How Multiple Planes Restrict a Spherical Region
A solid inside a ball and cut by several planes is described by several inequalities that must hold at once. In spherical coordinates, some planes restrict the directions that are allowed, while others limit how far a point can travel along an allowed direction. The key is to test the inequalities along each ray from the origin, then combine their restrictions.
Use the standard spherical coordinates \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), and \(z=\rho\cos\phi\), with \(\rho\ge 0\), \(0\le\phi\le\pi\), and \(0\le\theta<2\pi\). Here \(\phi\) is the angle from the positive \(z\)-axis and \(\theta\) is the azimuthal angle in the \(xy\)-plane.
Write a direction using the unit vector \(\mathbf{u}(\phi,\theta)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)\). Every point on its ray has the form \(\rho\mathbf{u}\). A ball centered at the origin with radius \(R\) contributes the simple radial bound \(0\le\rho\le R\). Each plane inequality can then be checked by substituting \(\rho\mathbf{u}\).
What a Plane Does Along a Ray
Consider the plane \(ax+by+cz=d\) and the half-space \(ax+by+cz\le d\). Define \(q(\phi,\theta)=a\sin\phi\cos\theta+b\sin\phi\sin\theta+c\cos\phi\). Along the ray in direction \(\mathbf{u}\), the plane expression is \(\rho q\), so the half-space condition becomes \(\rho q\le d\).
The cases follow directly from \(\rho q\le d\). When dividing by \(q\), the inequality direction reverses if \(q<0\). For \(d<0\), a negative \(q\) is necessary, and \(d/q\) is then positive. When a plane is written as \(ax+by+cz\ge d\), multiply the entire inequality by \(-1\) and apply the rule to the resulting \(\le\) inequality.
A plane through the origin, for which \(d=0\), restricts directions rather than stopping a ray at a positive distance. An offset plane can instead cap the radial distance or, for a negative offset, require the ray to begin outside a positive radial threshold. In a solid bounded by a ball, each ray must satisfy the ball limit and every plane condition simultaneously.
Combining the Constraints
For several planes, first find the angular directions that satisfy all direction-only conditions. For each remaining direction, gather every radial lower and upper bound. The allowed radial interval is the intersection of those bounds with the ball’s interval \(0\le\rho\le R\). If a required lower bound exceeds an upper bound, that direction contributes no points.
Include the ball boundary and every plane defining the cut solid, with the inequality signs matching the intended side of each plane.
For each plane, collect the angular terms into \(q(\phi,\theta)\), so its condition has the form \(\rho q\le d\).
Use the sign of \(q\) to determine whether a plane restricts angles, supplies a radial bound, or excludes a direction. Intersect all resulting bounds.
For a volume integral, once the region has been described, the spherical volume element is \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\). The bounds describe the solid; the factor \(\rho^2\sin\phi\) accounts for the change in volume in spherical coordinates.
Worked Example: A Ball in the First Octant
Worked Example: A Ball in the First Octant
Describe the solid inside the ball \(x^2+y^2+z^2\le4\) and satisfying \(x\ge0\), \(y\ge0\), and \(z\ge0\).
The ball condition becomes \(\rho^2\le4\), so \(0\le\rho\le2\). Since \(0\le\phi\le\pi\), \(\sin\phi\ge0\). Thus \(z=\rho\cos\phi\ge0\) requires \(\cos\phi\ge0\), giving \(0\le\phi\le\pi/2\). For these \(\phi\), \(x\ge0\) requires \(\cos\theta\ge0\), and \(y\ge0\) requires \(\sin\theta\ge0\). Together these give \(0\le\theta\le\pi/2\).
Check the radial endpoint \(\rho=2\): \(x^2+y^2+z^2=\rho^2(\sin^2\phi+\cos^2\phi)=4\), so it lies on the sphere. At \(\phi=\pi/3\) and \(\theta=\pi/4\), all three coordinate factors are nonnegative, so the point lies in the first octant. At \(\theta=0\), \(y=0\); at \(\theta=\pi/2\), \(x=0\). These endpoints correctly include the two bounding coordinate planes. The azimuth is not unique on the positive \(z\)-axis, where \(\phi=0\), but allowing every \(\theta\) there describes the same axis and causes no error.
Worked Example: Two Offset Planes in a Ball
Worked Example: Two Offset Planes in a Ball
Describe the part of the ball \(x^2+y^2+z^2\le16\) in the first octant that also satisfies \(x\le1\) and \(y\le2\).
The first-octant restrictions give \(0\le\phi\le\pi/2\) and \(0\le\theta\le\pi/2\), as in the previous example. The ball gives \(0\le\rho\le4\). In this angular range, \(\sin\phi\cos\theta\ge0\), so \(x\le1\) gives \(\rho\le1/(\sin\phi\cos\theta)\) when the denominator is positive. Likewise, \(y\le2\) gives \(\rho\le2/(\sin\phi\sin\theta)\) when its denominator is positive. Thus the combined radial bound is
A quotient whose denominator is zero means that the corresponding plane imposes no radial upper bound on that ray: there the relevant coordinate is zero and already satisfies its inequality. This convention applies, for example, to \(x\le1\) when \(\cos\theta=0\), and to \(y\le2\) when \(\sin\theta=0\). At \(\phi=0\), both \(x\) and \(y\) are zero, so neither plane restricts the positive \(z\)-axis.
Check a direction with \(\phi=\pi/3\), \(\theta=\pi/6\). Then \(\sin\phi=\sqrt3/2\), \(\cos\theta=\sqrt3/2\), and \(\sin\theta=1/2\). The two plane limits are \(1/((\sqrt3/2)(\sqrt3/2))=4/3\) and \(2/((\sqrt3/2)(1/2))=8/\sqrt3\). Since \(4/3<4\) and \(4/3<8/\sqrt3\), the ray ends at \(\rho=4/3\). Substitution gives \(x=(4/3)(3/4)=1\) and \(y=(4/3)(\sqrt3/4)=\sqrt3/3<2\), while \(\rho=4/3<4\). The active plane is \(x=1\), and the other constraints hold.
Worked Example: An Angular Cut and an Offset Cut
Worked Example: An Angular Cut and an Offset Cut
Describe the region inside the ball of radius \(3\) satisfying \(x\ge0\), \(y\ge0\), \(x-y\ge0\), \(z\ge0\), and \(x+y\le2\).
The conditions \(x\ge0\) and \(y\ge0\) put \(\theta\) in the first quadrant. The additional inequality \(x-y\ge0\) becomes \(\rho\sin\phi(\cos\theta-\sin\theta)\ge0\). For directions with \(\sin\phi>0\), this requires \(\cos\theta\ge\sin\theta\), which within the first quadrant gives \(0\le\theta\le\pi/4\). The condition \(z\ge0\) gives \(0\le\phi\le\pi/2\). The positive \(z\)-axis, where \(\phi=0\), also satisfies the original inequalities; its azimuth is arbitrary, though it is represented by the endpoint \(\phi=0\) for every allowed \(\theta\).
The ball gives \(\rho\le3\). In the permitted angular region, \(x+y=\rho\sin\phi(\cos\theta+\sin\theta)\), whose angular coefficient is nonnegative. The plane \(x+y\le2\) therefore gives the radial upper limit \(2/(\sin\phi(\cos\theta+\sin\theta))\) whenever that denominator is positive. The spherical bounds are
When the denominator is zero, as on the positive \(z\)-axis, the plane \(x+y\le2\) imposes no radial restriction. To check a direction away from that axis, take \(\phi=\pi/6\) and \(\theta=\pi/6\). Then \(\sin\phi=1/2\), \(\cos\theta=\sqrt3/2\), and \(\sin\theta=1/2\). The plane limit is \(2/((1/2)(\sqrt3/2+1/2))=8/(\sqrt3+1)=4(\sqrt3-1)\). This is less than \(3\), since \(\sqrt3<7/4\), so it is the active radial limit.
At that endpoint, \(x=\rho\sqrt3/4=\sqrt3(\sqrt3-1)\), \(y=\rho/4=\sqrt3-1\), and \(z=\rho\sqrt3/2=2\sqrt3(\sqrt3-1)\). These values give \(x+y=4(\sqrt3-1)(\sqrt3+1)/4=2\), \(x-y=4-2\sqrt3=2(2-\sqrt3)>0\), and \(z>0\). Also \(x^2+y^2+z^2=\rho^2=16(\sqrt3-1)^2<9\), as required by the ball. The chosen endpoint satisfies the plane boundary and all other inequalities.
Why the Minimum Bound Is Essential
In the second example, it would be incorrect to use only the ball bound \(\rho\le4\) or only one of the plane bounds. A point on a ray belongs to the solid only if it satisfies every defining inequality. Therefore, when several inequalities produce upper radial bounds, the permitted endpoint is the smallest of those bounds. Similarly, all lower radial bounds must be respected, and the largest lower bound controls the beginning of the allowed interval.
A common mistake is to divide by an angular coefficient without checking its sign. For example, the condition \(x\le1\) becomes \(\rho\sin\phi\cos\theta\le1\). If \(\sin\phi\cos\theta>0\), then \(\rho\le1/(\sin\phi\cos\theta)\). But if that coefficient is negative, dividing by it reverses the inequality and would give a lower bound instead. In the first octant the coefficient is nonnegative, which is why the upper-bound formula works there. At coefficient zero, division is not allowed; the original inequality reads \(0\le1\) and supplies no restriction.
Using the Bounds to Set Up an Integral
After the bounds are established, they can be inserted into an iterated integral in an order consistent with the described region. For example, the first-octant portion of the ball of radius \(2\) has volume setup
For the regions with competing offset-plane limits, the radial endpoint is the minimum of the ball radius and the plane limits. One may keep that minimum in the setup when the purpose is to describe the bounds. Determining where one formula becomes smaller than another can require splitting the angular domain; the important task here is to express each constraint correctly before deciding whether a split is needed for further evaluation.
Check Your Understanding
Use the sign of each angular coefficient and the original plane inequality to justify your answers.
- For a ray with direction \(\mathbf{u}\), what quantity replaces \(ax+by+cz\) after writing a point as \(\rho\mathbf{u}\)?
- When does the half-space \(ax+by+cz\le d\), with \(d>0\), give an upper radial bound?
- Why does the inequality \(x\le1\) impose no restriction on a ray along which \(x=0\)?
- What angular ranges describe the first-octant directions?
- If two planes and a ball produce three upper radial bounds on the same ray, how do you combine them?