Centroids Of Standard Planar Shapes

Centroids as Coordinate Averages

A centroid locates the geometric balance point of a planar region when its area is treated as uniformly distributed. The region’s shape determines how much area lies at each coordinate, so locating the centroid amounts to averaging the coordinates over the entire region—not simply choosing a visually central point.

For a region \(R\) of positive area \(A\), the area element \(dA\) represents a small piece of area. The first moment about the \(y\)-axis adds up each piece’s \(x\)-coordinate weighted by its area; the first moment about the \(x\)-axis does the same with its \(y\)-coordinate. Dividing each moment by total area gives the corresponding average coordinate.

These formulas apply to the geometric centroid, which treats area uniformly. If the region is a uniform lamina with constant density, its center of mass has the same coordinates. The calculation requires finite, positive area and integrable coordinate functions over the region.

Why the First-Moment Formulas Work

Imagine partitioning \(R\) into many small pieces. A piece of area \(\Delta A_i\), located near \((x_i,y_i)\), contributes approximately \(x_i\Delta A_i\) to the total \(x\)-coordinate weighted by area, and \(y_i\Delta A_i\) to the corresponding \(y\)-coordinate total. Adding these contributions and taking the limit gives \(\iint_R x\,dA\) and \(\iint_R y\,dA\). Dividing by the total area forms the average coordinate. Thus the first-moment formulas are the continuous versions of weighted averages.

This also explains a useful check: multiplying a computed centroid coordinate by the area must recover the corresponding first moment. A coordinate shift changes the numerical centroid coordinates, but it does not change the underlying point in the plane.

Disk: Symmetry and Direct Integration

Consider first a disk of radius \(R\) centered at the origin. Its vertical cross-section at coordinate \(x\) runs from \(y=-\sqrt{R^2-x^2}\) to \(y=\sqrt{R^2-x^2}\). Integrating the cross-section lengths gives the area:

For this disk, the first moment about the \(y\)-axis is the integral of \(x\) over each vertical slice. The integrand \(2x\sqrt{R^2-x^2}\) is odd over the symmetric interval \([-R,R]\), so the moment is zero. The first moment about the \(x\)-axis is also zero: each vertical slice has equal area above and below the \(x\)-axis. Therefore the centroid is the origin.

For a disk centered at \((h,k)\), use local coordinates \(u=x-h\) and \(v=y-k\). The centered disk has zero first moments in \(u\) and \(v\), so its averages satisfy \(\bar{u}=\bar{v}=0\). Since \(x=u+h\) and \(y=v+k\), the centroid in the original coordinates is \((h,k)\).

Semicircle: A Centroid Away from the Center

A semicircle has symmetry across its vertical diameter, which places its centroid on that diameter. But it does not have symmetry across its horizontal diameter: the area lies entirely on one side. For the upper semicircle of radius \(R\) centered at the origin, a horizontal slice at height \(y\), where \(0\le y\le R\), has width \(2\sqrt{R^2-y^2}\).

If this semicircle is translated so its center is \((h,k)\) and its straight edge lies on \(y=k\), the centroid becomes \((h,k+4R/(3\pi))\). The offset is measured from the center along the symmetry axis; it is not measured from the lowest point of the semicircle.

Triangle: Direct Integration and the Vertex Formula

For a right triangle with vertices \((0,0)\), \((a,0)\), and \((0,b)\), where \(a,b>0\), the sloping side has equation \(y=b(1-x/a)\). Integrating between the horizontal bounds \(0\le y\le b(1-x/a)\) gives $$ A=\int_0^a b\left(1-\frac{x}{a}\right)\,dx=\frac{ab}{2}. $$

The first moment about the \(y\)-axis is $$ \iint_R x\,dA =\int_0^a xb\left(1-\frac{x}{a}\right)\,dx =b\left(\frac{a^2}{2}-\frac{a^2}{3}\right) =\frac{ba^2}{6}. $$ The first moment about the \(x\)-axis is $$ \iint_R y\,dA =\int_0^a \frac{1}{2}b^2\left(1-\frac{x}{a}\right)^2\,dx =\frac{ab^2}{6}. $$ Dividing these moments by \(ab/2\) gives \(\bar{x}=a/3\) and \(\bar{y}=b/3\). The centroid is one-third of the way along each leg from the right-angle vertex.

The corresponding formula for any triangle follows by direct integration in coordinates adapted to its vertices. Let its vertices be \(P_0=(x_0,y_0)\), \(P_1=(x_1,y_1)\), and \(P_2=(x_2,y_2)\). Every point in the triangle can be written $$ P(s,t)=P_0+s(P_1-P_0)+t(P_2-P_0), \qquad s\ge 0,\quad t\ge 0,\quad s+t\le 1. $$ The absolute value \(J\) of the transformation’s determinant is constant, so \(dA=J\,ds\,dt\). The parameter triangle has area $$ \int_0^1\int_0^{1-s}1\,dt\,ds =\int_0^1(1-s)\,ds =\frac12, $$ and therefore the physical area is \(J/2\). Its average \(s\)-coordinate is $$ \frac{\int_0^1\int_0^{1-s}s\,dt\,ds}{1/2} =\frac{\int_0^1s(1-s)\,ds}{1/2} =\frac{1/6}{1/2} =\frac13. $$ By the same calculation, the average \(t\)-coordinate is \(1/3\). Since \(P(s,t)\) depends linearly on \(s\) and \(t\), its average is $$ \bar{P}=P_0+\frac13(P_1-P_0)+\frac13(P_2-P_0) =\frac{P_0+P_1+P_2}{3}. $$ Thus a triangle’s centroid is the arithmetic mean of its three vertex coordinates.

Rectangle: The Midpoint of Each Side Pair

For a rectangle \(a\le x\le b\), \(c\le y\le d\), its area is \((b-a)(d-c)\). Its first moments separate into one-dimensional integrals: $$ \iint_R x\,dA=(d-c)\int_a^b x\,dx, \qquad \iint_R y\,dA=(b-a)\int_c^d y\,dy. $$ Evaluating and dividing by the area gives \(\bar{x}=(a+b)/2\) and \(\bar{y}=(c+d)/2\). These are the midpoints of the rectangle’s horizontal and vertical coordinate intervals.

Coordinate Shifts and a Common Pitfall

Suppose local coordinates are defined by \(u=x-h\) and \(v=y-k\). For a region of area \(A\), $$ \bar{x}=\frac{1}{A}\iint_R(u+h)\,dA =\bar{u}+h, \qquad \bar{y}=\frac{1}{A}\iint_R(v+k)\,dA =\bar{v}+k. $$ The centroid is a point independent of the coordinate description, but its coordinate pair changes when the origin changes.

For a quick reasonableness check, the centroid of a region with nonnegative area density must lie within its convex hull. A disk’s centroid should be its center, and a rectangle’s centroid should be its midpoint. For a semicircle, symmetry fixes the coordinate along its straight edge, while the other coordinate must lie strictly inside the curved region. Such checks can reveal a sign error or an omitted translation, but they do not replace the moment calculations.