Worked Spherical Coordinate Conversions

From Formulas to Reliable Conversions

Spherical coordinates describe a point by its distance from the origin and two angles. The formulas are compact, but a correct conversion requires more than substitution: the angle must match the point's quadrant, the coordinate ranges must be respected, and special points such as the origin need care.

This tutorial works through representative points and surfaces. Each conversion will be checked by substituting back into the spherical formulas or comparing the resulting equation with the original Cartesian geometry.

Definition: In spherical coordinates, \(\rho\) is the distance from the origin, \(\theta\) is the angle in the \(xy\)-plane measured from the positive \(x\)-axis, and \(\phi\) is the angle measured down from the positive \(z\)-axis. The standard formulas and ranges are

\begin{aligned} x&=\rho\sin\phi\cos\theta,\\ y&=\rho\sin\phi\sin\theta,\\ z&=\rho\cos\phi, \end{aligned} \qquad \rho\ge 0,\quad 0\le\theta<2\pi,\quad 0\le\phi\le\pi.

For a Cartesian point other than the origin, its spherical coordinates can be found from

\rho=\sqrt{x^2+y^2+z^2},\qquad \phi=\arccos\left(\frac{z}{\rho}\right).

The angle \(\theta\) is determined by the direction of \((x,y)\) in the \(xy\)-plane. It must be selected using the signs of both \(x\) and \(y\), not just the quotient \(y/x\). On the \(z\)-axis, \(\theta\) can be chosen arbitrarily because \(x=y=0\). At the origin, \(\rho=0\), and neither angle is determined by the point.

A Conversion Strategy

When converting a point, first find \(\rho\), then identify the correct quadrant for \(\theta\), and finally use \(z/\rho\) to select \(\phi\) in the required interval. When converting a surface equation, replace \(x,y,z\) using the defining formulas and simplify. Keep track of any divisions: dividing by \(\rho\), for example, excludes the origin and may require checking it separately.

Record the geometry.
Note the Cartesian equation or the coordinates of the point, including signs and any special cases.

Convert carefully.
Use the spherical definitions and the stated angle ranges; do not infer an angle from a ratio without checking its quadrant.

Check the result.
Substitute the spherical coordinates back into the Cartesian formulas, or simplify the converted equation back to the stated geometry.

Worked Examples: Points and Surfaces

Worked Example: Converting a Point in the Second Quadrant

Convert the point \((-2,2\sqrt{3},4)\) to spherical coordinates.

First calculate the distance from the origin:

\rho=\sqrt{(-2)^2+(2\sqrt{3})^2+4^2}=\sqrt{4+12+16}=\sqrt{32}=4\sqrt{2}.

Since \(x<0\) and \(y>0\), the point's projection into the \(xy\)-plane is in quadrant II. Its reference angle satisfies \(\tan\alpha=|2\sqrt{3}/(-2)|=\sqrt{3}\), so \(\alpha=\pi/3\), giving \(\theta=2\pi/3\).

For the other angle,

\cos\phi=\frac{z}{\rho}=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}},\qquad \phi=\frac{\pi}{4}.

Thus the coordinates are \((4\sqrt{2},2\pi/3,\pi/4)\), in the order \((\rho,\theta,\phi)\). To check, \(\rho\sin\phi=4\), so the formulas give \(x=4\cos(2\pi/3)=-2\), \(y=4\sin(2\pi/3)=2\sqrt{3}\), and \(z=4\sqrt{2}\cos(\pi/4)=4\). These are the original coordinates.

Worked Example: A Sphere Centered on the \(z\)-Axis

Convert the sphere \(x^2+y^2+(z-2)^2=4\) to spherical coordinates, and identify its angular extent.

Substitute \(x^2+y^2=\rho^2\sin^2\phi\) and \(z=\rho\cos\phi\):

\rho^2\sin^2\phi+(\rho\cos\phi-2)^2=4.

Expand the square and use \(\sin^2\phi+\cos^2\phi=1\):

\rho^2\sin^2\phi+\rho^2\cos^2\phi-4\rho\cos\phi+4=4,

\rho^2-4\rho\cos\phi=0.

Factoring gives \(\rho(\rho-4\cos\phi)=0\). The origin is on the sphere, so it must not be discarded when dividing by \(\rho\). For its other points, the equation is \(\rho=4\cos\phi\). Since \(\rho\ge0\), this requires \(\cos\phi\ge0\), or \(0\le\phi\le\pi/2\). The sphere includes the origin at the endpoint \(\phi=\pi/2,\rho=0\).

Check the north pole \((0,0,4)\): its spherical coordinates are \((4,\theta,0)\), with any \(\theta\), and \(4=4\cos 0\). Also check \((2,0,2)\): it lies on the Cartesian sphere because \(2^2+(2-2)^2=4\). Its spherical coordinates have \(\rho=\sqrt{8}=2\sqrt{2}\) and \(\phi=\pi/4\); indeed \(4\cos(\pi/4)=2\sqrt{2}\). Both checks agree.

Worked Example: Converting a Double Cone

Convert the double cone \(z^2=x^2+y^2\) into spherical coordinates, then distinguish its two nappes.

Substitute \(z=\rho\cos\phi\) and \(x^2+y^2=\rho^2\sin^2\phi\):

\rho^2\cos^2\phi=\rho^2\sin^2\phi.

For \(\rho>0\), divide by \(\rho^2\) and rearrange:

\cos^2\phi=\sin^2\phi,\qquad \tan^2\phi=1.

In the allowed interval \(0\le\phi\le\pi\), these equations give \(\phi=\pi/4\) or \(\phi=3\pi/4\). The first angle has \(z>0\) and describes the upper nappe; the second has \(z<0\) and describes the lower nappe. The vertex \(\rho=0\) is also part of the cone, even though it was excluded when dividing by \(\rho^2\).

For a check, the point \((1,1,\sqrt{2})\) satisfies the original equation because \((\sqrt{2})^2=1^2+1^2\). Its distance is \(\rho=2\), and \(z/\rho=\sqrt{2}/2\), so \(\phi=\pi/4\), as required for the upper nappe. Replacing \(z\) by \(-\sqrt{2}\) gives \(\phi=3\pi/4\), confirming the lower nappe.

Worked Examples: Plane Sections

Worked Example: Horizontal Cuts of a Sphere

Find the spherical equations of the cuts made by the planes \(z=1\) and \(z=-1\) on the sphere \(x^2+y^2+z^2=4\).

The sphere is centered at the origin with radius \(2\), so its spherical equation is \(\rho=2\): substituting gives \(x^2+y^2+z^2=\rho^2=4\). A horizontal plane \(z=c\) becomes \(\rho\cos\phi=c\). For \(z=1\), the plane equation is \(2\cos\phi=1\), hence \(\phi=\pi/3\). For \(z=-1\), it is \(2\cos\phi=-1\), hence \(\phi=2\pi/3\).

Each cut is a circle, with \(\theta\) taking all values from \(0\) to \(2\pi\) and \(\rho=2\). Check the upper cut: its height is \(z=2\cos(\pi/3)=1\), and its distance from the \(z\)-axis is \(2\sin(\pi/3)=\sqrt{3}\). Thus \(x^2+y^2=3\), and \(x^2+y^2+z^2=3+1=4\), as required. For the lower cut, \(z=2\cos(2\pi/3)=-1\) and the same radial distance gives \(3+(-1)^2=4\).

Worked Example: An Inclined Plane

Convert the plane \(x+z=2\) to spherical coordinates, then verify the equation using the point \((0,0,2)\).

Substitute \(x=\rho\sin\phi\cos\theta\) and \(z=\rho\cos\phi\):

\rho\sin\phi\cos\theta+\rho\cos\phi=2,

or, equivalently, \(\rho(\sin\phi\cos\theta+\cos\phi)=2\). For the specified point, \(\rho=2\), \(\phi=0\), and \(\theta\) is arbitrary because the point lies on the \(z\)-axis. The left side is \(2(0+1)=2\), so the converted equation is satisfied. Directly, the original equation also gives \(x+z=0+2=2\). This check confirms that the angular ambiguity on the axis does not affect the plane equation.

Why the Substitutions Work

The conversion rules follow directly from the geometry encoded in the spherical definitions. The \(xy\)-projection has length \(\rho\sin\phi\), and its components are that length multiplied by \(\cos\theta\) and \(\sin\theta\). The vertical component is \(\rho\cos\phi\). Therefore,

\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}

Because \(\rho\ge0\), this yields \(\rho=\sqrt{x^2+y^2+z^2}\). Also, \(z=\rho\cos\phi\) gives \(\cos\phi=z/\rho\) whenever \(\rho>0\). The \(xy\)-components locate \(\theta\), but their signs determine which quadrant contains it. These facts justify the point-conversion procedure and explain why it needs special treatment at the origin and on the \(z\)-axis.

A Common Angle Error

A frequent mistake is to set \(\theta=\arctan(y/x)\) and accept the calculator's principal value without checking the quadrant. For the point \((-1,\sqrt{3},2)\), the quotient is \(-\sqrt{3}\), so this shortcut gives \(\theta=-\pi/3\). That angle points into quadrant IV, not quadrant II. Using \(\rho=2\sqrt{2}\) and \(\phi=\pi/4\), this incorrect angle would produce \(x=\rho\sin\phi\cos\theta=2(1/2)=1\), which has the wrong sign.

The correction is to use the signs of \(x\) and \(y\). Here \(x<0\) and \(y>0\), so \(\theta=2\pi/3\). The spherical formulas then give \(x=2\cos(2\pi/3)=-1\), \(y=2\sin(2\pi/3)=\sqrt{3}\), and \(z=2\sqrt{2}\cos(\pi/4)=2\). The corrected coordinates reproduce the point exactly.

Key takeaway: A converted equation is not fully checked until its coordinate ranges and exceptional points have been considered. In particular, do not lose the origin by dividing by \(\rho\), and do not choose \(\theta\) without checking the quadrant.

Check Your Understanding

Use the standard spherical ranges, and check each result by substituting into the defining formulas.

Convert \((1,-1,\sqrt{2})\) to spherical coordinates. Which quadrant determines \(\theta\)?
Convert \(x^2+y^2+z^2=9\) to a spherical equation and state the value of \(\rho\) on the sphere.
For the cone \(z^2=x^2+y^2\), which value of \(\phi\) describes its lower nappe?
On the sphere \(\rho=3\), what value of \(\phi\) corresponds to the horizontal cut \(z=3/2\)?
Why must the origin be checked separately when a surface equation has been divided by \(\rho\)?