Planes In Cylindrical And Spherical Coordinates

Why Plane Equations Change with Coordinates

A plane is a flat surface, but its equation can look quite different depending on how points are described. Cartesian coordinates use \(x\), \(y\), and \(z\). Cylindrical coordinates replace \(x\) and \(y\) by a distance from the \(z\)-axis and an angle around it. Spherical coordinates describe a point by its distance from the origin and two angles.

Changing coordinates does not change the plane. It changes how membership in that plane is expressed. Horizontal planes and vertical planes containing the \(z\)-axis have especially direct coordinate descriptions. A general plane also converts by substitution, but radial and angular factors must be kept.

Coordinate conventions: Cylindrical coordinates are \((r,\theta,z)\), with \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\). Spherical coordinates are \((\rho,\theta,\phi)\), with \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), and \(z=\rho\cos\phi\). Here \(r\geq 0\), \(\rho\geq 0\), \(0\leq\theta<2\pi\), and \(0\leq\phi\leq\pi\).

At \(r=0\), the cylindrical angle \(\theta\) is not determined. At \(\rho=0\), neither spherical angle is determined; at the positive and negative \(z\)-axis, \(\theta\) is also undetermined. These coordinate ambiguities describe the same geometric points in multiple ways. They matter when interpreting equations that contain angles.

Horizontal Planes

A horizontal plane has the form \(z=c\), where \(c\) is a fixed real number. In cylindrical coordinates, \(z\) is already a coordinate, so the equation remains \(z=c\). In spherical coordinates, substitute \(z=\rho\cos\phi\).

\text{Cylindrical: }z=c \qquad\qquad \text{Spherical: }\rho\cos\phi=c.

The spherical equation describes the same plane, not just a single point or a ray. For example, when \(c>0\), points in the plane have \(\rho\cos\phi=c\), while \(\theta\) can vary around the full circle. If \(c=0\), the equation is \(\rho\cos\phi=0\): it describes the plane \(z=0\), including the origin, even though the origin itself has no unique spherical angles.

Worked Example: A Horizontal Plane at Height Three

Convert \(z=3\) to cylindrical and spherical coordinates, then check the equations using the point \((4,0,3)\).

In cylindrical coordinates, \(z\) is unchanged, so the equation is \(z=3\). For the chosen point, \(r=\sqrt{4^2+0^2}=4\), \(\theta=0\), and \(z=3\); it satisfies the equation.

In spherical coordinates, \(z=\rho\cos\phi\), so the equation is \(\rho\cos\phi=3\). At \((4,0,3)\), \(\rho=\sqrt{4^2+0^2+3^2}=5\), \(\cos\phi=z/\rho=3/5\), and \(\theta=0\). Therefore \(\rho\cos\phi=5(3/5)=3\), as required.

Substituting these coordinates back into Cartesian form gives \(z=3\), so the point lies on the original plane as well.

Vertical Planes Through the \(z\)-Axis

A plane containing the \(z\)-axis contains every point \((0,0,z)\). Write a general Cartesian plane as \(ax+by+cz=d\). Substituting \((0,0,z)\) gives \(cz=d\) for every real \(z\), which is possible only if \(c=0\) and \(d=0\). Thus a plane containing the \(z\)-axis has an equation \(ax+by=0\), where \(a\) and \(b\) are not both zero.

Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) to obtain \(r(a\cos\theta+b\sin\theta)=0\). For points off the \(z\)-axis, \(r>0\), so the angle must satisfy \(a\cos\theta+b\sin\theta=0\). The axis itself is included in the plane regardless of \(\theta\), because \(r=0\).

ax+by=0 \quad\Longleftrightarrow\quad r(a\cos\theta+b\sin\theta)=0 \quad\Longleftrightarrow\quad a\cos\theta+b\sin\theta=0\ \text{when }r>0.

In spherical coordinates the same substitution gives \(\rho\sin\phi(a\cos\theta+b\sin\theta)=0\). Away from the \(z\)-axis, \(\rho\sin\phi>0\), and the angular condition is again \(a\cos\theta+b\sin\theta=0\). The resulting equation describes two opposite angular directions, which together give the full plane. An angular equation alone should be understood along with the axis points, where the angle may be undefined.

Worked Example: The Vertical Plane \(x-y=0\)

Convert the plane \(x-y=0\) into cylindrical and spherical coordinates. Check a point away from the \(z\)-axis and a point on the axis.

Here \(a=1\) and \(b=-1\). In cylindrical coordinates, substitution gives \(r\cos\theta-r\sin\theta=0\), or \(r(\cos\theta-\sin\theta)=0\). For \(r>0\), this requires \(\cos\theta=\sin\theta\), so \(\theta=\pi/4\) or \(5\pi/4\) in the stated range.

In spherical coordinates, the equation becomes \(\rho\sin\phi(\cos\theta-\sin\theta)=0\). For points off the \(z\)-axis, it has the same two angular solutions. Points on the \(z\)-axis also belong to the plane because their Cartesian coordinates have \(x=y=0\).

For the check, take \((1,1,2)\). It satisfies \(x-y=1-1=0\). Its cylindrical coordinates are \(r=\sqrt{1^2+1^2}=\sqrt2\), \(\theta=\pi/4\), and \(z=2\); indeed, \(r(\cos\theta-\sin\theta)=\sqrt2(1/\sqrt2-1/\sqrt2)=0\). Its spherical coordinates have \(\rho=\sqrt{1+1+4}=\sqrt6\), \(\theta=\pi/4\), and \(\sin\phi=r/\rho=\sqrt{2/6}=1/\sqrt3\). The spherical expression is \(\rho\sin\phi(\cos\theta-\sin\theta)=\sqrt6(1/\sqrt3)(0)=0\).

As an axis check, \((0,0,2)\) satisfies \(x-y=0\). It has \(r=0\), so the cylindrical product equation is zero for any assigned angle; in spherical coordinates \(\sin\phi=0\), so the product equation is also zero.

Converting a General Plane

Let \(ax+by+cz=d\), where \(a,b,c,d\) are constants and at least one of \(a,b,c\) is nonzero. This condition ensures the equation defines a plane rather than a degenerate equation. Substitute the coordinate formulas directly.

General plane in cylindrical coordinates: The equation \(ax+by+cz=d\) becomes \(r(a\cos\theta+b\sin\theta)+cz=d\).

General plane in spherical coordinates: The equation \(ax+by+cz=d\) becomes \(\rho(a\sin\phi\cos\theta+b\sin\phi\sin\theta+c\cos\phi)=d\).

The coefficients \(a,b,c\) are the components of a normal vector to the plane: they determine which direction is perpendicular to it. The coordinate equations follow by replacing each Cartesian coordinate, so the equivalence is exact. Every point represented by cylindrical or spherical coordinates has Cartesian coordinates given by the convention above; inserting those coordinates into the original plane equation yields precisely the displayed converted equation. Conversely, if the converted equation holds, the corresponding Cartesian point satisfies \(ax+by+cz=d\).

Keep the plane coefficients.
Start with \(ax+by+cz=d\); do not alter the plane while changing coordinates.

Substitute the coordinate identities.
For cylindrical coordinates use \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\).

Collect terms without dividing prematurely.
In particular, keep the factor \(r\) in the cylindrical equation, since points with \(r=0\) may be part of the plane.

Use the spherical identities separately.
Substitute all three spherical formulas, including the \(\sin\phi\) factors in \(x\) and \(y\).

Worked Example: Converting \(2x-y+2z=4\)

Write the plane in cylindrical and spherical coordinates, then verify both equations using the point \((2,0,0)\).

For cylindrical coordinates, substitute \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\):

2(r\cos\theta)-r\sin\theta+2z=4,

so the cylindrical equation is \(2r\cos\theta-r\sin\theta+2z=4\). At \((2,0,0)\), \(r=\sqrt{2^2+0^2}=2\), \(\theta=0\), and \(z=0\). The left side is \(2(2)(1)-2(0)+2(0)=4\), which equals the right side.

For spherical coordinates, substitute \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), and \(z=\rho\cos\phi\):

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

or, after factoring out \(\rho\), \(\rho(2\sin\phi\cos\theta-\sin\phi\sin\theta+2\cos\phi)=4\). The point \((2,0,0)\) has \(\rho=2\), \(\theta=0\), and \(\phi=\pi/2\). Its spherical left side is \(2(2(1)(1)-1(0)(0)+2(0))=4\). Substituting the point into Cartesian form also gives \(2(2)-0+2(0)=4\), confirming the conversion.

A Common Mistake: Losing a Radial Factor

When converting a plane to cylindrical coordinates, both \(x\) and \(y\) contain \(r\). Forgetting this factor changes the surface. For example, an incorrect conversion of \(2x-y+2z=4\) might be written as \(2\cos\theta-\sin\theta+2z=4\), omitting \(r\) from the first two terms.

Test this incorrect equation at the point \((2,0,0)\), which is on the original plane. Its cylindrical coordinates are \(r=2\), \(\theta=0\), \(z=0\). The incorrect left side is \(2(1)-0+0=2\), not \(4\). The correct left side is \(2r\cos\theta-r\sin\theta+2z=2(2)(1)-2(0)+0=4\).

A related error is dividing by \(r\) in a plane equation and then treating the resulting angular condition as valid on the axis. Division by \(r\) is permitted only when \(r>0\). The original product equation must be retained to account for \(r=0\).

Why These Forms Are Useful

The coordinate equation often reveals how a plane is positioned. A horizontal plane remains simple in cylindrical coordinates because \(z\) is unchanged. A vertical plane through the axis is controlled by angle, because every point off the axis lies along one of two opposite radial directions. A general plane shows the roles of the radial, angular, and vertical components explicitly.

These equations also provide a direct way to check whether a proposed point lies on a plane: convert the point to the chosen coordinates, substitute into the converted equation, and compare both sides. The check is reliable only when the coordinate convention is used consistently and all factors are retained.

Check Your Understanding

Use the coordinate conventions stated above. Keep track of points on the \(z\)-axis when an angular equation is involved.

What are the cylindrical and spherical equations of the horizontal plane \(z=-2\)?
Convert \(3x+4y=0\) into cylindrical coordinates. Which angular values in \(0\leq\theta<2\pi\) describe points off the \(z\)-axis?
Convert \(x+2y-3z=6\) into cylindrical and spherical coordinates.
Why must one be careful before dividing a cylindrical plane equation by \(r\)?
For the plane \(2x-y+2z=4\), verify whether \((0,0,2)\) lies on the plane using its spherical coordinates.