The Main Idea 

Finding the center of mass of a flat plate (lamina) means locating its balance point—where it would perfectly balance on a pin. For objects with variable density, this requires calculating three key quantities using double integrals: the mass and two moments that measure how mass is distributed relative to each axis.

The Core Concept

A lamina occupying region [latex]R[/latex] with density function [latex]\rho(x,y)[/latex] (mass per unit area) has:

• Mass: [latex]m = \iint_R \rho(x,y) \, dA[/latex]
• Moment about the [latex]x[/latex]-axis: [latex]M_x = \iint_R y\rho(x,y) \, dA[/latex]
• Moment about the [latex]y[/latex]-axis: [latex]M_y = \iint_R x\rho(x,y) \, dA[/latex]

The center of mass is then located at [latex](\overline{x}, \overline{y})[/latex] where:

[latex]\overline{x} = \frac{M_y}{m} \quad \text{and} \quad \overline{y} = \frac{M_x}{m}[/latex]

Understanding Moments

Moments measure how mass is distributed. The moment [latex]M_x[/latex] uses [latex]y[/latex] because it measures how far mass sits from the [latex]x[/latex]-axis. Similarly, [latex]M_y[/latex] uses [latex]x[/latex] to measure distance from the [latex]y[/latex]-axis. Dividing each moment by total mass gives the average position in that direction.

Centroid vs. Center of Mass

When density is constant (or you’re finding the geometric center regardless of mass), you’re finding the centroid. The density constant cancels out, giving simpler formulas:

[latex]x_c = \dfrac{\iint_R x \, dA}{\iint_R dA} \quad \text{and} \quad y_c = \dfrac{\iint_R y \, dA}{\iint_R dA}[/latex]

For uniform density, the center of mass and centroid coincide. For variable density, the center of mass shifts toward regions of higher density.

Problem-Solving Strategy

1. Sketch the region and identify its boundaries
2. Set up the double integral for mass using the given density function
3. Set up the double integrals for [latex]M_x[/latex] and [latex]M_y[/latex], including the appropriate coordinate ([latex]y[/latex] or [latex]x[/latex]) with the density
4. Evaluate all three integrals
5. Divide moments by mass to find [latex]\overline{x}[/latex] and [latex]\overline{y}[/latex]

The Main Idea 

Moments of inertia measure how mass is distributed relative to an axis of rotation—they describe an object’s resistance to rotational acceleration. While ordinary moments use distance from an axis, moments of inertia use the square of that distance, making them more sensitive to mass located far from the rotation axis.

The Core Formulas

For a lamina occupying region [latex]R[/latex] with density [latex]\rho(x,y)[/latex]:

• About the [latex]x[/latex]-axis: [latex]I_x = \iint_R y^2 \rho(x,y) \, dA[/latex]
• About the [latex]y[/latex]-axis: [latex]I_y = \iint_R x^2 \rho(x,y) \, dA[/latex]
• About the origin (polar moment): [latex]I_0 = I_x + I_y = \iint_R (x^2 + y^2) \rho(x,y) \, dA[/latex]

Notice the squared distance terms. For rotation about the [latex]x[/latex]-axis, we use [latex]y^2[/latex] because [latex]y[/latex] measures perpendicular distance from that axis. Similarly, [latex]x[/latex] measures distance from the [latex]y[/latex]-axis, so we use [latex]x^2[/latex].

Moments vs. Moments of Inertia

Compare the structure of these integrals:

• Moment: [latex]M_x = \iint_R y\rho(x,y) \, dA[/latex] (uses [latex]y[/latex])
• Moment of inertia: [latex]I_x = \iint_R y^2\rho(x,y) \, dA[/latex] (uses [latex]y^2[/latex])

The squared distance in moments of inertia means mass farther from the axis contributes much more to rotational resistance.

Polar Moment of Inertia

When rotating about the origin rather than a coordinate axis, use the polar moment [latex]I_0[/latex]. The distance from [latex](x,y)[/latex] to the origin is [latex]\sqrt{x^2 + y^2}[/latex], and squaring gives [latex]x^2 + y^2[/latex]. Conveniently, [latex]I_0 = I_x + I_y[/latex], so you can compute it either way. For regions with circular symmetry, polar coordinates simplify the calculation significantly.

Radius of Gyration

The radius of gyration tells you the effective distance where all the mass could be concentrated while maintaining the same moment of inertia:

[latex]R_x = \sqrt{\frac{I_x}{m}}, \quad R_y = \sqrt{\frac{I_y}{m}}, \quad R_0 = \sqrt{\frac{I_0}{m}}[/latex]

This gives intuition about mass distribution—a larger radius of gyration means mass is spread farther from the axis.

Problem-Solving Strategy

1. Set up the double integral with [latex]x^2[/latex] or [latex]y^2[/latex] depending on the axis
2. Evaluate the moment of inertia integral
3. For radius of gyration, also calculate mass and take [latex]\sqrt{I/m}[/latex]
4. Consider polar coordinates when the region or density has circular symmetry

The Main Idea 

The formulas for mass, moments, center of mass, and moments of inertia extend naturally from two dimensions to three dimensions. Instead of integrating over an area with [latex]dA[/latex], you integrate over a volume with [latex]dV[/latex]. The key is understanding which coordinate measures distance from each plane or axis.

Mass and Moments

For a solid [latex]Q[/latex] with density [latex]\rho(x,y,z)[/latex]:

• Mass: [latex]m = \iiint_Q \rho(x,y,z) \, dV[/latex]
• Moments about planes: [latex]M_{xy} = \iiint_Q z\rho(x,y,z) \, dV[/latex], [latex]M_{xz} = \iiint_Q y\rho(x,y,z) \, dV[/latex], [latex]M_{yz} = \iiint_Q x\rho(x,y,z) \, dV[/latex]

The moment subscript tells you which plane, and the missing coordinate is what you multiply by. For [latex]M_{xy}[/latex] (the [latex]xy[/latex]-plane), [latex]z[/latex] measures distance from that plane, so multiply by [latex]z[/latex].

Center of Mass

The center of mass coordinates are:

[latex]\overline{x} = \frac{M_{yz}}{m}, \quad \overline{y} = \frac{M_{xz}}{m}, \quad \overline{z} = \frac{M_{xy}}{m}[/latex]

Each coordinate equals its corresponding moment divided by mass. For uniform density (homogeneous solids), this gives the centroid—the geometric center.

Moments of Inertia

For rotation about each axis, use the sum of squares of the two perpendicular coordinates:

• About the [latex]x[/latex]-axis: [latex]I_x = \iiint_Q (y^2 + z^2) \rho(x,y,z) \, dV[/latex]
• About the [latex]y[/latex]-axis: [latex]I_y = \iiint_Q (x^2 + z^2) \rho(x,y,z) \, dV[/latex]
• About the [latex]z[/latex]-axis: [latex]I_z = \iiint_Q (x^2 + y^2) \rho(x,y,z) \, dV[/latex]

The pattern: for [latex]I_x[/latex], use everything except [latex]x[/latex] (namely [latex]y^2 + z^2[/latex]), because this gives the squared distance from the [latex]x[/latex]-axis.

Problem-Solving Strategy

1. Sketch the solid and determine integration bounds
2. Calculate mass using [latex]\iiint_Q \rho(x,y,z) , dV[/latex]
3. Calculate the three moments, multiplying by the appropriate coordinate
4. Divide moments by mass to find center of mass coordinates
5. For moments of inertia, use the sum of squared perpendicular coordinates with density