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Cones and Angular Bounds

Translate a cone’s geometry into polar-angle limits and use those limits to set up and evaluate spherical integrals over cone-shaped regions.

Intermediate 10 min read

What You'll Learn

  • Relate a cone’s semi-angle to its equation in cylindrical and spherical coordinates.
  • Identify the spherical bounds for the interior of a cone about the positive \(z\)-axis.
  • Describe the exterior of that cone within a ball using angular limits.
  • Set up spherical volume integrals for bounded conical regions.
  • Avoid confusing the cone’s slope with its polar-angle bound.

From a Cone to an Angular Bound

A cone with its vertex at the origin has a useful feature in spherical coordinates: its surface is described by a constant angle. Instead of expressing the cone boundary as a relationship between height and distance from the axis, we can specify the same boundary by fixing the polar angle.

This makes cones especially convenient when setting up spherical integrals. Once the cone’s angle is identified, its interior and exterior can be described by angular limits. The region still needs radial bounds to make it finite, but its conical side is handled directly by the polar-angle bound.

Spherical Coordinates and the Cone Angle

We use spherical coordinates \((\rho,\phi,\theta)\), where \(\rho\) is the distance from the origin, \(\phi\) is the polar angle measured down from the positive \(z\)-axis, and \(\theta\) is the azimuthal angle in the \(xy\)-plane. The coordinate relations and volume element are

$$ x=\rho\sin\phi\cos\theta,\qquad y=\rho\sin\phi\sin\theta,\qquad z=\rho\cos\phi, \qquad dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

Here \(0\leq\phi\leq\pi\), \(0\leq\theta<2\pi\), and \(\rho\geq0\). The cylindrical distance from the \(z\)-axis is \(r=\sqrt{x^2+y^2}=\rho\sin\phi\).

Consider a right circular cone whose axis is the positive \(z\)-axis and whose vertex is at the origin. Let \(\alpha\), with \(0<\alpha<\pi/2\), be its semi-angle: the angle between the axis and a line on the cone. A point on the cone satisfies \(r=z\tan\alpha\), with \(z\geq0\).

Definition: The solid interior of this cone consists of points whose direction from the origin makes a polar angle no greater than \(\alpha\) with the positive \(z\)-axis. In spherical coordinates, the cone surface is \(\phi=\alpha\), and its interior is \(0\leq\phi\leq\alpha\).

The corresponding equation in Cartesian or cylindrical coordinates is \(z=r\cot\alpha\). These equations agree because \(r=\rho\sin\phi\) and \(z=\rho\cos\phi\). On a cone ray, their ratio gives \(\tan\phi=r/z=\tan\alpha\), so the polar angle is constant along the ray.

$$ z=r\cot\alpha \quad\Longleftrightarrow\quad \rho\cos\phi=\rho\sin\phi\cot\alpha \quad\Longleftrightarrow\quad \phi=\alpha. $$

The equivalence describes the cone surface away from its vertex; the vertex itself is also on the surface. The angle \(\alpha\) is sometimes called the cone’s half-angle. It is not the angle between the cone’s side and the horizontal plane.

Why the Angular Bound Works

For a point other than the origin, \(\phi\) measures the direction of its position vector relative to the positive \(z\)-axis. Changing \(\rho\) moves along that direction but does not change \(\phi\). Thus, every ray from the vertex either lies inside the cone or outside it, according to its polar angle.

Geometric rule: For a cone about the positive \(z\)-axis with semi-angle \(\alpha\), the cone interior has \(0\leq\phi\leq\alpha\). The portion outside that interior has \(\alpha\leq\phi\leq\pi\), if all directions below the cone are included.

The exterior bound refers to the complement of the entire upward cone interior, not just the part of the exterior above the \(xy\)-plane. If only the upper half-space is included, the exterior angular range is instead \(\alpha\leq\phi\leq\pi/2\).

A double cone consists of two nappes, one about each direction of the \(z\)-axis. If the upper nappe has semi-angle \(\alpha\), its opposite nappe has polar angles \(\pi-\alpha\leq\phi\leq\pi\). The solid between the two nappes’ surfaces and around the axis has angular ranges \(0\leq\phi\leq\alpha\) and \(\pi-\alpha\leq\phi\leq\pi\); these are two separate intervals.

Worked Example: Volume Inside a Cone and a Ball

Worked Example: Volume Inside a Cone and a Ball

Find the volume inside the cone \(z=\sqrt{3}\,r\) and the ball \(\rho\leq2\).

First determine the semi-angle. Since \(z=r\cot\alpha\), we have \(\cot\alpha=\sqrt{3}\), so \(\alpha=\pi/6\). The region is inside the cone, so \(0\leq\phi\leq\pi/6\). It fills all azimuthal directions, and the ball supplies the radial bound \(0\leq\rho\leq2\). Therefore,

$$ V=\int_0^{2\pi}\int_0^{\pi/6}\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

Evaluate the radial integral first:

$$ \int_0^2\rho^2\,d\rho =\left[\frac{\rho^3}{3}\right]_0^2 =\frac{8}{3}. $$

The polar-angle integral is

$$ \int_0^{\pi/6}\sin\phi\,d\phi =\left[-\cos\phi\right]_0^{\pi/6} =1-\frac{\sqrt{3}}{2}. $$

The azimuthal integral is \(\int_0^{2\pi}d\theta=2\pi\). Multiplying the three factors gives

$$ V =\frac{8}{3}\left(1-\frac{\sqrt{3}}{2}\right)(2\pi) =\frac{8\pi}{3}(2-\sqrt{3}). $$

As a check, the radial factor is \(2^3/3=8/3\), the angular factor is positive because \(0<\pi/6\), and the result is smaller than the ball’s volume \(32\pi/3\), as it must be. The calculation uses the cone’s interior angle, not its exterior.

Setting Up Integrals Over Cone Regions

For a function \(f(x,y,z)\), a triple integral over a conical region is obtained by writing the integrand in spherical coordinates, multiplying by the volume element, and using the cone angle in the \(\phi\)-limits. The angular limits alone do not make an unbounded cone finite; a finite region also needs suitable radial bounds, such as a bounding sphere.

Setup rule: If a region has \(a\leq\rho\leq b\), \(0\leq\phi\leq\alpha\), and \(0\leq\theta\leq2\pi\), then $$ \iiint_E f(x,y,z)\,dV = \int_0^{2\pi}\int_0^\alpha\int_a^b f(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi) \,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$ Here \(0\leq a<b<\infty\), and the region includes every azimuthal direction.

When \(f=1\), the integral gives volume. For other integrands, the same bounds describe the region, but the function must also be converted using the spherical coordinate relations. The cone controls the polar-angle limit; it does not by itself determine the radial limit.

Worked Example: The Exterior of a Cone Within a Ball

Worked Example: The Exterior of a Cone Within a Ball

Find the volume inside \(\rho\leq3\) but outside the cone \(z=r/\sqrt{3}\), including the directions below the \(xy\)-plane.

The cone equation \(z=r\cot\alpha\) gives \(\cot\alpha=1/\sqrt{3}\), hence \(\alpha=\pi/3\). The exterior consists of directions with \(\pi/3\leq\phi\leq\pi\). The ball supplies \(0\leq\rho\leq3\), and all values of \(\theta\) are included:

$$ V=\int_0^{2\pi}\int_{\pi/3}^{\pi}\int_0^3 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

The radial factor is

$$ \int_0^3\rho^2\,d\rho =\left[\frac{\rho^3}{3}\right]_0^3 =9. $$

The angular factors are

$$ \int_{\pi/3}^{\pi}\sin\phi\,d\phi =\left[-\cos\phi\right]_{\pi/3}^{\pi} =1+\frac12 =\frac32, \qquad \int_0^{2\pi}d\theta=2\pi. $$

Thus, \(V=9\cdot(3/2)\cdot2\pi=27\pi\). To check, the ball has volume \(4\pi(3^3)/3=36\pi\). The cone interior within that ball has volume \(9\pi\), obtained by integrating over \(0\leq\phi\leq\pi/3\). Their difference is \(36\pi-9\pi=27\pi\), in agreement with the exterior integral.

Worked Example: A Finite Conical Shell

Worked Example: A Finite Conical Shell

Find the volume inside the cone with semi-angle \(\pi/4\), between the spheres \(\rho=1\) and \(\rho=2\).

“Inside the cone” gives \(0\leq\phi\leq\pi/4\), while the spherical shell gives \(1\leq\rho\leq2\). Since the region goes all the way around the axis, \(0\leq\theta\leq2\pi\). Hence

$$ V=\int_0^{2\pi}\int_0^{\pi/4}\int_1^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. $$

Compute the radial factor:

$$ \int_1^2\rho^2\,d\rho =\left[\frac{\rho^3}{3}\right]_1^2 =\frac{8-1}{3} =\frac{7}{3}. $$

The polar and azimuthal factors are

$$ \int_0^{\pi/4}\sin\phi\,d\phi =1-\cos\left(\frac{\pi}{4}\right) =1-\frac{\sqrt{2}}{2}, \qquad \int_0^{2\pi}d\theta=2\pi. $$

Therefore,

$$ V=\frac{7}{3}\left(1-\frac{\sqrt{2}}{2}\right)(2\pi) =\frac{7\pi}{3}(2-\sqrt{2}). $$

As a check, this must equal the volume inside the cone and \(\rho\leq2\) minus the volume inside the same cone and \(\rho\leq1\). Those volumes are respectively \(\frac{16\pi}{3}(1-\frac{\sqrt{2}}{2})\) and \(\frac{2\pi}{3}(1-\frac{\sqrt{2}}{2})\). Their difference is \(\frac{14\pi}{3}(1-\frac{\sqrt{2}}{2})=\frac{7\pi}{3}(2-\sqrt{2})\), as calculated.

A Common Angle Error

A frequent mistake is to read the cone’s coefficient as the tangent of its polar angle without checking which ratio the coefficient represents. For example, for \(z=\sqrt{3}\,r\), the incorrect calculation might be

$$ \tan\alpha=\sqrt{3} \quad\Longrightarrow\quad \alpha=\frac{\pi}{3}. $$

This uses the wrong ratio: the equation gives \(z/r=\sqrt{3}\), while \(\tan\phi=r/z\). The correction is

$$ \tan\alpha=\frac{r}{z}=\frac{1}{\sqrt{3}} \quad\Longrightarrow\quad \alpha=\frac{\pi}{6}, $$

or, equivalently, \(\cot\alpha=z/r=\sqrt{3}\). A reliable check is to substitute the angle into \(z=r\cot\alpha\): \(\cot(\pi/6)=\sqrt{3}\), which reproduces the given cone equation.

Choosing the Limits Carefully

Before writing an integral, identify three separate features of the region: whether it is inside or outside the cone, how far it extends from the origin, and whether it includes all directions around the axis. For the cones considered here, the cone determines the \(\phi\)-range, a ball or shell determines the \(\rho\)-range, and a complete rotation around the axis uses the full \(\theta\)-range.

  • For the interior of an upward cone, use \(0\leq\phi\leq\alpha\).
  • For the full exterior within a ball, use \(\alpha\leq\phi\leq\pi\).
  • For a finite conical shell, combine the cone’s angular limit with finite radial limits.

These bounds describe the region before the integrand is chosen. Keeping the geometry and the volume element distinct helps prevent a missing factor of \(\rho^2\sin\phi\) or an angle that describes the wrong side of the cone.

Check Your Understanding

Use the cone’s equation and the stated geometric region to determine the relevant polar-angle limits.

  1. For \(z=2r\), what is the cone’s semi-angle \(\alpha\)?
  2. Within a ball, what polar-angle interval describes the interior of a cone with semi-angle \(\pi/5\)?
  3. Within a ball, what interval describes the exterior of that same cone if directions below the \(xy\)-plane are included?
  4. Write a spherical-coordinate integral for the volume inside a cone of semi-angle \(\pi/4\) with \(2\leq\rho\leq5\), including all azimuthal directions.
  5. For the cone \(z=r/\sqrt{3}\), explain why its semi-angle is \(\pi/3\), rather than \(\pi/6\).

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