Why planes change spherical bounds
A plane has a simple equation in Cartesian coordinates, but its spherical-coordinate equation usually involves both angles and the radius. As a result, a plane may change the lower or upper radial limit, restrict the polar angle, restrict the azimuthal angle, or do several of these at once.
Using the course convention, spherical coordinates are
where \(\rho\ge 0\), \(0\le\phi\le\pi\), and \(0\le\theta<2\pi\). Here \(\phi\) is measured down from the positive \(z\)-axis, and \(\theta\) is measured in the \(xy\)-plane from the positive \(x\)-axis.
The key is to translate the plane constraint before deciding which variable should have which limit. For example, the horizontal plane \(z=c\) becomes \(\rho\cos\phi=c\), while the vertical plane \(x=a\) becomes \(\rho\sin\phi\cos\theta=a\).
Horizontal planes
For the horizontal plane \(z=c\), substitute \(z=\rho\cos\phi\). The plane and its associated half-spaces are described by
Suppose \(c>0\). To satisfy \(z\ge c\), a point must have \(\cos\phi>0\), so \(0\le\phi<\pi/2\). Dividing by \(\cos\phi\), which is positive on this angular range, gives a radial lower bound:
If the region also lies inside a sphere \(\rho\le R\), then the radial interval can be nonempty only when \(c\sec\phi\le R\). For \(0<c\le R\), this requires \(\cos\phi\ge c/R\), or \(0\le\phi\le\arccos(c/R)\). The azimuthal angle is unrestricted by a horizontal plane, because the equation contains no \(\theta\).
Worked Example: A horizontal plane inside a sphere
Describe the spherical bounds for the part of \(\rho\le 3\) satisfying \(z\ge 1\). The plane inequality becomes \(\rho\cos\phi\ge 1\). Since the right side is positive, \(\cos\phi\) must be positive, and therefore \(0\le\phi<\pi/2\). Dividing by \(\cos\phi>0\) gives \(\rho\ge\sec\phi\). The sphere supplies \(\rho\le3\), so the radial interval is \(\sec\phi\le\rho\le3\).
For this interval to exist, its lower endpoint cannot exceed its upper endpoint:
There is no restriction on \(\theta\), so \(0\le\theta<2\pi\). The bounds are thus \(0\le\theta<2\pi\), \(0\le\phi\le\arccos(1/3)\), and \(\sec\phi\le\rho\le3\). At \(\phi=0\), the radial interval is \(1\le\rho\le3\). At \(\phi=\arccos(1/3)\), \(\cos\phi=1/3\), so the radial interval reduces to the single value \(\rho=3\). Substituting the lower radial bound into \(z\) gives \((\sec\phi)\cos\phi=1\), as required by the plane.
The sign of the plane’s height matters. For \(z\ge0\), the condition is \(\rho\cos\phi\ge0\), which restricts the polar angle to \(0\le\phi\le\pi/2\), but creates no positive radial lower bound. For \(z\ge c\) with \(c<0\), do not automatically divide by \(\cos\phi\): that factor can be positive, zero, or negative, and division by a negative number reverses an inequality. Keeping the substituted inequality \(\rho\cos\phi\ge c\) visible prevents a sign error.
Vertical planes
For the vertical plane \(x=a\), substitution gives
For \(a>0\), the inequality \(x\ge a\) requires \(\sin\phi\cos\theta>0\). Since \(\sin\phi\ge0\) for \(0\le\phi\le\pi\), admissible points must have \(0<\phi<\pi\) and \(\cos\theta>0\). On those angles, division is valid and gives
Unlike the horizontal-plane bound, this radial lower bound depends on both angles. In the standard azimuth range, \(\cos\theta>0\) on \(0\le\theta<\pi/2\) and \(3\pi/2<\theta<2\pi\). A plane \(y=b\) works in the same way, using \(y=\rho\sin\phi\sin\theta\); its azimuthal factor is \(\sin\theta\) instead of \(\cos\theta\).
Worked Example: A vertical plane through the origin
Consider the region inside \(\rho\le4\) and on the side \(x\ge0\) of the plane \(x=0\). Substitution gives \(\rho\sin\phi\cos\theta\ge0\). Away from the \(z\)-axis, \(\rho\sin\phi>0\), so the condition is \(\cos\theta\ge0\). In \(0\le\theta<2\pi\), this gives \(0\le\theta\le\pi/2\) or \(3\pi/2\le\theta<2\pi\).
The plane through the origin creates no positive lower bound on \(\rho\): every radial distance from zero out to the sphere is allowed. Thus \(0\le\rho\le4\), and \(\phi\) remains in its full range \(0\le\phi\le\pi\). As a check, when \(\theta=\pi\), \(\cos\theta=-1\), so \(x=-\rho\sin\phi\le0\), which is on the excluded side except on the plane itself. When \(\theta=0\), \(x=\rho\sin\phi\ge0\), as required. On the \(z\)-axis, where \(\sin\phi=0\), \(x=0\) for every \(\theta\); the multiple azimuth descriptions of an axis point do not change the region.
Angular feasibility and radial bounds
A shifted vertical plane can restrict the polar and azimuthal angles as well as the radial coordinate. Suppose \(x\ge a>0\) and the region is inside \(\rho\le R\). The plane inequality gives \(\rho\sin\phi\cos\theta\ge a\), while the sphere gives \(\rho\le R\). Because \(\rho\) cannot exceed \(R\), there can be a point satisfying both only if
This condition describes the angular projection of the allowed region. Once an angular pair satisfies it, the radial interval begins at \(a/(\sin\phi\cos\theta)\) and ends at \(R\). In particular, a feasible polar angle must satisfy \(\sin\phi\ge a/R\); for \(0<a\le R\), this means \(\arcsin(a/R)\le\phi\le\pi-\arcsin(a/R)\). The azimuthal restriction still depends on \(\phi\), so it should be written with that dependence rather than treated as an independent fixed interval.
Worked Example: A shifted vertical plane inside a sphere
Describe the bounds for the part of \(\rho\le2\) satisfying \(x\ge1\). Substitution gives \(\rho\sin\phi\cos\theta\ge1\). Since the left side must be positive, \(\sin\phi>0\) and \(\cos\theta>0\). The radial lower bound is \(\rho\ge1/(\sin\phi\cos\theta)\), while the sphere gives \(\rho\le2\).
For this radial interval to be nonempty,
Since \(\cos\theta\le1\), this requires \(\sin\phi\ge1/2\), giving \(\pi/6\le\phi\le5\pi/6\). For each such \(\phi\), define \(\alpha(\phi)=\arccos(1/(2\sin\phi))\). The angular condition is equivalent to \(\cos\theta\ge1/(2\sin\phi)\), so the standard azimuth intervals are \(0\le\theta\le\alpha(\phi)\) and \(2\pi-\alpha(\phi)\le\theta<2\pi\). For these angles, the radial bounds are \(1/(\sin\phi\cos\theta)\le\rho\le2\).
At \(\phi=\pi/2\), \(\alpha=\arccos(1/2)=\pi/3\). Taking \(\theta=0\) gives the radial interval \(1\le\rho\le2\); at its lower endpoint, \(x=1\cdot1\cdot1=1\). At \(\theta=\pi/3\), the lower bound becomes \(1/(\sin(\pi/2)\cos(\pi/3))=2\), so only \(\rho=2\) is allowed. This checks both the plane boundary and the point where the radial interval closes.
Why the bound procedure works
The rules above follow directly from the coordinate equations and from checking whether the radial lower limit is compatible with the radial upper limit. For a positive horizontal cut inside a sphere, this reasoning is:
The condition \(z\ge c\) becomes \(\rho\cos\phi\ge c\).
When \(c>0\), the inequality forces \(\cos\phi>0\), so division by \(\cos\phi\) does not reverse the inequality.
The plane gives \(\rho\ge c/\cos\phi\), and the sphere gives \(\rho\le R\).
The interval exists only when \(c/\cos\phi\le R\). Solving this condition gives the polar-angle range.
The same logic applies to \(x\ge a>0\), except that the coefficient of \(\rho\) is \(\sin\phi\cos\theta\). It must be positive before division, and its maximum allowed value under the radial upper bound determines which angles are feasible. This is why a radial formula alone is not a complete description: its denominator must be positive, and the lower bound must not exceed the upper one.
Common mistake: dropping the angular factor
A frequent error is to treat a plane’s Cartesian offset as the spherical radial bound. For example, from \(z\ge1\), one might incorrectly write \(\rho\ge1\). But the actual substitution gives \(\rho\cos\phi\ge1\), so when \(\phi=\pi/3\), the correct requirement is \(\rho\ge2\), not merely \(\rho\ge1\).
Thus the incorrect bound would include a point that does not satisfy the original half-space constraint. For a vertical plane the corresponding mistake is to write \(\rho\ge a\) from \(x\ge a\); the correct factor is \(\sin\phi\cos\theta\), and it can be small or even negative. Always substitute first, check signs, and verify that the proposed limits imply the original inequality.
Using plane bounds in an integral
Once the region has been described correctly, its spherical limits can be placed around the spherical volume element \(\rho^2\sin\phi\). A horizontal plane often changes the radial lower bound and the polar range, while a vertical plane may make the radial bound depend on both angles. The plane equation determines the constraint; the other surfaces determine the remaining limits. A useful final check is to substitute a boundary value into \(z=\rho\cos\phi\), \(x=\rho\sin\phi\cos\theta\), or \(y=\rho\sin\phi\sin\theta\) and confirm that it gives the stated plane.
Check Your Understanding
For each question, identify the sign and feasibility conditions before writing limits.
- Translate \(z\ge2\) into spherical coordinates. What must be true of \(\phi\) before dividing by \(\cos\phi\)?
- Inside \(\rho\le5\), what polar-angle range is feasible for \(z\ge2\)?
- For \(x\ge3\), which angles must have positive trigonometric factors before the constraint can be written as a radial lower bound?
- Inside \(\rho\le4\), what condition on \(\phi\) and \(\theta\) makes the radial interval for \(x\ge1\) nonempty?
- Why can the plane \(z=0\) restrict \(\phi\) for \(z\ge0\) without creating a positive lower bound on \(\rho\)?