Why Polar Area Uses the Square of the Radius
A polar curve describes points by a distance from the pole and an angle. When the radius changes as the angle changes, a small piece of the region looks approximately like a thin circular sector. Its area depends on the square of its radius, which is why the polar area formula contains \(r^2\).
The goal here is to derive that formula. The central idea is to divide an angular interval into many small pieces, estimate the area of each piece with a sector, and take the limit of the resulting sums.
The Area of a Circular Sector
Start with a circle of radius \(\rho\). A sector with central angle \(\Delta\theta\) is the fraction \(\Delta\theta/(2\pi)\) of the full circle, provided the angle is in radians. Since the full circle has area \(\pi\rho^2\), the sector has area
This is an exact formula for a circular sector, not an approximation. The approximation enters when a piece of a polar region is treated as a sector even though its radius may change across that piece.
From Thin Sectors to an Integral
Divide \([a,b]\) into \(n\) subintervals. Write their endpoints as \(a=\theta_0<\theta_1<\cdots<\theta_n=b\), and let \(\Delta\theta_i=\theta_i-\theta_{i-1}\). Within the \(i\)th angular slice, the radius varies between a minimum \(m_i\) and a maximum \(M_i\).
The slice contains a sector of radius \(m_i\) and is contained within a sector of radius \(M_i\). Therefore its area lies between the corresponding sector areas:
Adding these inequalities over all slices bounds the whole region’s area between two sums. As the partition is refined, the angles \(\Delta\theta_i\) become small. Because \(r\) is continuous, its values on each sufficiently narrow slice are close together. Thus \(M_i^2-m_i^2\) becomes small, and the difference between the upper and lower area estimates tends to zero.
Choose any sample angle \(\theta_i^*\) in the \(i\)th slice. Using its radius to estimate that slice gives the sum
As the largest slice width tends to zero, these sums converge to the Riemann integral of \(\frac12r(\theta)^2\). The bounds above show that the actual area has the same limit.
Why the Limit Gives the Formula
The argument can be made precise by applying the upper- and lower-sum idea to the continuous function \(g(\theta)=r(\theta)^2\). A continuous function on a closed interval is Riemann integrable. For a partition \(P\), the difference between the upper and lower sums for \(\frac12g\) is
Continuity on \([a,b]\) implies uniform continuity. Hence, as the largest subinterval width tends to zero, the greatest difference between values of \(r\) within any one subinterval tends to zero. Since \(r\) is bounded, the differences between the corresponding squared values also tend uniformly to zero: \(M_i^2-m_i^2\to0\) uniformly across the slices. The sum of the widths is fixed, \(\sum_i\Delta\theta_i=b-a\), so the displayed difference tends to zero.
Each slice’s area is bounded by its inner and outer sector estimates. Consequently, the total area is bounded by sums whose difference tends to zero. Any tagged sector sum between those bounds therefore converges to the same value as the Riemann sums for \(\frac12r^2\). By the definition of the definite integral, that shared limit is
This proves the formula: thin sectors estimate the slices, continuity makes the estimates arbitrarily close, and the limiting sum is the integral.
Worked Examples: Checking the Sector Construction
Worked Example: One sector with a fixed radius
Find the area of a sector with radius \(3\) and angle \(\frac{\pi}{6}\) radians using the sector formula.
Substitute \(\rho=3\) and \(\Delta\theta=\frac{\pi}{6}\):
Check using the fraction of a full circle: the angle is \((\pi/6)/(2\pi)=1/12\) of a full turn. The circle’s area is \(9\pi\), and \(\frac1{12}(9\pi)=\frac{3\pi}{4}\), as required.
Worked Example: Adding equal sectors for a constant radius
Consider a constant radius \(r(\theta)=2\) over a quarter-turn, from \(0\) to \(\frac{\pi}{2}\). Divide the interval into four equal angular slices and add their sector areas.
Each slice has angle
Since the radius is \(2\), each slice is already an exact circular sector. Its area is
Adding four slices gives \(\displaystyle 4\left(\frac{\pi}{4}\right)=\pi\). Check against the geometry: this is one-quarter of a circle of radius \(2\), whose area is \(\frac14\pi(2)^2=\pi\). The sector sum agrees exactly, even before taking a finer limit, because the radius does not vary.
Worked Example: Taking a limiting sector sum for a changing radius
Let \(r(\theta)=1+\theta\) on \([0,1]\). Use left endpoints of \(n\) equal slices to form sector sums, then find their limit. This calculation illustrates the Riemann-sum step in the derivation.
Each slice has width \(\Delta\theta=\frac1n\), and its left endpoint is \(\theta_i=\frac{i}{n}\), for \(i=0,\ldots,n-1\). The sum of the sector estimates is
Expand the square and separate the sums:
Use \(\sum_{i=0}^{n-1}1=n\), \(\sum_{i=0}^{n-1}i=\frac{n(n-1)}2\), and \(\sum_{i=0}^{n-1}i^2=\frac{(n-1)n(2n-1)}6\):
As \(n\to\infty\), the three terms inside the parentheses tend to \(1\), \(1\), and \(\frac13\), respectively. Thus
Check the limit using the integral represented by the Riemann sums. An antiderivative of \(\frac12(1+\theta)^2\) is \(\frac16(1+\theta)^3\), since its derivative is \(\frac12(1+\theta)^2\). Evaluating at the endpoints gives \(\frac16(2^3-1^3)=\frac76\), matching the limiting sum.
Conditions and Common Errors
The angle must be measured in radians. The sector-area formula uses the fraction \(\Delta\theta/(2\pi)\), which represents the fraction of a full turn only when the angle is in radians. The derivation also assumes a continuous, nonnegative radial boundary and a sweep that does not count the same part of the region multiple times. If the curve crosses the pole or retraces part of the region, the geometry must be examined before using a single area integral over the entire parameter interval.
A frequent error is to omit the factor \(\frac12\). For example, a sector with radius \(4\) and angle \(\frac{\pi}{3}\) does not have area \(4^2\cdot\frac{\pi}{3}=\frac{16\pi}{3}\). The correct calculation is
The incorrect value is twice the correct one because it treats the sector as though its area were \(\rho^2\Delta\theta\), rather than half that quantity.
Check Your Understanding
Use the sector construction and its assumptions to answer these questions.
- Why must the angle in the sector-area formula be measured in radians?
- What is the area of a sector with radius \(5\) and angle \(\frac{\pi}{4}\)?
- In the derivation, why do the upper and lower sector estimates approach one another as the slices become narrower?
- Which factor is missing from the incorrect expression \(A=\int_a^b r(\theta)^2\,d\theta\) for a single radial sweep?
- Why can a curve that retraces part of a region require more care than a single nonnegative radial boundary?
