Annuli and Concentric Circular Bands

Reading a Circular Band in Polar Coordinates

A circular band is the region between two circles with the same center. When the center is the origin, its geometry is especially direct in polar coordinates: the distance \(r\) from the origin lies between two fixed radii, while the angle \(\theta\) ranges over the full circle or over a chosen interval.

This makes a band different from a disk, which extends from the origin to an outer radius. An annulus has a positive inner radius; allowing the inner radius to be zero gives a disk. The same radial-bound description also handles a sector of an annulus by restricting the angular interval.

Definition: Let \(0\le a<b\). The full circular band with inner radius \(a\) and outer radius \(b\), centered at the origin, consists of points whose polar radius satisfies \(a\le r\le b\), with angle covering one full turn. If \(a>0\), the band is an annulus. If \(a=0\), it is a disk. A sector of the band restricts the angle to an interval \(\alpha\le\theta\le\beta\), where \(0\le\beta-\alpha\le 2\pi\).

Here \(a\) and \(b\) are distances, so they are nonnegative, and \(a<b\). Full-turn bounds can be written as \(0\le\theta\le2\pi\). A different interval of length \(2\pi\), such as \(-\pi\le\theta\le\pi\), also covers the full band. The boundaries at the radii and angular endpoints do not change a double integral over the region.

The Polar Bounds and Area Element

The radial bounds are constant throughout a concentric band: \(a\le r\le b\). The angle bounds describe how much of the band is included. For an integrable function \(f(x,y)\), the polar-coordinate form uses \(x=r\cos\theta\), \(y=r\sin\theta\), and the area element \(dA=r\,dr\,d\theta\). Thus, over a sector of the band,

\iint_R f(x,y)\,dA = \int_{\alpha}^{\beta}\int_a^b f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

The factor \(r\) is part of the area element; it is not an additional feature of the integrand. For the full band, choose any angular interval of length \(2\pi\), for example \([0,2\pi]\). For an annular sector, use the desired smaller interval \([\alpha,\beta]\). These bounds apply because the inner and outer circles are centered at the origin and the radial limits do not vary with angle.

Key formula: The area of a sector of a concentric band with radii \(a,b\) and angular width \(\beta-\alpha\) is \(\frac{\beta-\alpha}{2}(b^2-a^2)\), with angles measured in radians. A full band has angular width \(2\pi\), so its area is \(\pi(b^2-a^2)\).

Why the Band-Area Formula Works

The formula follows directly by integrating the polar area element. For a sector, take \(f(x,y)=1\), since integrating \(1\) over a region gives its area. The radial and angular bounds are constant, so the iterated integral separates into an angular factor and a radial integral:

\begin{aligned} \operatorname{Area}(R) &=\int_{\alpha}^{\beta}\int_a^b r\,dr\,d\theta\\ &=\int_{\alpha}^{\beta}\left[\frac{r^2}{2}\right]_a^b\,d\theta\\ &=\int_{\alpha}^{\beta}\frac{b^2-a^2}{2}\,d\theta\\ &=\frac{\beta-\alpha}{2}(b^2-a^2). \end{aligned}

For a full band, set \(\beta-\alpha=2\pi\). Then the area is \(\pi(b^2-a^2)\). This is also the area of the outer disk minus the area of the inner disk: \(\pi b^2-\pi a^2\). The agreement is a useful independent check on the polar calculation.

Worked Examples

Worked Example: Area of a Full Annulus

Find the area of the annulus centered at the origin with inner radius \(2\) and outer radius \(5\).

The full angular interval is \(0\le\theta\le2\pi\), and the radial bounds are \(2\le r\le5\). Integrate the polar area element:

\begin{aligned} A &=\int_0^{2\pi}\int_2^5 r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^2}{2}\right]_2^5\,d\theta\\ &=\int_0^{2\pi}\left(\frac{25}{2}-\frac{4}{2}\right)\,d\theta\\ &=\int_0^{2\pi}\frac{21}{2}\,d\theta\\ &=\frac{21}{2}(2\pi)=21\pi. \end{aligned}

Check by subtracting disk areas: \(\pi(5^2)-\pi(2^2)=25\pi-4\pi=21\pi\), which agrees.

Worked Example: Area of an Annular Sector

Find the area of the part of the annulus \(1\le r\le3\) lying between the rays \(\theta=\frac{\pi}{6}\) and \(\theta=\frac{\pi}{2}\).

The radial bounds are \(1\le r\le3\), and the angular width is \(\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}\). Integrate over those bounds:

\begin{aligned} A &=\int_{\pi/6}^{\pi/2}\int_1^3 r\,dr\,d\theta\\ &=\int_{\pi/6}^{\pi/2}\left[\frac{r^2}{2}\right]_1^3\,d\theta\\ &=\int_{\pi/6}^{\pi/2}\frac{9-1}{2}\,d\theta\\ &=\int_{\pi/6}^{\pi/2}4\,d\theta\\ &=4\left(\frac{\pi}{2}-\frac{\pi}{6}\right) =4\left(\frac{\pi}{3}\right) =\frac{4\pi}{3}. \end{aligned}

Check using the sector formula: \(\frac{1}{2}\cdot\frac{\pi}{3}\cdot(3^2-1^2)=\frac{1}{2}\cdot\frac{\pi}{3}\cdot8=\frac{4\pi}{3}\).

Worked Example: A Constant Integral Over Two Concentric Bands

Let \(f(x,y)=2\) on the inner disk \(0\le r\le2\), and let \(f(x,y)=5\) on the surrounding band \(2\le r\le4\). Find the integral of \(f\) over the full disk \(0\le r\le4\). The shared circle \(r=2\) has area zero, so its assignment to either band does not affect the integral.

Split the region into the two bands and integrate the constant value on each:

\begin{aligned} \iint_R f(x,y)\,dA &=\int_0^{2\pi}\int_0^2 2r\,dr\,d\theta +\int_0^{2\pi}\int_2^4 5r\,dr\,d\theta\\ &=\int_0^{2\pi}2\left[\frac{r^2}{2}\right]_0^2\,d\theta +\int_0^{2\pi}5\left[\frac{r^2}{2}\right]_2^4\,d\theta\\ &=\int_0^{2\pi}4\,d\theta+\int_0^{2\pi}30\,d\theta\\ &=4(2\pi)+30(2\pi)\\ &=8\pi+60\pi=68\pi. \end{aligned}

Check by area times constant on each part. The inner disk has area \(4\pi\), contributing \(2(4\pi)=8\pi\). The outer band has area \(\pi(4^2-2^2)=12\pi\), contributing \(5(12\pi)=60\pi\). The sum is \(68\pi\), as found by integration.

Concentric Bands as Pieces of a Region

A region can be divided into multiple concentric bands by listing its radii in increasing order. For radii \(0\le r_0<r_1<\cdots<r_n\), the successive pieces have radial bounds \(r_0\le r\le r_1\), \(r_1\le r\le r_2\), and so on. If the integrand has different constant values on different pieces, integrate each piece with its own value and add the results.

The pieces should cover the intended region without gaps, and their interiors should not overlap. Shared circular boundaries are harmless for area integrals. When the entire region is a full set of concentric bands, every piece has the same full angular bounds; when only sectors are included, use the specified angular interval for the corresponding piece.

Identify the radii.
Record the inner and outer radius of each band, in increasing order.

Set the angular interval.
Use a width of \(2\pi\) for a full band, or the given interval for a sector.

Write the polar integral.
Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\), and include the factor \(r\).

Integrate each piece and check.
Evaluate the pieces separately, then compare the area calculation with disk differences or sector proportions when possible.

A Common Error: Omitting the Factor \(r\)

The most consequential setup mistake is to replace the area element \(dA\) with \(dr\,d\theta\). Polar coordinates do not preserve area uniformly: a small angular change covers a longer arc at a larger radius. The area element is therefore \(r\,dr\,d\theta\).

For example, omitting \(r\) when finding the area of the annulus \(2\le r\le5\) gives

\int_0^{2\pi}\int_2^5 1\,dr\,d\theta =2\pi(5-2)=6\pi.

This is incorrect. It fails the disk-difference check, since the actual area is \(\pi(25-4)=21\pi\). The correction is to integrate the polar area element:

\int_0^{2\pi}\int_2^5 r\,dr\,d\theta =21\pi.

Another useful check is qualitative: with the same radial thickness, a band farther from the origin has greater area because its circles have larger circumference. A formula that assigns equal areas to equally thick bands at different radii has likely lost the radial factor.

Check Your Understanding

Use radial bounds, angular bounds, and the polar area element to answer the following.

Write the polar bounds for the full annulus with inner radius \(3\) and outer radius \(7\).
Find the area of that annulus using a polar integral, then check it by subtracting disk areas.
Write the bounds for the sector \(2\le r\le5\), \(\frac{\pi}{4}\le\theta\le\frac{3\pi}{4}\), and find its area.
What factor must multiply \(dr\,d\theta\) in the polar area element, and what geometric feature explains its presence?
A circular region is divided at radii \(1\) and \(3\). Describe the radial bounds of the two resulting concentric pieces.