15.3 Applications of spherical triple integrals

Volume and Mass Calculation

Calculating Volume with Triple Integrals

In spherical coordinates, the volume element picks up a factor that accounts for the geometry of the coordinate system. That factor is what makes spherical coordinates so natural for regions bounded by spheres, cones, or spherical shells.

The volume element is:

$dV = \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi$

To find the volume of a region $E$, you integrate this element over the region:

$V = \iiint_E \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi$

Setting up the bounds correctly is the main challenge. Here's a reliable approach:

1. Identify the radial bounds. Determine how $\rho$ varies. For a full sphere of radius $R$, that's $0 \leq \rho \leq R$. For a spherical shell, it's $a \leq \rho \leq b$.
2. Identify the polar angle bounds. $\phi$ measures the angle down from the positive $z$-axis. A full sphere uses $0 \leq \phi \leq \pi$. A cone of half-angle $\alpha$ restricts this to $0 \leq \phi \leq \alpha$.
3. Identify the azimuthal bounds. $\theta$ sweeps around the $z$-axis. A full revolution is $0 \leq \theta \leq 2\pi$. A wedge-shaped region restricts this range.
4. Integrate in the order $d\rho \, d\phi \, d\theta$ (or whichever order suits the bounds), always including the $\rho^2 \sin\phi$ Jacobian factor.

Determining Mass and Density

Mass connects geometry to physics. If a solid has a mass density function $\delta(\rho, \phi, \theta)$ (mass per unit volume at each point), the total mass is:

$m = \iiint_E \delta(\rho, \phi, \theta)\, \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi$

Note the notation: since $\rho$ is already used for the radial coordinate in spherical coordinates, many texts use $\delta$ for the density function to avoid confusion. Watch for this in your course materials.

• Constant density $\delta_0$: the integral reduces to $m = \delta_0 \cdot V$, so you just need the volume.
• Variable density: you must integrate the full expression. The density function often depends on position, such as $\delta = \delta(\rho)$ for a radially varying material.

Leveraging Spherical Symmetry

When the density depends only on the distance from the origin, i.e., $\delta = \delta(\rho)$, the angular integrals separate out and can be evaluated independently. The $\theta$ integral gives $2\pi$, and the $\phi$ integral gives $\int_0^{\pi}\sin\phi\,d\phi = 2$. This collapses the triple integral into a single integral:

$m = 4\pi \int_0^R \rho^2 \, \delta(\rho) \, d\rho$

This is a huge simplification. For example, if a solid sphere of radius $R$ has density $\delta(\rho) = \delta_0 e^{-\rho}$, you only need to evaluate one integral instead of three nested ones. Always check whether your problem has this radial symmetry before grinding through a full triple integral.

Moments and Centers

Moment of Inertia Calculations

The moment of inertia quantifies how difficult it is to spin an object about a given axis. It depends not just on total mass, but on how that mass is distributed relative to the axis.

For rotation about the $z$-axis, the moment of inertia is:

$I_z = \iiint_E \delta(\rho, \phi, \theta)\, d_{\perp}^2 \, dV$

where $d_{\perp}$ is the perpendicular distance from the $z$-axis. In spherical coordinates, the distance from the $z$-axis is $\rho\sin\phi$, so $d_{\perp}^2 = \rho^2\sin^2\phi$. Substituting the spherical volume element:

$I_z = \iiint_E \delta(\rho, \phi, \theta)\, \rho^2\sin^2\phi \cdot \rho^2\sin\phi \, d\rho\, d\phi\, d\theta = \iiint_E \delta\, \rho^4\sin^3\phi \, d\rho\, d\phi\, d\theta$

For a uniform solid sphere of mass $m$ and radius $R$, this evaluates to $I_z = \frac{2}{5}mR^2$. That's a standard result worth remembering as a sanity check.

To compute moment of inertia about a different axis (say the $x$-axis), you'd replace $d_{\perp}^2$ with the squared distance from the $x$-axis: $d_{\perp}^2 = y^2 + z^2 = \rho^2(\sin^2\phi\sin^2\theta + \cos^2\phi)$.

Locating the Center of Mass

The center of mass is the average position of all the mass in the object. Its coordinates are:

$\bar{x} = \frac{1}{m}\iiint_E x\,\delta\,dV, \quad \bar{y} = \frac{1}{m}\iiint_E y\,\delta\,dV, \quad \bar{z} = \frac{1}{m}\iiint_E z\,\delta\,dV$

In spherical coordinates, recall the conversions:

• $x = \rho\sin\phi\cos\theta$
• $y = \rho\sin\phi\sin\theta$
• $z = \rho\cos\phi$

So, for example, the $z$-coordinate of the center of mass becomes:

$\bar{z} = \frac{1}{m}\iiint_E (\rho\cos\phi)\,\delta(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$

Symmetry shortcut: If the object and its density function are symmetric about the $xy$-plane, then $\bar{z} = 0$ without computing anything. Similarly, full rotational symmetry about the $z$-axis forces $\bar{x} = \bar{y} = 0$. Always exploit symmetry before integrating.

For a solid with radial symmetry and a density that depends only on $\rho$, the center of mass sits at the origin. The interesting cases arise when the region or the density breaks some of that symmetry (e.g., a hemisphere, or a sphere with density that depends on $\phi$).

Fields and Potentials

Gravitational Potential Energy

The gravitational potential at a point due to a continuous mass distribution is found by summing contributions from every small mass element. For a mass distribution with density $\delta$, the gravitational potential at the origin is:

$\Phi = -G \iiint_E \frac{\delta(\rho, \phi, \theta)}{\rho} \cdot \rho^2\sin\phi \, d\rho\, d\phi\, d\theta = -G\iiint_E \delta\,\rho\sin\phi\,d\rho\,d\phi\,d\theta$

More generally, the potential at an arbitrary field point $\vec{r}$ due to a source distribution is:

$\Phi(\vec{r}) = -G \iiint_E \frac{\delta(\vec{r}')}{|\vec{r} - \vec{r}'|} \, dV'$

where the integration is over the source coordinates $\vec{r}'$.

For a spherically symmetric mass distribution, the shell theorem provides a powerful shortcut:

• Outside the distribution ($r > R$): the potential is $\Phi = -\frac{Gm}{r}$, as if all the mass were concentrated at the center.
• Inside a uniform spherical shell: the potential is constant (no net gravitational force from the shell).

The gravitational self-energy (the energy required to assemble the mass from infinity) for a uniform sphere of mass $m$ and radius $R$ is:

$U = -\frac{3Gm^2}{5R}$

This result comes from integrating shell-by-shell, bringing each infinitesimal shell in from infinity onto the already-assembled mass.

Electric Field and Charge Distribution

The electric field due to a continuous charge distribution follows the same mathematical structure as gravity, with charge density $\rho_c$ replacing mass density and $\frac{1}{4\pi\epsilon_0}$ replacing $G$:

$\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \iiint_E \frac{\rho_c(\vec{r}')(\vec{r} - \vec{r}')}{|\vec{r} - \vec{r}'|^3} \, dV'$

This is a vector integral, so in practice you often compute each component separately.

For spherically symmetric charge distributions, the calculation simplifies dramatically through Gauss's law:

$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$

By choosing a spherical Gaussian surface of radius $r$, symmetry forces $\vec{E}$ to point radially with constant magnitude on the surface, giving:

$E(r) = \frac{Q_{\text{enc}}(r)}{4\pi\epsilon_0 r^2}$

where $Q_{\text{enc}}(r) = \int_0^r 4\pi {\rho'}^2 \rho_c(\rho')\,d\rho'$ is the total charge enclosed within radius $r$. This turns a difficult vector triple integral into a single scalar integral for the enclosed charge, which is exactly the kind of simplification that makes spherical coordinates so valuable.