Why Intersection Bounds Need Care
When a region lies between two polar curves, its angular bounds usually come from points where the curves meet. Finding those points is not always as simple as setting their radial formulas equal at the same angle. Polar coordinates can represent one point in more than one way, negative radii reverse direction, and a curve can reach the pole at an angle different from the angle at which another curve reaches it.
The goal is to find the actual shared points of the curves first, then express those points using angles that match the region being considered. A sketch can help identify the intended region, but the algebra and coordinate checks establish the bounds.
Equivalent Polar Coordinates
A polar coordinate pair \((r,\theta)\) represents the Cartesian point \((r\cos\theta,r\sin\theta)\). Consequently, angles differing by a full turn represent the same direction, and a negative radius points in the direction opposite to the angle.
For example, \((-1,\pi)\) and \((1,0)\) both represent \((1,0)\). This matters when a curve uses negative radial values: it may pass through a point that another curve describes using a positive radius at an angle shifted by \(\pi\).
A Reliable Intersection Check
If two curves are given by \(r=f(\theta)\) and \(r=g(\theta)\), solving \(f(\theta)=g(\theta)\) finds intersections represented at the same angle with the same radial value. This is often the quickest first check, especially on an interval where both radii are nonnegative. It is not automatically a complete check if either curve uses negative radii.
For additional intersections, compare the points in Cartesian coordinates, or use the equivalent-coordinate rule. For instance, a point \((f(\theta),\theta)\) can match a point on the second curve represented at \(\theta+\pi\) with the opposite radius. The pole must also be checked separately: the pole is an intersection only if each curve reaches \(r=0\), not necessarily at the same angle.
Set the radial expressions equal and solve on the angle interval relevant to the curves and region.
Ask whether a negative radius on one curve describes a point with positive radius on the other curve at an angle shifted by \(\pi\).
Determine whether each curve reaches \(r=0\), even if the two zero-radius values occur at different angles.
Substitute candidate points into both curves, then identify which intersection angles bound the particular region in question.
Worked Example: Intersections of Two Circles
Worked Example: Find angular bounds for a region between two circles
Find the intersections of \(r=2\cos\theta\) and \(r=1\), and identify the angular interval for the region inside the first circle but outside the unit circle.
For the same-angle intersections, set the radii equal:
The solutions around the right side of the plane are \(\theta=-\frac{\pi}{3}\) and \(\theta=\frac{\pi}{3}\). At both angles, the common radius is \(1\). The first circle has radius \(2\cos\theta\), which is at least \(1\) between these angles, so the requested region has angular bounds \(-\frac{\pi}{3}\leq\theta\leq\frac{\pi}{3}\).
Check the points in Cartesian coordinates. At \(\theta=\frac{\pi}{3}\), the point is \((\frac12,\frac{\sqrt3}{2})\). It lies on the unit circle because \(x^2+y^2=\frac14+\frac34=1\). It also satisfies the Cartesian equation of the first circle, \(x^2+y^2=2x\), because \(1=2(\frac12)\). At \(\theta=-\frac{\pi}{3}\), the point is \((\frac12,-\frac{\sqrt3}{2})\), and the same checks give \(x^2+y^2=1=2x\).
The pole is not a common point: \(2\cos\theta=0\) at \(\theta=\frac{\pi}{2}\) and \(\frac{3\pi}{2}\), while the second curve has radius \(1\) at every angle. Thus the two verified intersections are the endpoints for this region.
Worked Example: Negative Radius Creates Another Intersection
Worked Example: Compare a limaçon with the unit circle
Find all intersections of \(r=1+2\cos\theta\) and \(r=1\). The unit circle has distance \(1\) from the pole at every point. A point on the limaçon has distance \(|r|=|1+2\cos\theta|\), so intersection requires \(|1+2\cos\theta|=1\).
There are two cases. If \(1+2\cos\theta=1\), then \(\cos\theta=0\), giving \(\theta=\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). The radius is \(1\), so these are the points \((0,1)\) and \((0,-1)\).
If \(1+2\cos\theta=-1\), then \(2\cos\theta=-2\), so \(\cos\theta=-1\) and \(\theta=\pi\). The limaçon has coordinate \((-1,\pi)\), which represents \((1,0)\). The unit circle represents this same point as \((1,0)\), with angle \(0\).
For verification, \((1,0)\) is on the unit circle and is represented on the limaçon by \(r=1+2\cos\pi=1-2=-1\). At \((0,1)\), the limaçon has \(r=1+2\cos(\frac{\pi}{2})=1\); at \((0,-1)\), it has \(r=1+2\cos(\frac{3\pi}{2})=1\). All three points are therefore shared. The pole is not shared: the limaçon reaches it when \(1+2\cos\theta=0\), but the unit circle never has radius zero.
The angular locations are \(\frac{\pi}{2}\), \(\pi\), and \(\frac{3\pi}{2}\) on the limaçon's parameterization. The intersection at \((1,0)\), however, is described on the unit circle at angle \(0\). A region's bounds must be read from the relevant arcs and coordinate representations, rather than assuming every shared point has one common angle on both formulas.
Worked Example: The Pole Can Occur at Different Angles
Worked Example: Find intersections of two circles through the pole
Consider \(r=2\sin\theta\) and \(r=2\cos\theta\). Find their common points and describe the angles that bound the arcs between them.
First convert each circle to Cartesian form. Since \(x=r\cos\theta\), \(y=r\sin\theta\), and \(r^2=x^2+y^2\), the first equation gives \(r^2=2r\sin\theta=2y\), hence \(x^2+y^2=2y\). The second gives \(x^2+y^2=2x\). At a shared point, subtracting the equations gives \(2y=2x\), so \(y=x\). Substituting \(y=x\) into \(x^2+y^2=2x\) gives \(2x^2=2x\), or \(2x(x-1)=0\). Thus \(x=0\) or \(x=1\), and the shared points are \((0,0)\) and \((1,1)\).
The point \((1,1)\) has polar coordinates \((\sqrt2,\frac{\pi}{4})\). Substitution confirms that both equations give \(r=2\sin(\frac{\pi}{4})=\sqrt2\) and \(r=2\cos(\frac{\pi}{4})=\sqrt2\). The pole is also shared: the first curve has \(r=0\) at \(\theta=0\), and the second has \(r=0\) at \(\theta=\frac{\pi}{2}\). Both represent \((0,0)\), even though the zero radii occur at different angles.
For the arcs from the pole to \((1,1)\), the first curve uses angles from \(0\) to \(\frac{\pi}{4}\). The second curve uses angles from \(\frac{\pi}{2}\) down to \(\frac{\pi}{4}\). These are different parameter intervals for arcs with the same endpoints. This is why identifying the actual shared points and then selecting the desired arcs is essential when setting angular bounds.
Why the Coordinate Check Works
The equivalence rule follows directly from the Cartesian coordinates. For any integer \(k\), adding \(2k\pi\) leaves sine and cosine unchanged, so \((r,\theta)\) and \((r,\theta+2k\pi)\) give identical coordinates. For a negative radius, use \(\cos(\theta+\pi)=-\cos\theta\) and \(\sin(\theta+\pi)=-\sin\theta\):
Thus \((-r,\theta+\pi)\) and \((r,\theta)\) are the same point. Conversely, checking equality of Cartesian coordinates tests whether two representations name the same location, regardless of their angles or radius signs. This provides a direct verification whenever the same-angle equation is not enough.
From Intersections to Bounds
Intersections provide candidate endpoints, but the intended region determines which candidates to use. A pair of curves can meet more than twice, trace the pole in different ways, or have several arcs connecting the same points. Use a sketch or known graph features to select the boundary arcs of the particular region. Then express their endpoint angles consistently with the chosen representation and angle interval.
For a region whose curves have a fixed radial order over one angular interval, the relevant endpoint angles are typically the angles at which the two bounding arcs meet. The earlier material on setting up polar area between curves with fixed radial order applies once those bounds and the radial order are established. Here the essential task is the reliable identification of the endpoints; when radial order changes, the bounds alone do not settle the integral setup.
Check Your Understanding
For each question, distinguish the point itself from the particular polar coordinates used to describe it.
- What Cartesian point is represented by \((-3,\frac{\pi}{6})\), and give an equivalent representation with positive radius.
- For \(r=2\cos\theta\) and \(r=2\), what equation gives their same-angle intersection candidates?
- Why should the pole be checked separately when two polar curves are compared?
- For \(r=1+2\cos\theta\), what angle represents the point \((1,0)\) using a negative radius?
- Can two curves meet at the pole even if their zero-radius values occur at different angles? Explain briefly.
