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Composite Regions And Holes

Use signed areas or masses and their first moments to calculate the centroid of composite planar regions, including regions with holes.

Intermediate 10 min read

What You'll Learn

  • Combine component areas and first moments to find the centroid of a union.
  • Subtract a hole’s area and first moments from those of the surrounding region.
  • Use signed mass and first moments when component densities differ.
  • Recognize when overlapping pieces must be partitioned to avoid double counting.
  • Check a composite centroid by recovering its total first moments.

Why Composite Regions Need Careful Bookkeeping

A region with a cutout, or a shape assembled from several simpler pieces, may not have a convenient centroid formula of its own. But if its pieces have known areas or masses and known centroids, its centroid can be found by combining their contributions.

The key is to combine first moments as well as areas or masses. A large piece far from the origin contributes more to a first moment than a small piece nearby. A hole contributes negatively because its material has been removed.

Key idea: Add the areas or masses of pieces that are present, subtract those of holes, and apply the same signed bookkeeping to their first moments. Divide the resulting first moments by the resulting total area or mass.

Area and First-Moment Bookkeeping

For a planar region with area \(A\) and centroid \((\bar{x},\bar{y})\), its first moments about the coordinate axes are \(Q_y=A\bar{x}\) and \(Q_x=A\bar{y}\). Thus, \(Q_y\) tracks horizontal position and \(Q_x\) tracks vertical position. Equivalently, for a region \(R\), \(Q_y=\iint_R x\,dA\) and \(Q_x=\iint_R y\,dA\).

Definition: If a region is assembled from nonoverlapping pieces \(R_i\), assign each piece a sign \(s_i=+1\) when it is present and \(s_i=-1\) when it is removed as a hole. If \(A_i\) and \((\bar{x}_i,\bar{y}_i)\) are the area and centroid of piece \(i\), then $$ A=\sum_i s_iA_i,\qquad Q_y=\sum_i s_iA_i\bar{x}_i,\qquad Q_x=\sum_i s_iA_i\bar{y}_i. $$ Provided \(A>0\), the composite centroid is $$ \bar{x}=\frac{Q_y}{A},\qquad \bar{y}=\frac{Q_x}{A}. $$

For a union, all pieces have positive signs if their interiors do not overlap. For a region with a hole, the hole has a negative sign. The pieces and holes must be chosen so that the signed combination describes the intended region without counting any part twice.

Why Signed First Moments Work

The formulas follow from additivity of integrals. If \(R_1\) and \(R_2\) have disjoint interiors, their union contains each point of each piece exactly once. Therefore, integrating \(1\), \(x\), or \(y\) over the union gives the sum of the corresponding integrals over the pieces. These integrals are area, \(Q_y\), and \(Q_x\), respectively.

If a hole \(H\) lies inside an outer region \(R\), the remaining region is \(R\setminus H\). The integral over \(R\) is the sum of the integrals over \(R\setminus H\) and \(H\), so subtracting the integral over \(H\) gives the integral over the material that remains. Applying this to each of \(1\), \(x\), and \(y\) proves the signed formulas.

$$ \iint_{R\setminus H} f\,dA = \iint_R f\,dA-\iint_H f\,dA, \qquad f=1,\ x,\ \text{or }y. $$

This argument also explains an important limitation: simply adding the areas of overlapping pieces counts the overlap more than once. Before using the formulas, split the shape into pieces with disjoint interiors, or otherwise account for the overlap. A negative sign represents removed material in the bookkeeping; it does not mean that the physical hole has negative area.

Worked Example: A Rectangular Plate with a Circular Hole

Worked Example: A Rectangular Plate with a Circular Hole

A uniform plate occupies the rectangle \(0\le x\le10\), \(0\le y\le8\). A circular hole of radius \(2\), centered at \((6,4)\), is cut from it. Find the area and centroid of the remaining region.

The rectangle has area \(A_1=10\cdot8=80\), and its centroid is \((5,4)\). The circular hole has area \(A_2=\pi(2)^2=4\pi\), and its centroid is its center, \((6,4)\). Give the rectangle a positive sign and the hole a negative sign.

The remaining area is

$$ A=80-4\pi. $$

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

$$ Q_y=(80)(5)-(4\pi)(6)=400-24\pi. $$

Its first moment about the \(x\)-axis is

$$ Q_x=(80)(4)-(4\pi)(4)=320-16\pi. $$

Since \(80-4\pi>0\), the centroid is defined. Its coordinates are

$$ \bar{x}=\frac{400-24\pi}{80-4\pi}, \qquad \bar{y}=\frac{320-16\pi}{80-4\pi}=4. $$

To check, multiply the centroid coordinates by the net area. This recovers \(A\bar{x}=400-24\pi=Q_y\) and \(A\bar{y}=320-16\pi=Q_x\), as required. Also, \(\bar{x}<5\): removing area to the right of the rectangle’s center shifts the centroid left. The unchanged value \(\bar{y}=4\) agrees with symmetry about the horizontal line \(y=4\).

Worked Example: A Region Built from Two Rectangles

Worked Example: A Region Built from Two Rectangles

An L-shaped region consists of a horizontal rectangle \(0\le x\le6,\ 0\le y\le2\) and a vertical rectangle \(0\le x\le2,\ 2\le y\le6\). Their interiors do not overlap. Find the area and centroid.

The horizontal rectangle has area \(A_1=6\cdot2=12\) and centroid \((3,1)\). The vertical rectangle has area \(A_2=2\cdot4=8\) and centroid \((1,4)\). Both pieces are present, so both signs are positive.

The total area is

$$ A=12+8=20. $$

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

$$ Q_y=(12)(3)+(8)(1)=36+8=44. $$

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

$$ Q_x=(12)(1)+(8)(4)=12+32=44. $$

Therefore,

$$ \bar{x}=\frac{44}{20}=\frac{11}{5}, \qquad \bar{y}=\frac{44}{20}=\frac{11}{5}. $$

For the check, \((20)(11/5)=44\) for each coordinate, recovering both calculated first moments. The equal coordinates are consistent with the component areas and locations, although the L-shaped region itself is not symmetric about the line \(y=x\).

Mass and Centroid for Different Materials

The same method applies when pieces have different uniform densities. For a piece with mass \(m_i\) and centroid \((\bar{x}_i,\bar{y}_i)\), its first moments are \(m_i\bar{x}_i\) and \(m_i\bar{y}_i\). If pieces have densities \(\rho_i\), their masses are \(m_i=\rho_iA_i\). A hole is subtracted using the mass and first moments of the material that would have occupied the removed part.

Formula: For disjoint pieces of uniform density, with signed contributions \(s_i\), the total mass and mass first moments are $$ M=\sum_i s_i m_i,\qquad Q_y=\sum_i s_i m_i\bar{x}_i,\qquad Q_x=\sum_i s_i m_i\bar{y}_i. $$ If \(M>0\), the center of mass is $$ \bar{x}=\frac{Q_y}{M},\qquad \bar{y}=\frac{Q_x}{M}. $$

Worked Example: Two Joined Pieces with Different Densities

Worked Example: Two Joined Pieces with Different Densities

A rectangular piece occupies \(0\le x\le4,\ 0\le y\le2\) and has uniform density \(3\). A circular disk of radius \(1\), density \(2\), and center \((5,1)\) is joined to it. The disk touches the rectangle at its boundary, so their interiors do not overlap. Find the center of mass.

The rectangle has area \(4\cdot2=8\), so its mass is \(m_1=3\cdot8=24\). Its centroid is \((2,1)\). The disk has area \(\pi(1)^2=\pi\), so its mass is \(m_2=2\pi\). Its centroid is \((5,1)\).

The total mass is \(M=24+2\pi\). The first moment about the \(y\)-axis is

$$ Q_y=(24)(2)+(2\pi)(5)=48+10\pi. $$

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

$$ Q_x=(24)(1)+(2\pi)(1)=24+2\pi. $$

Thus the center of mass is

$$ \bar{x}=\frac{48+10\pi}{24+2\pi}, \qquad \bar{y}=\frac{24+2\pi}{24+2\pi}=1. $$

Checking gives \(M\bar{x}=48+10\pi=Q_y\) and \(M\bar{y}=24+2\pi=Q_x\). The vertical coordinate is \(1\) because both pieces have that vertical centroid coordinate. The different densities matter through the masses, not just through the component areas.

A Common Sign Error

When subtracting a hole, its area and both first moments must be subtracted. A frequent error is to subtract the hole’s area but add its first moment. For the plate in the first example, that incorrect calculation would give

$$ \bar{x}_{\mathrm{wrong}} = \frac{(80)(5)+(4\pi)(6)}{80-4\pi} = \frac{400+24\pi}{80-4\pi}. $$

This treats the removed material as if it still contributed to the horizontal first moment, even though it has been removed from the area total. The consistent calculation subtracts both contributions: \(\bar{x}=(400-24\pi)/(80-4\pi)\). As a quick reasonableness check, removing a hole to the right of the rectangle’s centroid should shift the centroid left, not right.

Before accepting a result, check that the net area or mass is positive, that the signs represent the material actually present, and that multiplying the proposed centroid coordinates by the net area or mass recovers the calculated first moments. These checks catch many bookkeeping errors.

Check Your Understanding

For each question, identify the signed areas or masses and first moments that should be combined.

  1. A region is formed from three rectangles with disjoint interiors. How would you calculate its total area and its two first moments from the rectangles’ areas and centroids?
  2. A hole is cut from a region. Why must the hole’s first moments be subtracted along with its area?
  3. A plate consists of two joined pieces with different uniform densities. Should their centroid coordinates be weighted by area or by mass? Explain.
  4. What can go wrong if two overlapping pieces are simply added as if their interiors were disjoint?
  5. What calculation verifies that a proposed composite centroid has the correct first moments?

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