From Pole Crossings to an Enclosed Region
A polar curve can form a closed boundary by leaving the pole, tracing an arc, and returning to the pole. The area of the region enclosed by that arc and the pole is found by integrating one-half the squared radius over the angle interval that traces the boundary.
The important work is not only evaluating an integral. You must identify the correct angle bounds and check how the curve is traced. In particular, the polar area integral counts repeated tracing with multiplicity. It gives ordinary area only when the interval traces the region without overlap.
If two consecutive pole crossings bound an arc that encloses a region with the pole, the area setup is generally an integral from the first crossing to the second. Before treating the result as ordinary area, verify that the curve traces the relevant boundary once and does not retrace a portion of the region.
The Area Formula for a Region Bounded by the Pole
For a polar curve \(r=f(\theta)\) on an interval \([\alpha,\beta]\), the area swept out is represented by \(\frac12\int_\alpha^\beta [f(\theta)]^2\,d\theta\). If the curve traces the enclosed region exactly once, with no overlapping interior, this integral equals its ordinary area.
For a region bounded by the pole and a curve arc, the endpoints \(\alpha\) and \(\beta\) are often consecutive pole crossings, so \(r(\alpha)=r(\beta)=0\). The radius need not be zero at every angle between them. It describes how far the boundary is from the pole as the angle changes.
The square matters. If a polar equation uses negative values of \(r\), those values still contribute nonnegative area because \(r^2\geq 0\). But the geometry must still be checked: a signed radius places a point in the direction opposite the angle, and different angle-radius pairs can represent the same point.
Why the Squared-Radius Integral Works
The formula follows by approximating the region with thin sectors. Over a small angle change \(\Delta\theta\), if the radius is nearly constant at \(r\), the area is approximately the area of a circular sector with radius \(|r|\). A sector with angle \(\Delta\theta\) measured in radians has area \(\frac12 r^2\Delta\theta\).
Divide \([\alpha,\beta]\) into small subintervals. On each one, the radius changes only a little if \(r\) is continuous.
At a sample angle \(\theta_i\), the sector area is approximately \(\frac12 r(\theta_i)^2\Delta\theta_i\).
The total approximation is \(\sum_i \frac12 r(\theta_i)^2\Delta\theta_i\). If the trace overlaps itself, this sum includes the area again whenever that portion is swept again.
As the largest subinterval width approaches zero, the sum becomes the Riemann integral \(\frac12\int_\alpha^\beta r(\theta)^2\,d\theta\).
This gives a useful reason for both parts of the formula: the squared radius comes from sector area, and the angle interval records how the curve is swept. The latter is also why checking the trace matters.
A Reliable Setup
For a region bounded by a polar curve and the pole, use this sequence:
- Solve \(r(\theta)=0\) to find pole crossings.
- Choose consecutive crossings that bound the particular arc and region of interest.
- Check the trace on that interval. Confirm that it encloses the intended region and does not retrace part or all of it.
- Substitute the radius into \(A=\frac12\int_\alpha^\beta r^2\,d\theta\).
- Evaluate the integral and check the result, for example by verifying the antiderivative or comparing with a known geometric area when possible.
The word “consecutive” is a useful guide to candidate bounds, not a substitute for examining the curve. The graph and the angle-radius representation together determine which arc is being traced.
Worked Example: A Circle Above the Pole
Find the area enclosed by \(r=3\sin\theta\) and the pole, using the interval between consecutive pole crossings from \(0\) to \(\pi\).
First find the crossings. Setting \(r=0\) gives \(3\sin\theta=0\), so the relevant consecutive angles are \(\theta=0\) and \(\theta=\pi\). Between them, \(\sin\theta\geq 0\), and the curve traces the circle once, returning to the pole at the second endpoint.
Set up and evaluate the area:
Check the result geometrically. Since \(r=3\sin\theta\), multiplication by \(r\) gives \(r^2=3r\sin\theta\). Using \(r^2=x^2+y^2\) and \(r\sin\theta=y\), the curve satisfies \(x^2+y^2=3y\), or \(x^2+(y-\frac32)^2=(\frac32)^2\). This is a circle of radius \(\frac32\), whose area is \(\pi(\frac32)^2=\frac{9\pi}{4}\), matching the integral.
Worked Example: A Circle to the Right of the Pole
Find the area enclosed by \(r=4\cos\theta\) and the pole over the interval \([-\frac{\pi}{2},\frac{\pi}{2}]\).
The pole crossings satisfy \(4\cos\theta=0\), giving the endpoints \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). On the open interval, \(\cos\theta>0\). As \(\theta\) increases from the first endpoint to the second, the curve traces the circle once.
The antiderivative checks because \(\frac{d}{d\theta}\left(\theta+\frac12\sin(2\theta)\right)=1+\cos(2\theta)=2\cos^2\theta\). The Cartesian equation gives a second check: \(r=4\cos\theta\) implies \(r^2=4r\cos\theta=4x\), so \((x-2)^2+y^2=4\). The circle has radius \(2\), and its area is \(4\pi\).
A Curve with One Pole Crossing per Revolution
Not every region bounded by the pole has two different crossing angles within one period. A curve may return to the pole at the same direction after a full turn. In that case, use successive occurrences of the crossing in the unwrapped angle interval.
Worked Example: The Region Enclosed by a Cardioid
Find the area enclosed by \(r=1+\cos\theta\). The radius is zero when \(1+\cos\theta=0\), which occurs at \(\theta=\pi+2k\pi\), where \(k\) is an integer. Thus \(\pi\) and \(3\pi\) are consecutive pole crossings. On \((\pi,3\pi)\), the curve is positive and traces the cardioid once.
For an antiderivative check, differentiating \(\frac{3\theta}{4}+\sin\theta+\frac{\sin(2\theta)}{8}\) gives \(\frac34+\cos\theta+\frac14\cos(2\theta)\). Since \(\cos^2\theta=\frac{1+\cos(2\theta)}2\), this equals \(\frac12(1+2\cos\theta+\cos^2\theta)\), the integrand. The result is therefore \(\frac{3\pi}{2}\) square units.
Repeated Tracing: A Necessary Check
A common mistake is to assume that every polar area integral automatically gives the ordinary area of the set of points in the plane. It does not. The integral follows the parametrization as \(\theta\) changes, so an area swept twice is counted twice.
For example, \(r=1\) on \([0,4\pi]\) traces the unit circle twice. The incorrect interpretation is to calculate \(\frac12\int_0^{4\pi}1\,d\theta=2\pi\) and call \(2\pi\) the area of the unit disk. But the disk is covered twice, and its ordinary area is \(\pi\). The integral gives the swept area counted with multiplicity.
The correction is to use an interval that traces the disk once, such as \([0,2\pi]\):
This example has no pole crossings, but it makes the same tracing issue clear. For a region bounded by the pole, consecutive crossings help identify an interval, while a graph or careful tracing check confirms whether the boundary is covered once. Endpoints alone do not establish that there is no overlap in the interior of the interval.
What the Bounds and the Answer Tell You
The angle bounds describe the part of the curve used in the integral. If the bounds are too short, the curve may not return to the pole; if they are too long, it may trace a boundary again. A useful first check is to substitute each endpoint into \(r(\theta)\) and confirm that the result is zero when the region is supposed to close at the pole.
The integrand is nonnegative because it contains \(r^2\), so the computed swept area cannot be negative. If an antiderivative calculation produces a negative value, check the order of the bounds and the evaluation at the endpoints. Also retain exact values such as \(\pi\) until the end unless a decimal approximation is specifically requested.
Check Your Understanding
For each question, identify the pole crossings or tracing issue before evaluating or interpreting an area.
- For \(r=2\sin\theta\), what consecutive pole crossings bound the circle, and what integral gives its enclosed area?
- For \(r=2\cos\theta\) on \([-\frac{\pi}{2},\frac{\pi}{2}]\), set up the polar area integral and evaluate it exactly.
- For \(r=1+\cos\theta\), identify two consecutive pole crossings that enclose one full trace of the curve.
- Why does \(\frac12\int_0^{4\pi}1\,d\theta\) not give the ordinary area of the unit disk?
- What geometric condition must be checked before interpreting a polar area integral as ordinary area?
