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Spherical Triple Integrals · Tutorial 663 of 999

Radial Integrals on Spherical Regions

Reduce triple integrals of functions that depend only on distance from the origin to radial integrals, while accounting correctly for the spherical volume factor and the region’s angular extent.

Intermediate 9 min read

What You'll Learn

  • Identify a radial integrand as a function of distance from the origin.
  • Use the spherical volume element with its required squared-radius factor.
  • Separate radial and angular integrals when the region has independent radial and angular bounds.
  • Evaluate radial integrals over a ball, a shell, and an angular sector.
  • Recognize when a singularity at the origin requires an improper integral.
  • Avoid omitting the spherical volume factor in a triple integral.

Why Radial Integrands Simplify

A function of three variables is called radial about the origin when its value depends only on the distance of the point from the origin. This special dependence can simplify a triple integral: although the region may be three-dimensional, the function itself does not distinguish one direction from another.

For example, the distance from \((x,y,z)\) to the origin is \(r=\sqrt{x^2+y^2+z^2}\). Thus the function \(\sqrt{x^2+y^2+z^2}\) is radial; in spherical coordinates it is simply \(r\). To use this simplification correctly, we must still account for how volume is distributed in spherical coordinates.

Definition: A function \(g\) on a region in \(\mathbb{R}^3\) is radial about the origin if there is a function \(f\) such that \(g(x,y,z)=f(r)\), where \(r=\sqrt{x^2+y^2+z^2}\). In spherical coordinates, the same function is written \(f(r)\).

The Spherical Volume Factor

We use the spherical-coordinate conventions \(x=r\sin\phi\cos\theta\), \(y=r\sin\phi\sin\theta\), and \(z=r\cos\phi\), with \(r\geq 0\), \(0\leq\phi\leq\pi\), and \(0\leq\theta<2\pi\). Here \(\phi\) is the angle measured down from the positive \(z\)-axis, and \(\theta\) is the azimuthal angle in the \(xy\)-plane.

The spherical volume element is \(dV=r^2\sin\phi\,dr\,d\phi\,d\theta\). Therefore the triple integral of a radial function has integrand \(f(r)r^2\sin\phi\), not merely \(f(r)\). The factor \(r^2\) is the radial part of the volume scaling, while \(\sin\phi\) accounts for the angular coordinate scaling.

Suppose a region has fixed radial bounds \(a\leq r\leq b\) and an angular domain \(A\) that does not depend on \(r\). Then the radial and angular bounds are independent. If \(f\) is continuous on \([a,b]\), the spherical-coordinate formula gives the following separation.

Theorem: For \(D=\{(r,\phi,\theta):a\leq r\leq b,\;(\phi,\theta)\in A\}\), where \(0\leq a<b\), and a continuous radial integrand \(f(r)\), $$ \iiint_D f(r)\,dV = \left(\int_a^b f(r)r^2\,dr\right) \left(\iint_A \sin\phi\,d\phi\,d\theta\right). $$ The angular factor is determined by the region, and the radial factor is determined by the integrand and the radial bounds.

Why the Integral Separates

The separation follows directly from the spherical volume element and the independence of the bounds. Write the triple integral as an iterated integral, with \(r\) ranging from \(a\) to \(b\) for every angle in \(A\). The integrand is a product: one factor depends only on \(r\), and the other depends only on \(\phi\).

1
Insert the volume element.
The integrand becomes \(f(r)r^2\sin\phi\).
2
Use the independent bounds.
The radial interval is the same at each angle, so the radial integration range does not change with \(\phi\) or \(\theta\).
3
Separate the factors.
Since \(f(r)r^2\) has no angular dependence, its integral can be taken independently of the angular integral of \(\sin\phi\).

In detail, integrating the product over the independent radial and angular ranges yields

$$ \int_a^b\int\!\!\int_A f(r)r^2\sin\phi\,d\phi\,d\theta\,dr = \left(\int_a^b f(r)r^2\,dr\right) \left(\iint_A\sin\phi\,d\phi\,d\theta\right). $$

This argument is the reason the reduction works. It also shows its limitation: if the radial bounds vary with the angles, the factors do not have independent integration ranges, so this particular product formula does not apply as written.

Full Spherical Regions

For the entire sphere of directions, the angular factor is

$$ \int_0^{2\pi}\int_0^\pi \sin\phi\,d\phi\,d\theta = \left(\int_0^{2\pi}d\theta\right) \left(\int_0^\pi\sin\phi\,d\phi\right) = 2\pi\cdot 2 = 4\pi. $$

Consequently, when a radial function is integrated over a ball or a spherical shell centered at the origin, the triple integral reduces to \(4\pi\) times a one-dimensional radial integral. The bounds determine whether the region includes the origin or begins at a positive radius.

Worked Example: Integrating Distance over a Ball

Evaluate \(\iiint_B \sqrt{x^2+y^2+z^2}\,dV\), where \(B\) is the ball \(x^2+y^2+z^2\leq 4\).

Step 1: Identify the radial function and bounds. Since \(\sqrt{x^2+y^2+z^2}=r\), the integrand is \(f(r)=r\). The ball has \(0\leq r\leq 2\), and it includes every direction.

Step 2: Include the volume factor. Multiplying \(f(r)=r\) by the radial part \(r^2\) of the volume element gives \(r^3\). The full angular factor is \(4\pi\), so

$$ \iiint_B \sqrt{x^2+y^2+z^2}\,dV = 4\pi\int_0^2 r^3\,dr. $$

Step 3: Evaluate the radial integral. An antiderivative of \(r^3\) is \(r^4/4\), because \(\frac{d}{dr}(r^4/4)=r^3\). Thus

$$ 4\pi\left[\frac{r^4}{4}\right]_0^2 = 4\pi\left(\frac{2^4}{4}-\frac{0^4}{4}\right) = 4\pi(4-0) = 16\pi. $$

Check: Differentiating the antiderivative \(r^4/4\) returns \(r^3\), the radial integrand after including the volume factor. Substituting the endpoints gives \(16\pi\), as above.

Spherical Shells

A spherical shell centered at the origin has a positive inner radius and an outer radius. If the integrand has a negative power of \(r\), the inner radius can matter: a shell with inner radius greater than zero avoids the origin, where such an integrand might not be defined. On a shell with fixed radii, the same full angular factor \(4\pi\) applies.

Worked Example: A Reciprocal-Square Integrand on a Shell

Evaluate \(\iiint_S \frac{1}{x^2+y^2+z^2}\,dV\), where \(S\) is the shell \(1\leq x^2+y^2+z^2\leq 9\).

Step 1: Translate the region and integrand. Since \(x^2+y^2+z^2=r^2\), the shell has \(1\leq r\leq 3\), and the integrand is \(1/r^2\). The origin is outside the shell, so the integrand is defined throughout the region.

Step 2: Multiply by the volume factor. The radial integrand in the spherical integral becomes

$$ \frac{1}{r^2}r^2=1. $$

The angular factor is \(4\pi\), so the integral is

$$ 4\pi\int_1^3 1\,dr = 4\pi[r]_1^3 = 4\pi(3-1) = 8\pi. $$

Check: The derivative of the antiderivative \(r\) is \(1\), as required. Evaluating at \(3\) and \(1\) gives \(3-1=2\); multiplying by \(4\pi\) gives \(8\pi\).

Angular Sectors

A spherical region need not include every direction. For example, it may occupy only an octant or a wedge. In that case, do not automatically use \(4\pi\): evaluate the angular integral over the actual angular bounds. The radial factor is still \(\int_a^b f(r)r^2\,dr\) when the radial interval is fixed throughout that angular domain.

Worked Example: A Radial Integrand in an Octant

Evaluate \(\iiint_D r\,dV\), where \(D\) is the part of the ball \(r\leq 2\) in the first octant, defined by \(x\geq0\), \(y\geq0\), and \(z\geq0\).

Step 1: Set the spherical bounds. The radius ranges from \(0\) to \(2\). The condition \(z\geq0\) gives \(0\leq\phi\leq\pi/2\); the conditions \(x\geq0\) and \(y\geq0\) give \(0\leq\theta\leq\pi/2\). The integrand is \(r\).

Step 2: Include the volume factor and integrate. The radial part of the integrand becomes \(r\cdot r^2=r^3\). Therefore

$$ \begin{aligned} \iiint_D r\,dV &= \int_0^{\pi/2}\int_0^{\pi/2}\int_0^2 r^3\sin\phi\,dr\,d\phi\,d\theta\\ &= \left(\int_0^2 r^3\,dr\right) \left(\int_0^{\pi/2}\sin\phi\,d\phi\right) \left(\int_0^{\pi/2}d\theta\right)\\ &= \left[\frac{r^4}{4}\right]_0^2 \left[-\cos\phi\right]_0^{\pi/2} \left[\theta\right]_0^{\pi/2}\\ &= (4)(1)\left(\frac{\pi}{2}\right) = 2\pi. \end{aligned} $$

Check: The derivative of \(r^4/4\) is \(r^3\), the derivative of \(-\cos\phi\) is \(\sin\phi\), and the derivative of \(\theta\) is \(1\). The endpoint values are \(4\), \(1\), and \(\pi/2\), whose product is \(2\pi\).

A Common Mistake: Omitting \(r^2\)

The most common error in a radial spherical integral is to replace the original integrand by \(f(r)\) and then forget the \(r^2\) in \(dV\). This changes the integral, not just its notation. For the ball example, omitting \(r^2\) would produce the following incorrect calculation:

$$ 4\pi\int_0^2 r\,dr = 4\pi\left[\frac{r^2}{2}\right]_0^2 = 8\pi. $$

This is wrong because the spherical integrand should be \(r\cdot r^2=r^3\), not \(r\). Restoring the volume factor gives \(4\pi\int_0^2 r^3\,dr=16\pi\), as computed above. A reliable setup is to write the original radial function, multiply it by \(r^2\), and only then evaluate the radial integral.

When the Origin Requires Care

If a radial integrand is undefined or unbounded at \(r=0\), a region that includes the origin may require an improper integral. The relevant radial expression is \(f(r)r^2\), not \(f(r)\) alone. For a full ball of radius \(b\), the question of convergence is therefore determined by the behavior of \(\int_0^b f(r)r^2\,dr\). If that integral is finite, the radial reduction gives a finite triple integral; if it diverges, the triple integral does not have a finite value.

For instance, the factor \(r^2\) can offset some singular behavior in \(f(r)\), but it does not guarantee convergence for every radial function. Check the resulting radial integral at the origin rather than assuming that a singularity is harmless.

Using the Reduction Effectively

For a radial integrand, the calculation has two distinct parts: the radial integral records how the function varies with distance, and the angular factor records which directions belong to the region. For a full ball or shell, the latter is \(4\pi\). For a restricted sector, compute it from the angular bounds. This separation is useful only when the radial interval is the same in every included direction; otherwise the radial and angular bounds are linked and must be handled together.

Before evaluating, check three items: whether the region is described correctly in spherical coordinates, whether the integrand is expressed as a function of \(r\), and whether the volume factor \(r^2\sin\phi\) is present. These checks prevent the most common setup errors and make the remaining one-dimensional integration straightforward.

Check Your Understanding

Use the radial reduction and show the role of the spherical volume factor in each setup.

  1. For a ball of radius \(3\), what are the radial bounds and the full angular factor?
  2. Write the reduced radial integral for integrating \(r^2\) over the ball \(r\leq 1\). Do not evaluate it.
  3. For a shell \(2\leq r\leq 5\) with all directions included, what radial integral corresponds to the integrand \(1/r\)?
  4. Why is \(4\pi\) not the correct angular factor for a region restricted to the first octant?
  5. When the region includes the origin and \(f(r)\) is singular there, which radial expression must be checked for convergence?

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