Spherical Caps and Cut Balls

Why a Horizontal Cut Changes the Bounds

A ball centered at the origin has a simple spherical description: its radius runs from zero to the ball’s radius, and its angles cover all directions. A horizontal plane removes part of that ball. The resulting bounds depend on whether the plane lies above, below, or outside the center and on how the plane meets the sphere.

In this tutorial, \(R>0\) is the radius of a ball centered at the origin, and the cutting plane is \(z=h\). We use spherical coordinates with \(r\) the distance from the origin, \(\phi\) the angle measured down from the positive \(z\)-axis, and \(\theta\) the azimuthal angle. In particular, \(z=r\cos\phi\), and the spherical volume element is \(r^2\sin\phi\,dr\,d\phi\,d\theta\).

Definition: The portion of the ball \(x^2+y^2+z^2\le R^2\) above the plane \(z=h\) is the set of points satisfying \(r\le R\) and \(r\cos\phi\ge h\). The portion below the plane satisfies \(r\le R\) and \(r\cos\phi\le h\).

Writing the plane condition as \(r\cos\phi\ge h\) or \(r\cos\phi\le h\) is the key step. Depending on which variable is integrated first, it may give a radial bound or an angular bound.

Where the Plane Meets the Sphere

On the sphere \(r=R\), the plane equation \(z=h\) becomes \(R\cos\phi=h\). When \(-R\le h\le R\), the intersection circle therefore has polar angle \(\phi_0=\arccos(h/R)\). This angle lies in \([0,\pi]\). It is a transition angle: at the sphere, directions on one side of \(\phi_0\) meet the plane before reaching the sphere, while directions on the other side do not.

\phi_0=\arccos\left(\frac{h}{R}\right), \qquad -R\le h\le R.

For an upper cap with \(0<h<R\), a ray at angle \(\phi\) reaches the plane at \(r=h/\cos\phi\). The ray intersects the cap only when \(\phi\le\phi_0<\pi/2\), so \(\cos\phi\) is positive and the radial lower bound is well-defined. For \(h<0\), the transition angle is greater than \(\pi/2\) only when \(-R<h<0\). The cases \(h\le -R\) must be handled separately; in particular, \(\arccos(h/R)\) is not real when \(h<-R\).

Takeaway: A plane height determines an angle on the sphere only when \(-R\le h\le R\). Always compare \(h\) with \(R\) before using \(\arccos(h/R)\).

Bounds for an Upper Spherical Cap

Suppose \(0<h<R\), and consider the part of the ball above the plane. On a ray with polar angle \(\phi\), the plane condition is \(r\cos\phi\ge h\). To satisfy it, the ray must reach at least \(r=h/\cos\phi\). The outer edge of the ball is at \(r=R\). For these radial limits to form a nonempty interval, \(h/\cos\phi\le R\), or \(\cos\phi\ge h/R\). Thus \(0\le\phi\le\phi_0\).

0\le\theta\le2\pi,\qquad 0\le\phi\le\arccos(h/R),\qquad \frac{h}{\cos\phi}\le r\le R.

The cap height, measured from the plane to the top of the ball, is \(a=R-h\). The plane lies above the center when \(h>0\), so this is a cap smaller than a hemisphere. When \(h=0\), the region above the plane is a hemisphere; its spherical bounds simplify to \(0\le r\le R\), \(0\le\phi\le\pi/2\), and \(0\le\theta\le2\pi\).

Worked Example: A Cap Above \(z=3\) in a Ball of Radius \(5\)

Find spherical bounds and the volume of the part of \(x^2+y^2+z^2\le25\) above \(z=3\). Here \(R=5\), \(h=3\), and \(\phi_0=\arccos(3/5)\). The radial lower bound is \(3/\cos\phi\), so the bounds are

0\le\theta\le2\pi,\qquad 0\le\phi\le\arccos(3/5),\qquad \frac{3}{\cos\phi}\le r\le5.

Integrate the volume element over these bounds. First integrate with respect to \(r\):

\int_{3/\cos\phi}^{5}r^2\,dr =\left[\frac{r^3}{3}\right]_{3/\cos\phi}^{5} =\frac{125-27\sec^3\phi}{3}.

Then integrate over \(\theta\), giving a factor of \(2\pi\). Since \(\int\sin\phi\,d\phi=-\cos\phi\) and \(\int\sec^3\phi\sin\phi\,d\phi=\tfrac12\tan^2\phi\), the remaining integral is

\begin{aligned} V &=\frac{2\pi}{3}\int_0^{\arccos(3/5)} \left(125-27\sec^3\phi\right)\sin\phi\,d\phi\\ &=\frac{2\pi}{3}\left[ 125(1-\cos\phi_0)-\frac{27}{2}\tan^2\phi_0 \right]. \end{aligned}

Here \(\cos\phi_0=3/5\), and \(\tan^2\phi_0=(4/3)^2=16/9\). Therefore the bracket is \(125(2/5)-(27/2)(16/9)=50-24=26\), and \(V=52\pi/3\). As a check, the cap height is \(a=5-3=2\); the cap-volume formula derived below gives \(\pi a^2(3R-a)/3=\pi(4)(15-2)/3=52\pi/3\), in agreement.

Radial-First Bounds for a Cut Ball

When the plane is below the center, the region above it contains the origin. Along some downward-pointing rays, a point begins inside the region and eventually crosses the plane. For such a region, it is often simpler to integrate radially from the origin and let the plane determine an angular upper limit.

For \(-R<h<0\), consider the portion above \(z=h\). At fixed \(r\), the plane condition becomes \(\cos\phi\ge h/r\). If \(0\le r<-h\), then even the bottom of the sphere of radius \(r\) has \(z=-r>h\), so the entire spherical shell is above the plane. For \(-h\le r\le R\), the allowed polar angles end at \(\arccos(h/r)\). Thus one valid set of bounds is

\begin{cases} 0\le r\le -h,\quad 0\le\phi\le\pi,\\ -h\le r\le R,\quad 0\le\phi\le\arccos(h/r), \end{cases} \qquad 0\le\theta\le2\pi.

The threshold \(r=-h\) is where a sphere of radius \(r\) first touches the plane. For \(h=0\), this description reduces to the upper hemisphere bounds. For \(h=-R\), the portion above the plane is the entire ball except for its bottom point, which has zero volume. For \(h<-R\), the whole ball is above the plane. These endpoint and exterior cases do not require an inverse cosine.

Worked Example: The Ball of Radius \(4\) Above \(z=-2\)

Set up and evaluate the volume of the part of \(x^2+y^2+z^2\le16\) above \(z=-2\). Since \(-h=2\), the radial-first bounds are \(0\le r\le2\) with all polar angles, and \(2\le r\le4\) with \(0\le\phi\le\arccos(-2/r)\). The azimuthal angle runs from \(0\) to \(2\pi\).

For the first radial interval, \(\int_0^\pi\sin\phi\,d\phi=2\). For the second,

\int_0^{\arccos(-2/r)}\sin\phi\,d\phi =1-\cos(\arccos(-2/r)) =1+\frac{2}{r}.

Therefore

\begin{aligned} V &=2\pi\left[ \int_0^2 2r^2\,dr+ \int_2^4 r^2\left(1+\frac{2}{r}\right)\,dr \right]\\ &=2\pi\left[ \left(\frac{2r^3}{3}\right)_0^2+ \left(\frac{r^3}{3}+r^2\right)_2^4 \right]\\ &=2\pi\left[ \frac{16}{3} +\left(\frac{64}{3}+16-\frac{8}{3}-4\right) \right]\\ &=2\pi\left(\frac{16}{3}+\frac{92}{3}\right) =72\pi. \end{aligned}

Check by subtraction: the full ball has volume \(4\pi(4^3)/3=256\pi/3\). The portion below \(z=-2\) is a cap of height \(2\), whose volume is \(40\pi/3\). Their difference is \(256\pi/3-40\pi/3=72\pi\), as obtained from the integral.

Lower Caps and the Other Side of the Plane

A lower cap is the portion below a plane \(z=h\) with \(-R<h<0\). Its angle on the sphere is \(\phi_0=\arccos(h/R)\), which lies between \(\pi/2\) and \(\pi\). A ray can reach the cap only when \(\phi\ge\phi_0\), and the plane condition \(r\cos\phi\le h\) requires \(r\ge h/\cos\phi\). Both \(h\) and \(\cos\phi\) are negative over this angular range, so the radial lower bound is positive.

0\le\theta\le2\pi,\qquad \arccos(h/R)\le\phi\le\pi,\qquad \frac{h}{\cos\phi}\le r\le R, \qquad -R<h<0.

Worked Example: A Lower Cap Below \(z=-1\) in a Ball of Radius \(3\)

Find the volume of the portion of \(x^2+y^2+z^2\le9\) below \(z=-1\). The transition angle is \(\phi_0=\arccos(-1/3)\), so the bounds are \(0\le\theta\le2\pi\), \(\phi_0\le\phi\le\pi\), and \(-1/\cos\phi\le r\le3\). Integrating first in \(r\) gives

\int_{-1/\cos\phi}^{3}r^2\,dr =\frac{27+1/\cos^3\phi}{3}.

Consequently, using \(\cos\phi_0=-1/3\), \(\sin\phi/\cos^3\phi\) as the derivative of \(\tfrac12\sec^2\phi\), and \(\sec^2\phi_0=9\), we obtain

\begin{aligned} V &=\frac{2\pi}{3}\int_{\phi_0}^{\pi} \left(27+\sec^3\phi\right)\sin\phi\,d\phi\\ &=\frac{2\pi}{3}\left[ 27(1+\cos\phi_0) +\frac12\left(\sec^2\pi-\sec^2\phi_0\right) \right]\\ &=\frac{2\pi}{3}\left[ 27\left(\frac23\right)+\frac12(1-9) \right] =\frac{2\pi}{3}(18-4) =\frac{28\pi}{3}. \end{aligned}

For a check, the cap height is \(a=3-1=2\). The cap formula gives \(\pi a^2(3R-a)/3=\pi(4)(9-2)/3=28\pi/3\), confirming the integral.

Why the Spherical Cap Formula Works

The spherical bounds describe a cap, but its volume also has a simple expression in terms of its height \(a\). A direct cross-section argument explains the formula. Measure vertical distance \(s\) downward from the top of a sphere of radius \(R\). At height \(s\), the cross-section is a disk. The sphere equation implies that the squared disk radius is \(R^2-(R-s)^2=2Rs-s^2\). Integrating disk area from the top (\(s=0\)) to the cutting plane (\(s=a\)) gives

\begin{aligned} V &=\int_0^a \pi(2Rs-s^2)\,ds\\ &=\pi\left[Rs^2-\frac{s^3}{3}\right]_0^a\\ &=\frac{\pi a^2(3R-a)}{3}. \end{aligned}

This derivation applies when \(0\le a\le2R\), so the plane is between the top and bottom of the sphere. If \(a=0\), the cap is a single point and has zero volume; if \(a=2R\), it is the full ball. The formula is a useful independent check on a spherical-coordinate setup, not a substitute for identifying the correct bounds.

Common Mistake: Using an Impossible Transition Angle

A tempting but invalid step is to write \(\phi_0=\arccos(h/R)\) for every negative \(h\). If \(h=-2R\), then \(h/R=-2\), which is outside the domain \([-1,1]\) of the real inverse cosine. There is no intersection circle: the plane lies below the ball, so the portion below it is empty and the portion above it is the entire ball.

The correction is to check the height first. Use \(\arccos(h/R)\) only when \(-R\le h\le R\). For \(-R<h<0\), it gives a real angle greater than \(\pi/2\). At \(h=-R\), the portion below is just the bottom point; for \(h<-R\), it is empty. Similarly, for \(h=R\), the portion above is just the top point, and for \(h>R\), it is empty.

Bound-checking rule: Before writing a transition angle, compare the plane height with both \(R\) and \(-R\). Then check that the proposed radial interval is nonempty and that its endpoints satisfy the plane and sphere equations.

How to Choose and Check Bounds

The same cut ball may be described by integrating in different orders. For an upper cap with \(h>0\), ray-based bounds put the plane in the radial lower limit and are often efficient. For a cut ball above a negative-height plane, radial-first bounds can represent the full inner shells and the partial outer shells cleanly. In either case, the inequalities \(r\le R\) and \(r\cos\phi\ge h\) (or \(r\cos\phi\le h\)) determine the region; the picture is a useful check, but the inequalities are decisive.

Check the height.
Compare \(h\) with \(-R\), \(0\), and \(R\) to determine whether the cut is empty, tangent, a cap, or a larger portion of the ball.

Write the plane condition.
For the portion above, use \(r\cos\phi\ge h\); for the portion below, use \(r\cos\phi\le h\).

Choose the variable order.
Let the inequality give either a radial endpoint or an angular endpoint, and include the full azimuthal range \(0\le\theta\le2\pi\).

Verify the limits.
Check that the plane boundary satisfies \(r\cos\phi=h\), the sphere boundary satisfies \(r=R\), and all bounds describe points on the intended side of the plane.

These checks matter because a sign error in \(h/\cos\phi\) can make a radial lower bound negative, while using an inverse cosine outside its domain can invent a transition angle where no plane-sphere intersection exists. A final volume comparison with the cap formula or with the full-ball volume can reveal such errors.

Check Your Understanding

Use the plane condition and the ball radius to justify each answer.

For the portion above \(z=2\) in a ball of radius \(6\), what is the transition angle, and what are the radial and polar limits along a ray?
For a ball of radius \(5\) cut by \(z=-2\), why do shells with \(0\le r\le2\) have all polar angles in the portion above the plane?
What is the portion below \(z=-R\), and why should no inverse cosine be used for \(h<-R\)?
Give radial-first bounds for the portion above \(z=-1\) in a ball of radius \(3\).
For a cap of height \(a\) in a ball of radius \(R\), state the cap-volume formula and the range of \(a\) for which it applies.

What You'll Learn