Introduction
In spherical coordinates, a boundary is described using distance from the origin and two angles. To convert a solid’s boundary, substitute the spherical-coordinate expressions for \(x\), \(y\), and \(z\) into its Cartesian equation, then simplify. The resulting equation describes the same surface in a different coordinate system.
This procedure applies to spheres, cones, and planes, but the algebra can look quite different from one surface to another. It is important to keep track of the coordinate ranges, whether an equation describes one or both parts of a surface, and whether a simplification has divided away a special point such as a cone’s apex.
The Substitution Rule for Boundaries
A surface equation in Cartesian coordinates becomes a spherical-coordinate equation by replacing each occurrence of \(x\), \(y\), and \(z\) with its expression above. The substitution does not change the surface; it changes how points on that surface are described.
The coordinate formulas also imply the identity \(x^2+y^2+z^2=\rho^2\). Indeed, substitution gives \[ x^2+y^2+z^2 =\rho^2\sin^2\phi\cos^2\theta +\rho^2\sin^2\phi\sin^2\theta +\rho^2\cos^2\phi =\rho^2. \] In the middle expression, \(\cos^2\theta+\sin^2\theta=1\), and then \(\sin^2\phi+\cos^2\phi=1\). This identity makes spheres centered at the origin especially simple.
Spheres
A sphere of radius \(R>0\), centered at the origin, has Cartesian equation \(x^2+y^2+z^2=R^2\). Using the identity just derived, its spherical equation is \(\rho^2=R^2\). Because \(\rho\ge0\), its boundary is \(\rho=R\), not both \(\rho=R\) and \(\rho=-R\).
Worked Example: A Sphere Centered at the Origin
Convert the sphere \(x^2+y^2+z^2=25\) to spherical coordinates. First substitute all three coordinate expressions: \[ (\rho\sin\phi\cos\theta)^2+(\rho\sin\phi\sin\theta)^2+(\rho\cos\phi)^2=25. \] Expanding the squares gives \[ \rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta+\rho^2\cos^2\phi=25. \] Factor the terms: \[ \rho^2\sin^2\phi(\cos^2\theta+\sin^2\theta)+\rho^2\cos^2\phi=25. \] Apply the identity \(\cos^2\theta+\sin^2\theta=1\): \[ \rho^2\sin^2\phi+\rho^2\cos^2\phi=25. \] Then \(\sin^2\phi+\cos^2\phi=1\), so \(\rho^2=25\). Since \(\rho\ge0\), the boundary is \(\rho=5\). To check, any point on this spherical surface has \(x^2+y^2+z^2=\rho^2=25\), which is the original equation.
For a sphere not centered at the origin, the equation generally depends on the angles as well as on \(\rho\). If the center is \((a,b,c)\), the Cartesian equation is \[ (x-a)^2+(y-b)^2+(z-c)^2=R^2. \] Expanding the squares first, or substituting directly and collecting terms, gives \[ \rho^2-2\rho(a\sin\phi\cos\theta+b\sin\phi\sin\theta+c\cos\phi) +a^2+b^2+c^2=R^2. \] The terms involving \(a\), \(b\), and \(c\) record how far the center is displaced from the origin.
Worked Example: A Shifted Sphere
Convert the sphere \((x-1)^2+y^2+z^2=4\). Its center is \((1,0,0)\) and its radius is \(2\). Substitute: \[ (\rho\sin\phi\cos\theta-1)^2 +(\rho\sin\phi\sin\theta)^2 +(\rho\cos\phi)^2=4. \] Expand the first square: \[ \rho^2\sin^2\phi\cos^2\theta -2\rho\sin\phi\cos\theta+1 +\rho^2\sin^2\phi\sin^2\theta +\rho^2\cos^2\phi=4. \] Collect the three terms quadratic in \(\rho\): \[ \rho^2\bigl(\sin^2\phi\cos^2\theta+ \sin^2\phi\sin^2\theta+\cos^2\phi\bigr) -2\rho\sin\phi\cos\theta+1=4. \] The expression in parentheses is \[ \sin^2\phi(\cos^2\theta+\sin^2\theta)+\cos^2\phi =\sin^2\phi+\cos^2\phi=1. \] Therefore the spherical equation is \[ \rho^2-2\rho\sin\phi\cos\theta=3. \] For a check, the point \((3,0,0)\) lies on the original sphere because \((3-1)^2=4\). Its spherical coordinates are \(\rho=3\), \(\phi=\pi/2\), and \(\theta=0\). The transformed left-hand side is \(3^2-2(3)(1)(1)=3\), as required.
Cones
A cone with axis along the \(z\)-axis is often described using \(z\) and the distance from the \(z\)-axis, \(\sqrt{x^2+y^2}\). The coordinate substitution gives \[ \sqrt{x^2+y^2}=\rho\sin\phi,\qquad z=\rho\cos\phi, \] because \(\sin\phi\ge0\) on \(0\le\phi\le\pi\). Thus a relation between height and distance from the axis can often be rewritten as a relation involving \(\phi\) alone, apart from the apex \(\rho=0\).
Worked Example: The Upper Napple of a Cone
Convert the upper cone \(z=\sqrt{3}\sqrt{x^2+y^2}\). Substituting gives \[ \rho\cos\phi=\sqrt{3}\sqrt{ \rho^2\sin^2\phi\cos^2\theta+ \rho^2\sin^2\phi\sin^2\theta}. \] Inside the square root, factor \(\rho^2\sin^2\phi\): \[ \rho\cos\phi=\sqrt{3}\sqrt{\rho^2\sin^2\phi (\cos^2\theta+\sin^2\theta)} =\sqrt{3}\rho\sin\phi. \] For points other than the apex, \(\rho>0\), so divide by \(\rho\): \[ \cos\phi=\sqrt{3}\sin\phi. \] The upper nappe has \(0\le\phi<\pi/2\), and this equation gives \(\tan\phi=1/\sqrt{3}\), hence \(\phi=\pi/6\). The apex \(\rho=0\) also belongs to the original cone, even though it was set aside when dividing by \(\rho\). To check the angle, take \((x,y,z)=(1,0,\sqrt{3})\). Then \(\rho=2\), \(\cos\phi=\sqrt{3}/2\), and \(\phi=\pi/6\), so the transformed equation holds.
A related equation, \(z^2=3(x^2+y^2)\), describes a double cone rather than only its upper nappe. Substitution yields \[ \rho^2\cos^2\phi=3\rho^2\sin^2\phi. \] For \(\rho>0\), this reduces to \(\cos^2\phi=3\sin^2\phi\). The square is important: it allows both positive and negative values of \(z\), and therefore represents both nappes. The origin is again included in the original equation and should be retained when describing the whole surface.
Planes
A general plane can be written as \(Ax+By+Cz=D\), where \(A\), \(B\), and \(C\) are not all zero. Substituting gives \[ \rho(A\sin\phi\cos\theta+B\sin\phi\sin\theta+C\cos\phi)=D. \] This is the spherical-coordinate equation of the plane. When \(D=0\), the plane passes through the origin. In that case the equation has a factor of \(\rho\); for points with \(\rho>0\), the remaining angular expression describes the plane’s directions from the origin. The origin itself is also on the plane.
Worked Example: A Plane with All Three Coordinate Terms
Convert \(2x-y+2z=6\). Substitute the coordinate expressions: \[ 2\rho\sin\phi\cos\theta-\rho\sin\phi\sin\theta +2\rho\cos\phi=6. \] Factor out \(\rho\): \[ \rho(2\sin\phi\cos\theta-\sin\phi\sin\theta+2\cos\phi)=6. \] This is the spherical equation of the plane. To check it, the point \((3,0,0)\) lies on the Cartesian plane because \(2(3)-0+0=6\). Its spherical coordinates are \(\rho=3\), \(\phi=\pi/2\), \(\theta=0\). Substituting them into the transformed left-hand side gives \[ 3\left(2(1)(1)-(1)(0)+2(0)\right)=3(2)=6. \] Thus the transformed equation is satisfied at the point.
A horizontal plane \(z=h\) is a useful special case. Substitution gives \(\rho\cos\phi=h\). The equation should not be divided by \(\cos\phi\) without considering where \(\cos\phi=0\): at those angles, the left side is zero, so such points cannot lie on \(z=h\) unless \(h=0\). This illustrates why retaining the substituted equation is often safer than immediately solving for one variable.
Why the Substitution Works
The spherical-coordinate formulas assign Cartesian coordinates to each allowed triple \((\rho,\phi,\theta)\). Therefore, a triple represents a point on a Cartesian surface exactly when the Cartesian coordinates produced by the formulas satisfy that surface’s equation. Replacing \(x\), \(y\), and \(z\) in the equation is precisely this test.
For example, a plane is defined by \(Ax+By+Cz=D\).
Use \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), and \(z=\rho\cos\phi\).
The resulting equation holds exactly for the spherical-coordinate descriptions of points on the original surface.
This reasoning also applies to solid regions described by inequalities. For example, \(x^2+y^2+z^2\le R^2\) becomes \(\rho^2\le R^2\), or \(0\le\rho\le R\). For more complicated inequalities, the substitution remains valid, but rearranging the result may require attention to signs and zero factors.
Common Pitfall: Losing the Apex
A common error is to divide a cone equation by \(\rho\) and then forget that the original surface includes \(\rho=0\). For example, starting from \(z=\sqrt{3}\sqrt{x^2+y^2}\), substitution gives \(\rho\cos\phi=\sqrt{3}\rho\sin\phi\). Dividing by \(\rho\) and stating only \(\cos\phi=\sqrt{3}\sin\phi\) omits the apex, where \(\rho=0\).
The corrected description is: the apex \(\rho=0\) is on the cone, and every other point on the upper nappe satisfies \(\phi=\pi/6\). In a solid-region problem the apex may be included automatically by the radial range, but when describing the boundary equation itself, it should not be silently discarded.
Another frequent error is to replace \(z\) by \(\rho\sin\phi\). Under the convention used here, \(z=\rho\cos\phi\), while the distance from the \(z\)-axis is \(\rho\sin\phi\). Keeping these two expressions distinct prevents many incorrect plane and cone equations.
Summary
To convert a sphere, cone, or plane, substitute the spherical-coordinate expressions directly into its Cartesian equation and simplify using \(\sin^2 u+\cos^2 u=1\). A sphere centered at the origin simplifies to \(\rho=R\); a shifted sphere generally involves both \(\rho\) and the angles; cones often reduce to angular equations away from the apex; and planes take the form \(\rho(A\sin\phi\cos\theta+B\sin\phi\sin\theta+C\cos\phi)=D\).
Always check the coordinate ranges and whether an algebraic step, such as dividing by \(\rho\), has excluded a point that belongs to the original surface. Substituting a known point on the boundary back into the transformed equation is a direct way to catch errors.
Check Your Understanding
For each question, perform the substitution before simplifying.
- What is the spherical-coordinate equation of the sphere \(x^2+y^2+z^2=16\)?
- Convert the plane \(x+z=2\) into spherical coordinates.
- Why does \(z^2=4(x^2+y^2)\) describe both nappes of a cone?
- What point might be lost if a cone equation is divided by \(\rho\), and why?
- What are the spherical-coordinate expressions for \(z\) and the distance \(\sqrt{x^2+y^2}\) from the \(z\)-axis?