^2}=5. $$

The point lies on the negative \(y\)-axis. Its angle is \(3\pi/2\) in \(0\le\theta<2\pi\), or \(-\pi/2\) in \(-\pi<\theta\le\pi\). Check using either representative:

$$ 5\cos\left(\frac{3\pi}{2}\right)=0,\qquad 5\sin\left(\frac{3\pi}{2}\right)=-5, $$ $$ 5\cos\left(-\frac{\pi}{2}\right)=0,\qquad 5\sin\left(-\frac{\pi}{2}\right)=-5. $$

Including \(z=4\), the two descriptions are \((5,3\pi/2,4)\) and \((5,-\pi/2,4)\). Each gives the original Cartesian point \((0,-5,4)\).

Why a Specified Interval Gives One Representative

Any real angle can be shifted by whole turns until it lies in \(0\le\theta<2\pi\). For instance, for an angle \(\alpha\), choose the integer \(k\) so that \(k\le\alpha/(2\pi)<k+1\). Then \(\alpha-2\pi k\) lies in that interval. The representative is unique: if two angles in the interval differed by \(2\pi k\) for a nonzero integer \(k\), their difference would have absolute value at least \(2\pi\), even though two members of the interval differ by less than \(2\pi\). This is impossible.

The same idea applies to \(-\pi<\theta\le\pi\): shift by full turns until the angle is in that interval. Its length is \(2\pi\), and exactly one endpoint is included, so every direction again has exactly one representative. For any point with \(r>0\), choosing one of these intervals therefore removes the ambiguity in the angle.

Theorem: If \(r>0\) and a convention specifies either \(0\le\theta<2\pi\) or \(-\pi<\theta\le\pi\), each horizontal direction has exactly one allowed angle representative. Without a specified interval, all representatives \(\theta+2\pi k\), \(k\in\mathbb{Z}\), describe the same direction.

The Special Case \(r=0\)

When \(r=0\), the point lies on the \(z\)-axis. The horizontal coordinates are \(x=0\) and \(y=0\) for every angle, because multiplying either sine or cosine by zero gives zero. Consequently, \(\theta\) does not identify a direction and is not determined by the point. Even after an interval is specified, every angle in that interval represents the same point on the axis for the same \(z\).

Worked Example: Nonuniqueness on the \(z\)-Axis

Consider the point \((0,0,-2)\). Its cylindrical radius is

$$ r=\sqrt{0^2+0^2}=0. $$

Using \(\theta=0\) gives \(x=0\cos 0=0\) and \(y=0\sin 0=0\). Using \(\theta=\pi/3\) also gives \(x=0\cos(\pi/3)=0\) and \(y=0\sin(\pi/3)=0\). Both choices have \(z=-2\), so both \((0,0,-2)\) and \((0,\pi/3,-2)\) represent the same Cartesian point. In fact, every real value of \(\theta\) gives \((0,0,-2)\) when \(r=0\).

A Common Mistake: Using Tangent Without Checking the Quadrant

Suppose \(x=-3\) and \(y=-3\sqrt{3}\). A tempting calculation is \(\tan\theta=y/x=\sqrt{3}\), followed by \(\theta=\pi/3\). That answer is wrong for this point: \(\pi/3\) lies in the first quadrant, where both coordinates would be positive.

Checking the signs corrects the calculation. Since both coordinates are negative, the point is in the third quadrant, and the angle in \(0\le\theta<2\pi\) is \(\pi+\pi/3=4\pi/3\). Substitution confirms the correction: \(6\cos(4\pi/3)=-3\) and \(6\sin(4\pi/3)=-3\sqrt{3}\). The lesson is to use the quadrant as well as a tangent ratio; the ratio alone does not select the direction.

Why the Convention Is Useful

Consistent angle ranges make cylindrical descriptions easier to compare and prevent an equivalent angle from being mistaken for a different direction. When converting a point, first identify the direction and then adjust its angle to the interval required by the problem. When checking a proposed representation, substitute into \(x=r\cos\theta\) and \(y=r\sin\theta\). Finally, check whether \(r=0\): if it is, the angle is unrestricted, not a uniquely determined coordinate.

Summary: Adding \(2\pi k\) to \(\theta\) preserves the direction. The intervals \(0\le\theta<2\pi\) and \(-\pi<\theta\le\pi\) each select one representative for every direction when \(r>0\). If \(r=0\), every angle represents the same point on the \(z\)-axis.

Check Your Understanding

Use exact angles and state which interval you are using where relevant.

  1. Convert \(7\pi/4\) to an equivalent angle in \(-\pi<\theta\le\pi\).
  2. Give the angle for the negative \(x\)-axis in each of the two standard intervals. Why is \(-\pi\) excluded from the symmetric interval?
  3. A point has \(r=3\) and \(\theta=5\pi/3\). Give an equivalent angle in \(-\pi<\theta\le\pi\), and explain why the direction is unchanged.
  4. For the Cartesian point \((0,6,2)\), find \(r\) and an allowed angle in each standard interval.
  5. Why does a point with \(r=0\) have no uniquely determined value of \(\theta\)?