Spherical Wedges and Azimuthal Bounds

Why Azimuthal Bounds Describe Wedges

A vertical wedge is bounded by vertical half-planes whose common edge is the \(z\)-axis. Each boundary half-plane points in a particular horizontal direction. In spherical coordinates, that direction is recorded by the azimuthal angle, so the wedge can often be described with a single interval for that angle.

This is useful because the three spherical variables describe different parts of a region: the radial coordinate \(\rho\) controls distance from the origin, the polar angle \(\phi\) controls direction from the positive \(z\)-axis, and the azimuthal angle \(\theta\) controls direction around that axis. A wedge restriction is therefore naturally combined with radial and polar bounds rather than substituted for them.

Coordinate convention: In this tutorial, \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), and \(z=\rho\cos\phi\), with \(\rho\geq 0\), \(0\leq\phi\leq\pi\), and azimuth \(\theta\) measured from the positive \(x\)-axis in the \(xy\)-plane. Angles differing by \(2\pi\) describe the same azimuth.

From a Boundary Half-Plane to an Angle

A vertical half-plane through the \(z\)-axis can be specified by the horizontal direction it contains. If that direction makes angle \(\alpha\) with the positive \(x\)-axis, then the half-plane consists of points whose horizontal projection points in direction \(\alpha\), together with every possible height. Except on the \(z\)-axis, this means \(\theta=\alpha\).

Two such half-planes, with azimuths \(\alpha\) and \(\beta\), bound a wedge. Once an orientation and the intended side are specified, the wedge is described by an interval from \(\alpha\) to \(\beta\). The width of that interval is the wedge's azimuthal angle. Because azimuth is periodic, the interval might need to cross \(2\pi\), rather than run from the smaller listed angle to the larger one.

Definition: An azimuthal interval is a range of \(\theta\)-values describing the directions of horizontal projections in a region. A wedge bounded by vertical half-planes through the \(z\)-axis is described by choosing the interval between the azimuths of its two boundary half-planes. The \(z\)-axis itself has no unique azimuth, but it lies on the limiting boundaries and does not affect a volume description.

For instance, the first-quadrant wedge in the \(xy\)-plane has \(x\geq 0\) and \(y\geq 0\), so its azimuths run from \(0\) to \(\pi/2\). The same conclusion follows from \(x=\rho\sin\phi\cos\theta\) and \(y=\rho\sin\phi\sin\theta\): away from the \(z\)-axis, both coordinates are nonnegative precisely when \(\theta\) is in that quadrant.

Deriving Angle Restrictions from Horizontal Inequalities

Horizontal inequalities can be converted into angle conditions by substituting the spherical formulas for \(x\) and \(y\). Since \(\rho\sin\phi\geq 0\), this common factor can be removed when it is positive. The remaining trigonometric inequalities determine the permitted azimuths. At points on the \(z\)-axis, where \(\rho\sin\phi=0\), the horizontal inequalities hold with equality or reduce to zero; those points are handled as boundary points.

Derivation: Suppose a point satisfies \(x\geq 0\) and \(y\geq 0\). Away from the \(z\)-axis, substituting the spherical coordinate formulas gives \(\rho\sin\phi\cos\theta\geq 0\) and \(\rho\sin\phi\sin\theta\geq 0\). Dividing by the positive factor \(\rho\sin\phi\) yields \(\cos\theta\geq 0\) and \(\sin\theta\geq 0\). In one full turn, both inequalities hold exactly for \(0\leq\theta\leq\pi/2\). Thus the first-quadrant restriction is the azimuthal interval \([0,\pi/2]\), including its two boundary half-planes.

The same procedure works for other linear inequalities involving \(x\) and \(y\). A useful check is to test a direction clearly inside the proposed wedge and another clearly outside it. If their inequalities do not match the proposed interval, the angle range or its orientation needs correction.

Combining the Three Spherical Bounds

A complete spherical description states the permitted ranges of all three coordinates. For a simple wedge-shaped solid, these can be written as a radial interval, a polar interval, and an azimuthal interval. With the spherical volume element \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\), the corresponding integral setup follows the same bounds. If an azimuthal interval crosses the zero-angle direction, it can be written as a single unwrapped interval or as two intervals within \([0,2\pi]\).

Takeaway: The azimuthal interval specifies how far around the \(z\)-axis the region extends; it does not specify how far from the origin the region extends or how high or low it lies. Those restrictions must still be stated using \(\rho\) and \(\phi\).

Worked Example: A First-Quadrant Wedge in the Upper Half of a Ball

Describe and find the volume of the part of the ball \(\rho\leq 3\) that satisfies \(x\geq 0\), \(y\geq 0\), and \(z\geq 0\).

First, \(x\geq 0\) and \(y\geq 0\) give \(0\leq\theta\leq\pi/2\), as derived above. The condition \(z\geq 0\) gives \(\rho\cos\phi\geq 0\), hence \(0\leq\phi\leq\pi/2\). The ball supplies \(0\leq\rho\leq 3\). Thus the full spherical description is

0\leq\rho\leq 3,\qquad 0\leq\phi\leq\frac{\pi}{2},\qquad 0\leq\theta\leq\frac{\pi}{2}.

The wedge occupies one quarter of the azimuthal directions, and the condition \(z\geq0\) selects the upper hemisphere. Together they give one eighth of the ball. The ball of radius \(3\) has volume \(\frac{4}{3}\pi(3)^3=36\pi\), so the requested volume is \(36\pi\cdot\frac18=\frac{9\pi}{2}\).

Check: the proposed volume is positive and is one eighth of \(36\pi\), as required by the two independent restrictions. A point such as \((1,1,1)\) satisfies all three inequalities and has radius \(\sqrt{3}<3\), confirming that the described region is nonempty.

Worked Example: A Spherical Shell in a Quarter-Turn Wedge

Describe and find the volume of the solid between the spheres \(\rho=1\) and \(\rho=2\), inside the wedge \(x\geq0,\ y\geq x\).

For a point away from the \(z\)-axis, \(x\geq0\) means \(\cos\theta\geq0\), so \(\theta\) lies between \(-\pi/2\) and \(\pi/2\), modulo \(2\pi\). The second inequality becomes \(\sin\theta\geq\cos\theta\). In that right-half-plane interval, this selects \(\pi/4\leq\theta\leq\pi/2\). Equivalently, the wedge lies between the rays of azimuth \(\pi/4\) and \(\pi/2\). The full bounds are

1\leq\rho\leq2,\qquad 0\leq\phi\leq\pi,\qquad \frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}.

The azimuthal width is \(\pi/2-\pi/4=\pi/4\), one eighth of a full turn of \(2\pi\). Since the radial shell and the full range of \(\phi\) do not favor any azimuth, the wedge contains one eighth of the entire spherical shell. The shell volume is the radius-\(2\) ball minus the radius-\(1\) ball:

\frac{4}{3}\pi(2)^3-\frac{4}{3}\pi(1)^3 =\frac{32\pi}{3}-\frac{4\pi}{3} =\frac{28\pi}{3}.

Therefore the wedge volume is \(\frac18\cdot\frac{28\pi}{3}=\frac{7\pi}{6}\). Check: a point with \(\rho=3/2,\phi=\pi/2,\theta=3\pi/8\) has horizontal coordinates with \(x>0\) and \(y>x\), and its radius is strictly between \(1\) and \(2\). It lies inside the stated bounds.

Intervals That Cross the Zero-Angle Direction

Angles are periodic, so a geometrically connected wedge need not correspond to a simple interval from a smaller number to a larger number in the standard range \(0\leq\theta<2\pi\). For example, the directions from \(3\pi/2\) through \(2\pi\) to \(\pi/4\) make one connected wedge. In the standard range, that set is two intervals. Alternatively, an unwrapped interval from \(3\pi/2\) to \(9\pi/4\) describes it continuously, since \(9\pi/4=2\pi+\pi/4\).

Worked Example: Intersecting a Wrapped Wedge with a Polar Band

Describe the part of the region \(1\leq\rho\leq4\), \(\pi/3\leq\phi\leq2\pi/3\) that also satisfies \(x\geq0\) and \(y\leq x\).

First, \(x\geq0\) requires \(\cos\theta\geq0\), so in \([0,2\pi)\) the azimuth lies in \([0,\pi/2]\cup[3\pi/2,2\pi)\). The inequality \(y\leq x\) becomes \(\sin\theta\leq\cos\theta\). Intersecting these conditions leaves \([0,\pi/4]\cup[3\pi/2,2\pi)\). The endpoint \(2\pi\) is the same direction as \(0\), so it can be included when describing this closed wedge. Equivalently, the azimuthal range is the unwrapped interval \([3\pi/2,9\pi/4]\), whose width is \(9\pi/4-3\pi/2=3\pi/4\).

Combining the bounds gives

1\leq\rho\leq4,\qquad \frac{\pi}{3}\leq\phi\leq\frac{2\pi}{3},\qquad \theta\in\left[\frac{3\pi}{2},2\pi\right]\cup \left[0,\frac{\pi}{4}\right].

To check the intersection, take \(\rho=2,\phi=\pi/2,\theta=0\). This gives \((x,y,z)=(2,0,0)\), which satisfies \(x\geq0\) and \(y\leq x\), and all three coordinate values lie within the listed ranges. In contrast, at the same \(\rho\) and \(\phi\) but \(\theta=\pi/2\), the point is \((0,2,0)\); it fails \(y\leq x\). The excluded direction is therefore correctly absent from the azimuthal intervals.

A Common Mistake: Choosing the Wrong Side

Knowing the two boundary angles is not always enough: the inequalities determine which side of each boundary is intended. A frequent error is to convert each inequality into a familiar-looking interval and then forget to intersect the intervals.

For example, \(x\geq0\) alone permits azimuths in \([0,\pi/2]\cup[3\pi/2,2\pi]\), not all azimuths from \(0\) to \(\pi/2\) if the intended region also includes directions near the negative \(y\)-axis. If the additional condition is \(y\leq x\), it cuts out some directions from the first interval and retains directions from the second. Replacing the intersection with just \([0,\pi/2]\) would incorrectly discard part of the wedge.

Correction: Translate each horizontal inequality into its own condition on \(\theta\), intersect the resulting sets on the circle, and test a direction from each proposed interval. If the interval crosses \(0\), write it as two standard intervals or use a clearly identified unwrapped interval.

Why the Bounds Matter

Correct azimuthal bounds make the spherical description match the geometry, and they prevent a common source of missing or duplicated regions in a triple-integral setup. The three bounds have distinct jobs: radial limits determine the distance range, polar limits determine the range above and below the \(xy\)-plane, and azimuthal limits determine the wedge around the \(z\)-axis. Writing them separately makes it easier to see whether a solid has been fully described.

Read the wedge boundaries.
Record the horizontal direction, or azimuth, of each vertical half-plane through the \(z\)-axis.

Determine the intended side.
Use the inequalities or a test point to decide which angular interval lies inside the wedge.

Handle periodicity.
Represent a wrapped interval as two pieces in \([0,2\pi]\), or as one unwrapped interval, and state which convention is being used.

Add the other bounds.
Specify the radial and polar ranges independently so that all parts of the solid are accounted for.

Check Your Understanding

For each question, identify the azimuthal interval or explain how it should be determined.

What azimuthal interval describes the region \(x\geq0,\ y\geq0\) in one full turn?
Why must the conditions \(x\geq0\) and \(y\leq x\) be intersected rather than treated as unrelated angle ranges?
Write the azimuthal interval from \(3\pi/2\) through \(2\pi\) to \(\pi/4\) as two intervals in \([0,2\pi]\).
What do the radial, polar, and azimuthal bounds each specify in a spherical description?
Give one reason to test a point from a proposed azimuthal interval in the original inequalities.