Nonuniform Density on Triangular Laminae

Why Triangles with Varying Density Need Care

A triangular lamina is a thin, flat object occupying a triangular region in the plane. If its density is constant, its mass is distributed uniformly. If density varies from point to point, however, parts of the triangle contribute different amounts to the total mass and to its first moments. The centroid therefore need not be the geometric center of the triangle.

For an affine density, the density changes linearly with the coordinates. The computations use the same mass and first-moment integrals used for other laminae, but the triangle’s sloping sides must be translated into correct strip bounds. A reliable solution therefore connects three things: the region, the density, and the moments.

Mass, First Moments, and Centroid

Definition: Let a lamina occupy a bounded planar region $R$ and have density $\rho(x,y)$, measured as mass per unit area. Its mass $m$, first moment about the $x$-axis $M_x$, and first moment about the $y$-axis $M_y$ are $$ m=\iint_R \rho(x,y)\,dA,\qquad M_x=\iint_R y\rho(x,y)\,dA,\qquad M_y=\iint_R x\rho(x,y)\,dA. $$ If $m>0$, its centroid is $$ (\bar{x},\bar{y})=\left(\frac{M_y}{m},\frac{M_x}{m}\right). $$

The coordinate order in the centroid formula matters. The moment about the $y$-axis uses the distance $x$, so it determines $\bar{x}$. The moment about the $x$-axis uses the distance $y$, so it determines $\bar{y}$.

An affine density has the form $\rho(x,y)=a+bx+cy$, where $a,b,c$ are constants. For a physical lamina, density should be nonnegative throughout the region, and its total mass must be positive. On a triangle, checking the density at the three vertices is enough to establish nonnegativity throughout: every point of the triangle is a weighted average of its vertices with nonnegative weights summing to one, and an affine function takes the same weighted average of its vertex values. Thus, if the three vertex values are nonnegative, so is every value inside the triangle.

Turning a Triangle into Strip Bounds

For vertical strips, describe the region by $x$ between its left and right limits, then give the lower and upper $y$-bounds as functions of $x$. A vertical strip has thickness $dx$, so its area element is $dA=dy\,dx$. For horizontal strips, give the left and right $x$-bounds as functions of $y$; then $dA=dx\,dy$.

Once the bounds are set, integrate the density for mass. For the first moment about the $x$-axis, include the additional factor $y$; for the first moment about the $y$-axis, include $x$. The bounds may change if a triangle cannot be described by one set of strips in the chosen direction. For a single triangle with one vertex on an axis and the opposite side a straight line, one strip description is often especially direct.

Key setup: For vertical bounds $x_0\le x\le x_1$ and $g(x)\le y\le h(x)$, $$ m=\int_{x_0}^{x_1}\int_{g(x)}^{h(x)}\rho(x,y)\,dy\,dx,\quad M_x=\int_{x_0}^{x_1}\int_{g(x)}^{h(x)}y\rho(x,y)\,dy\,dx,\quad M_y=\int_{x_0}^{x_1}\int_{g(x)}^{h(x)}x\rho(x,y)\,dy\,dx. $$

These formulas follow by dividing the region into thin vertical strips. A small piece at $(x,y)$ with area $dA$ has mass approximately $\rho(x,y)dA$. Its contributions to the first moments are approximately $y\rho(x,y)dA$ and $x\rho(x,y)dA$, respectively. Adding those contributions over the region and taking the limit gives the area integrals above. The same reasoning with horizontal strips gives the corresponding integrals in the opposite order.

Worked Examples

Worked Example: Vertical Strips on a Right Triangle

A triangular lamina occupies the triangle with vertices $(0,0)$, $(4,0)$, and $(0,3)$, with coordinates measured in meters. Its density is $\rho(x,y)=2+x$ kilograms per square meter. Find its mass and centroid.

The sloping side joins $(4,0)$ and $(0,3)$, so its equation is $y=3-\frac{3}{4}x$. Vertical bounds are $0\le x\le4$ and $0\le y\le3-\frac{3}{4}x$. The density values at the vertices are $2$, $6$, and $2$, so the density is positive throughout.

For the mass, first integrate in $y$:

\begin{aligned} m&=\int_0^4\int_0^{3-3x/4}(2+x)\,dy\,dx\\ &=\int_0^4(2+x)\left(3-\frac{3x}{4}\right)\,dx\\ &=\int_0^4\left(6+\frac{3x}{2}-\frac{3x^2}{4}\right)\,dx\\ &=\left[6x+\frac{3x^2}{4}-\frac{x^3}{4}\right]_0^4\\ &=24+12-16=20\text{ kg}. \end{aligned}

As a check on the integration, the derivative of the antiderivative is $6+\frac{3x}{2}-\frac{3x^2}{4}$, the integrand above.

For the first moment about the $y$-axis, multiply by $x$:

\begin{aligned} M_y&=\int_0^4\int_0^{3-3x/4}x(2+x)\,dy\,dx\\ &=\int_0^4x\left(6+\frac{3x}{2}-\frac{3x^2}{4}\right)\,dx\\ &=\int_0^4\left(6x+\frac{3x^2}{2}-\frac{3x^3}{4}\right)\,dx\\ &=\left[3x^2+\frac{x^3}{2}-\frac{3x^4}{16}\right]_0^4\\ &=48+32-48=32\text{ kg}\cdot\text{m}. \end{aligned}

The derivative of the displayed antiderivative is the stated integrand. For the first moment about the $x$-axis, integrate $y(2+x)$ over each strip:

\begin{aligned} M_x&=\int_0^4\frac{2+x}{2}\left(3-\frac{3x}{4}\right)^2\,dx\\ &=\frac{9}{2}\int_0^4\left(2-\frac{3x^2}{8}+\frac{x^3}{16}\right)\,dx\\ &=\frac{9}{2}\left[2x-\frac{x^3}{8}+\frac{x^4}{64}\right]_0^4\\ &=\frac{9}{2}(8-8+4)=18\text{ kg}\cdot\text{m}. \end{aligned}

Differentiating the antiderivative gives $2-\frac{3x^2}{8}+\frac{x^3}{16}$, as required. Therefore,

\bar{x}=\frac{32}{20}=\frac85=1.6\text{ m},\qquad \bar{y}=\frac{18}{20}=\frac9{10}=0.9\text{ m}.

The point satisfies $0\le 1.6\le4$, $0\le0.9$, and $0.9\le3-\frac34(1.6)=1.8$, so it lies inside the triangle. The larger density at larger $x$-values is consistent with the centroid lying to the right of the triangle’s geometric centroid.

Worked Example: Horizontal Strips on a Triangle

A triangle has vertices $(0,0)$, $(4,0)$, and $(4,3)$. Let its density be $\rho(x,y)=1+y$ kilograms per square meter. Find the mass and centroid using horizontal strips.

The sloping side is $x=\frac43y$, so the bounds are $0\le y\le3$ and $\frac43y\le x\le4$. The density at the vertices is $1$, $1$, and $4$, so it is positive throughout.

The strip width is $4-\frac43y$. Thus the mass is

\begin{aligned} m&=\int_0^3(1+y)\left(4-\frac43y\right)\,dy\\ &=\int_0^3\left(4+\frac83y-\frac43y^2\right)\,dy\\ &=\left[4y+\frac43y^2-\frac49y^3\right]_0^3\\ &=12+12-12=12\text{ kg}. \end{aligned}

The derivative of the antiderivative is the integrand. The moment about the $x$-axis is

\begin{aligned} M_x&=\int_0^3y(1+y)\left(4-\frac43y\right)\,dy\\ &=\int_0^3\left(4y+\frac83y^2-\frac43y^3\right)\,dy\\ &=\left[2y^2+\frac89y^3-\frac13y^4\right]_0^3\\ &=18+24-27=15\text{ kg}\cdot\text{m}. \end{aligned}

The derivative of this antiderivative is $4y+\frac83y^2-\frac43y^3$. To calculate the other moment, integrate $x$ across each strip:

\begin{aligned} M_y&=\int_0^3\frac{1+y}{2}\left(4^2-\left(\frac43y\right)^2\right)\,dy\\ &=8\int_0^3\left(1+y-\frac{y^2}{9}-\frac{y^3}{9}\right)\,dy\\ &=8\left[y+\frac{y^2}{2}-\frac{y^3}{27}-\frac{y^4}{36}\right]_0^3\\ &=8\left(3+\frac92-1-\frac94\right)=34\text{ kg}\cdot\text{m}. \end{aligned}

The derivative of the antiderivative inside the brackets is $1+y-\frac{y^2}{9}-\frac{y^3}{9}$. The centroid is therefore

\bar{x}=\frac{34}{12}=\frac{17}{6}\text{ m},\qquad \bar{y}=\frac{15}{12}=\frac54\text{ m}.

It lies within the triangle: $0\le\frac54\le3$, $\frac{17}{6}\le4$, and $\frac{17}{6}\ge\frac43\cdot\frac54=\frac53$. The density increases with $y$, and the centroid’s $y$-coordinate exceeds the triangle’s uniform-density value of $1$ meter.

Worked Example: An Affine Density Depending on Both Coordinates

Consider the triangle $0\le x\le2$, $0\le y\le x$, with density $\rho(x,y)=4-x-y$ kilograms per square meter. Find its mass and centroid.

The vertices are $(0,0)$, $(2,0)$, and $(2,2)$. The density there is $4$, $2$, and $0$, so it is nonnegative throughout. Using vertical strips,

\begin{aligned} m&=\int_0^2\int_0^x(4-x-y)\,dy\,dx\\ &=\int_0^2\left((4-x)x-\frac{x^2}{2}\right)\,dx\\ &=\int_0^2\left(4x-\frac32x^2\right)\,dx\\ &=\left[2x^2-\frac{x^3}{2}\right]_0^2=8-4=4\text{ kg}. \end{aligned}

The antiderivative differentiates to $4x-\frac32x^2$. For the two first moments,

\begin{aligned} M_x&=\int_0^2\int_0^x y(4-x-y)\,dy\,dx\\ &=\int_0^2\left(\frac{(4-x)x^2}{2}-\frac{x^3}{3}\right)\,dx\\ &=\int_0^2\left(2x^2-\frac56x^3\right)\,dx\\ &=\left[\frac23x^3-\frac5{24}x^4\right]_0^2 =\frac{16}{3}-\frac{10}{3}=2\text{ kg}\cdot\text{m},\\[4pt] M_y&=\int_0^2\int_0^x x(4-x-y)\,dy\,dx\\ &=\int_0^2\left(4x^2-\frac32x^3\right)\,dx\\ &=\left[\frac43x^3-\frac38x^4\right]_0^2 =\frac{32}{3}-6=\frac{14}{3}\text{ kg}\cdot\text{m}. \end{aligned}

In each case, differentiating the displayed antiderivative gives the integrand immediately above it. Hence

\bar{x}=\frac{14/3}{4}=\frac76\text{ m},\qquad \bar{y}=\frac24=\frac12\text{ m}.

The centroid satisfies $0\le\frac76\le2$ and $0\le\frac12\le\frac76$, so it is within the triangle. Since the density is greatest near the origin and decreases toward the other vertices, the centroid is shifted toward the origin relative to the uniform-density centroid $(\frac43,\frac23)$.

A Common Moment-Assignment Error

A common mistake is to use the moment about the $x$-axis to calculate $\bar{x}$. For the first example, that would give the incorrect calculation $\bar{x}=M_x/m=18/20=0.9$ meters. The correction is to match each moment with the coordinate it measures: $M_x$ contains the factor $y$, while $M_y$ contains the factor $x$. Thus the correct values are $\bar{x}=M_y/m=32/20=1.6$ meters and $\bar{y}=M_x/m=18/20=0.9$ meters.

Another useful check is geometric. A computed centroid of a positive-density triangle should be consistent with the triangle’s boundaries. Testing its coordinates in the triangle’s inequalities can reveal a bounds error, a missing coordinate factor in a moment, or an arithmetic mistake. Passing this test does not by itself prove the integrals are correct, but failing it signals that something needs to be revisited.

Check Your Understanding

Use the definitions and strip setups above to answer these questions.

For vertical strips, what additional factor belongs in the integrand for $M_y$, and what factor belongs in the integrand for $M_x$?
A triangle has vertices $(0,0)$, $(3,0)$, and $(0,2)$. What are suitable vertical bounds?
Why is checking the density at the three vertices sufficient to establish that an affine density is nonnegative throughout a triangle?
If a lamina has mass $10$, $M_x=12$, and $M_y=25$, what are its centroid coordinates?
What inequality would you check to confirm that a point lies below the sloping side joining $(4,0)$ and $(0,3)$?