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Area, Mass, and Moments · Tutorial 582 of 999

Symmetry Tests For Centroids

Use reflection tests on a region and its density to identify centroid coordinates, and see why a symmetric shape need not have a centered mass distribution.

Intermediate 10 min read

What You'll Learn

  • State the mass and first-moment conditions needed for a centroid.
  • Test symmetry of both the region and the density under reflection.
  • Use symmetry about coordinate axes or a horizontal line to determine a centroid coordinate.
  • Compute a centroid coordinate when symmetry determines only the other coordinate.
  • Identify how nonuniform density can shift the centroid of a symmetric region.

Why Symmetry Helps

A centroid is a balance point for a region carrying mass. Computing it directly can require several double integrals, but reflection symmetry can make one first moment vanish without evaluating that integral. The important qualification is that symmetry must apply to the mass distribution, not merely to the outline of the region.

For a lamina occupying a region \(R\) with nonnegative density \(\rho(x,y)\), the mass at each point is weighted by that density. If one side of a geometrically symmetric region is denser than its reflected side, the mass may not balance on the symmetry axis. The symmetry tests below make this distinction precise.

Centroid Formulas and Their Conditions

Assume \(R\) is a planar region, \(\rho\) is integrable and nonnegative on \(R\), and the mass and first moments are finite. The total mass must be positive for a centroid to be defined. The first moment about the \(x\)-axis weights each mass element by its \(y\)-coordinate; the first moment about the \(y\)-axis weights it by its \(x\)-coordinate.

Definition: The mass \(M\), first moments \(M_x\) and \(M_y\), and centroid \((\bar{x},\bar{y})\) are
$$ M=\iint_R \rho(x,y)\,dA,\qquad M_x=\iint_R y\rho(x,y)\,dA,\qquad M_y=\iint_R x\rho(x,y)\,dA, $$
with
$$ \bar{x}=\frac{M_y}{M},\qquad \bar{y}=\frac{M_x}{M}. $$

The reversed-looking subscripts in the coordinate formulas are intentional: \(M_y\), the moment about the \(y\)-axis, determines the \(x\)-coordinate. If \(\rho\) is constant and positive, the centroid is also the geometric centroid of the region. With variable density, it is the center of mass and need not coincide with the geometric centroid.

The Reflection Tests

Reflection in the \(y\)-axis sends \((x,y)\) to \((-x,y)\). For a symmetry test to force \(\bar{x}=0\), the reflected region must still be the same region and the density at each point must match its reflected value. Reflection in the \(x\)-axis is treated in the same way, with \((x,y)\) sent to \((x,-y)\).

Theorem: Suppose \(R\) has finite positive mass and finite first moments. If \(R\) is invariant under reflection in the \(y\)-axis and \(\rho(-x,y)=\rho(x,y)\) throughout \(R\) (up to boundary sets of area zero), then \(M_y=0\), so \(\bar{x}=0\). If \(R\) is invariant under reflection in the \(x\)-axis and \(\rho(x,-y)=\rho(x,y)\), then \(M_x=0\), so \(\bar{y}=0\).

Why the Reflection Tests Work

Consider reflection in the \(y\)-axis, and define \(f(x,y)=x\rho(x,y)\), the integrand in \(M_y\). Under reflection, \(f(-x,y)=-x\rho(-x,y)=-x\rho(x,y)=-f(x,y)\). Since the region is unchanged, changing variables by this reflection leaves the value of the integral over \(R\) unchanged. But the integrand changes sign, so the same integral is also its negative.

$$ M_y=\iint_R f(x,y)\,dA =\iint_R f(-x,y)\,dA =-\iint_R f(x,y)\,dA=-M_y. $$

Therefore \(M_y=-M_y\), which implies \(M_y=0\). Since \(M>0\), the centroid formula gives \(\bar{x}=M_y/M=0\). The proof for reflection in the \(x\)-axis is parallel: \(y\rho(x,y)\) changes sign under \((x,y)\mapsto(x,-y)\), so \(M_x=0\) and \(\bar{y}=0\).

The same reasoning gives a useful shifted version. If both the region and density are invariant under reflection across the horizontal line \(y=c\), then \((y-c)\rho(x,y)\) changes sign under that reflection. Its integral is zero, so \(M_x-cM=0\), and consequently \(\bar{y}=c\). The centroid lies on the symmetry line even if the line is not an axis.

Worked Example: Symmetry Determines One Coordinate

Worked Example: Symmetry Determines One Coordinate

Let \(R=[-2,2]\times[0,3]\), with density \(\rho(x,y)=5+y\). Find the centroid.

The region and density are both symmetric about the \(y\)-axis: replacing \(x\) by \(-x\) changes neither the bounds nor the density. Thus \(M_y=0\). We calculate the mass and the other first moment to find \(\bar{y}\):

$$ M=\int_{-2}^{2}\int_0^3(5+y)\,dy\,dx =\int_{-2}^{2}\left[5y+\frac{y^2}{2}\right]_0^3dx =\int_{-2}^{2}\frac{39}{2}\,dx =4\cdot\frac{39}{2}=78. $$

For the first moment about the \(x\)-axis, integrate first with respect to \(y\):

$$ M_x=\int_{-2}^{2}\int_0^3 y(5+y)\,dy\,dx =\int_{-2}^{2}\left[\frac{5y^2}{2}+\frac{y^3}{3}\right]_0^3dx =\int_{-2}^{2}\frac{63}{2}\,dx =126. $$

For completeness, the moment about the \(y\)-axis is zero by symmetry; directly, its integrand factors into an \(x\)-integral and a \(y\)-integral, and \(\int_{-2}^{2}x\,dx=0\). Therefore

$$ \bar{x}=\frac{M_y}{M}=\frac{0}{78}=0,\qquad \bar{y}=\frac{M_x}{M}=\frac{126}{78}=\frac{21}{13}. $$

Check: \(78(21/13)=6\cdot21=126=M_x\), and \(78(0)=0=M_y\). The centroid is \((0,21/13)\). The density varies with height, so symmetry gives no reason for \(\bar{y}\) to be halfway between \(0\) and \(3\).

A Symmetric Region Can Have an Off-Center Centroid

A region symmetric about both axes has a geometric centroid at the origin when its density is uniform. But if the density is not invariant under both reflections, those symmetry tests no longer apply. The next example keeps a symmetric region and changes only its density.

Worked Example: A Disk with a Density That Shifts the Centroid

Let \(R\) be the unit disk \(x^2+y^2\leq1\), and let \(\rho(x,y)=4+x\). Find its centroid. The density is positive throughout the disk because \(x\geq-1\), so \(\rho\geq3\).

The region is symmetric about both axes, but the density is not symmetric about the \(y\)-axis: \(\rho(-x,y)=4-x\), which generally differs from \(\rho(x,y)=4+x\). We must calculate the \(x\)-coordinate rather than infer it from the disk’s shape. In polar coordinates, \(x=r\cos\theta\) and \(dA=r\,dr\,d\theta\):

$$ M=\int_0^{2\pi}\int_0^1(4+r\cos\theta)r\,dr\,d\theta =4\int_0^{2\pi}\int_0^1r\,dr\,d\theta +\int_0^{2\pi}\cos\theta\,d\theta\int_0^1r^2\,dr =4\pi+0=4\pi. $$

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

$$ \begin{aligned} M_y &=\iint_R x(4+x)\,dA\\ &=\iint_R (4x+x^2)\,dA\\ &=4\iint_R x\,dA+\iint_R x^2\,dA\\ &=0+\int_0^{2\pi}\cos^2\theta\,d\theta\int_0^1r^3\,dr\\ &=\pi\cdot\frac14=\frac{\pi}{4}. \end{aligned} $$

Here \(\iint_R x\,dA=0\) because the region is symmetric about the \(y\)-axis, and \(\int_0^{2\pi}\cos^2\theta\,d\theta=\pi\). For the other first moment, the terms \(4y\) and \(xy\) each integrate to zero: the first is odd under reflection in the \(x\)-axis, and the second is odd under reflection in either coordinate axis. Thus \(M_x=0\). The centroid is

$$ \bar{x}=\frac{M_y}{M}=\frac{\pi/4}{4\pi}=\frac{1}{16}, \qquad \bar{y}=\frac{M_x}{M}=0. $$

Check: \(4\pi(1/16)=\pi/4=M_y\), and \(4\pi(0)=0=M_x\). The centroid is \((1/16,0)\), shifted to the right because the density is greater on the right side. This is why geometric symmetry alone cannot determine a center of mass.

Worked Example: Symmetry About a Horizontal Line

Worked Example: Symmetry About a Horizontal Line

Let \(R=[-1,1]\times[0,2]\) and \(\rho(x,y)=1+(y-1)^2\). Find the centroid. The region and density are symmetric about \(y=1\), so the shifted reflection test gives \(\bar{y}=1\). The region and density are also symmetric about the \(y\)-axis, so \(\bar{x}=0\). We can verify both coordinates directly.

The mass is

$$ M=2\int_0^2\bigl(1+(y-1)^2\bigr)\,dy =2\left[y+\frac{(y-1)^3}{3}\right]_0^2 =2\left(\frac{8}{3}\right)=\frac{16}{3}. $$

Symmetry in \(x\) gives \(M_y=0\). For \(M_x\), substitute \(u=y-1\), so \(y=u+1\) and the limits are \(-1\) and \(1\):

$$ \begin{aligned} M_x &=2\int_0^2 y\bigl(1+(y-1)^2\bigr)\,dy\\ &=2\int_{-1}^{1}(u+1)(1+u^2)\,du\\ &=2\int_{-1}^{1}(u+u^3+1+u^2)\,du\\ &=2\left(0+0+2+\frac{2}{3}\right)=\frac{16}{3}. \end{aligned} $$

The odd terms \(u\) and \(u^3\) integrate to zero over \([-1,1]\). Therefore \(\bar{x}=0/(16/3)=0\) and \(\bar{y}=(16/3)/(16/3)=1\). Check: \(M\bar{x}=0=M_y\) and \(M\bar{y}=16/3=M_x\), as required.

A Common Mistake

A frequent error is to conclude that the centroid of a symmetric shape must lie on every axis of symmetry, even when the density varies. For the disk with \(\rho=4+x\), the incorrect calculation would be: “The disk is symmetric about the \(y\)-axis, so \(\bar{x}=0\).” That conclusion uses symmetry of \(R\) but ignores the density.

The correction is to check both conditions. Although the disk reflects onto itself, its density does not satisfy \(\rho(-x,y)=\rho(x,y)\). The actual first moment is \(M_y=\pi/4\), so \(\bar{x}=(\pi/4)/(4\pi)=1/16\), not zero. A symmetry test is a sufficient way to establish that a first moment vanishes only when the mass distribution shares that symmetry.

How to Apply the Test

1
Identify the proposed reflection.
For the \(y\)-axis, compare \((x,y)\) with \((-x,y)\). For the \(x\)-axis, compare \((x,y)\) with \((x,-y)\).
2
Check the region.
Confirm that reflecting the region maps it onto itself, apart from boundary differences that do not affect the integrals.
3
Check the density.
Confirm that the density at a point equals the density at its reflected point. If not, geometry alone does not establish a zero first moment.
4
Use the result carefully.
Matching reflection symmetry forces the coordinate perpendicular to the symmetry axis to be zero, or to equal the axis location for a shifted symmetry line. Calculate any coordinate the symmetry does not determine.

Symmetry is useful because it can remove an integral from a centroid calculation or provide a check on an integral evaluated directly. It does not say that the centroid is the origin whenever the region is centered there, and it does not determine a coordinate parallel to the symmetry axis. The density and the direction of reflection both matter.

Check Your Understanding

For each question, distinguish what follows from the region’s shape from what requires symmetry of the density.

  1. A region is symmetric about the \(x\)-axis, and its density satisfies \(\rho(x,-y)=\rho(x,y)\). Which first moment must be zero, and which centroid coordinate follows?
  2. A disk is symmetric about both coordinate axes, but its density is \(\rho(x,y)=3+y\). Which coordinate can be determined from reflection symmetry alone?
  3. What two conditions must hold for reflection in the \(y\)-axis to force \(\bar{x}=0\)?
  4. A region and density are symmetric about \(y=4\). What value does symmetry give for \(\bar{y}\)?
  5. Why is it incorrect to infer the center of mass of a nonuniform lamina from the shape of its boundary alone?

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