Lamina in Calculus (Mass & Density Formulas)
A lamina is a thin, flat object modeled by a region (R) in the plane and an areal density (\rho(x,y)). Its mass is (m=\iint_R\rho\,dA), and its center of mass is the density-weighted average of its coordinates. Moments of inertia use the same density, weighted by squared distance from an axis.
When a problem gives you a shape and a density function, the main challenge is turning them into the right double integrals. A careful sketch, sensible coordinates, and a units check make the work easier to follow and help catch mistakes before they affect the answer.
Lamina Models, Density, and Core Assumptions
A lamina is an idealized thin sheet whose mass spreads across a two-dimensional region. We describe the sheet’s shape by (R), a bounded region in the (xy)-plane, and its areal density by (\rho(x,y)). The model assumes the sheet has no thickness for the calculation.
Areal density measures mass per unit area. For example, if (\rho) is measured in kilograms per square meter, multiplying it by a small area in square meters gives an approximate mass in kilograms. The density may be constant, or it may change from one point to another.
For a small patch of area (dA), the mass is approximately (\rho(x,y)dA). Adding the masses of all such patches leads to the double-integral formulas below. In this model, (\rho) should be nonnegative, and the total mass should be finite.
- Constant density: (\rho(x,y)=c), where (c) is a fixed value. Then (m=c\,\text{Area}(R)).
- Variable density: (\rho) depends on position, so some parts of the lamina may be heavier per unit area than others.
- Region (R): the part of the plane occupied by the lamina. Its boundaries determine the integration limits.
I find it useful to separate the shape from the density before integrating. The shape tells you where the material is; the density tells you how much mass is packed into each unit of that shape.
Mass, Center of Mass, and Moment Formulas
Mass, center of mass, and moments of inertia are found by integrating density across the lamina. The mass totals the material, the center of mass gives its balance point, and each moment of inertia measures how strongly the mass is spread away from a chosen axis.
Mass and center of mass
The total mass is
[ m=\iint_R \rho(x,y)\,dA. ]
The first moments about the coordinate axes are
[ M_y=\iint_R x\rho(x,y)\,dA,\qquad M_x=\iint_R y\rho(x,y)\,dA. ]
For (m>0), the center of mass is
[ \bar{x}=\frac{M_y}{m} =\frac{1}{m}\iint_R x\rho(x,y)\,dA,\qquad \bar{y}=\frac{M_x}{m} =\frac{1}{m}\iint_R y\rho(x,y)\,dA. ]
The coordinate names can be easy to mix up: (M_y) uses (x), the distance from the (y)-axis, while (M_x) uses (y), the distance from the (x)-axis. Each centroid coordinate is a weighted average, with denser parts carrying more weight.
Moments of inertia
A moment of inertia describes how mass is distributed relative to an axis. For a lamina in the plane, the moments about the coordinate axes are
[ I_x=\iint_R y^2\rho(x,y)\,dA,\qquad I_y=\iint_R x^2\rho(x,y)\,dA. ]
The squared distance matters: mass farther from an axis contributes more. The polar moment about the origin is
[ I_0=\iint_R (x^2+y^2)\rho(x,y)\,dA=I_x+I_y. ]
If mass is in kilograms and distance is in meters, a moment of inertia has units of (\text{kg}\cdot\text{m}^2). It is not a mass or a length; it describes mass distribution relative to an axis.
| Quantity | Formula | Units when (\rho) is in kg/m² and lengths are in m |
|---|---|---|
| Mass | (\iint_R \rho\,dA) | kg |
| (x)-coordinate of center of mass | (\frac{1}{m}\iint_R x\rho\,dA) | m |
| (y)-coordinate of center of mass | (\frac{1}{m}\iint_R y\rho\,dA) | m |
| Moment about the (x)-axis | (\iint_R y^2\rho\,dA) | kg·m² |
| Moment about the (y)-axis | (\iint_R x^2\rho\,dA) | kg·m² |
Setting Up Integrals and Checking Results
A correct calculation starts with the region and density, not with an antiderivative. Sketch (R), choose coordinates that describe it cleanly, and make sure the bounds cover every point exactly once. Then compute mass before using it to find the center of mass.
A worked Cartesian example
Suppose a rectangular lamina occupies (0\le x\le2), (0\le y\le1), with density (\rho(x,y)=3+x) kg/m². Since the boundaries are straight and aligned with the axes, Cartesian coordinates give direct bounds:
[ m=\int_0^2\int_0^1(3+x)\,dy\,dx =\int_0^2(3+x)\,dx=8\text{ kg}. ]
The first moments are
[ M_y=\int_0^2\int_0^1x(3+x)\,dy\,dx =\frac{26}{3}\text{ kg}\cdot\text{m}, ]
[ M_x=\int_0^2\int_0^1y(3+x)\,dy\,dx =4\text{ kg}\cdot\text{m}. ]
Thus,
[ (\bar{x},\bar{y}) =\left(\frac{M_y}{m},\frac{M_x}{m}\right) =\left(\frac{13}{12},\frac12\right)\text{ m}. ]
The density increases as (x) increases, so the center of mass lies to the right of the rectangle’s geometric center, at (x=1). This check agrees with the calculation.
The moments of inertia for the same lamina are
[ I_x=\int_0^2\int_0^1 y^2(3+x)\,dy\,dx=\frac83\text{ kg}\cdot\text{m}^2, ]
[ I_y=\int_0^2\int_0^1 x^2(3+x)\,dy\,dx=12\text{ kg}\cdot\text{m}^2. ]
The units provide a useful check: the density contributes kg/m², (dA) contributes m², and the squared distance in an inertia integral contributes m² more.
Choosing Cartesian or polar coordinates
Cartesian coordinates use (dA=dx\,dy) and work well for rectangles, many triangles, and regions described by (x)- or (y)-bounds. Polar coordinates are often simpler for disks, sectors, or regions bounded by circles. In polar form,
[ x=r\cos\theta,\qquad y=r\sin\theta,\qquad dA=r\,dr\,d\theta. ]
That final factor, (r), is the Jacobian. It must be included because a small polar patch has area approximately (r\,dr\,d\theta), not just (dr\,d\theta).
For a uniform disk of radius (a) and constant density (\rho_0), the mass is
[ m=\int_0^{2\pi}\int_0^a \rho_0\,r\,dr\,d\theta =\rho_0\pi a^2. ]
Symmetry places its center of mass at the origin. Its moments are (I_x=I_y=\rho_0\pi a^4/4), so (I_0=\rho_0\pi a^4/2). This example also shows why leaving out the polar factor (r) gives an incorrect mass.
| Region or density pattern | Usually helpful coordinates | Setup detail to watch |
|---|---|---|
| Rectangle with axis-aligned edges | Cartesian | Use bounds matching the four edges |
| Disk or circular sector | Polar | Include (r) in (dA) |
| Density depending on distance from origin | Polar | Replace (x,y) using (r\cos\theta,r\sin\theta) |
| Region between two curves | Often Cartesian | Split the region if one set of bounds cannot cover it cleanly |
Before accepting an answer, check that the bounds match your sketch, that the mass is positive when (\rho) is nonnegative and the region has positive area, and that the units fit the quantity. For a nonnegative density, the center of mass should lie within the convex hull of the lamina. Symmetry can provide another check: a shape and density symmetric about the (y)-axis must have (\bar{x}=0).
Applications, Interpretation, and Common Errors
These formulas turn a shape and a density rule into measurable properties of a thin object. In a math problem, that may mean finding a balance point or comparing how mass is spread about an axis. The same setup applies whether density is constant or varies across the region.
Common mistakes often come from the model rather than the integration:
Quick diagnostic checklist
- Does the region (R) match the physical or stated shape?
- Is (\rho(x,y)\ge0), with units of mass per area?
- Do the bounds cover the region exactly once?
- If using polar coordinates, is (r) included?
- Does the final mass have units of mass, and does each inertia have units of mass times length squared?
- Is the center of mass consistent with the density pattern and symmetry?
Frequently Asked Questions
These answers address common points of confusion when setting up planar mass integrals. Keep the region, density, and coordinate system distinct: the region determines where to integrate, density weights each point, and the coordinates determine how to express the area element and bounds.
What is a lamina in calculus?
A lamina is a thin, flat object represented by a bounded plane region with mass spread across it.
What is the mass formula for a lamina?
The mass is (m=\iint_R\rho(x,y)\,dA), where (\rho) is areal density and (R) is the region.
What units should areal density use?
Use units of mass per area, such as kg/m² or g/cm². Keep length and density units consistent.
How do I find the center of mass?
Calculate (m), then divide each first moment by (m): (\bar{x}=\frac{1}{m}\iint_R x\rho\,dA) and (\bar{y}=\frac{1}{m}\iint_R y\rho\,dA).
What is the difference between center of mass and centroid?
For uniform density, the center of mass matches the geometric centroid. With variable density, denser parts pull the center of mass toward them.
Why is there an (r) in polar area integrals?
The polar area element is (dA=r\,dr\,d\theta). The factor (r) accounts for the fact that the width of an angular slice grows with its distance from the origin.
What is the moment of inertia about the (x)-axis?
It is (I_x=\iint_R y^2\rho(x,y)\,dA). The squared distance from the (x)-axis is (y^2).
What is the polar moment of inertia about the origin?
It is (I_0=\iint_R(x^2+y^2)\rho\,dA=I_x+I_y).
Can a center of mass lie outside the lamina?
For a nonnegative density over a bounded region, it lies within the region’s convex hull. It need not lie inside a region that has a hole or a concave shape.
When should I use polar coordinates?
Use them when the region or density is naturally circular or depends on distance from the origin. Remember to include the Jacobian (r).
(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)
