Identify inner and outer boundaries
Which curve is farther out
The solid curve is the circle \(r = 4\cos\theta\), and the dashed curve is the circle \(r = 2\). The slider turns the faint ray.
The ray meets each curve at an orange dot. The outer curve has the dot farther from the pole, and the inner curve has the nearer one. The teal region fills the space between the dots on every ray from 0 to θ.
Turn the ray to \(\theta = 0\). The solid circle has \(r = 4\) and the dashed circle has \(r = 2\), so the solid circle is the outer curve.
Turn the ray to \(\theta = \frac{\pi}{2}\). The solid circle has \(r = 0\), so the dashed circle is the outer curve.
The dots meet at \(\theta = \frac{\pi}{3}\), where \(4\cos\theta = 2\), and the outer curve changes there.
Outer and inner at every angle
The curves are now \(r = f(\theta)\) and \(r = g(\theta)\). On each ray, the region between them starts at the nearer dot and ends at the farther dot.
The radius of the nearer dot is \(r_1\), and the radius of the farther dot is \(r_2\), so \(r_1 \le r_2\) on every ray.
Drag the angle across a crossing, marked by a white ring. The labels \(r_1\) and \(r_2\) move to the other curves, but the curves do not change.
The order changes only where the curves meet, so between two meeting angles the same curve stays outer.
Notation
Find the angles where the curves meet by solving \(f(\theta) = g(\theta)\), as in the previous skill. These angles cut the range of \(\theta\) into smaller intervals, called stretches.
\[f(\theta) = g(\theta)\]Choose a test angle \(\theta_t\) inside one stretch. Find \(r\) on each curve at that angle.
\[f(\theta_t) \quad\text{and}\quad g(\theta_t)\]The curve with the larger \(r\) is the outer curve \(r_2\) on the whole stretch, and the other curve is the inner curve \(r_1\). The order cannot change inside the stretch, because the curves do not meet there.
\[0 \le \cy{r_1} \le \cy{r_2}\]Repeat for every stretch.
This works when both values of \(r\) are 0 or positive on the stretch.
When it fails
Compare the curve \(r = 1 + 2\cos\theta\) with the circle \(r = \frac12\). At \(\theta = \pi\), the curve has \(r = -1\), and since \(-1 < \frac12\), the curve seems to be the inner curve.
A negative \(r\) puts the dot on the opposite side of the pole, at distance 1, so the ray at \(\theta = \pi\) meets only the circle. Compare values of \(r\) only where both are 0 or positive.
Example 1: a curve inside a circle
Find the outer and inner curves of \(r = 7\) and \(r = 4 + 2\sin\theta\), for \(0 \le \theta \le 2\pi\).
Look for angles where the curves meet. Solve \(4 + 2\sin\theta = 7\). Then \(\sin\theta = \frac32\), which no angle satisfies. The curves never meet, so the order cannot change.
\[4 + 2\sin\theta = 7\]Test one angle, \(\theta = \frac{\pi}{2}\). The curve has \(r = 6\), and the circle has \(r = 7\).
\[4 + 2\sin\tfrac{\pi}{2} = 6 < 7\]The circle is the outer curve at every angle. The other curve is the inner curve.
\[\cy{r_2 = 7} \qquad \cy{r_1 = 4 + 2\sin\theta}\]
Example 2: a curve that crosses a circle
For \(0 \le \theta \le \frac{\pi}{2}\), find the outer and inner curves of \(r = 2 + 2\sin\theta\) and \(r = 3\).
Find where the curves meet. Solve \(2 + 2\sin\theta = 3\). Then \(\sin\theta = \frac12\), so \(\theta = \frac{\pi}{6}\) in this range.
\[2 + 2\sin\theta = 3\]A tempting wrong move: The curve reaches \(r = 4\), which is more than 3. It is the outer curve for the whole range.
\[r_2 = 2 + 2\sin\theta\]At \(\theta = 0\), the curve has \(r = 2\), and the circle has \(r = 3\). In the graph, the circle's dot is farther out near the x-axis.
Test one angle in each stretch, \(\theta = 0\) and \(\theta = \frac{\pi}{2}\).
\[\begin{aligned} 2 + 2\sin 0 &= 2 < 3 \\ 2 + 2\sin\tfrac{\pi}{2} &= 4 > 3 \end{aligned}\]The circle is outer for \(0 \le \theta \le \frac{\pi}{6}\). The other curve is outer for \(\frac{\pi}{6} \le \theta \le \frac{\pi}{2}\).
Practice
For \(0 \le \theta \le \frac{\pi}{2}\), find the outer curve on each stretch for \(r = 4\sin 2\theta\) and \(r = 2\).
Find the meeting angles first. Then turn the ray into each stretch and see which dot is farther from the pole.
Solution
Set the radii equal. Then \(\sin 2\theta = \frac12\), so \(2\theta = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\). The curves meet at \(\theta = \frac{\pi}{12}\) and \(\theta = \frac{5\pi}{12}\).
\[4\sin 2\theta = 2\]These angles cut the range into three stretches. Test one angle in each, \(\theta = 0\), \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\), and compare each value with the circle's \(r = 2\).
\[\begin{aligned} 4\sin(2\cdot 0) &= 0 \\ 4\sin(2\cdot\tfrac{\pi}{4}) &= 4 \\ 4\sin(2\cdot\tfrac{\pi}{2}) &= 0 \end{aligned}\]The circle is outer for \(0 \le \theta \le \frac{\pi}{12}\) and for \(\frac{5\pi}{12} \le \theta \le \frac{\pi}{2}\). The other curve is outer for \(\frac{\pi}{12} \le \theta \le \frac{5\pi}{12}\).
Questions
Questions 1 and 2 use the curve \(r = 3 - 2\cos\theta\) and the circle \(r = 2\), for \(0 \le \theta \le \pi\).
Which curve is the outer curve at \(\theta = \pi\)?
- The curve \(r = 3 - 2\cos\theta\)
- The circle \(r = 2\)
- Neither, because the curves meet.
Answer
The curve \(r = 3 - 2\cos\theta\)
For which angles is the circle the outer curve?
- \(0 \le \theta < \frac{\pi}{3}\)
- \(\frac{\pi}{3} < \theta \le \pi\)
- \(0 \le \theta < \frac{2\pi}{3}\)
Answer
\(0 \le \theta < \frac{\pi}{3}\)
At \(\theta = \frac{\pi}{6}\), which curve is the outer curve: \(r = 6\sin\theta\) or \(r = 4\cos\theta\)?
- \(r = 4\cos\theta\)
- \(r = 6\sin\theta\)
- Neither, because they meet at \(\frac{\pi}{6}\).
Answer
\(r = 4\cos\theta\)
Summary
Between two angles where the curves meet, the curve with the larger r at one test angle is the outer curve for the whole stretch.
This works because the order of two curves can change only where they meet.
You need it to find the area between two polar curves, because that area subtracts the inner radius squared from the outer one.
This is one of four skills in topic 9.9 of Which & Why: AP Calculus.