Mass Of A Lamina With Variable Density

From Surface Density to Total Mass

A lamina is a thin, flat sheet occupying a region $R$ in the plane. If material is spread uniformly, its mass is the product of its surface density and its area. When the material is not spread uniformly, different parts of the lamina contribute different amounts of mass. A double integral accounts for that variation across the region.

The aim here is to find the total mass of a lamina. The density may be constant, may depend on the Cartesian coordinates, or may depend only on distance from a central point. In each case, the setup must reflect the region and the density, and the result must have units of mass.

Definition: A lamina with region $R$ and surface density $\sigma(x,y)$ has total mass $$ m=\iint_R \sigma(x,y)\,dA, $$ provided $\sigma$ is integrable over $R$. For a physical mass distribution, the density must also be nonnegative on $R$.

Surface density measures mass per unit area. For example, if mass is measured in kilograms and length in metres, then $\sigma$ has units of $\text{kg}/\text{m}^2$. Multiplying density by a small area gives an amount of mass; integrating adds those contributions over the whole lamina.

Why the Mass Integral Works

Divide the region into many small pieces. On a small piece with area $\Delta A_i$, the density is approximately constant, so its mass is approximately $\sigma(x_i,y_i)\Delta A_i$ for a sample point $(x_i,y_i)$ in that piece. Adding these estimates gives an approximation to the total mass. As the pieces become finer, these sums approach the double integral, when the density is integrable.

m\approx \sum_i \sigma(x_i,y_i)\,\Delta A_i \quad\longrightarrow\quad \iint_R \sigma(x,y)\,dA.

This explains why density is the integrand and area is the element being integrated. It also shows why the density must be evaluated throughout the region, rather than at a single representative point, unless the density is constant.

Existence condition: A bounded region and a density that is continuous only in the interior do not, by themselves, guarantee a finite mass. A sufficient condition is that $R$ be a compact Jordan-measurable region and that $\sigma$ be continuous on $R$. More generally, the mass formula applies when $\sigma$ is integrable over $R$. A nonnegative density can still have a divergent integral if it is unbounded near the boundary.

Constant Surface Density

If the surface density has the same value $\sigma_0$ everywhere on $R$, it can be taken outside the integral. The remaining integral is the area of $R$, so total mass is density times area.

m=\iint_R \sigma_0\,dA =\sigma_0\iint_R dA =\sigma_0\,\operatorname{Area}(R).

Worked Example: A Uniform Rectangular Lamina

A rectangular sheet occupies $0\leq x\leq 2\,\text{m}$, $0\leq y\leq 1\,\text{m}$, and has constant density $3\,\text{kg}/\text{m}^2$. Find its mass.

The density is positive and constant. The area element in Cartesian coordinates is $dA=dx\,dy$, so

m=\int_0^{2\,\text{m}}\int_0^{1\,\text{m}} 3\,\frac{\text{kg}}{\text{m}^2}\,dy\,dx.

First integrate with respect to $y$:

\int_0^{1\,\text{m}}3\,\frac{\text{kg}}{\text{m}^2}\,dy = 3\,\frac{\text{kg}}{\text{m}^2}(1\,\text{m}) = 3\,\frac{\text{kg}}{\text{m}}.

Then integrate with respect to $x$:

m=\int_0^{2\,\text{m}}3\,\frac{\text{kg}}{\text{m}}\,dx = 3\,\frac{\text{kg}}{\text{m}}(2\,\text{m}) = 6\,\text{kg}.

As a check, the area is $(2\,\text{m})(1\,\text{m})=2\,\text{m}^2$, and density times area is $(3\,\text{kg}/\text{m}^2)(2\,\text{m}^2)=6\,\text{kg}$, agreeing with the integral.

Density Depending on Both Coordinates

A coordinate-dependent density assigns different values to different locations. In Cartesian coordinates, the mass is found by integrating that function over the actual region. Before evaluating, check that the density is nonnegative throughout the region. This is a physical requirement; a formula that becomes negative somewhere does not describe an ordinary nonnegative mass distribution there.

Worked Example: A Density That Increases Across a Rectangle

Let $R$ be the rectangle $0\leq x\leq 2\,\text{m}$, $0\leq y\leq 1\,\text{m}$, and let $\sigma(x,y)=\left(2+\frac{x}{1\,\text{m}}+\frac{y}{1\,\text{m}}\right)\text{kg}/\text{m}^2$. Find the mass.

On the given rectangle, $x/(1\,\text{m})\geq 0$ and $y/(1\,\text{m})\geq 0$, so $\sigma(x,y)\geq 2\,\text{kg}/\text{m}^2>0$. To make the numerical limits transparent, set $X=x/(1\,\text{m})$ and $Y=y/(1\,\text{m})$. Then $0\leq X\leq 2$, $0\leq Y\leq 1$, and $dA=(1\,\text{m}^2)dX\,dY$. Thus

m=(1\,\text{m}^2)\int_0^2\int_0^1(2+X+Y)\,\frac{\text{kg}}{\text{m}^2}\,dY\,dX.

Evaluate the inner integral:

\int_0^1(2+X+Y)\,dY = \left[(2+X)Y+\frac{Y^2}{2}\right]_0^1 = 2+X+\frac12 = \frac52+X.

Now evaluate the outer integral:

\int_0^2\left(\frac52+X\right)\,dX = \left[\frac52X+\frac{X^2}{2}\right]_0^2 = 5+2 = 7.

The factor $1\,\text{m}^2$ cancels the inverse-area unit in the density. Therefore $m=7\,\text{kg}$. As an independent check, the constant part of the density contributes $2(2\,\text{m}^2)=4\,\text{kg}$, the $X$ part contributes $2\,\text{kg}$, and the $Y$ part contributes $1\,\text{kg}$. Their sum is $7\,\text{kg}$, as found by integration.

Radial Density and Polar Coordinates

Sometimes density depends only on distance from the origin. Write $r=\sqrt{x^2+y^2}$; a radial density has the form $\sigma(r)$. For a disk of radius $a$, polar coordinates use $0\leq r\leq a$ and $0\leq\theta\leq 2\pi$. The polar area element is $dA=r\,dr\,d\theta$, so the radial distance factor must appear in the mass integral.

m=\int_0^{2\pi}\int_0^a \sigma(r)\,r\,dr\,d\theta.

For other regions, the angular and radial limits must be chosen to describe that region. The density alone does not determine the bounds. If the density is nonnegative and integrable, the resulting mass is nonnegative and finite.

Worked Example: A Radially Increasing Density on a Disk

A disk of radius $2\,\text{m}$ has density $\sigma(r)=\left(1+\frac{r}{1\,\text{m}}\right)\text{kg}/\text{m}^2$. Find its mass.

Throughout the disk, $0\leq r\leq 2\,\text{m}$, so the factor $1+r/(1\,\text{m})$ is at least $1$. The density is therefore nonnegative. Write $R=r/(1\,\text{m})$, so $0\leq R\leq 2$, $r\,dr=(1\,\text{m}^2)R\,dR$, and the angular variable ranges from $0$ to $2\pi$. The mass is

m=(1\,\text{m}^2)\int_0^{2\pi}\int_0^2(1+R)R\,\frac{\text{kg}}{\text{m}^2}\,dR\,d\theta.

First integrate in $R$:

\int_0^2(1+R)R\,dR = \int_0^2(R+R^2)\,dR = \left[\frac{R^2}{2}+\frac{R^3}{3}\right]_0^2 = 2+\frac83 = \frac{14}{3}.

Then integrate over the full angle:

m=(1\,\text{m}^2)(2\pi)\left(\frac{14}{3}\right)\frac{\text{kg}}{\text{m}^2} = \frac{28\pi}{3}\,\text{kg}.

Numerically, this is approximately $29.32\,\text{kg}$ to two decimal places. For a check, the constant part of the density contributes $\int_0^{2\pi}\int_0^2 r\,dr\,d\theta=4\pi\,\text{m}^2$, hence $4\pi\,\text{kg}$. The part proportional to $r/(1\,\text{m})$ contributes $\frac{16\pi}{3}\,\text{kg}$. Their sum is $4\pi+\frac{16\pi}{3}=\frac{28\pi}{3}\,\text{kg}$, matching the calculation.

A Common Error: Omitting the Polar Area Factor

In polar coordinates, $dr\,d\theta$ alone is not the area element. The change from Cartesian to polar coordinates has area factor $r$, so every polar mass integral must use $r\,dr\,d\theta$. Omitting it changes how area is counted: a small angular wedge farther from the origin covers more area than an equally wide wedge near the origin.

For the disk example, the incorrect setup would be

\int_0^{2\pi}\int_0^2 (1+r)\,dr\,d\theta = 2\pi\left[r+\frac{r^2}{2}\right]_0^2 = 8\pi.

This is wrong because it omits the factor $r$, and its units are also inconsistent: density times $dr\,d\theta$ has units of $\text{kg}/\text{m}$, not $\text{kg}$. The correct integral includes the area element:

\int_0^{2\pi}\int_0^2 (1+r)r\,dr\,d\theta = \frac{28\pi}{3}\,\text{kg},

with the length scale understood as $1\,\text{m}$ in the density expression. Checking both the Jacobian factor and units is an effective way to catch this mistake.

Setting Up a Mass Calculation Reliably

Before evaluating, make the mathematical and physical checks explicit. A useful sequence is:

Specify the region.
Write bounds that cover the lamina exactly once, using Cartesian or polar coordinates as appropriate.

Check the density.
Verify that $\sigma\geq 0$ throughout the region for a physical mass distribution, and that the density is integrable there.

Choose the area element.
Use $dx\,dy$ in Cartesian coordinates and $r\,dr\,d\theta$ in polar coordinates.

Evaluate and check.
Keep units attached, verify the arithmetic, and confirm that the final result has units of mass.

Total mass is a scalar quantity: it records how much material the entire lamina contains. Later planar applications use density-weighted integrals for other quantities, but those require different integrands. Here the essential operation is always to accumulate the nonnegative surface density over area.

Check Your Understanding

Use the density, region, and area element together in each question.

A rectangular lamina has constant density $4\,\text{kg}/\text{m}^2$ and area $3\,\text{m}^2$. What is its mass?
What condition on $\sigma(x,y)$ makes it a nonnegative physical surface density over $R$?
For a radial density on a disk of radius $a$, write the polar-coordinate mass integral, including the area factor.
Why does continuity on a bounded region alone not always guarantee a finite mass?
What units should result when a density measured in $\text{kg}/\text{m}^2$ is integrated over area measured in $\text{m}^2$?