Cones as Constant-Angle Surfaces
A circular cone with its vertex at the origin consists of rays that make a fixed angle with its axis. In spherical coordinates, directions are recorded by angles. That makes a cone about the \(z\)-axis especially direct to describe: its surface has a fixed polar angle \(\phi\).
We use the spherical-coordinate convention in which \(\rho\) is the distance from the origin, \(\theta\) is the angle in the \(xy\)-plane, and \(\phi\) is measured down from the positive \(z\)-axis. Thus \(\rho\ge 0\), \(0\le\theta<2\pi\), and \(0\le\phi\le\pi\). The coordinate formulas needed here are
In particular, the distance from a point to the \(z\)-axis is \(r=\sqrt{x^2+y^2}=\rho\sin\phi\). For points other than the origin, \(\phi\) tells us the direction of the point relative to the positive \(z\)-axis.
The word nappe means one of the two parts of a double cone: the upper part points in the positive \(z\)-direction, and the lower part points in the negative \(z\)-direction. Because \(\phi\) is measured from the positive \(z\)-axis, the upper nappe has a small polar angle, while the lower nappe has an angle greater than \(\frac{\pi}{2}\).
From a Cone's Angle to Its Cartesian Equation
For a point on the upper nappe, \(\phi=\alpha\). Using \(r=\rho\sin\phi\) and \(z=\rho\cos\phi\), its height and distance from the axis satisfy \(z/r=\cot\alpha\). Consequently, the upper nappe can also be written \(z=(\cot\alpha)\sqrt{x^2+y^2}\). The lower nappe has the opposite sign for \(z\). Squaring combines these two cases into the equation of the double cone.
This equation describes both nappes, not just the upper one. At the origin, both sides are zero, so the vertex is included. Away from the origin, the equation fixes the ratio between height and distance from the axis.
Worked Examples
Worked Example: Convert a Cone Equation to Spherical Coordinates
Rewrite \(z^2=3(x^2+y^2)\) in spherical coordinates, and identify its nappes.
Use \(z=\rho\cos\phi\) and \(x^2+y^2=\rho^2\sin^2\phi\). Substitution gives
For \(\rho>0\), divide by \(\rho^2\). The resulting equation is \(\cos^2\phi=3\sin^2\phi\), or \(\cot^2\phi=3\). In the permitted range \(0\le\phi\le\pi\), this gives \(\phi=\frac{\pi}{6}\) or \(\phi=\frac{5\pi}{6}\). The first is the upper nappe; the second is the lower nappe. The vertex \(\rho=0\) is also on the original cone, even though it was excluded when dividing by \(\rho^2\).
Check: At either angle, \(\cot^2\phi=3\), so \(\cos^2\phi=3\sin^2\phi\). Multiplying by \(\rho^2\) recovers \(z^2=3(x^2+y^2)\). The two angles give the two nappes, and \(\rho=0\) gives the vertex.
Worked Example: Find the Angle of an Upper Nappe
Express the upper nappe \(z=\sqrt{3}\sqrt{x^2+y^2}\) as a spherical-coordinate equation.
Replace \(z\) by \(\rho\cos\phi\) and \(\sqrt{x^2+y^2}\) by \(\rho\sin\phi\). Since \(\sin\phi\ge0\) for \(0\le\phi\le\pi\), the equation becomes
For \(\rho>0\), divide by \(\rho\). Because the equation has a positive right-hand side, its solutions must have \(\cos\phi\ge0\); the nonzero points on the upper nappe therefore have \(0\le\phi<\frac{\pi}{2}\). Dividing by \(\cos\phi\) gives \(\tan\phi=\frac{1}{\sqrt{3}}\), so \(\phi=\frac{\pi}{6}\). The vertex is included separately at \(\rho=0\).
Check: With \(\phi=\frac{\pi}{6}\), \(\cos\phi=\frac{\sqrt{3}}{2}\) and \(\sin\phi=\frac12\). Thus \(z=\frac{\sqrt{3}}{2}\rho\), while \(\sqrt{x^2+y^2}=\frac12\rho\); multiplying the latter by \(\sqrt{3}\) gives the same height.
Angle Inequalities Describe Conical Regions
An equation \(\phi=\alpha\) picks out a cone surface. An inequality such as \(0\le\phi\le\alpha\) instead describes all directions within that angle of the positive \(z\)-axis. Allowing \(\rho\) to vary fills out the corresponding solid region. The bound on \(\theta\) allows every direction around the axis; restricting \(\theta\) would keep only part of the region.
More generally, when \(0<\alpha<\frac{\pi}{2}\), the bounds \(\alpha\le\phi\le\pi-\alpha\) describe directions between the upper and lower nappes, around the plane \(\phi=\frac{\pi}{2}\). They include the horizontal directions and exclude the narrower regions near both parts of the \(z\)-axis. These are statements about angular directions; bounds on \(\rho\) can then restrict how far the region extends from the vertex.
Worked Example: Interpret an Inequality in \(\phi\)
Describe the region \(z\ge\sqrt{3}\sqrt{x^2+y^2}\) using spherical coordinates, allowing all \(\rho\ge0\) and all \(\theta\).
Substitute \(z=\rho\cos\phi\) and \(\sqrt{x^2+y^2}=\rho\sin\phi\):
At \(\rho=0\), the inequality holds and includes the origin. For \(\rho>0\), divide by the positive number \(\rho\), obtaining \(\cos\phi\ge\sqrt{3}\sin\phi\). Since \(\sin\phi\ge0\), this inequality requires \(\cos\phi\ge0\), so \(0\le\phi\le\frac{\pi}{2}\). In that range, \(\cos\phi\ge\sqrt{3}\sin\phi\) is equivalent to \(\tan\phi\le\frac{1}{\sqrt{3}}\). Therefore \(0\le\phi\le\frac{\pi}{6}\).
The region is the solid cone around the positive \(z\)-axis, including its boundary and vertex. In spherical bounds, it is \(0\le\rho<\infty\), \(0\le\theta<2\pi\), and \(0\le\phi\le\frac{\pi}{6}\).
Check: For any point in these bounds, \(\cos\phi-\sqrt{3}\sin\phi\ge0\). Multiplying by \(\rho\ge0\) gives \(\rho\cos\phi\ge\sqrt{3}\rho\sin\phi\), which is precisely the original inequality after substitution.
Derivation: Why a Cone Has a Fixed Polar Angle
Here is the geometric reason the polar-angle description works. Consider a point \((x,y,z)\ne(0,0,0)\) on a cone whose axis is the positive \(z\)-axis and whose angle to that axis is \(\alpha\). Its distance from the axis is \(r=\sqrt{x^2+y^2}\). In the right triangle formed by the point, its projection onto the \(z\)-axis, and the origin, the angle at the origin is \(\alpha\). Thus \(\tan\alpha=r/z\) on the upper nappe, or \(z=r\cot\alpha\).
They are \(r=\rho\sin\phi\) and \(z=\rho\cos\phi\).
For \(\rho>0\), \(z/r=(\rho\cos\phi)/(\rho\sin\phi)=\cot\phi\). The geometric condition \(z/r=\cot\alpha\) therefore gives \(\cot\phi=\cot\alpha\).
For the upper nappe, \(0<\phi<\frac{\pi}{2}\), so \(\phi=\alpha\). Reflection across the \(xy\)-plane changes \(z\) to \(-z\), taking \(\phi\) to \(\pi-\phi\). The lower nappe consequently has \(\phi=\pi-\alpha\).
This reasoning also explains why \(\rho\) and \(\theta\) are unrestricted on a complete cone surface: changing \(\rho\) moves along a ray, and changing \(\theta\) selects a different ray around the axis. Neither change alters the fixed angle with the axis.
Common Mistake: Keeping Only One Nappe
A frequent error is to solve a double-cone equation and report only the acute polar angle. For example, from \(z^2=3(x^2+y^2)\), the calculation \(\cot^2\phi=3\) might be followed by the incorrect conclusion “\(\phi=\frac{\pi}{6}\), so the cone is \(\phi=\frac{\pi}{6}\).” That describes only the upper nappe.
The square in \(\cot^2\phi=3\) allows both \(\cot\phi=\sqrt{3}\) and \(\cot\phi=-\sqrt{3}\). Over \(0\le\phi\le\pi\), these give \(\phi=\frac{\pi}{6}\) and \(\phi=\frac{5\pi}{6}\), respectively. Equivalently, direct substitution shows that both angles have \(\cos^2\phi=3\sin^2\phi\). The correction is to include both angular values for a double cone, unless the original equation or inequality separately restricts the sign of \(z\).
Check the Meaning Before Using the Bounds
Before accepting a spherical-coordinate description, check whether it specifies a surface or a solid region, and whether it includes the upper nappe, the lower nappe, or both. At the vertex, \(\rho=0\), the angle coordinates are not unique; this does not remove the origin from a cone or its solid interior. Keeping these distinctions clear makes angle bounds a reliable way to describe conical regions.
Check Your Understanding
For each question, use the convention that \(\phi\) is measured from the positive \(z\)-axis.
- What polar angle describes the upper nappe of a cone whose semi-vertical angle is \(\frac{\pi}{4}\)? What angle describes its lower nappe?
- Convert \(z^2=x^2+y^2\) to a condition on \(\phi\), and state whether it describes one nappe or both.
- Describe the conical region around the negative \(z\)-axis whose boundary makes an angle of \(\frac{\pi}{6}\) with that axis. Give its \(\phi\)-bounds.
- Why must the vertex be considered separately when dividing a spherical-coordinate cone equation by \(\rho\) or \(\rho^2\)?
- Does \(0\le\phi\le\frac{\pi}{4}\) describe a cone surface or a solid conical region? Explain the distinction.