Sectors with Restricted Angular Bounds

Reading a Sector as Polar Bounds

A circular sector is the part of a disk swept out as a radius turns between two rays. In polar coordinates, those rays determine the angular limits, while the circular arc determines the outer radial limit. Once these limits are identified, the region can be used directly in a double integral.

The order of the angular limits matters: the angle is taken to increase counterclockwise. An interval may begin at a negative angle or at an angle greater than \(2\pi\); what matters is that it sweeps through the intended sector exactly once.

Definition: For an outer radius \(R>0\) and angles \(\alpha<\beta\) with \(0<\beta-\alpha\leq 2\pi\), the circular sector \(S\) is the region \(S=\{(r\cos\theta,r\sin\theta):0\leq r\leq R,\ \alpha\leq\theta\leq\beta\}\). Its straight boundaries are the rays at angles \(\alpha\) and \(\beta\); its curved boundary is the arc \(r=R\).

The restriction on the angular width means the interval describes one sweep, not multiple coverings of the same points. When the width is \(2\pi\), the sector is the full disk; the repeated boundary ray does not affect the integral.

The Polar Integral over a Sector

In polar coordinates, the area element is \(dA=r\,dr\,d\theta\). The factor \(r\) is essential: it accounts for the fact that a small angular change covers a wider arc farther from the origin. For an integrable function \(f\) on the sector, the integral is therefore

\iint_S f(x,y)\,dA = \int_{\alpha}^{\beta}\int_0^R f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

This form uses radial integration first and angular integration second. The bounds say that for every angle from \(\alpha\) to \(\beta\), the radius runs from the origin out to the circular edge. If \(f=1\), the double integral gives the area of the sector. If \(f\) is a constant density, it gives the mass of a sector-shaped lamina with that density.

Rule: For a sector with one circular arc centered at the origin, use \(\alpha\leq\theta\leq\beta\) and \(0\leq r\leq R\). Include the polar area factor \(r\) in the integrand.

Why the Sector Area Formula Works

The general polar integral formula follows from the polar area element and the stated bounds. Applying it to the constant function \(f=1\) gives a direct derivation of the sector’s area. The radial integral contributes half the square of the outer radius, and the angular integral contributes the sector’s angle in radians.

Set the integrand to one.
The integral of \(1\) over a region is its area, so use the sector bounds with the polar area factor \(r\).

Integrate with respect to radius.
For each fixed angle, \(\int_0^R r\,dr=R^2/2\).

Integrate over the angle.
The value \(R^2/2\) is constant in \(\theta\), so integrating it from \(\alpha\) to \(\beta\) multiplies it by \(\beta-\alpha\).

\begin{aligned} \operatorname{Area}(S) &=\int_{\alpha}^{\beta}\int_0^R r\,dr\,d\theta\\ &=\int_{\alpha}^{\beta}\left[\frac{r^2}{2}\right]_0^R\,d\theta\\ &=\int_{\alpha}^{\beta}\frac{R^2}{2}\,d\theta\\ &=\frac{R^2}{2}(\beta-\alpha). \end{aligned}

This also explains why the angle must be measured in radians: the polar area element and the resulting formula use the angular width \(\beta-\alpha\) as a radian measure.

Theorem: A sector of radius \(R\) and angular width \(\beta-\alpha\), measured in radians, has area \(\frac{1}{2}R^2(\beta-\alpha)\).

Worked Examples

Worked Example: Area of a Narrow Sector

Find the area of the sector with radius \(3\) and bounding rays at \(\theta=\frac{\pi}{6}\) and \(\theta=\frac{\pi}{2}\).

The angular interval is already in increasing order. Its width is

\frac{\pi}{2}-\frac{\pi}{6} = \frac{3\pi}{6}-\frac{\pi}{6} = \frac{\pi}{3}.

The polar bounds are \(\frac{\pi}{6}\leq\theta\leq\frac{\pi}{2}\) and \(0\leq r\leq3\). Integrate \(1\) over the region to find its area:

\begin{aligned} \operatorname{Area}(S) &=\int_{\pi/6}^{\pi/2}\int_0^3 r\,dr\,d\theta\\ &=\int_{\pi/6}^{\pi/2}\left[\frac{r^2}{2}\right]_0^3\,d\theta\\ &=\int_{\pi/6}^{\pi/2}\frac{9}{2}\,d\theta\\ &=\frac{9}{2}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)\\ &=\frac{9}{2}\cdot\frac{\pi}{3} =\frac{3\pi}{2}. \end{aligned}

Check with the sector area formula: \(\frac{1}{2}(3)^2\cdot\frac{\pi}{3}=\frac{3\pi}{2}\), the same result.

Worked Example: Area from Two Nonstandard Ray Angles

Find the area of the sector with radius \(4\), starting ray at \(\theta=\frac{3\pi}{4}\), and ending ray at \(\theta=\frac{7\pi}{6}\), sweeping counterclockwise between them.

First check the order and calculate the angular width using a common denominator:

\frac{3\pi}{4}=\frac{9\pi}{12}, \qquad \frac{7\pi}{6}=\frac{14\pi}{12}, \qquad \Delta\theta=\frac{14\pi}{12}-\frac{9\pi}{12} =\frac{5\pi}{12}.

The region is described by \(\frac{3\pi}{4}\leq\theta\leq\frac{7\pi}{6}\) and \(0\leq r\leq4\). Thus

\begin{aligned} \operatorname{Area}(S) &=\int_{3\pi/4}^{7\pi/6}\int_0^4 r\,dr\,d\theta\\ &=\int_{3\pi/4}^{7\pi/6}\left[\frac{r^2}{2}\right]_0^4\,d\theta\\ &=\int_{3\pi/4}^{7\pi/6}8\,d\theta\\ &=8\left(\frac{7\pi}{6}-\frac{3\pi}{4}\right)\\ &=8\left(\frac{14\pi}{12}-\frac{9\pi}{12}\right)\\ &=8\cdot\frac{5\pi}{12} =\frac{10\pi}{3}. \end{aligned}

Check: the sector formula gives \(\frac{1}{2}(4)^2\cdot\frac{5\pi}{12}=8\cdot\frac{5\pi}{12}=\frac{10\pi}{3}\). The bounds use the specified counterclockwise sweep; reversing the endpoints without changing the intended region would not describe the same interval.

Worked Example: A Sector Crossing the Zero-Angle Ray

A sector has radius \(2\), with rays at \(\frac{5\pi}{3}\) and \(\frac{\pi}{4}\). The intended sector is the counterclockwise sweep from the ray at \(\frac{5\pi}{3}\) through the positive horizontal axis to the ray at \(\frac{\pi}{4}\). Find its mass if its density is the constant \(5\).

Since \(\frac{5\pi}{3}>\frac{\pi}{4}\), those endpoints in that order would not give the desired increasing angular interval. Replace \(\frac{5\pi}{3}\) by the equivalent angle \(-\frac{\pi}{3}\). The intended sweep is then \(-\frac{\pi}{3}\leq\theta\leq\frac{\pi}{4}\), with \(0\leq r\leq2\). Its width is

\frac{\pi}{4}-\left(-\frac{\pi}{3}\right) = \frac{3\pi}{12}+\frac{4\pi}{12} = \frac{7\pi}{12}.

For constant density \(5\), the mass integral includes both the density and the polar area factor:

\begin{aligned} \operatorname{Mass}(S) &=\int_{-\pi/3}^{\pi/4}\int_0^2 5r\,dr\,d\theta\\ &=\int_{-\pi/3}^{\pi/4}5\left[\frac{r^2}{2}\right]_0^2\,d\theta\\ &=\int_{-\pi/3}^{\pi/4}10\,d\theta\\ &=10\left(\frac{\pi}{4}+\frac{\pi}{3}\right)\\ &=10\cdot\frac{7\pi}{12} =\frac{35\pi}{6}. \end{aligned}

Check the area first: \(\frac{1}{2}(2)^2\cdot\frac{7\pi}{12}=\frac{7\pi}{6}\). Multiplying by the constant density gives \(5\cdot\frac{7\pi}{6}=\frac{35\pi}{6}\), as in the integral.

Common Mistake: Losing the Half or the Angular Width

A frequent error is to omit the factor \(\frac{1}{2}\) when integrating the polar area factor \(r\). For example, for the first example one might incorrectly calculate the area as \(3^2\cdot\frac{\pi}{3}=3\pi\). That is twice the correct value, because \(\int_0^3 r\,dr\) is not \(3^2\); it is \(\left[\frac{r^2}{2}\right]_0^3=\frac{9}{2}\).

The corrected calculation is \(\frac{9}{2}\cdot\frac{\pi}{3}=\frac{3\pi}{2}\). A reliable check is to compare the integral with \(\frac{1}{2}R^2(\beta-\alpha)\). Also check that the interval’s width matches the intended counterclockwise sweep and is measured in radians.

Choosing and Checking the Bounds

When a sector is described in words or by a diagram, identify its boundaries before writing the integral. The outer circular arc gives \(r=R\); because the sector starts at the origin, the radial interval begins at \(r=0\). The two bounding rays give the angular endpoints. If the sector crosses the positive horizontal axis, an equivalent negative angle can make the increasing interval clear, as in the third example.

For a sector with constant density, the integral is also an effective check on the geometry: its mass must equal its area multiplied by the density. If an angular interval has width \(\Delta\theta\), its area is the same fraction \(\frac{\Delta\theta}{2\pi}\) of the full disk’s area \(\pi R^2\). This gives \(\frac{\Delta\theta}{2\pi}\pi R^2=\frac{1}{2}R^2\Delta\theta\), agreeing with the derived formula.

Check Your Understanding

For each question, identify the bounds or use the polar area element carefully. Keep angular widths in radians.

Write the polar bounds for the sector of radius \(5\) between \(\theta=\frac{\pi}{3}\) and \(\theta=\frac{2\pi}{3}\).
Find the area of a sector of radius \(6\) with angular width \(\frac{\pi}{4}\).
A sector sweeps counterclockwise from \(\frac{7\pi}{4}\) through the positive horizontal axis to \(\frac{\pi}{6}\). Give an increasing angular interval for it.
Explain why the polar area integral over a sector includes a factor of \(r\).
For a sector of radius \(3\), angular width \(\frac{\pi}{2}\), and constant density \(2\), set up the mass integral and evaluate it.