Common Area of Two Polar Regions

Why Common Area Requires Two Comparisons

When two polar regions overlap, their common area is not generally the area inside either one of the curves alone. A point belongs to both regions only if its direction is allowed by both angular ranges and its distance from the pole is within both radial bounds. Thus, finding common area requires two decisions: which angles occur in the overlap, and which radial boundary is nearer on each of those angles.

The polar area formula gives the area swept out by a radial bound \(r=R(\theta)\) over an angular interval. For common area, the relevant radial bound is the smaller of the two boundaries, but only on the angles shared by both regions. If the smaller boundary changes, the integral must be split at the change.

Key idea: On a shared ray, the common region extends from the pole only as far as the nearer of the two radial boundaries. The common area is found by integrating that nearer boundary over the shared angular range.

Radial Descriptions of the Regions

Suppose a region is described by an angular interval \(I\) and a nonnegative radial bound \(R(\theta)\). Its points can be written as \(0\le r\le R(\theta)\) for angles \(\theta\in I\). We will use intervals that describe each region once, with \(R(\theta)\ge 0\) throughout. Endpoints do not affect area, so including or excluding a single endpoint makes no difference to the integral.

Definition: If two polar regions have angular intervals \(I_1\) and \(I_2\), with radial bounds \(R_1(\theta)\) and \(R_2(\theta)\), their common angular range is \(I_1\cap I_2\). For an angle in this intersection, the common radial extent is \(0\le r\le \min(R_1(\theta),R_2(\theta))\).

Here the angular intervals are assumed to be written so that the same direction has the same angle representative (angles that differ by a multiple of \(2\pi\) describe the same direction, so shift an interval by \(2\pi\) if necessary). If the intervals then do not overlap, the regions have no shared rays and their common area is zero. When they do overlap, the shared interval may still need to be divided into smaller pieces: the two radial bounds can exchange which one is smaller. If both bounds are continuous, such an exchange can occur only where \(R_1(\theta)=R_2(\theta)\), or at an endpoint of the shared angular range; if a bound has jumps, treat each jump as an additional breakpoint and check the ordering on each piece.

The Common-Area Formula

Let \(J=I_1\cap I_2\) be the shared angular range. If the same direction is represented once on the chosen intervals, the common area is

A_{\mathrm{common}} = \frac12\int_J \min\!\bigl(R_1(\theta)^2,R_2(\theta)^2\bigr)\,d\theta.

Since both radial bounds are nonnegative, the smaller squared radius is the square of the smaller radius. If the radial bounds exchange order at angles \(\alpha_1,\ldots,\alpha_k\) inside \(J\), split the integral at those angles and use the boundary that is smaller on each resulting interval.

Important condition: The formula assumes the regions are described by radial segments from the pole, \(0\le r\le R_i(\theta)\), on their chosen angular intervals. For regions that are not represented this way, first divide or describe the regions using angular intervals on which such a radial description is valid.

Why the Smaller Boundary Gives the Common Area

Fix an angle \(\theta\) that belongs to both angular ranges. The first region contains points on that ray whose radii satisfy \(0\le r\le R_1(\theta)\); the second contains points satisfying \(0\le r\le R_2(\theta)\). A radius belongs to both sets of points exactly when it satisfies both inequalities. Therefore it must obey

0\le r\le \min(R_1(\theta),R_2(\theta)).

By the polar area formula, the area contributed by this ray direction as the angle varies is one half the integral of the square of its radial extent. Substituting the common radial extent gives the common-area formula. This also explains why the angular range must be intersected first: outside that range, at least one region contains no points on the ray, so there is no common area to count.

When a boundary crossing lies inside the shared angular range, solve \(R_1(\theta)=R_2(\theta)\) to locate it. Then determine which boundary is smaller on each side. The equality identifies where the choice can change; a test value in each interval establishes which boundary limits the overlap there.

Worked Examples

Worked Example: Two Constant Radial Bounds with Different Angular Ranges

Let the first region have radial bound \(R_1(\theta)=3\) for \(0\le\theta\le\frac{\pi}{2}\). Let the second have radial bound \(R_2(\theta)=2\) for \(\frac{\pi}{3}\le\theta\le\frac{2\pi}{3}\). Find their common area.

The shared angular range is the intersection of the two intervals:

\left[0,\frac{\pi}{2}\right] \cap \left[\frac{\pi}{3},\frac{2\pi}{3}\right] = \left[\frac{\pi}{3},\frac{\pi}{2}\right].

Throughout that interval, \(R_2=2<3=R_1\), so the common radial boundary is \(2\). The angle width is \(\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}\). Hence

\begin{aligned} A_{\mathrm{common}} &=\frac12\int_{\pi/3}^{\pi/2}2^2\,d\theta\\ &=\frac12\cdot4\left(\frac{\pi}{2}-\frac{\pi}{3}\right)\\ &=2\cdot\frac{\pi}{6}\\ &=\frac{\pi}{3}. \end{aligned}

Check: The second region’s full angular interval has width \(\frac{2\pi}{3}-\frac{\pi}{3}=\frac{\pi}{3}\), so its area is \(\frac12\cdot2^2\cdot\frac{\pi}{3}=\frac{2\pi}{3}\). The common area \(\frac{\pi}{3}\) is less than that full-region area. The first region’s sector on the shared interval has area \(\frac12\cdot3^2\cdot\frac{\pi}{6}=\frac{3\pi}{4}\), also greater than the common area.

Worked Example: Splitting at a Boundary Crossing

Suppose the first region has \(R_1(\theta)=2\) and the second has \(R_2(\theta)=4\cos\theta\), both for \(0\le\theta\le\frac{\pi}{2}\). Find their common area.

The angular intervals are identical. To find where the radial boundaries meet, solve

2=4\cos\theta \quad\Longrightarrow\quad \cos\theta=\frac12 \quad\Longrightarrow\quad \theta=\frac{\pi}{3}

in the given interval. For \(0\le\theta\le\frac{\pi}{3}\), the value \(4\cos\theta\) is at least \(2\), so the smaller boundary is \(2\). For \(\frac{\pi}{3}\le\theta\le\frac{\pi}{2}\), \(4\cos\theta\le2\), so the smaller boundary is \(4\cos\theta\). Thus

\begin{aligned} A_{\mathrm{common}} &=\frac12\int_0^{\pi/3}2^2\,d\theta +\frac12\int_{\pi/3}^{\pi/2}(4\cos\theta)^2\,d\theta\\ &=2\cdot\frac{\pi}{3} +8\int_{\pi/3}^{\pi/2}\cos^2\theta\,d\theta. \end{aligned}

Use \(\cos^2\theta=\frac{1+\cos(2\theta)}{2}\), so an antiderivative is \(\frac{\theta}{2}+\frac{\sin(2\theta)}{4}\). Therefore,

\begin{aligned} A_{\mathrm{common}} &=\frac{2\pi}{3} +8\left[ \frac{\theta}{2}+\frac{\sin(2\theta)}{4} \right]_{\pi/3}^{\pi/2}\\ &=\frac{2\pi}{3} +8\left[ \left(\frac{\pi}{4}+0\right) -\left(\frac{\pi}{6}+\frac{\sqrt3}{8}\right) \right]\\ &=\frac{2\pi}{3} +8\left(\frac{\pi}{12}-\frac{\sqrt3}{8}\right)\\ &=\frac{4\pi}{3}-\sqrt3. \end{aligned}

Check: The derivative of \(\frac{\theta}{2}+\frac{\sin(2\theta)}{4}\) is \(\frac12+\frac12\cos(2\theta)=\cos^2\theta\), as required. The first region’s full quarter-sector has area \(\frac12\cdot2^2\cdot\frac{\pi}{2}=\pi\), and \(\frac{4\pi}{3}-\sqrt3<\pi\) because \(\frac{\pi}{3}<\sqrt3\). The common area is also less than the second region’s full area: that area is \(8\int_0^{\pi/2}\cos^2\theta\,d\theta=2\pi\), which is greater than \(\frac{4\pi}{3}-\sqrt3\).

Worked Example: A Shared Range Where One Boundary Is Always Smaller

Let the first region have \(R_1(\theta)=2\) for \(0\le\theta\le\pi\), and let the second have \(R_2(\theta)=1+\cos\theta\) for \(\frac{\pi}{3}\le\theta\le\frac{2\pi}{3}\). Find the common area.

The common angular range is \([\frac{\pi}{3},\frac{2\pi}{3}]\). On this interval, \(-\frac12\le\cos\theta\le\frac12\), so \(\frac12\le1+\cos\theta\le\frac32<2\). The second boundary is smaller throughout, so there is no interior crossing to split at.

\begin{aligned} A_{\mathrm{common}} &=\frac12\int_{\pi/3}^{2\pi/3}(1+\cos\theta)^2\,d\theta\\ &=\frac12\int_{\pi/3}^{2\pi/3} (1+2\cos\theta+\cos^2\theta)\,d\theta. \end{aligned}

An antiderivative of the integrand is \(\frac{3\theta}{2}+2\sin\theta+\frac{\sin(2\theta)}{4}\). Evaluating at the endpoints gives

\begin{aligned} A_{\mathrm{common}} &=\frac12\left[ \frac{3\theta}{2}+2\sin\theta+\frac{\sin(2\theta)}{4} \right]_{\pi/3}^{2\pi/3}\\ &=\frac12\left[ \frac{\pi}{2}+0-\frac{\sqrt3}{4} \right]\\ &=\frac{\pi}{4}-\frac{\sqrt3}{8}. \end{aligned}

Check: Differentiating the displayed antiderivative gives \(\frac32+2\cos\theta+\frac12\cos(2\theta)=1+2\cos\theta+\cos^2\theta=(1+\cos\theta)^2\). The answer is positive since \(2\pi-\sqrt3>0\). It is less than the area of the second region itself, as it must be: the common region uses exactly the same radial bound but only over the second region’s angular range, and its radial bound is smaller than \(2\) throughout.

A Common Mistake: Choosing the Larger Boundary

A frequent error is to identify where the curves intersect but then use the larger radial value in the common-area integral. The larger boundary describes points that lie in one region but extend beyond the other region, so those points are not shared.

For the first worked example, using radius \(3\) instead of radius \(2\) over the shared interval would give

\frac12\int_{\pi/3}^{\pi/2}3^2\,d\theta = \frac12\cdot9\cdot\frac{\pi}{6} = \frac{3\pi}{4}.

This is not the common area: it uses the larger boundary, \(3\), where the second region ends at radius \(2\). In fact, \(\frac{3\pi}{4}>\frac{2\pi}{3}\), the area of the entire second region in that example. A proposed common area cannot exceed the area of either original region, so this comparison warns that the setup is wrong. The correct common area, using radius \(2\), is \(\frac{\pi}{3}\).

A Reliable Setup

For two polar regions given by radial bounds, organize the calculation in this order:

Find the shared angular range.
Intersect the angular intervals that describe the two regions. If there is no overlap, the common area is zero.

Find possible changes in the limiting boundary.
Solve \(R_1(\theta)=R_2(\theta)\) within the shared range, and include the range’s endpoints as breakpoints.

Choose the smaller radius on each piece.
Test an angle in each interval between consecutive breakpoints, or compare the expressions directly.

Integrate and check.
Use one half the integral of the squared smaller radius on each piece, then compare the result with the areas of the original regions.

This order keeps angular restrictions separate from radial comparisons. That distinction matters: an intersection of the boundary curves does not by itself determine the common area. The curves may intersect outside the shared angular range, and even within it the boundary that limits the overlap must be identified on both sides of every crossing.

Check Your Understanding

For each question, identify the common angular range before deciding which radial bound to integrate.

One region has \(0\le r\le4\) for \(0\le\theta\le\frac{\pi}{2}\), and another has \(0\le r\le2\) for \(\frac{\pi}{4}\le\theta\le\frac{3\pi}{4}\). What is the shared angular range and which boundary limits the common area there?
If two radial bounds cross at \(\theta=\alpha\) inside the common angular range, what additional check is needed before writing the split integral?
For regions with no shared angular directions, what is their common area?
Why does using the larger radial boundary on a shared ray count points outside the common region?
What comparison can help identify an incorrectly large common-area result?