Horizontal Cuts in Spherical Coordinates
A horizontal plane has equation \(z=c\), where \(c\) is constant. A spherical-coordinate point has height \(z=\rho\cos\phi\), so a horizontal cut is described by the relation \(\rho\cos\phi=c\). The challenge is to turn that relation into bounds that describe the desired part of a ball without dividing by zero or including points outside the ball.
We will work with the ball centered at the origin with radius \(R>0\). The spherical-coordinate convention is \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), and \(z=\rho\cos\phi\), with \(\rho\ge0\), \(0\le\phi\le\pi\), and \(0\le\theta<2\pi\). Thus, the ball is given by \(0\le\rho\le R\).
Upper Spherical Caps
Fix a height \(h\) with \(0\le h\le R\). The upper cap at or above the plane \(z=h\) consists of points in the ball satisfying \(\rho\cos\phi\ge h\). When \(h>0\), this inequality forces \(\cos\phi>0\), so it is valid to divide by \(\cos\phi\). The ray at angle \(\phi\) reaches height \(h\) at radius \(h/\cos\phi\).
The angular limit follows from requiring the lower radial bound to be no larger than \(R\). For \(h>0\), this means \(h/\cos\phi\le R\), or \(\cos\phi\ge h/R\). Since \(0\le\phi\le\pi/2\) in the upper half-space, this is equivalent to \(\phi\le\arccos(h/R)\).
Worked Example: An Upper Cap Above \(z=3\)
Describe the part of the ball of radius \(5\) lying at or above \(z=3\). Here \(R=5\) and \(h=3\), so \(0<h<R\). The angular limit is \(\arccos(3/5)\), and the radial coordinate must begin at the point where the ray reaches height \(3\):
For a check, take \(\cos\phi=4/5\), so the radial bounds are \(3/(4/5)=15/4\) and \(5\). Choosing \(\rho=4\) gives \(z=4(4/5)=16/5\), which is at least \(3\). At the limiting angle, where \(\cos\phi=3/5\), the lower radial bound is \(3/(3/5)=5\). The only permitted radius on that ray is \(5\), the point where the cap meets the sphere.
Why the Cap Bounds Work
The bounds come directly from the height inequality and the ball constraint; they do not rely on a sketch alone. For \(h>0\), the condition \(\rho\cos\phi\ge h\) first requires \(\cos\phi>0\). Dividing by this positive quantity gives \(\rho\ge h/\cos\phi\). The ball requires \(\rho\le R\), so a point is possible on a ray only if \(h/\cos\phi\le R\).
For \(z\ge h>0\), require \(\rho\cos\phi\ge h\), which forces \(\cos\phi>0\).
Dividing by \(\cos\phi\) gives \(\rho\ge h/\cos\phi\); being in the ball also requires \(\rho\le R\).
The radial interval is nonempty exactly when \(h/\cos\phi\le R\), or \(\phi\le\arccos(h/R)\).
This also explains why the case \(h=0\) must be handled separately. At the equatorial angle \(\phi=\pi/2\), the expression \(h/\cos\phi\) would be \(0/0\), which is undefined. But the original inequality at that angle is simply \(0\ge0\), so the full radial interval \(0\le\rho\le R\) is allowed. Using the separate half-ball bounds includes that equatorial radial segment correctly.
Lower Spherical Caps
The lower cap at or below \(z=-h\) is the reflection of the upper cap across the \(xy\)-plane. In the lower half-space, \(\cos\phi<0\), and \(z\le-h\) means \(\rho\cos\phi\le-h\). Equivalently, \(\rho(-\cos\phi)\ge h\). For \(h>0\), division by the positive quantity \(-\cos\phi\) gives the radial lower bound.
Worked Example: A Lower Cap Below \(z=-2\)
Inside the ball of radius \(5\), describe the points satisfying \(z\le-2\). Here \(h=2\), so the angular range is from \(\pi-\arccos(2/5)\) to \(\pi\), and the radial lower bound is \(2/(-\cos\phi)\):
For a check, choose \(\phi=2\pi/3\), for which \(\cos\phi=-1/2\). The radial range is \(2/(1/2)=4\le\rho\le5\). At \(\rho=4\), \(z=4(-1/2)=-2\); at \(\rho=5\), \(z=-5/2\). Both satisfy \(z\le-2\). At the angular endpoint, \(\cos\phi=-2/5\), and the radial lower bound is \(2/(2/5)=5\), where the cap meets the sphere.
A Slab Between Two Positive Heights
A region between two horizontal planes is a slab. Inside the ball, the positive-height slab \(a\le z\le b\), where \(0<a<b<R\), has two different radial patterns as \(\phi\) changes. On rays close to the positive \(z\)-axis, the ray reaches the upper plane before leaving the ball. On more slanted rays, it leaves the ball before reaching that upper plane.
To derive the split, the height inequalities give \(a/\cos\phi\le\rho\le b/\cos\phi\), provided \(\cos\phi>0\). The upper radial limit must also obey \(\rho\le R\). The upper plane's radial intersection \(b/\cos\phi\) is inside the ball when \(\cos\phi\ge b/R\), which is \(\phi\le\alpha\). For \(\alpha\le\phi\le\beta\), that intersection lies at or beyond the sphere, so the ball's boundary \(R\) becomes the radial upper bound. No point in the slab is possible for \(\phi>\beta\), because even the sphere's outer edge is below height \(a\).
Worked Example: The Slab \(1\le z\le3\) in a Ball
Let \(R=5\), \(a=1\), and \(b=3\). The angle thresholds are \(\alpha=\arccos(3/5)\) and \(\beta=\arccos(1/5)\). Therefore the region is the union of
and
On the axis, \(\phi=0\), the first interval gives \(1\le\rho\le3\), and \(z=\rho\) lies between \(1\) and \(3\). For a ray in the second interval, take \(\cos\phi=1/2\), or \(\phi=\pi/3\). Its radial range is \(2\le\rho\le5\): the lower bound gives \(z=2(1/2)=1\), while the outer edge gives \(z=5(1/2)=5/2\le3\). At the split angle, \(\cos\phi=3/5\), the first interval's upper radial bound is \(3/(3/5)=5\), agreeing with the second interval's bound.
When a Slab Begins at the Equator
If \(a=0\), the positive slab reaches the equatorial plane. The angle \(\phi=\pi/2\) is included, so formulas containing \(a/\cos\phi\) cannot be used there: they give \(0/0\). The original condition \(0\le z\le b\), together with \(z=\rho\cos\phi\), instead gives a radial lower bound of zero throughout the allowed upper half-space.
The first angular interval ends where the upper plane reaches the sphere. In the second interval, the sphere's boundary is at height no greater than \(b\), so the entire radius from \(0\) to \(R\) is allowed. In particular, at \(\phi=\pi/2\), every radius satisfies \(z=0\), and the equatorial radial segment is included.
Common Pitfall: Dividing at the Equator
A tempting shortcut is to use \(\rho\ge h/\cos\phi\) for all \(h\ge0\) and all upper-half angles. For \(h=0\) and \(\phi=\pi/2\), this substitutes into \(0/\cos(\pi/2)=0/0\), which is undefined. Replacing that undefined expression with zero without explanation can hide an incorrect or incomplete description.
The correction is to return to the original inequality at the exceptional angle. For the upper half-ball, \(z\ge0\) becomes \(\rho\cos(\pi/2)=0\ge0\), true for every \(0\le\rho\le R\). Thus the correct bounds at \(h=0\) use the full radial interval and include \(\phi=\pi/2\). The same check is essential for a slab whose lower height is zero.
Geometric Meaning of the Cut
Where a plane \(z=h\) meets the sphere \(\rho=R\), substitution gives \(R\cos\phi=h\). For \(0\le h\le R\), this intersection occurs at the polar angle \(\phi=\arccos(h/R)\). The boundary circle lies at that fixed angle while \(\theta\) runs around its full range. This is why the cap's angular limit is determined by the sphere-plane intersection, while its radial bound determines where each ray first reaches the plane.
For a cap with \(h=R\), the upper cap reduces to the north pole, \(\rho=R\), \(\phi=0\); the full \(\theta\)-range merely repeats that point. For \(h=0\), it is a hemisphere, and the equator must be retained. These limiting cases are useful checks on any proposed bounds.
Check Your Understanding
For each question, identify the allowed angles and radii, and check any endpoint where a denominator could vanish.
- Describe the part of the ball \(\rho\le4\) satisfying \(z\ge2\).
- Why must the upper half-ball at \(z\ge0\) be stated separately from the positive-height cap formula?
- Describe the lower cap \(z\le-1\) inside the ball \(\rho\le3\).
- For the slab \(2\le z\le4\) inside \(\rho\le6\), which angle separates the two radial-bound patterns?
- In the region \(0\le z\le2\) inside \(\rho\le5\), what radial interval is allowed at \(\phi=\pi/2\)?