Centroids Of Sectors And Annuli

Why These Centroids Are Useful

A circular sector is bounded by two radii and a circular arc. An annular sector has the same angular bounds but lies between two concentric circular arcs. Their curved boundaries can make Cartesian integration awkward, while polar coordinates describe both regions directly.

The centroid is the balance point of a uniform lamina. For these shapes, its location depends on two features: how far the material extends from the origin and how the angular interval is positioned. Polar first moments separate those contributions, giving formulas that are also easy to check using symmetry.

Definition and Setup

Take the origin to be the common centre of the circular arcs. Let \(0\leq a<b\), and let the angular interval run from \(\alpha\) to \(\beta\), with \(0<\Delta=\beta-\alpha\leq 2\pi\). The region is described by \(a\leq r\leq b\) and \(\alpha\leq\theta\leq\beta\). A sector has \(a=0\); an annular sector has \(a>0\). Angles are measured in radians. The angular bounds can be chosen as a continuous interval even if the region crosses a conventional \(0\)-angle boundary.

Assume the lamina has constant positive density \(\rho\). In polar coordinates, \(x=r\cos\theta\), \(y=r\sin\theta\), and \(dA=r\,dr\,d\theta\). Its mass is \(m=\rho A\), where \(A\) is its area. The first moment about the \(y\)-axis is \(M_y=\iint_R x\rho\,dA\), and the first moment about the \(x\)-axis is \(M_x=\iint_R y\rho\,dA\). The centroid coordinates are \(\bar{x}=M_y/m\) and \(\bar{y}=M_x/m\).

Deriving the General Formula

The area and mass are

For the first moment about the \(y\)-axis, multiply the horizontal coordinate \(x=r\cos\theta\) by the mass element \(\rho\,dA\). For the moment about the \(x\)-axis, use \(y=r\sin\theta\). This gives

Dividing each moment by the mass cancels the constant density. The resulting coordinates can be written using the angular midpoint \(\theta_c=(\alpha+\beta)/2\) and width \(\Delta=\beta-\alpha\).

Why the Formula Works

The proof follows from the first-moment definition, but the midpoint form also explains the geometric direction of the centroid. The trigonometric identities

rewrite the two moment integrals in terms of \(\theta_c\) and \(\Delta\). Dividing by \(m=\rho(b^2-a^2)\Delta/2\) yields the stated coordinates. In particular, when \(0<\Delta<2\pi\), the centroid lies along the angular bisector, because its coordinates are a common nonnegative factor times \((\cos\theta_c,\sin\theta_c)\). The radial factor \(C\) accounts for the inner and outer radii; the factor \(2\sin(\Delta/2)/\Delta\) accounts for the angular spread.

For \(\Delta=2\pi\), the angular factor is zero. This is consistent with rotational symmetry: opposite parts of a full disk or annulus balance, placing the centroid at the origin. The centroid is not found by averaging the radii alone; each area element has polar area \(r\,dr\,d\theta\), and the extra coordinate in a first moment contributes another factor of \(r\).

Worked Examples

Symmetry and Special Cases

Symmetry provides a useful independent check on an integral. If a region and its uniform density are symmetric across the \(x\)-axis, contributions to \(M_x=\iint_R y\rho\,dA\) cancel in pairs, so \(\bar{y}=0\). Symmetry across the \(y\)-axis similarly gives \(\bar{x}=0\). A full disk and a full annulus have both symmetries, so their centroids are at the origin.

For a sector symmetric about a line through the origin at angle \(\theta_c\), the centroid must lie on that line. The midpoint formula makes the same fact explicit. For a semicircle of radius \(b\), take \(a=0\), \(\Delta=\pi\), and \(\theta_c\) along its symmetry axis. The centroid is \(4b/(3\pi)\) from the origin along that axis. For an annular sector the same angular factor applies, but the radial factor depends on both radii. Removing the central disk therefore changes the centroid's distance from the origin even when the angular bounds remain fixed.

A Common Error: Omitting the Polar Area Factor

A common mistake is to use \(dA=dr\,d\theta\) instead of \(dA=r\,dr\,d\theta\). For a quarter-disk of radius \(R\), omitting \(r\) would give the incorrect area \(\int_0^{\pi/2}\int_0^R1\,dr\,d\theta=\pi R/2\), which has units of length rather than area. The same mistake would give an incorrect first moment \(M_y/\rho=\int_0^{\pi/2}\int_0^R r\cos\theta\,dr\,d\theta=R^2/2\), and hence the false coordinate \(\bar{x}=R/\pi\).

With the correct area element, the area is \(\pi R^2/4\) and \(M_y/\rho=\int_0^{\pi/2}\int_0^R r^2\cos\theta\,dr\,d\theta=R^3/3\). Their ratio is \(\bar{x}=4R/(3\pi)\), as required. A dimensional check is a quick warning: area must have units of length squared, a first moment must have units of density times length cubed, and a centroid coordinate must have units of length.

Practical Checks Before Accepting a Result

• Confirm that the region has positive area: \(a<b\) and \(\Delta>0\).
• Use \(r\,dr\,d\theta\) for area and \(r^2\cos\theta\,dr\,d\theta\) or \(r^2\sin\theta\,dr\,d\theta\) for the corresponding first moment.
• Keep the moment labels straight: \(M_y\) uses \(x\), while \(M_x\) uses \(y\).
• Check the coordinate signs against the region's location and its symmetries.
• Verify the final coordinates by multiplying \(\bar{x}\) by \(m\) to recover \(M_y\), and \(\bar{y}\) by \(m\) to recover \(M_x\).