Spheres And Spherical Shells

Distance from the Origin in Spherical Coordinates

Spherical coordinates are especially useful when a surface or region is determined by distance from the origin. In the convention used here, \(\rho\) is the distance from the point to the origin, \(\phi\) is the angle measured down from the positive \(z\)-axis, and \(\theta\) is the angle in the \(xy\)-plane. The corresponding Cartesian coordinates are

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

Since \(\rho\) is a distance, it must be nonnegative. For any point \((x,y,z)\), its distance from the origin is \(\sqrt{x^2+y^2+z^2}\). This makes \(\rho\) the natural coordinate for describing spheres centered at the origin and the regions between them.

Definition: A sphere centered at the origin with radius \(c\), where \(c>0\), is the set of points whose distance from the origin is \(c\). In spherical coordinates, this surface is described by \(\rho=c\).

From \(\rho=c\) to a Cartesian Equation

To express a fixed-\(\rho\) surface in Cartesian coordinates, square the distance formula. Because \(\rho\ge 0\), the equation \(\rho=c\) is equivalent to saying that the distance from the origin is \(c\). Squaring gives the familiar Cartesian equation.

\rho=c \quad\Longleftrightarrow\quad \sqrt{x^2+y^2+z^2}=c \quad\Longleftrightarrow\quad x^2+y^2+z^2=c^2, \qquad c>0.

This equation describes a sphere, not a filled-in region: every point on it is exactly \(c\) units from the origin. The squared equation also makes the radius evident. For example, \(x^2+y^2+z^2=16\) has radius \(4\), because the radius is the nonnegative square root of \(16\).

Worked Example: Converting a Sphere to Cartesian Form

Describe the surface \(\rho=3\) in Cartesian coordinates, and check that the result represents points three units from the origin.

Use \(\rho^2=x^2+y^2+z^2\). Squaring \(\rho=3\) gives

\rho^2=3^2 \quad\Longrightarrow\quad x^2+y^2+z^2=9.

Thus the surface is the sphere of radius \(3\) centered at the origin. To check the equation at \((3,0,0)\), substitute its coordinates:

$$ 3^2+0^2+0^2=9. $$

Its distance from the origin is \(\sqrt{9}=3\), as required by \(\rho=3\). The point \((0,0,3)\) also satisfies the equation, since \(0^2+0^2+3^2=9\), and its distance is likewise \(3\).

Why the Cartesian Formula Works

The conversion follows directly from the spherical-coordinate formulas. Substitute the formulas for \(x\), \(y\), and \(z\) into \(x^2+y^2+z^2\):

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

The third equality uses \(\cos^2\theta+\sin^2\theta=1\), and the last uses \(\sin^2\phi+\cos^2\phi=1\). Since \(\rho\) is nonnegative, this identity gives \(\rho=\sqrt{x^2+y^2+z^2}\). In particular, fixing \(\rho\) fixes the distance from the origin, while leaving the angles free to vary over the directions represented by the spherical coordinates.

Result: For any point represented in spherical coordinates, \(\rho^2=x^2+y^2+z^2\) and \(\rho=\sqrt{x^2+y^2+z^2}\). Therefore \(\rho=c\), for \(c>0\), is equivalent to \(x^2+y^2+z^2=c^2\).

Spherical Shells as Bounds on \(\rho\)

A single equation \(\rho=c\) describes one spherical surface. An interval of values for \(\rho\) describes a region between spherical surfaces. For example, \(a\le\rho\le b\), with \(0\le a\le b\), includes the points whose distance from the origin is at least \(a\) and at most \(b\). When \(a<b\), this is a solid spherical shell with inner radius \(a\) and outer radius \(b\), including both boundary surfaces.

Because all quantities in the comparison are nonnegative, squaring preserves their order. Thus the same region can be described in Cartesian coordinates by squaring the endpoints and the distance:

a\le\rho\le b \quad\Longleftrightarrow\quad a^2\le x^2+y^2+z^2\le b^2, \qquad 0\le a\le b.

Strict inequalities work in the same way. An endpoint is included exactly when its inequality includes equality. For instance, \(a<\rho\le b\) excludes the inner sphere and includes the outer one. If both inequalities are strict, neither boundary sphere is included.

Worked Example: Translating a Closed Spherical Shell

Convert \(2\le\rho\le5\) into Cartesian coordinates and identify its boundary spheres.

The lower and upper distance bounds are nonnegative, so squaring all three quantities preserves the inequalities:

2\le\rho\le5 \quad\Longleftrightarrow\quad 2^2\le\rho^2\le5^2 \quad\Longleftrightarrow\quad 4\le x^2+y^2+z^2\le25.

This region lies between the sphere of radius \(2\), given by \(x^2+y^2+z^2=4\), and the sphere of radius \(5\), given by \(x^2+y^2+z^2=25\). Both surfaces are included. To check a point in the region, take \((1,2,2)\). Its squared distance is \(1^2+2^2+2^2=9\), so its distance is \(3\), which satisfies \(2\le3\le5\). The Cartesian inequality also holds, since \(4\le9\le25\). At \((1,1,1)\), the squared distance is \(3\), which fails \(4\le3\); its distance \(\sqrt{3}\) is less than \(2\), so it is correctly excluded.

Open and Half-Open Shells

Changing an inequality changes whether its boundary surface belongs to the region. The Cartesian form preserves that information; it does not turn an excluded boundary into an included one.

Worked Example: Keeping Track of Included Boundaries

Translate \(1<\rho\le3\) into Cartesian coordinates and check both a boundary point and an interior point.

All distances in the stated range are nonnegative. Squaring the lower strict inequality and the upper inclusive inequality gives

1<\rho\le3 \quad\Longleftrightarrow\quad 1<\rho^2\le9 \quad\Longleftrightarrow\quad 1<x^2+y^2+z^2\le9.

The inner sphere is excluded because the lower inequality is strict. The outer sphere is included because the upper inequality allows equality. The point \((1,0,0)\) has squared distance \(1\), so it fails the strict lower bound and is excluded. The point \((0,0,3)\) has squared distance \(9\), so it satisfies both Cartesian inequalities and is included on the outer boundary. Finally, \((0,0,2)\) has squared distance \(4\), and \(1<4\le9\), so it lies inside the shell. Its distance is \(2\), which also verifies \(1<2\le3\).

Special Cases and Common Errors

The radius must be nonnegative. If \(c=0\), then \(\rho=0\) describes only the origin, since a point has distance zero from the origin exactly when it is the origin. If \(c<0\), the equation \(\rho=c\) has no solutions because \(\rho\) cannot be negative. Likewise, a proposed interval \(a\le\rho\le b\) describes a shell of the stated form only when \(0\le a\le b\).

Common mistake: Replacing \(\rho\) by \(\sqrt{x^2+y^2}\) and then concluding that \(\rho=3\) is the cylinder \(x^2+y^2=9\). That expression measures distance from the \(z\)-axis, not distance from the origin. For example, \((0,0,3)\) has \(\sqrt{x^2+y^2}=0\), but its distance from the origin is \(3\). The correct Cartesian equation for \(\rho=3\) is \(x^2+y^2+z^2=9\).

A reliable translation keeps the geometric meaning in view: \(\rho\) measures distance from the origin in three dimensions, so its Cartesian expression involves all three coordinates. A fixed positive value gives a spherical surface; a range of values gives the points between the corresponding concentric spheres. Writing the inequalities carefully then determines which boundary surfaces are part of the region.

Identify the distance condition.
Read a value or interval for \(\rho\) as a condition on distance from the origin.

Use the Cartesian distance.
Replace \(\rho^2\) by \(x^2+y^2+z^2\).

Preserve the bounds.
Square nonnegative endpoints and retain each strict or inclusive inequality.

Check the geometry.
Confirm that the resulting equation or inequalities describe the intended distances and boundary spheres.

Check Your Understanding

For each question, use the fact that \(\rho\) is distance from the origin.

Write the Cartesian equation of the sphere \(\rho=6\), and state its radius.
Translate \(3\le\rho<7\) into Cartesian inequalities. Which boundary sphere is excluded?
What set of points is described by \(\rho=0\)? What happens if the equation is \(\rho=-2\)?
Does the point \((2,1,2)\) lie in the region \(2\le\rho\le4\)? Verify using both its distance and the Cartesian inequality.
Explain why \(x^2+y^2=9\) does not describe the sphere \(\rho=3\).

What You'll Learn