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Mass, Moments, and Centroids · Tutorial 484 of 998

Centroids of Regions Between Curves

Set up mass and moment integrals for a region between curves, and use them to find and verify its centroid.

Intermediate 10 min read

What You'll Learn

  • Identify the upper and lower curves on a specified interval
  • Set up mass and first-moment integrals using vertical slices
  • Derive the centroid formulas for a lamina of constant density
  • Evaluate polynomial integrals to find a centroid
  • Check calculations and avoid confusing the moment integrand with the squared slice height

From a Bounded Region to Its Centroid

A region between two curves can be divided into thin vertical strips. Each strip contributes some mass, and its position determines how much it contributes to the first moments about the coordinate axes. Adding these contributions across the region gives the mass and moments; dividing each moment by the mass gives the centroid.

The key is to keep the upper and lower boundaries in the correct order and to use the appropriate moment integrand. The height of a strip determines its mass, while its vertical position also matters when calculating the moment about the \(x\)-axis.

Definition and Formulas

Let \(a<b\), and let \(f\) and \(g\) be continuous on \([a,b]\), with \(f(x)\leq g(x)\) throughout the interval. The region \(R\) between their graphs and the vertical lines \(x=a\) and \(x=b\) consists of points satisfying \(a\leq x\leq b\) and \(f(x)\leq y\leq g(x)\). Assume this region has positive area, and suppose its lamina has constant areal density \(\rho>0\).

Definition: The mass \(m\), first moments \(M_y\) and \(M_x\) about the \(y\)-axis and \(x\)-axis, and centroid \((\bar{x},\bar{y})\) are given by
$$ m=\rho\int_a^b \bigl(g(x)-f(x)\bigr)\,dx, \qquad M_y=\rho\int_a^b x\bigl(g(x)-f(x)\bigr)\,dx, $$
$$ M_x=\frac{\rho}{2}\int_a^b \bigl(g(x)^2-f(x)^2\bigr)\,dx, \qquad \bar{x}=\frac{M_y}{m}, \qquad \bar{y}=\frac{M_x}{m}. $$
The positive-area assumption ensures \(m>0\), so the ratios defining the centroid exist.

Since \(g-f\) is continuous and nonnegative, the area is \(\int_a^b(g-f)\,dx\). Requiring positive area is important: if the two curves agree throughout the interval, then the region has zero area and, for positive density, zero mass. In that case the centroid ratios are undefined.

For constant density, the factor \(\rho\) appears in the mass and both moments, so it cancels from the centroid coordinates. The location of the centroid therefore depends on the shape of the region, not on the particular positive value of its uniform density.

Why the Moment Formulas Work

Consider a thin vertical strip at position \(x\), with width \(dx\) and vertical extent from \(f(x)\) to \(g(x)\). Its height is \(g(x)-f(x)\), so its area is approximately \((g(x)-f(x))\,dx\), and its mass is approximately \(\rho(g(x)-f(x))\,dx\). At this strip, the horizontal coordinate is \(x\), which gives its contribution to the moment about the \(y\)-axis.

For the moment about the \(x\)-axis, the mass in the strip is distributed over its full height. A thin horizontal element at height \(y\) has area \(dy\,dx\) and mass \(\rho\,dy\,dx\); its moment about the \(x\)-axis is its mass times its distance \(y\) from that axis. Integrating over the strip's height gives the strip's moment contribution.

1
Sum the strip masses.
The mass contribution is \(\rho(g(x)-f(x))\,dx\). Integrating from \(a\) to \(b\) gives \(m=\rho\int_a^b(g-f)\,dx\).
2
Weight each mass by its horizontal coordinate.
The moment contribution about the \(y\)-axis is \(x\,\rho(g(x)-f(x))\,dx\). Integrating gives \(M_y=\rho\int_a^b x(g-f)\,dx\).
3
Integrate the vertical coordinate within each strip.
The moment contribution about the \(x\)-axis is \(\rho\int_{f(x)}^{g(x)}y\,dy\,dx=\frac{\rho}{2}(g(x)^2-f(x)^2)\,dx\). Integrating across the interval gives the stated formula for \(M_x\).
4
Divide each moment by the mass.
By the definition of the centroid of a lamina, its coordinates are \(M_y/m\) and \(M_x/m\), respectively.

The inner integral in Step 3 uses \(\int y\,dy=y^2/2\), evaluated at the upper and lower limits. Thus the difference of squares is \(g(x)^2-f(x)^2\), not the square of the strip's height.

Worked Polynomial Examples

Worked Example: A Parabola and a Line

Find the mass, first moments, and centroid of the region between \(y=x^2\) and \(y=2x\) on \(0\leq x\leq2\), with constant density \(\rho\).

First check the order of the curves. Their difference is \(2x-x^2=x(2-x)\), which is nonnegative on \([0,2]\) and positive for \(0<x<2\). The region has positive area, so the centroid is defined. Here \(f(x)=x^2\) and \(g(x)=2x\).

The mass is

$$ m=\rho\int_0^2(2x-x^2)\,dx =\rho\left[x^2-\frac{x^3}{3}\right]_0^2 =\rho\left(4-\frac{8}{3}\right) =\frac{4\rho}{3}. $$

The moment about the \(y\)-axis is

$$ M_y=\rho\int_0^2x(2x-x^2)\,dx =\rho\int_0^2(2x^2-x^3)\,dx =\rho\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^2 =\rho\left(\frac{16}{3}-4\right) =\frac{4\rho}{3}. $$

The moment about the \(x\)-axis is

$$ \begin{aligned} M_x &=\frac{\rho}{2}\int_0^2\bigl((2x)^2-(x^2)^2\bigr)\,dx\\ &=\frac{\rho}{2}\int_0^2(4x^2-x^4)\,dx\\ &=\rho\left[\frac{2x^3}{3}-\frac{x^5}{10}\right]_0^2\\ &=\rho\left(\frac{16}{3}-\frac{16}{5}\right) =\frac{32\rho}{15}. \end{aligned} $$

Therefore

$$ \bar{x}=\frac{4\rho/3}{4\rho/3}=1, \qquad \bar{y}=\frac{32\rho/15}{4\rho/3}=\frac{8}{5}. $$

To check, the integrands used above have antiderivatives \(x^2-x^3/3\), \(2x^3/3-x^4/4\), and \(2x^3/3-x^5/10\), respectively; differentiating each recovers its integrand. Substituting the evaluated mass and moments into the centroid ratios gives \((1,8/5)\), as calculated. At \(x=1\), the region runs from \(y=1\) to \(y=2\), so the centroid's coordinates are also consistent with the region's vertical bounds.

Worked Example: Two Curves Meeting at Both Ends

Find the centroid of the region between \(y=x^2\) and \(y=x\) on \(0\leq x\leq1\), with constant density \(\rho\).

On this interval, \(x-x^2=x(1-x)\geq0\), and it is positive for \(0<x<1\). Thus \(f(x)=x^2\), \(g(x)=x\), and the region has positive area.

The mass is

$$ m=\rho\int_0^1(x-x^2)\,dx =\rho\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 =\rho\left(\frac12-\frac13\right) =\frac{\rho}{6}. $$

The moments are

$$ \begin{aligned} M_y &=\rho\int_0^1x(x-x^2)\,dx =\rho\int_0^1(x^2-x^3)\,dx\\ &=\rho\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1 =\rho\left(\frac13-\frac14\right) =\frac{\rho}{12}, \end{aligned} $$
$$ \begin{aligned} M_x &=\frac{\rho}{2}\int_0^1(x^2-x^4)\,dx =\rho\left[\frac{x^3}{6}-\frac{x^5}{10}\right]_0^1\\ &=\rho\left(\frac16-\frac1{10}\right) =\frac{\rho}{15}. \end{aligned} $$

Hence

$$ \bar{x}=\frac{\rho/12}{\rho/6}=\frac12, \qquad \bar{y}=\frac{\rho/15}{\rho/6}=\frac25. $$

For a direct check of the integrations, differentiating \(x^2/2-x^3/3\), \(x^3/3-x^4/4\), and \(x^3/6-x^5/10\) gives \(x-x^2\), \(x^2-x^3\), and \((x^2-x^4)/2\), respectively. The moment-to-mass ratios give the stated centroid.

Worked Example: A Linear Upper Boundary

Find the mass, moments, and centroid of the region between \(y=0\) and \(y=x+1\), from \(x=0\) to \(x=2\), with constant density \(\rho\).

The upper curve exceeds the lower curve throughout the interval because \(x+1\geq1\). In particular, the area is positive. Set \(f(x)=0\) and \(g(x)=x+1\).

The mass is

$$ m=\rho\int_0^2(x+1)\,dx =\rho\left[\frac{x^2}{2}+x\right]_0^2 =\rho(2+2) =4\rho. $$

The moment about the \(y\)-axis is

$$ \begin{aligned} M_y &=\rho\int_0^2x(x+1)\,dx =\rho\int_0^2(x^2+x)\,dx\\ &=\rho\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^2 =\rho\left(\frac83+2\right) =\frac{14\rho}{3}. \end{aligned} $$

The moment about the \(x\)-axis is

$$ \begin{aligned} M_x &=\frac{\rho}{2}\int_0^2(x+1)^2\,dx =\frac{\rho}{2}\int_0^2(x^2+2x+1)\,dx\\ &=\frac{\rho}{2}\left[\frac{x^3}{3}+x^2+x\right]_0^2 =\frac{\rho}{2}\left(\frac83+4+2\right) =\frac{13\rho}{3}. \end{aligned} $$

Therefore

$$ \bar{x}=\frac{14\rho/3}{4\rho}=\frac76, \qquad \bar{y}=\frac{13\rho/3}{4\rho}=\frac{13}{12}. $$

Differentiating the antiderivatives used for \(m\), \(M_y\), and \(M_x\) recovers \(x+1\), \(x^2+x\), and \((x+1)^2/2\), respectively. The computed mass and moments therefore yield the centroid \((7/6,13/12)\) by substitution into the defining ratios.

A Common Mistake: Squaring the Slice Height

The mass of a vertical strip depends on its height \(g-f\), but the moment about the \(x\)-axis is found by integrating the vertical coordinate over that strip. These are different calculations. For the region in the second example, a mistaken calculation might use

$$ M_x\ \text{(incorrect)} =\frac{\rho}{2}\int_0^1(x-x^2)^2\,dx =\frac{\rho}{2}\left[\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\right]_0^1 =\frac{\rho}{60}. $$

This incorrectly squares the strip height. The correct integral is \(\frac{\rho}{2}\int_0^1(x^2-x^4)\,dx=\rho/15\), obtained by integrating \(y\) from the lower curve to the upper curve. The incorrect moment would give \(\bar{y}=(\rho/60)/(\rho/6)=1/10\); the correct moment gives \(\bar{y}=(\rho/15)/(\rho/6)=2/5\).

Before evaluating the integrals, check the upper and lower curves on the entire interval, confirm that the region has positive area, and write down which moment each integrand represents. If the curves cross inside the interval, the upper and lower boundaries are not ordered as assumed here; the interval must first be split at the crossing points and the bounds assigned correctly on each part.

Check Your Understanding

Use the definitions and formulas to answer these questions. Assume constant positive density unless a question says otherwise.

  1. For a region between \(f(x)\) and \(g(x)\) on \([a,b]\), which curve belongs in the upper-bound position in the moment formula for \(M_x\)?
  2. Why does positive area matter when defining the centroid by ratios of moments to mass?
  3. Set up, but do not evaluate, the mass and first-moment integrals for the region between \(y=x^2\) and \(y=3x\) on \(0\leq x\leq2\).
  4. Explain why the integrand for \(M_x\) involves \(g(x)^2-f(x)^2\), rather than \((g(x)-f(x))^2\).
  5. For the region between \(y=0\) and \(y=2-x\) on \(0\leq x\leq1\), identify \(f\), \(g\), and the expression for the mass before integration.

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