13.3 Applications to volume and mass

Volume and Mass Calculation

Calculating Volume Using Triple Integrals

The volume of a solid region $D$ in $\mathbb{R}^3$ is given by the triple integral

$\text{Volume} = \iiint_D dV = \iiint_D dx\, dy\, dz$

The integrand is just 1, so you're summing up all the infinitesimal volume elements $dV$ throughout the region. The real work is in setting up the limits of integration correctly.

Setting up the integral:

1. Sketch or visualize the region $D$ and identify its bounding surfaces.
2. Choose an order of integration (e.g., $dz\, dy\, dx$). Pick the order that makes the innermost limits simplest to express.
3. Write the innermost limits as functions of the remaining variables, the middle limits as functions of the outermost variable, and the outermost limits as constants.
4. Evaluate from the inside out.

If the region has cylindrical or spherical symmetry, switching to cylindrical coordinates ($r, \theta, z$) or spherical coordinates ($\rho, \phi, \theta$) can dramatically simplify the integral. Remember to include the appropriate Jacobian factor: $dV = r\, dr\, d\theta\, dz$ for cylindrical, or $dV = \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$ for spherical.

Determining Mass and Density

When an object has a density function $\rho(x, y, z)$ describing mass per unit volume at each point, the total mass is

$\text{Mass} = \iiint_D \rho(x, y, z)\, dV$

• If density is uniform (constant throughout), then $\rho(x, y, z) = \rho_0$ and the mass integral reduces to $\rho_0 \cdot \text{Volume}$.
• For non-uniform density, $\rho$ varies with position. Common examples include density that increases linearly along an axis (e.g., $\rho = kz$) or density that depends on distance from the origin (e.g., $\rho = \rho_0 e^{-r^2}$).

The setup process is the same as for volume: identify the region, choose coordinates and integration order, then substitute the density function into the integrand before evaluating.

Applying Volume and Mass Calculations

For irregular shapes bounded by multiple surfaces, the key challenge is expressing the integration limits correctly. Break the region into simpler sub-regions if a single set of limits doesn't cover it.

You can also combine volume and mass to find the average density of an object:

$\bar{\rho} = \frac{\text{Mass}}{\text{Volume}} = \frac{\iiint_D \rho(x,y,z)\, dV}{\iiint_D dV}$

This is useful as a sanity check: your computed average density should fall between the minimum and maximum values of $\rho$ over the region.

Center of Mass and Moment of Inertia

Center of Mass Calculation

The center of mass $(x_c, y_c, z_c)$ is the weighted average position of all the mass in the object. Each coordinate is computed by integrating the corresponding position variable times the density, then dividing by total mass:

$x_c = \frac{\iiint_D x\, \rho(x, y, z)\, dV}{\iiint_D \rho(x, y, z)\, dV}$

$y_c = \frac{\iiint_D y\, \rho(x, y, z)\, dV}{\iiint_D \rho(x, y, z)\, dV}$

$z_c = \frac{\iiint_D z\, \rho(x, y, z)\, dV}{\iiint_D \rho(x, y, z)\, dV}$

The denominator in each formula is just the total mass $M$, so you typically compute that first and reuse it. The numerators $\iiint x\rho\, dV$, etc., are called the first moments of mass ($M_{yz}$, $M_{xz}$, $M_{xy}$ respectively).

For objects with uniform density, $\rho$ cancels from numerator and denominator, and the center of mass becomes the centroid, which depends only on geometry. Use symmetry to your advantage: if the region is symmetric about a plane and the density respects that symmetry, the center of mass lies on that plane, saving you an integral.

Moment of Inertia Determination

The moment of inertia quantifies how mass is distributed relative to an axis of rotation. More mass farther from the axis means greater resistance to angular acceleration.

$I = \iiint_D r^2\, \rho(x, y, z)\, dV$

Here $r$ is the perpendicular distance from the point $(x, y, z)$ to the axis of rotation. The specific expression for $r^2$ depends on which axis you're rotating about:

• About the $z$-axis: $r^2 = x^2 + y^2$
• About the $x$-axis: $r^2 = y^2 + z^2$
• About the $y$-axis: $r^2 = x^2 + z^2$

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

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

where $M$ is the total mass. This saves you from re-evaluating the full triple integral for every parallel axis.

Gravitational Potential

Calculating Gravitational Potential

The gravitational potential $V(x, y, z)$ at a point due to a continuous mass distribution is the potential energy per unit mass a test particle would have at that location. It's computed by summing contributions from every mass element in the body:

$V(x, y, z) = -G \iiint_D \frac{\rho(x', y', z')}{\sqrt{(x - x')^2 + (y - y')^2 + (z - z')^2}}\, dV'$

Here $G$ is the gravitational constant, and $(x', y', z')$ are the integration variables running over the mass distribution $D$. The point $(x, y, z)$ where you evaluate the potential is treated as a fixed parameter during integration.

The denominator is the distance between the field point and the source point, which makes this integral harder than typical mass or volume integrals. For bodies with symmetry, choosing the right coordinate system is critical. A solid sphere, for instance, is best handled in spherical coordinates, where the angular integrals often simplify or can be evaluated using known results.

Applying Gravitational Potential

Once you have $V$, you can extract the gravitational force on a test particle of mass $m$ by taking the negative gradient:

$\vec{F}(x, y, z) = -m\, \nabla V(x, y, z)$

This connects the scalar potential (one integral) to the vector force field (which would otherwise require three separate integrals, one per component). That's why working with the potential is often more efficient.

For classic symmetric distributions:

• Uniform sphere: Outside the sphere, the potential is identical to that of a point mass at the center: $V = -\frac{GM}{r}$. Inside, the potential varies as $V = -\frac{GM}{2R^3}(3R^2 - r^2)$, where $R$ is the sphere's radius.
• Uniform thin rod or plate: These typically require direct evaluation of the integral, but symmetry arguments reduce the number of nontrivial integration variables.

The gravitational potential energy of the entire mass distribution (the energy required to assemble it from infinity) involves a self-energy integral, which is a more advanced calculation where you integrate $\rho\, V$ over the body, being careful to avoid double-counting.