Set up and calculate the area

Outer sector minus inner sector

The dashed circles have radius 3 and radius 1. The shaded region lies between them, from the x-axis to the ray at the angle \(\beta\).

Drag \(\beta\). A sector with radius \(r\) and angle \(\beta\) has area \(\tfrac12 r^2 \beta\), so the outer sector has area \(\tfrac92\beta\) and the inner sector has area \(\tfrac12\beta\).

The shaded region is the outer sector with the inner sector removed, so its area is \(\tfrac92\beta - \tfrac12\beta = 4\beta\). At \(\beta = \frac{\pi}{2}\), the area is \(2\pi\).

Any two curves

The circles are now two curves: the solid outer curve \(\cy{r_2(\theta)}\) and the dashed inner curve \(\cy{r_1(\theta)}\). Over a thin angle \(d\theta\), each curve is almost an arc of a circle.

Press the button to show one thin wedge. It is a thin outer sector with a thin inner sector removed, so its area is \(\tfrac12 r_2^2\,d\theta - \tfrac12 r_1^2\,d\theta\).

Drag the ray. The shaded region is made of these wedges, and at every angle each wedge is the outer sector minus the inner sector.

Notation

1. Find the meeting angles, where the curves cross. They give the limits \(a\) and \(b\) of the angle range, or the angles where you split it.

2. On the range, name the outer curve \(\cy{r_2}\) and the inner curve \(\cy{r_1}\), with \(0 \le r_1 \le r_2\).

3. The area is the sum of the thin wedges.

   \[A = \cb{\tfrac12\int_a^b \big(r_2^2 - r_1^2\big)\,d\theta}\]

4. The area is half the integral of the outer radius squared minus the inner radius squared. Square each radius first. Then subtract.

   For the region inside both curves, the edge is the nearer curve. Use \(\tfrac12\int r^2\,d\theta\) with the nearer curve on each part of the range.

When it fails

Suppose you name the nearer circle \(r = 1\) as \(\cr{r_2}\) and the circle \(r = 3\) as \(\cr{r_1}\). Then \(r_2^2 - r_1^2\) is negative, so the integral from 0 to \(\frac{\pi}{2}\) is \(-2\pi\). The red lines mark wedges that count as negative.

An area cannot be negative, so the order is wrong. Check that \(r_2\) is the farther curve on the whole range.

Example 1: inside one circle, outside another

Find the area of the region inside \(r = 4\sin\theta\) and outside \(r = 2\).

1. The curves meet where \(4\sin\theta = 2\). Then \(\sin\theta = \tfrac12\), so \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

   \[4\sin\theta = 2\]

2. Between these angles, \(4\sin\theta \ge 2\), so \(r = 4\sin\theta\) is the outer curve and \(r = 2\) is the inner curve.

   \[\cy{r_2 = 4\sin\theta} \qquad \cy{r_1 = 2}\]

3. Square each radius. Then subtract.

   \[A = \tfrac12\int_{\pi/6}^{5\pi/6}\big(16\sin^2\theta - 4\big)\,d\theta\]

4. Use \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\), so \(\tfrac12\big(16\sin^2\theta - 4\big) = 2 - 4\cos 2\theta\).

   \[\begin{aligned} A &= \int_{\pi/6}^{5\pi/6}\big(2 - 4\cos 2\theta\big)\,d\theta \\ &= \Big[2\theta - 2\sin 2\theta\Big]_{\pi/6}^{5\pi/6} \end{aligned}\]

5. Substitute the limits.

   \[\cb{A = \tfrac{4\pi}{3} + 2\sqrt3}\]

6. As a check, \(A \approx 7.65\). The region lies inside the circle \(r = 4\sin\theta\), which has area \(4\pi \approx 12.57\), so the answer must be smaller.

Example 2: a loop outside a circle

Find the area of the region inside \(r = 4\cos 2\theta\) and outside \(r = 2\), for \(-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}\).

1. The curves meet where \(4\cos 2\theta = 2\). Then \(\cos 2\theta = \tfrac12\), so \(\theta = \pm\frac{\pi}{6}\). Between these angles, \(4\cos 2\theta \ge 2\), so \(r = 4\cos 2\theta\) is the outer curve.

   \[4\cos 2\theta = 2\]

2. A tempting wrong move: Subtract the radii first. Then square.

   \[A = \tfrac12\int_{-\pi/6}^{\pi/6}\big(4\cos 2\theta - 2\big)^2d\theta\]

   At \(\theta = 0\), the region runs from \(r = 2\) to \(r = 4\). The thin wedge there has area \(\frac12(16 - 4)\,d\theta = 6\,d\theta\). The squared difference gives only \(\frac12(4 - 2)^2\,d\theta = 2\,d\theta\), which is too small.

3. Square each radius. Then subtract.

   \[A = \tfrac12\int_{-\pi/6}^{\pi/6}\big(16\cos^2 2\theta - 4\big)\,d\theta\]

4. The region is symmetric about the x-axis, so double the integral from 0 to \(\frac{\pi}{6}\). Use \(\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}\).

   \[A = \int_0^{\pi/6}\big(4 + 8\cos 4\theta\big)\,d\theta\]

5. Integrate and substitute the limits.

   \[\begin{aligned} A &= \Big[4\theta + 2\sin 4\theta\Big]_0^{\pi/6} \\ &= \cb{\tfrac{2\pi}{3} + \sqrt3} \end{aligned}\]

6. As a check, \(A \approx 3.83\).

Practice

Find the area of the region inside \(r = 2 + 2\cos\theta\) and outside \(r = 2\).

Find the angles where the curves meet. Then drag the ray between them and see which dot is farther from the pole.

Questions

Questions 1 and 2 use the region inside the circle \(r = 4\) and outside the curve \(r = 2\cos 2\theta\), for \(-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}\).

1. What are the outer radius \(r_2\) and the inner radius \(r_1\) at \(\theta = \frac{\pi}{6}\)?

   • \(r_2 = 4\) and \(r_1 = 1\)
   • \(r_2 = 1\) and \(r_1 = 4\)
   • \(r_2 = 4\) and \(r_1 = 2\)
   Answer

   \(r_2 = 4\) and \(r_1 = 1\)

2. What is the area of the thin wedge at \(\theta = \frac{\pi}{6}\), divided by \(d\theta\)?

   • \(\tfrac{15}{2}\)
   • \(\tfrac92\)
   • \(\tfrac{17}{2}\)
   • 15
   Answer

   \(\tfrac{15}{2}\)

3. Find the area of the whole region, from \(\theta = -\frac{\pi}{4}\) to \(\theta = \frac{\pi}{4}\).

   • \(\tfrac{7\pi}{2}\)
   • \(4\pi\)
   • \(\tfrac{9\pi}{2} - 8\)
   Answer

   \(\tfrac{7\pi}{2}\)

Summary

You write half the integral of the outer radius squared minus the inner radius squared, from one meeting angle to the next.

This works because each thin wedge between the curves is a thin outer sector minus a thin inner sector.

You use it for any region between two polar curves, after you find the meeting angles and name the outer and inner curves.

This is one of four skills in topic 9.9 of Which & Why: AP Calculus.