Why Subtract Squared Radii?
When two polar curves bound a region over the same range of angles, the distance between their radii is not enough by itself to determine the area. A thin angular slice is a sector-like piece whose area depends on the square of its radius. Therefore, the area between the curves comes from subtracting the inner squared radius from the outer squared radius.
The key setup decision is to identify which curve is farther from the pole on every ray in the interval. This tutorial assumes that the shared angular interval and the radial order have already been established. If the curves exchange order inside the interval, one fixed-order formula cannot describe the whole region.
Conditions for the Fixed-Order Formula
Suppose the region is described over an angular interval \([\alpha,\beta]\) by two continuous, nonnegative radial functions \(R\) and \(r\). At each angle, the points in the region have radial distances between \(r(\theta)\) and \(R(\theta)\). For the formula below to give the area without overcounting, require all of the following:
- The interval is shared by both curves and describes the region being measured.
- The radii are nonnegative distances from the pole on that interval.
- The radial order is fixed: \(R(\theta)\ge r(\theta)\) for every \(\theta\) in the interval.
- The interval describes the region only once in angle. For example, an interval of width at most \(2\pi\) has no repeated interior ray; its two endpoint rays may coincide when the width is \(2\pi\).
Here \(R\) and \(r\) mean radial distances, not merely whichever expressions look larger by their coefficients. Before integrating, compare the functions over the entire stated interval. If an equation uses a negative signed radius, convert it to a representation by nonnegative radial distances before treating it as an outer or inner radius.
Why the Formula Works
For a curve with nonnegative radial function \(\rho(\theta)\), the polar area formula gives the area swept over a single angular interval as \(\frac12\int \rho(\theta)^2\,d\theta\). The region inside the outer curve and outside the inner curve is obtained by removing the inner region from the outer one, angle by angle. The fixed order ensures that this subtraction describes the intended radial band.
At an angle \(\theta\), the thin outer sector has area approximately \(\frac12 R(\theta)^2\Delta\theta\). The corresponding inner sector has area approximately \(\frac12 r(\theta)^2\Delta\theta\).
The radial band between the curves has approximate area \(\frac12\bigl(R(\theta)^2-r(\theta)^2\bigr)\Delta\theta\).
Summing these contributions over a partition gives an approximation to the region's area. As the partition widths approach zero, the sums approach the definite integral \(\frac12\int_{\alpha}^{\beta}(R^2-r^2)\,d\theta\).
This argument relies on a single angular tracing of the region. If a ray occurs repeatedly in the interval, its radial band is included repeatedly in the sum. The single-coverage condition is therefore part of the theorem's use, not a minor technical detail.
Worked Examples
Worked Example: A Constant Outer Radius
Find the area between \(R(\theta)=4\) and \(r(\theta)=2\cos\theta\) on \([0,\pi/3]\). On this interval, \(0\le 2\cos\theta\le2\), so \(4\ge2\cos\theta\ge0\). The angle interval has width \(\pi/3<2\pi\), so it covers no ray more than once.
Set up the area with the outer squared radius first:
Use \(\cos^2\theta=(1+\cos 2\theta)/2\). An antiderivative of \(8-2\cos^2\theta\) is \(7\theta-\frac12\sin 2\theta\), since its derivative is \(7-\cos 2\theta=8-2\cos^2\theta\). Thus
The result is positive, as expected because the outer radius is larger throughout. The antiderivative check confirms that the integrand is the required squared-radius difference with its factor of \(\frac12\).
Worked Example: Curves That Meet at an Endpoint
Find the area between \(R(\theta)=2\) and \(r(\theta)=1+\sin\theta\) on \([0,\pi/2]\). Since \(0\le\sin\theta\le1\) there, \(1\le1+\sin\theta\le2\). Thus \(R\ge r\ge0\), with equality at the endpoint \(\theta=\pi/2\). A shared boundary point does not change the area formula.
An antiderivative of the integrand including the outside factor \(\frac12\) is \(\frac54\theta+\cos\theta+\frac18\sin 2\theta\). Its derivative is \(\frac54-\sin\theta+\frac14\cos 2\theta=\frac32-\sin\theta-\frac12\sin^2\theta\), which is the integrand. Therefore
This area is positive: \(5\pi/8>1\). The radii become equal only at one endpoint, while the outer curve remains no smaller on the entire interval.
Worked Example: A Symmetric Angular Interval
Find the area between \(R(\theta)=2+\cos\theta\) and \(r(\theta)=1\) on \([-\pi/2,\pi/2]\). Since \(\cos\theta\ge0\) on this interval, \(R(\theta)\ge2>1=r(\theta)\). Both radii are nonnegative, and the interval has width \(\pi\), so the region is described once in angle.
An antiderivative of the integrand is \(\frac74\theta+2\sin\theta+\frac18\sin 2\theta\). Differentiating gives \(\frac74+2\cos\theta+\frac14\cos 2\theta=\frac32+2\cos\theta+\frac12\cos^2\theta\), equal to the integrand after expanding the outside factor. Evaluate at the bounds:
The two endpoint evaluations have equal and opposite signs because the antiderivative is odd. Their difference is therefore positive and gives the exact area.
Common Setup Errors
A frequent error is to square the difference of the radii. For example, if \(R=3\) and \(r=1\) on \([0,\pi/2]\), the incorrect setup \(\frac12\int_0^{\pi/2}(3-1)^2\,d\theta\) gives \(\frac12(4)(\pi/2)=\pi\). The correct setup is
The error occurs because each sector's area depends on radius squared. The radial width \(R-r\) is not itself a radius to square.
Another important error is integrating over a range that traces directions repeatedly. If \(R=2\), \(r=1\), and the interval is \([0,4\pi]\), direct use of the formula gives \(\frac12(4-1)(4\pi)=6\pi\). But the same annulus is traced twice: its area counted once is \(\pi(2^2-1^2)=3\pi\). Use an interval that covers the desired region once, or otherwise account for repeated coverage rather than applying the formula directly.
Finally, do not use a single fixed-order subtraction unless the radial order holds throughout the interval. The expression \(R^2-r^2\) represents the intended area only when \(R\) is consistently the outer radius and \(r\) consistently the inner radius.
Why the Setup Matters
For polar area problems, a correct integral depends first on a correct geometric description. The bounds specify which directions are included; the order specifies the outer and inner radial distances; and the single-coverage condition ensures that each part of the region is counted once. Only after these checks should the integral be evaluated.
Check Your Understanding
For each question, focus on the geometric setup before doing any integration.
- For \(R(\theta)=5\) and \(r(\theta)=2\) on \([0,\pi]\), write the area integral and state why the angle interval does not repeat interior rays.
- On \([0,\pi/2]\), which radius is outer: \(3\) or \(1+\sin\theta\)? Give the inequality that verifies your choice.
- Explain why \(\frac12\int_\alpha^\beta(R-r)^2\,d\theta\) is not the fixed-order polar area formula.
- What problem arises if the interval \([0,4\pi]\) is used to describe an annular region by constant radii?
- If the proposed outer radius is smaller than the proposed inner radius at some angle, which condition of the fixed-order formula has failed?
