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Lamina — Definition, Formula & Examples

A lamina is an idealized flat, thin plate that has mass spread across a two-dimensional region. In calculus, it serves as the physical model when you compute centers of mass, moments, and related quantities using double or single integrals.

A lamina is a two-dimensional body occupying a bounded region RR in the xyxy-plane, equipped with a density function ρ(x,y)\rho(x, y) (mass per unit area). When ρ\rho is constant, the lamina is called homogeneous; otherwise it is heterogeneous.

Key Formula

m=∬Rρ(x,y) dA,xˉ=Mym,yˉ=Mxmm = \iint_R \rho(x,y)\,dA, \qquad \bar{x} = \frac{M_y}{m}, \qquad \bar{y} = \frac{M_x}{m}
Where:
  • mm = Total mass of the lamina
  • ρ(x,y)\rho(x,y) = Density function (mass per unit area)
  • RR = Region in the xy-plane occupied by the lamina
  • MxM_x = Moment about the x-axis: $\iint_R y\,\rho(x,y)\,dA$
  • MyM_y = Moment about the y-axis: $\iint_R x\,\rho(x,y)\,dA$
  • xˉ,yˉ\bar{x}, \bar{y} = Coordinates of the center of mass

How It Works

You define a lamina by specifying a region RR and a density function ρ(x,y)\rho(x,y). The total mass is found by integrating ρ\rho over RR. Moments about the axes are computed by weighting position coordinates by ρ\rho and integrating. Dividing each moment by the total mass gives the coordinates of the center of mass (xˉ,yˉ)(\bar{x}, \bar{y}). For a homogeneous lamina, ρ\rho cancels and the center of mass depends only on the geometry of RR — this special case is called the centroid.

Worked Example

Problem: Find the mass and center of mass of a lamina occupying the region bounded by y=0y = 0, y=xy = x, and x=2x = 2, with density ρ(x,y)=3\rho(x,y) = 3 (constant).
Step 1: Compute the total mass by integrating the constant density over the triangular region.
m=∫02∫0x3 dy dx=∫023x dx=3x22∣02=6m = \int_0^2 \int_0^x 3\,dy\,dx = \int_0^2 3x\,dx = \frac{3x^2}{2}\Big|_0^2 = 6
Step 2: Compute the moment about the y-axis.
My=∫02∫0x3x dy dx=∫023x2 dx=x3∣02=8M_y = \int_0^2 \int_0^x 3x\,dy\,dx = \int_0^2 3x^2\,dx = x^3\Big|_0^2 = 8
Step 3: Compute the moment about the x-axis.
Mx=∫02∫0x3y dy dx=∫023x22 dx=x32∣02=4M_x = \int_0^2 \int_0^x 3y\,dy\,dx = \int_0^2 \frac{3x^2}{2}\,dx = \frac{x^3}{2}\Big|_0^2 = 4
Answer: The mass is m=6m = 6, and the center of mass is (xˉ,yˉ)=(86, 46)=(43, 23)(\bar{x}, \bar{y}) = \left(\frac{8}{6},\, \frac{4}{6}\right) = \left(\frac{4}{3},\, \frac{2}{3}\right).

Why It Matters

Lamina problems appear throughout Calculus II and III courses and are a standard topic on the AP Calculus BC exam. Engineers use the same framework to locate centroids of flat structural components, which determines how beams and plates respond to loads.

Common Mistakes

Mistake: Confusing MxM_x and MyM_y — using xx in the integrand for MxM_x.
Correction: MxM_x (moment about the x-axis) uses yy in the integrand, and MyM_y (moment about the y-axis) uses xx. The subscript names the axis, not the variable.