Density Proportional to Distance From an Edge

Why the Direction of the Strips Matters

A lamina can be denser in some places than in others. If its density increases in proportion to distance from one of its edges, points farther from that edge contribute more mass. To find the centroid, the density must be included in both the mass and the first moments.

The choice of strips is especially useful here. A strip parallel to the edge has the same perpendicular distance from the edge at every point along its length. Its density is therefore constant across the strip, so mass and moments can be calculated using one density value for the whole strip.

Key idea: If density depends on perpendicular distance from an edge, use strips parallel to that edge. The strip’s thickness is measured perpendicular to the edge, in the direction in which density changes.

Density, Mass, and First Moments

Let $\rho(x,y)$ denote mass per unit area at the point $(x,y)$. If $R$ is the lamina’s region, its mass and first moments are

m=\iint_R \rho(x,y)\,dA,\qquad M_y=\iint_R x\rho(x,y)\,dA,\qquad M_x=\iint_R y\rho(x,y)\,dA.

When $m>0$, the centroid coordinates are

\bar{x}=\frac{M_y}{m},\qquad \bar{y}=\frac{M_x}{m}.

For density proportional to distance $d$ from an edge, write $\rho=kd$, where $k>0$ is a constant. If density is measured in mass per area and distance in length, then $k$ has units of mass per length cubed. Density is zero at the specified edge and increases as the distance from it increases.

Rectangular Lamina: Density Increasing from a Horizontal Edge

Consider the rectangle $0\le x\le a,\ 0\le y\le b$, with its lower edge $y=0$ specified. The perpendicular distance from this edge is $y$, so the density is $\rho(x,y)=ky$. Horizontal strips are parallel to the edge and have thickness $dy$. On a strip at height $y$, density is constant along the strip.

Result for a rectangle: If $\rho=ky$ on $0\le x\le a,\ 0\le y\le b$, then $$ m=\frac{kab^2}{2},\qquad \bar{x}=\frac{a}{2},\qquad \bar{y}=\frac{2b}{3}. $$

To see why, a horizontal strip has area $a\,dy$, so its mass is $ky\,a\,dy$. Its horizontal centroid is $a/2$, by symmetry. Thus

m=ka\int_0^b y\,dy=\frac{kab^2}{2}, \qquad M_y=\frac{a}{2}m, \qquad \bar{x}=\frac{M_y}{m}=\frac{a}{2}.

For the vertical first moment, every point in the strip has height $y$, giving

M_x=ka\int_0^b y^2\,dy=\frac{kab^3}{3}, \qquad \bar{y}=\frac{M_x}{m} =\frac{kab^3/3}{kab^2/2} =\frac{2b}{3}.

The result places the centroid above the rectangle’s midpoint: the upper part is denser, so it contributes more mass.

Worked Example: A Rectangle Denser Toward Its Top Edge

A rectangular sheet occupies $0\le x\le6$ m and $0\le y\le3$ m. Its density is $\rho(x,y)=2y$ kg/m². Find its mass and centroid.

Because density depends on $y$, use horizontal strips. A strip at height $y$ has width $6$ m, area $6\,dy$, and density $2y$ kg/m². Its mass is $12y\,dy$ kg. Therefore,

m=\int_0^3 12y\,dy =12\left[\frac{y^2}{2}\right]_0^3 =12\cdot\frac{9}{2} =54\text{ kg}.

Each strip has horizontal centroid $x=3$ m. Hence

M_y=3m=162\text{ kg}\cdot\text{m}, \qquad \bar{x}=\frac{162}{54}=3\text{ m}.

For the first moment about the $x$-axis, multiply each strip’s mass by its height:

M_x=\int_0^3 y(12y)\,dy =12\left[\frac{y^3}{3}\right]_0^3 =12\cdot9 =108\text{ kg}\cdot\text{m}, \qquad \bar{y}=\frac{108}{54}=2\text{ m}.

Check: Multiplying the mass by the centroid coordinates gives $54(3)=162$ kg·m and $54(2)=108$ kg·m, the calculated first moments. The centroid is $(3,2)$ m; its height is greater than the rectangle’s midpoint height of $1.5$ m, as expected for density increasing upward.

Rectangular Lamina: Density Increasing from a Vertical Edge

If the specified edge is vertical, density depends on the horizontal coordinate. For example, on $0\le x\le a,\ 0\le y\le b$, density proportional to distance from the right edge $x=a$ is $\rho=k(a-x)$. Vertical strips are parallel to that edge, so density is constant across each strip. Reflection of the horizontal-edge calculation gives

m=\frac{kba^2}{2},\qquad \bar{x}=\frac{a}{3},\qquad \bar{y}=\frac{b}{2}.

The centroid is closer to the left edge, where density is greatest. The formula for $\bar{x}$ can also be checked directly: $M_y=bk\int_0^a x(a-x)\,dx=kba^3/6$, so $M_y/m=a/3$. The rectangle is symmetric in $y$, and its density does not vary with $y$, so $\bar y=b/2$.

Worked Example: Density Increasing Away from the Right Edge

A rectangle occupies $0\le x\le4$ m and $0\le y\le2$ m. Its density is $\rho(x,y)=3(4-x)$ kg/m². Find its mass and centroid.

Density depends on distance $4-x$ from the right edge, so use vertical strips. A strip at coordinate $x$ has height $2$ m and mass $6(4-x)\,dx$ kg. Its total mass is

m=6\int_0^4(4-x)\,dx =6\left[4x-\frac{x^2}{2}\right]_0^4 =6(16-8) =48\text{ kg}.

The first moment about the $y$-axis is

M_y=6\int_0^4 x(4-x)\,dx =6\left[2x^2-\frac{x^3}{3}\right]_0^4 =6\left(32-\frac{64}{3}\right) =6\cdot\frac{32}{3} =64\text{ kg}\cdot\text{m}.

Therefore $\bar{x}=64/48=4/3$ m. The density is independent of $y$, and each vertical strip is symmetric about $y=1$ m. Thus $M_x=48(1)=48$ kg·m and $\bar{y}=48/48=1$ m.

Check: The proposed centroid gives $m\bar{x}=48(4/3)=64$ kg·m and $m\bar{y}=48(1)=48$ kg·m, matching both first moments. Since density is greatest near $x=0$, a horizontal centroid coordinate of $4/3$ m, to the left of the midpoint $x=2$ m, is plausible.

Triangular Lamina: Density Increasing from the Base

Consider the right triangle with vertices $(0,0)$, $(L,0)$, and $(0,H)$, where $L,H>0$. Its base lies on $y=0$, and the sloping side is $y=H(1-x/L)$. The density $\rho=ky$ increases with distance above the base.

Horizontal strips are parallel to the base, and their density is constant along each strip. At height $y$, the strip extends from $x=0$ to $x=L(1-y/H)$, so its width is $w(y)=L(1-y/H)$. The mass and first moments follow from integrating over these strips.

Result for a triangle: For this right triangle with $\rho=ky$, $$ m=\frac{kLH^2}{6},\qquad \bar{x}=\frac{L}{4},\qquad \bar{y}=\frac{H}{2}. $$

Here is a derivation. A horizontal strip at height $y$ has mass $ky\,w(y)\,dy$. Therefore,

m=kL\int_0^H y\left(1-\frac{y}{H}\right)\,dy =kL\left(\frac{H^2}{2}-\frac{H^2}{3}\right) =\frac{kLH^2}{6}.

For $M_y$, the integral of $x$ across a strip from $0$ to $w$ is $w^2/2$. Thus

M_y=\frac{k}{2}\int_0^H y\,w(y)^2\,dy =\frac{kL^2}{2}\int_0^H y\left(1-\frac{y}{H}\right)^2\,dy.

Expanding the square and integrating gives

M_y=\frac{kL^2}{2} \left(\frac{H^2}{2}-\frac{2H^2}{3}+\frac{H^2}{4}\right) =\frac{kL^2H^2}{24}.

For $M_x$, multiply strip mass by its height:

M_x=kL\int_0^H y^2\left(1-\frac{y}{H}\right)\,dy =kL\left(\frac{H^3}{3}-\frac{H^3}{4}\right) =\frac{kLH^3}{12}.

Dividing by the mass gives $\bar{x}=(kL^2H^2/24)/(kLH^2/6)=L/4$ and $\bar{y}=(kLH^3/12)/(kLH^2/6)=H/2$. The increased density shifts the centroid away from the base. Because the triangle narrows toward its top, the density-weighted centroid also lies left of the geometric center in the horizontal direction.

Worked Example: A Triangular Sheet Denser Toward Its Apex

A triangular sheet has vertices $(0,0)$, $(6,0)$, and $(0,4)$, measured in metres. Its density is $\rho(x,y)=2y$ kg/m². Find its mass and centroid.

The base is $y=0$, so use horizontal strips. The sloping side is $y=4(1-x/6)$, which gives strip width $w(y)=6(1-y/4)$. A strip has mass $2y\,w(y)\,dy$. Hence

m=12\int_0^4 y\left(1-\frac{y}{4}\right)\,dy =12\left(\frac{16}{2}-\frac{16}{3}\right) =12\cdot\frac{8}{3} =32\text{ kg}.

The first moment about the $y$-axis uses the strip integral $\int_0^{w}x\,dx=w^2/2$:

M_y=\int_0^4 2y\,\frac{[6(1-y/4)]^2}{2}\,dy =36\int_0^4 y\left(1-\frac{y}{4}\right)^2\,dy =36\cdot\frac{16}{12} =48\text{ kg}\cdot\text{m}.

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

M_x=12\int_0^4 y^2\left(1-\frac{y}{4}\right)\,dy =12\left(\frac{64}{3}-16\right) =12\cdot\frac{16}{3} =64\text{ kg}\cdot\text{m}.

Consequently, $\bar{x}=48/32=1.5$ m and $\bar{y}=64/32=2$ m.

Check: The centroid coordinates reproduce the first moments: $32(1.5)=48$ kg·m and $32(2)=64$ kg·m. The centroid is halfway up the triangle, above the midpoint of its height, consistent with density increasing from the base.

A Common Strip-Orientation Mistake

For the triangular example, a tempting but incorrect shortcut is to multiply the triangle’s area by the density at its halfway height. The area is $6\cdot4/2=12$ m², and the density at $y=2$ m is $2(2)=4$ kg/m². This shortcut gives $12(4)=48$ kg, not the correct mass of $32$ kg. It is wrong because density varies throughout the triangle; the density at one height cannot represent the whole region.

Instead, horizontal strips keep density constant along each strip while allowing it to vary from strip to strip. The mass integral accounts for that variation. Alternatively, vertical strips could be used only if density is integrated along the strip as well; assigning one density value to an entire vertical strip would again be incorrect. The strip direction does not make a valid integral impossible, but it can make the setup simpler or more error-prone.

Practical check: Locate the specified edge, write the perpendicular distance from that edge, and choose strips parallel to it. Then check that the density has units of mass per area and that the centroid lies toward the more densely weighted part of the lamina.

Check Your Understanding

Use strip integrals and first moments to answer the following questions.

On a rectangle $0\le x\le a,\ 0\le y\le b$, what density function represents density proportional to distance from the edge $y=b$?
For a rectangle with density $\rho=k(a-x)$, which strip direction makes density constant across each strip, and why?
For the right triangle with vertices $(0,0)$, $(L,0)$, and $(0,H)$, write the width of a horizontal strip at height $y$.
Why is density at the midpoint height insufficient for calculating the mass of a triangular lamina with density $\rho=ky$?
For the triangle in the worked example, verify the centroid by multiplying its mass by each centroid coordinate to recover the corresponding first moment.