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Cylindrical and Spherical Coordinates · Tutorial 290 of 999

Spheres Centered On The Z-Axis

Translate the distance equation for a sphere into spherical coordinates, then read its center, radius, and interior from the result.

Intermediate 9 min read

What You'll Learn

  • Derive the spherical equation for a sphere centered at \((0,0,a)\)
  • Identify the sphere’s center and radius from its equation
  • Convert the equality into an inequality for the sphere’s interior
  • Interpret positive and negative values of the center parameter
  • Check a proposed equation using points on the sphere and inside it
  • Recognize why omitting the spherical angle changes the surface

Why the Center’s Position Matters

A sphere centered at the origin has a particularly simple equation in spherical coordinates: its distance from the origin is constant. If the center moves along the \(z\)-axis, the distance from a point on the sphere to the center is no longer just its spherical radial coordinate \(\rho\). It depends on the point’s height as well.

The key is to express that height using the spherical angle \(\phi\). This produces a compact equation that describes the sphere and makes its location and size visible.

Definition and Spherical Coordinate Facts

A sphere with center \((0,0,a)\) and radius \(R\) consists of all points whose distance from \((0,0,a)\) is \(R\). Here \(a\) may be positive, negative, or zero, while \(R\) is a nonnegative radius.

Under the spherical-coordinate convention used here, \(\rho\) is the distance from the origin and \(\phi\) is the angle measured from the positive \(z\)-axis. The coordinate relations needed below are \(z=\rho\cos\phi\) and \(x^2+y^2+z^2=\rho^2\).

Definition: The sphere centered at \((0,0,a)\) with radius \(R\) is the set of points satisfying \(x^2+y^2+(z-a)^2=R^2\). Its interior consists of the points satisfying the corresponding strict inequality \(x^2+y^2+(z-a)^2<R^2\).

Deriving the Spherical Equation

Start from the defining distance equation. Expand the square involving \(z-a\), then use the spherical-coordinate identities. This gives the equation for the sphere directly.

1
Write the distance equation.
A point lies on the sphere precisely when its squared distance from \((0,0,a)\) is \(R^2\): \(x^2+y^2+(z-a)^2=R^2\).
2
Expand the square.
Since \((z-a)^2=z^2-2az+a^2\), the left side becomes \(x^2+y^2+z^2-2az+a^2\).
3
Substitute spherical coordinates.
Use \(x^2+y^2+z^2=\rho^2\) and \(z=\rho\cos\phi\). The result is the spherical equation below.
$$ \rho^2-2a\rho\cos\phi+a^2=R^2. $$

This derivation also explains why the term \(\cos\phi\) is essential: it converts \(\rho\), the distance from the origin, into the point’s \(z\)-coordinate. The equation is not simply a condition on \(\rho\) when the center is displaced from the origin.

Theorem: The equation \(\rho^2-2a\rho\cos\phi+a^2=R^2\) describes the sphere of radius \(R\) centered at \((0,0,a)\), provided \(R\geq 0\). It follows by expanding the squared distance from \((0,0,a)\) and substituting \(z=\rho\cos\phi\) and \(x^2+y^2+z^2=\rho^2\).

Reading the Center and Radius

In the displayed form, the coefficient of \(\rho\cos\phi\) is \(-2a\), and the constant term on the left is \(a^2\). The right side is the squared radius. Thus \(a\) gives the signed \(z\)-coordinate of the center, so the center is \((0,0,a)\), and \(R\) is the nonnegative square root of the right side.

A negative \(a\) places the center below the \(xy\)-plane; it does not mean the radius is negative. If \(a=0\), the equation reduces to \(\rho^2=R^2\), the sphere centered at the origin. The case \(R=0\) is a degenerate sphere consisting of its center alone.

Worked Example: Write the Equation for a Sphere Above the Origin

Find the spherical equation for the sphere centered at \((0,0,3)\) with radius \(2\), and write the inequality for its interior.

Here \(a=3\) and \(R=2\). Substitute these values into the general equation:

$$ \rho^2-2(3)\rho\cos\phi+3^2=2^2. $$

Evaluate the powers and product:

$$ \rho^2-6\rho\cos\phi+9=4. $$

The surface is given by equality. For the interior, points must have distance less than \(2\) from the center, so replace the equality by a strict inequality:

$$ \rho^2-6\rho\cos\phi+9<4. $$

Check a surface point: \((0,0,5)\) is two units above the center. It has \(\rho=5\) and \(\phi=0\), so the left side is \(25-6(5)(1)+9=25-30+9=4\), as required. The center itself has \(\rho=3,\phi=0\), giving \(9-18+9=0<4\), which confirms that it is inside.

Identifying a Sphere from Its Equation

An equation may not initially look like the standard form. To identify the sphere, compare its terms with the canonical equation. If the coefficient of \(\rho\cos\phi\) is \(-2a\), that determines \(a\); then combine the constant terms to find \(R^2\). The resulting \(R^2\) must be nonnegative for a real sphere.

Worked Example: Identify the Center and Radius from an Expanded Equation

Identify the center and radius, and write the spherical-coordinate inequality for the interior:

$$ \rho^2-8\rho\cos\phi+12=0. $$

Match the coefficient \(-8\) with \(-2a\): \(-2a=-8\), so \(a=4\). Move the constant term to the right:

$$ \rho^2-8\rho\cos\phi=-12. $$

Add \(a^2=16\) to both sides to complete the spherical form:

$$ \rho^2-8\rho\cos\phi+16=-12+16, $$ $$ \rho^2-8\rho\cos\phi+16=4. $$

Therefore \(R^2=4\), so \(R=2\), and the center is \((0,0,4)\). The interior is described by replacing the equality with a strict inequality:

$$ \rho^2-8\rho\cos\phi+16<4. $$

Check the identification with a point on the sphere: \((0,0,6)\) is two units above the center. Since \(\rho=6\) and \(\phi=0\), the original left side is \(36-8(6)(1)+12=36-48+12=0\). Thus the point satisfies the original equation. At the center \((0,0,4)\), \(\rho=4,\phi=0\), and the original left side is \(16-32+12=-4\), which is not zero; that is consistent because the center is inside, not on the surface.

Centers Below the Origin and the Interior

The same formula works when \(a\) is negative. The sign change in the mixed term records the center’s direction along the \(z\)-axis. To describe the interior, use the same left-hand expression as for the surface, but require it to be less than \(R^2\). This follows from the distance definition: a point is inside exactly when its distance from the center is less than \(R\). Because distances and \(R\) are nonnegative, squaring preserves this comparison.

Worked Example: A Sphere Centered Below the Origin

Write the equation and interior inequality for the sphere centered at \((0,0,-2)\) with radius \(3\). Then check a point on its surface and its center.

Use \(a=-2\) and \(R=3\). Substitution gives:

$$ \rho^2-2(-2)\rho\cos\phi+(-2)^2=3^2, $$ $$ \rho^2+4\rho\cos\phi+4=9. $$

The interior inequality is:

$$ \rho^2+4\rho\cos\phi+4<9. $$

For a surface check, take \((0,0,1)\), which is three units above the center. Its coordinates are \(\rho=1,\phi=0\). Substitution gives \(1+4(1)(1)+4=9\), so it lies on the sphere. For the center \((0,0,-2)\), the coordinates are \(\rho=2,\phi=\pi\), so \(\cos\phi=-1\). The expression is \(4+4(2)(-1)+4=4-8+4=0<9\). The center lies inside, as expected.

A Common Mistake: Dropping the Angle

A frequent error is to replace \(\rho\cos\phi\) by \(\rho\), as though the radial distance \(\rho\) were always the height \(z\). That is only true for points on the positive \(z\)-axis. In general, \(z=\rho\cos\phi\), and confusing the two changes the surface.

Common mistake: For the sphere centered at \((0,0,3)\) with radius \(2\), writing \(\rho^2-6\rho+9=4\) drops the factor \(\cos\phi\). The correct equation is \(\rho^2-6\rho\cos\phi+9=4\).

For a concrete check, consider \((2,0,3)\). Its distance from \((0,0,3)\) is \(2\), so it lies on the sphere. Its spherical coordinates have \(\rho=\sqrt{13}\) and \(\cos\phi=3/\sqrt{13}\). The correct left side evaluates to \(13-6\sqrt{13}(3/\sqrt{13})+9=13-18+9=4\). The incorrect expression gives \(13-6\sqrt{13}+9=22-6\sqrt{13}\), which is not \(4\). The omitted angle therefore makes the proposed equation fail even for a known point on the sphere.

Why This Form Is Useful

The spherical equation preserves the geometric meaning of the sphere while using coordinates adapted to distance from the origin and direction from the \(z\)-axis. Its mixed term indicates how far the center is shifted along that axis, while the completed square form keeps the radius explicit.

When interpreting an equation, check all three features: the coefficient of \(\rho\cos\phi\), the constant term needed to complete the square, and the nonnegative value of \(R^2\). Then distinguish the surface, which uses equality, from its interior, which uses a strict inequality. Points on the boundary satisfy equality, while points outside have the expression greater than \(R^2\).

Check Your Understanding

Use the distance interpretation and the spherical-coordinate identities to answer these questions.

  1. What are the center and radius of the sphere \(\rho^2-10\rho\cos\phi+21=0\)?
  2. Write the spherical-coordinate equation for a sphere centered at \((0,0,-5)\) with radius \(2\).
  3. For a sphere centered at \((0,0,a)\), what inequality describes its interior?
  4. Why does the spherical equation contain \(\rho\cos\phi\), rather than just \(\rho\), in its mixed term?
  5. What does a negative value of \(a\) indicate, and what does it not indicate?

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