The Coordinate Surfaces Of Spherical Coordinates

How Fixing a Coordinate Creates a Surface

Spherical coordinates describe a point by its distance from the origin and two angles. Fixing one of these three values while allowing the other two to vary traces out a coordinate surface. Recognizing these surfaces makes it easier to read equations written in spherical coordinates and to understand which points they include.

We use the convention in which \(\rho\) is the distance from the origin, \(\theta\) is the azimuthal angle in the \(xy\)-plane measured from the positive \(x\)-axis, and \(\phi\) is the polar angle measured down from the positive \(z\)-axis. The conversion formulas are

x=\rho\sin\phi\cos\theta,\qquad y=\rho\sin\phi\sin\theta,\qquad z=\rho\cos\phi.

The coordinate ranges are \(\rho\geq 0\), \(0\leq\theta<2\pi\), and \(0\leq\phi\leq\pi\). At the origin and on the \(z\)-axis, some angle coordinates are not unique. That ambiguity is important when interpreting the surfaces below.

Definition: A coordinate surface is the set of points obtained by fixing one coordinate and allowing the other coordinates to vary over their permitted ranges. Its geometry must be interpreted together with those ranges and the coordinate ambiguities.

Fixing \(\rho\): Spheres Centered at the Origin

Let \(\rho=c\). Since \(\rho\) is the distance from the origin, a fixed positive value \(c\) means that every point on the surface is exactly \(c\) units from the origin. To see the Cartesian equation directly, square the coordinate formulas and add:

\begin{aligned} x^2+y^2+z^2 &=\rho^2\sin^2\phi\cos^2\theta +\rho^2\sin^2\phi\sin^2\theta +\rho^2\cos^2\phi\\ &=\rho^2\sin^2\phi(\cos^2\theta+\sin^2\theta) +\rho^2\cos^2\phi\\ &=\rho^2(\sin^2\phi+\cos^2\phi)\\ &=\rho^2. \end{aligned}

Thus, if \(\rho=c\), the Cartesian equation is \(x^2+y^2+z^2=c^2\). The allowed values of \(\rho\) determine which cases actually occur: \(c>0\) gives a sphere of radius \(c\), \(c=0\) gives just the origin, and \(c<0\) gives no points. The last case is empty because a distance cannot be negative.

Worked Example: The Surface \(\rho=3\)

Set \(\rho=3\) in the coordinate formulas and eliminate the angles by squaring and adding:

\begin{aligned} x^2+y^2+z^2 &=9\sin^2\phi\cos^2\theta +9\sin^2\phi\sin^2\theta +9\cos^2\phi\\ &=9\sin^2\phi+9\cos^2\phi\\ &=9. \end{aligned}

The surface is therefore a sphere of radius \(3\) centered at the origin. For a check, choose \(\phi=\pi/3\) and \(\theta=\pi/4\). The corresponding point is

(x,y,z)= \left(\frac{3\sqrt{6}}{4},\frac{3\sqrt{6}}{4},\frac{3}{2}\right).

Its squared distance from the origin is \(54/16+54/16+9/4=27/4+9/4=9\), so its distance is \(3\), as required.

Fixing \(\theta\): A Half-Plane Through the \(z\)-Axis

Let \(\theta=c\), where \(c\) is any azimuthal angle. The formulas become

x=\rho\sin\phi\cos c,\qquad y=\rho\sin\phi\sin c,\qquad z=\rho\cos\phi.

The horizontal distance from the \(z\)-axis is \(r=\sqrt{x^2+y^2}=\rho\sin\phi\), which is nonnegative because \(\rho\geq0\) and \(0\leq\phi\leq\pi\). Hence points on this surface satisfy \(x=r\cos c\), \(y=r\sin c\), with \(r\geq0\), while \(z\) may take any real value. These are precisely the points on a vertical half-plane whose boundary is the \(z\)-axis.

For angles with \(\cos c\ne0\), one description is \(y=x\tan c\), with the additional restriction that the horizontal direction points along \(c\), not in the opposite direction. That restriction matters: the full vertical plane also contains the opposite half-plane. When \(\cos c=0\), the plane equation is \(x=0\), but the intended half-plane is selected by the sign of \(y\). In every case the \(z\)-axis is the boundary shared by the two opposite half-planes.

The azimuth is periodic: \(c\) and \(c+2\pi k\) describe the same direction for any integer \(k\). But \(c\) and \(c+\pi\) describe opposite horizontal directions, so they give opposite half-planes. Both include the \(z\)-axis. The axis is included because the coordinate formulas give \(x=y=0\) when \(\phi=0\) or \(\phi=\pi\), regardless of \(\theta\); the origin is also included at \(\rho=0\).

Worked Example: The Surface \(\theta=\pi/6\)

With \(\theta=\pi/6\), the horizontal coordinates satisfy

x=\frac{\sqrt{3}}{2}\rho\sin\phi,\qquad y=\frac{1}{2}\rho\sin\phi.

Therefore \(y=x/\sqrt{3}\), with \(x\geq0\). The surface is the half-plane in the vertical plane \(y=x/\sqrt{3}\) lying in the horizontal direction \((\cos(\pi/6),\sin(\pi/6))\), together with its \(z\)-axis boundary.

For a check, take \(\rho=2\), \(\phi=\pi/3\), and \(\theta=\pi/6\). Then \(\rho\sin\phi=\sqrt{3}\), so \(x=3/2\), \(y=\sqrt{3}/2\), and \(z=1\). We have \(y=x/\sqrt{3}=\sqrt{3}/2\) and \(x\geq0\). The point lies on the stated half-plane. The axis is included separately by taking \(\phi=0\) or \(\phi=\pi\).

Fixing \(\phi\): A Cone or a Limiting Surface

Let \(\phi=c\), with \(0\leq c\leq\pi\). The angle \(c\) fixes the direction away from the positive \(z\)-axis while \(\rho\) changes the distance from the origin and \(\theta\) sweeps around that axis. For \(0<c<\pi\), the formulas imply

\sqrt{x^2+y^2}=\rho\sin c,\qquad z=\rho\cos c.

Because \(\sin c>0\) in this range, dividing gives \(z=\sqrt{x^2+y^2}\cot c\). This equation describes a single nappe: if \(c<\pi/2\), its points lie above the \(xy\)-plane; if \(c>\pi/2\), they lie below it. The origin is the vertex. At \(c=\pi/2\), the relation becomes \(z=0\), the \(xy\)-plane, which is a limiting, degenerate case rather than a cone with a nonzero slope.

The endpoints \(c=0\) and \(c=\pi\) are also limiting cases. When \(\phi=0\), the formulas give \(x=y=0\) and \(z=\rho\geq0\): the positive \(z\)-axis. When \(\phi=\pi\), they give \(x=y=0\) and \(z=-\rho\leq0\): the negative \(z\)-axis. Thus the two endpoint surfaces are rays, not ordinary cones. Values of \(c\) outside \([0,\pi]\) are not permitted polar angles in this convention and produce no coordinate surface.

Worked Example: The Surface \(\phi=2\pi/3\)

Here \(\cot(2\pi/3)=-1/\sqrt{3}\), so the surface equation is

z=-\frac{\sqrt{x^2+y^2}}{\sqrt{3}}.

This is the lower nappe of a cone, including its vertex. To check the description using a point on the surface, choose \(\rho=4\) and \(\theta=\pi/3\). Since \(\sin(2\pi/3)=\sqrt{3}/2\) and \(\cos(2\pi/3)=-1/2\), the coordinates are

x=\sqrt{3},\qquad y=3,\qquad z=-2.

Then \(\sqrt{x^2+y^2}=\sqrt{3+9}=2\sqrt{3}\), and \(-\sqrt{x^2+y^2}/\sqrt{3}=-2=z\). The point satisfies the equation and lies below the \(xy\)-plane, as required.

Why the Cone Equation Needs Care

For \(0<\phi<\pi\), the cone equation follows by eliminating \(\rho\) from \(z=\rho\cos\phi\) and \(\sqrt{x^2+y^2}=\rho\sin\phi\). Keeping the sign of the resulting relation preserves which nappe is selected. Squaring both sides can lose that information.

Important: For a fixed interior angle \(c\), use \(z=\sqrt{x^2+y^2}\cot c\) to describe the coordinate surface. Squaring gives \(z^2=(x^2+y^2)\cot^2 c\), which generally includes an extra nappe. The endpoint cases \(c=0,\pi\) must be read from the coordinate formulas, not by substituting into a formula involving \(\cot c\).

For instance, at \(c=\pi/3\), the correct relation is \(z=\sqrt{x^2+y^2}/\sqrt{3}\), so \(z\geq0\). Squaring produces \(z^2=(x^2+y^2)/3\), which also admits points with \(z<0\). The squared equation therefore describes both nappes and is not equivalent to the fixed-angle surface unless the required sign condition is retained.

Comparing the Three Coordinate Surfaces

Fixed coordinate	Geometric surface	Important limiting or range case
\(\rho=c\)	Sphere of radius \(c\) centered at the origin, when \(c>0\)	\(c=0\): origin; \(c<0\): empty
\(\theta=c\)	Vertical half-plane through the \(z\)-axis	Angle is periodic; the \(z\)-axis is the boundary
\(\phi=c\)	One nappe of a cone, when \(0<c<\pi\) and \(c\ne\pi/2\)	\(c=\pi/2\): \(xy\)-plane; \(c=0,\pi\): positive or negative \(z\)-axis

The coordinate formulas justify each description: fixing \(\rho\) fixes distance, fixing \(\theta\) fixes horizontal direction, and fixing \(\phi\) fixes the angle to the positive \(z\)-axis. These interpretations are useful when sketching a spherical equation and when deciding whether an algebraic equation represents all, part, or more than the intended coordinate surface.

Check Your Understanding

For each question, attend to both the geometry and the permitted coordinate range.

What geometric set is described by \(\rho=0\), and why does \(\rho=-2\) describe no points?
For \(\theta=5\pi/4\), what is the boundary of the half-plane, and which horizontal direction does it select?
What surface is described by \(\phi=\pi/2\)? Explain why it is a limiting case of the cone description.
For \(\phi=\pi/4\), write the unsquared Cartesian relation and state which side of the \(xy\)-plane contains the surface.
Why does squaring the cone relation risk adding points that are not on the fixed-\(\phi\) coordinate surface?