14.3 Applications of cylindrical triple integrals

Volume and Mass

Calculating Volume with Triple Integrals

The volume of a solid region $D$ in $\mathbb{R}^3$ is computed by integrating the volume element over that region. In cylindrical coordinates, the volume element picks up the Jacobian factor $r$:

$V = \iiint_D dV = \iiint_D r \, dr \, d\theta \, dz$

That extra factor of $r$ is not optional. It accounts for the fact that cylindrical "slices" farther from the $z$-axis sweep out more area. Forgetting it is one of the most common mistakes on exams.

Setting up the integral:

1. Sketch the region $D$ and identify its symmetry.
2. Determine the bounds for $r$, $\theta$, and $z$. These may be constants or functions of the other variables.
3. Choose an integration order that matches the region. Common choices are $dz \, dr \, d\theta$ or $dr \, dz \, d\theta$, but any valid order works.
4. Evaluate from the inside out, applying standard integration techniques as needed.

Example: Find the volume of a right circular cylinder with radius $a$ and height $h$.

$V = \int_0^{2\pi} \int_0^a \int_0^h r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^a rh \, dr \, d\theta = \int_0^{2\pi} \frac{a^2 h}{2} \, d\theta = \pi a^2 h$

Determining Mass and Fluid Pressure

Mass of a solid object with a spatially varying density function $\rho(r, \theta, z)$ (mass per unit volume) is:

$M = \iiint_D \rho(r, \theta, z) \, r \, dr \, d\theta \, dz$

Don't forget: the $r$ factor from the cylindrical volume element multiplies whatever integrand you already have.

Example: Find the mass of a solid cone with base radius $R$, height $h$, and density $\rho(z) = kz$ (density increases linearly with height). The cone has its apex at the origin and opens upward, so at height $z$ the radial bound is $r = Rz/h$.

$M = \int_0^{2\pi} \int_0^h \int_0^{Rz/h} (kz) \, r \, dr \, dz \, d\theta = \frac{1}{4} k \pi R^2 h^2$

Fluid pressure at the base of a container can be found by integrating the weight of the fluid column. For a cylindrical tank of radius $a$, height $h$, and constant density $\rho$, the total weight of the fluid (which acts on the base area) is:

$W = \iiint_D \rho g \, r \, dr \, d\theta \, dz = \rho g h \pi a^2$

Note that this gives the total gravitational force on the base, not the pressure at a single point. The hydrostatic pressure at depth $d$ below the surface is simply $P = \rho g d$. The triple integral approach is useful when density varies spatially or the geometry is irregular.

Probability Density Functions

Triple integrals can compute probabilities when a continuous random variable is distributed over a 3D region. If $f(x, y, z)$ is a joint probability density function (PDF), the probability that a random point falls in region $D$ is:

$P(X \in D) = \iiint_D f(x, y, z) \, dV$

A valid PDF must satisfy two conditions:

• $f(x, y, z) \geq 0$ everywhere
• $\iiint_{\mathbb{R}^3} f(x, y, z) \, dV = 1$

Example: A uniform PDF over a cylinder of radius $a$ and height $h$ is $f = \frac{1}{\pi a^2 h}$ inside the cylinder and 0 outside. Integrating over the entire cylinder:

$\iiint_D \frac{1}{\pi a^2 h} \, r \, dr \, d\theta \, dz = \frac{1}{\pi a^2 h} \cdot \pi a^2 h = 1$

This confirms it's a valid PDF. To find the probability of landing in a sub-region, you'd integrate the same PDF over that smaller region with cylindrical limits.

Center of Mass and Moment of Inertia

Calculating Center of Mass

The center of mass is the average position of all the mass in an object, weighted by density. For a solid with density $\rho$, the coordinates of the center of mass are:

$\bar{x} = \frac{1}{M}\iiint_D x \, \rho \, dV, \quad \bar{y} = \frac{1}{M}\iiint_D y \, \rho \, dV, \quad \bar{z} = \frac{1}{M}\iiint_D z \, \rho \, dV$

where $M = \iiint_D \rho \, dV$ is the total mass.

In cylindrical coordinates, remember that $x = r\cos\theta$ and $y = r\sin\theta$. For any solid that's symmetric about the $z$-axis (full $2\pi$ rotation, density independent of $\theta$), the integrals for $\bar{x}$ and $\bar{y}$ vanish because $\cos\theta$ and $\sin\theta$ integrate to zero over $[0, 2\pi]$. This means you only need to compute $\bar{z}$.

For objects with uniform density, the center of mass is the same as the centroid (purely geometric center), since $\rho$ cancels from numerator and denominator.

Example: For a solid hemisphere of radius $R$ and uniform density, symmetry gives $\bar{x} = \bar{y} = 0$. Computing $\bar{z}$:

$\bar{z} = \frac{\int_0^{2\pi}\int_0^R \int_0^{\sqrt{R^2 - r^2}} z \, r \, dz \, dr \, d\theta}{\frac{2}{3}\pi R^3} = \frac{3}{8}R$

The center of mass sits 3/8 of the way up from the flat base, which makes intuitive sense since there's more mass near the base than near the top.

Determining Moment of Inertia

Moment of inertia quantifies how much an object resists angular acceleration about a given axis. The farther the mass is from the axis, the larger the moment of inertia.

$I = \iiint_D r_{\perp}^2 \, \rho \, dV$

Here $r_{\perp}$ is the perpendicular distance from the point to the axis of rotation. For rotation about the $z$-axis, $r_{\perp}^2 = x^2 + y^2 = r^2$ in cylindrical coordinates, which makes the setup especially clean.

Example: Moment of inertia of a solid cylinder (radius $R$, height $h$, uniform density $\rho$) about its central axis:

$I_z = \int_0^{2\pi}\int_0^R \int_0^h r^2 \cdot \rho \cdot r \, dz \, dr \, d\theta = \rho \int_0^{2\pi} d\theta \int_0^R r^3 \, dr \int_0^h dz = \frac{1}{2}\rho \pi R^4 h$

Since the total mass is $M = \rho \pi R^2 h$, this simplifies to the familiar $I_z = \frac{1}{2}MR^2$.

Watch the integrand carefully: you get $r^2$ from the moment of inertia formula and another $r$ from the volume element, giving $r^3$ in the integrand.

Parallel axis theorem: If you know the moment of inertia $I_{CM}$ about an axis through the center of mass, the moment about any parallel axis a distance $d$ away is:

$I = I_{CM} + Md^2$

This saves you from re-doing the full integral when you shift axes.

Gravitational and Heat Distribution

Gravitational Potential and Force

The gravitational potential at point $(x, y, z)$ due to a continuous mass distribution is:

$V(x, y, z) = -G \iiint_D \frac{\rho(x', y', z')}{\|\vec{r} - \vec{r}'\|} \, dV'$

where $G$ is the gravitational constant, $(x', y', z')$ are coordinates of the mass element, and $\|\vec{r} - \vec{r}'\| = \sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}$.

The gravitational force is the negative gradient of the potential:

$\vec{F} = -\nabla V$

For a uniform sphere of radius $R$, mass $M = \frac{4}{3}\pi R^3 \rho$, and a field point at distance $r$ from the center:

• Outside ($r > R$): The sphere acts like a point mass.
  • $V = -\frac{GM}{r}$
  • $\vec{F} = -\frac{GM}{r^2}\hat{r}$
• Inside ($r < R$): Only the mass enclosed within radius $r$ contributes.
  • $V = -\frac{GM}{2R^3}(3R^2 - r^2)$
  • $\vec{F} = -\frac{GM r}{R^3}\hat{r}$

Cylindrical coordinates are particularly useful for computing potentials of cylindrical shells, solenoid-like mass distributions, or any geometry where the mass has rotational symmetry about an axis.

Heat Distribution and Flux

A temperature function $T(r, \theta, z)$ describes how temperature varies throughout a 3D region. The average temperature over region $D$ is:

$\bar{T} = \frac{\iiint_D T \, dV}{\iiint_D dV}$

This is the same structure as a center-of-mass calculation, but with temperature replacing density times a coordinate.

Heat flux describes the rate and direction of heat flow. By Fourier's law, the heat flux vector is $\vec{q} = -k \nabla T$, where $k$ is the thermal conductivity. The total heat flow through a surface $S$ is:

$\Phi = \iint_S \vec{q} \cdot \vec{n} \, dS$

where $\vec{n}$ is the outward unit normal. By the divergence theorem, the total flux through a closed surface equals:

$\Phi = \iiint_D \nabla \cdot \vec{q} \, dV = -k \iiint_D \nabla^2 T \, dV$

This connects the surface integral back to a triple integral, which is often easier to evaluate in cylindrical coordinates when the region has the right symmetry.

Example: For $T(x,y,z) = x^2 + y^2 + z^2$ over a cube of side $L$ (from 0 to $L$ in each direction):

• Average temperature: $\bar{T} = \frac{1}{L^3}\iiint_D (x^2 + y^2 + z^2)\,dV = \frac{L^2}{4}$
• Since $\nabla T = (2x, 2y, 2z)$ and $\nabla^2 T = 6$, the total outward heat flux through the cube's surface is $\Phi = -k \cdot 6 \cdot L^3$ (using the divergence theorem), or equivalently $6kL^3$ outward if you define flux as positive outward with $\vec{q} = -k\nabla T$.