Identify the region and its angle interval
A ray that sweeps a petal
The curve \(r = 4\cos 2\theta\) is a rose with four petals. The goal is the region inside the petal along the positive x-axis.
A ray from the pole turns counterclockwise, and its orange part reaches from the pole to the curve. As the ray turns, the teal region shows the area that this orange part covers.
At \(\theta = -\frac{\pi}{4}\), \(r = 4\cos(-\frac{\pi}{2}) = 0\), so the curve is at the pole and the teal region is empty. Drag \(\theta\) to the right, and the region grows.
At \(\theta = \frac{\pi}{4}\), \(r = 0\) again, so the petal is complete. Its angle interval is \(-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}\).
Two rays bound the region
The curve and two rays from the pole, at the start angle \(\alpha\) and the end angle \(\beta\), bound the region.
At each angle between the rays, the orange part of the ray reaches from the pole to the curve, a distance \(r = f(\theta)\).
If the question names a sector, the part between two given rays, use those rays. For a petal or a loop, the rays are at the angles where the curve is at the pole, the solutions of \(f(\theta) = 0\).
Drag the end ray \(\beta\) to the right. The teal region fills the petal when \(\alpha\) and \(\beta\) are two such angles.
Notation
A point is in the region if its angle is between \(\alpha\) and \(\beta\) and its distance from the pole is between 0 and \(f(\theta)\).
\[\alpha \le \theta \le \beta, \qquad 0 \le r \le f(\theta)\]For a sector, use the given rays, here 0 and \(\frac{\pi}{6}\).
\[0 \le \theta \le \tfrac{\pi}{6}\]For a petal or a loop, solve \(f(\theta) = 0\), because the curve is at the pole at each solution. Choose two solutions next to each other.
\[f(\theta) = 0\]Test one angle between them to see where the loop lies. If \(f(\theta) < 0\) there, the loop is on the side of the pole opposite to the ray.
Here \(f(0) = 4\) is positive, so the petal lies along the ray.
The interval must not be longer than one full turn, because a longer interval covers some points twice.
\[\beta - \alpha \le 2\pi\]
When it fails
Move the end ray past \(\frac{\pi}{4}\). There \(r\) is negative, so the curve is drawn on the side of the pole opposite to the faint ray. A second teal petal grows along the negative y-axis.
The interval \(-\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}\) covers two petals. To describe one petal, stop at the next solution of \(f(\theta) = 0\).
Example 1: a sector of a cardioid
Describe the part of the region inside \(r = 4 - 4\sin\theta\) that lies to the right of the y-axis.
The right side of the y-axis lies between the ray that points down and the ray that points up. These two rays are the bounds.
\[-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}\]Because \(\sin\theta \le 1\), the radius \(4 - 4\sin\theta\) is never negative, so each ray reaches the curve. At \(\theta = 0\), the orange part of the ray has length 4.
The bounding rays meet the curve at the pole and at \((0, -8)\), where the white rings are.
\[r(\tfrac{\pi}{2}) = 0, \qquad r(-\tfrac{\pi}{2}) = 8\]The teal region is the right half of the cardioid.
\[\begin{aligned} -\tfrac{\pi}{2} &\le \theta \le \tfrac{\pi}{2} \\ 0 &\le r \le 4 - 4\sin\theta \end{aligned}\]
Example 2: the inner loop of a limaçon
Find the angle interval for the inner loop of \(r = 1 - 2\sin\theta\).
The loop starts and ends at the pole, so solve \(r = 0\) for \(0 \le \theta \le 2\pi\). The dashed rays show the two solutions.
\[1 - 2\sin\theta = 0 \;\Rightarrow\; \theta = \tfrac{\pi}{6} \text{ or } \tfrac{5\pi}{6}\]A tempting wrong move: A loop is a closed path, so use one full turn.
\[0 \le \theta \le 2\pi\]A full turn draws both loops. The small loop lies inside the large one, so the ray covers it twice and the teal there is darker.
A loop runs from one visit to the pole to the next. The solutions give two intervals, and the teal region shows the second one.
\[\begin{aligned} \tfrac{\pi}{6} \le \theta &\le \tfrac{5\pi}{6} \\ \tfrac{5\pi}{6} \le \theta &\le \tfrac{13\pi}{6} \end{aligned}\]Test one angle in each interval. At \(\frac{\pi}{2}\), \(r = -1\) gives the point \((0, -1)\) on the small loop. At \(\frac{3\pi}{2}\), \(r = 3\) gives \((0, -3)\) on the large loop. The inner loop uses the first interval.
\[\tfrac{\pi}{6} \le \theta \le \tfrac{5\pi}{6}\]
Practice
Find the angle interval for the petal of \(r = 5\sin 3\theta\) that has its tip on the ray \(\theta = \frac{\pi}{6}\).
The petal starts and ends where the curve is at the pole. Find the solutions of \(r = 0\) on each side of \(\frac{\pi}{6}\).
Solution
The curve is at the pole where \(\sin 3\theta = 0\), so \(3\theta\) is a multiple of \(\pi\). The dashed rays show two solutions.
\[\theta = 0, \ \tfrac{\pi}{3}, \ \tfrac{2\pi}{3}, \ \dots\]The solutions next to \(\frac{\pi}{6}\) are 0 and \(\frac{\pi}{3}\). At \(\frac{\pi}{6}\), \(r = 5\) is positive, so the petal between them has its tip on that ray.
The petal is the region for \(0 \le \theta \le \frac{\pi}{3}\).
\[0 \le \theta \le \tfrac{\pi}{3}\]
Questions
Questions 1 and 2 use the rose \(r = 4\sin 2\theta\), which has four petals. The grid rays are \(\frac{\pi}{4}\) apart. When you answer, the orange ray turns through the petal and the teal region shows it.
Which interval of \(\theta\) traces the petal in the first quadrant?
- \(0 \le \theta \le \frac{\pi}{2}\)
- \(0 \le \theta \le \pi\)
- \(0 \le \theta \le \frac{\pi}{4}\)
Answer
\(0 \le \theta \le \frac{\pi}{2}\)
Which interval of \(\theta\) traces the petal in the second quadrant?
- \(\frac{3\pi}{2} \le \theta \le 2\pi\)
- \(\frac{\pi}{2} \le \theta \le \pi\)
- \(0 \le \theta \le \frac{\pi}{2}\)
- \(\pi \le \theta \le \frac{3\pi}{2}\)
Answer
\(\frac{3\pi}{2} \le \theta \le 2\pi\)
Which interval of \(\theta\) traces the inner loop of \(r = 3 - 6\cos\theta\)?
- \(-\frac{\pi}{3} \le \theta \le \frac{\pi}{3}\)
- \(\frac{\pi}{3} \le \theta \le \frac{5\pi}{3}\)
- \(0 \le \theta \le \frac{\pi}{3}\)
Answer
\(-\frac{\pi}{3} \le \theta \le \frac{\pi}{3}\)
Summary
You find the angle interval from the rays that the question names, or, for a petal or a loop, from two neighboring solutions of \(f(\theta) = 0\).
This works because the turning ray covers the region between the pole and the curve, and a loop begins and ends at the pole.
You use it before you write an area integral, so that the bounds cover the requested region and no other.
This is one of four skills in topic 9.8 of Which & Why: AP Calculus.