Area Between Two Polar Curves — AP® Calculus BC Notes

Topic 9.8 found the area swept by one polar curve. Now there are two curves, and the region lies between them: inside one and outside the other. The idea is the same as the washer in a volume problem: a big piece with a small piece removed.

Outer squared minus inner squared

Cut the region into thin wedges from the pole. Each piece is a wedge of the outer curve $R(\theta)$ with a wedge of the inner curve $r(\theta)$ removed, so its area is

$\frac{1}{2}R^2\,\Delta\theta - \frac{1}{2}r^2\,\Delta\theta = \frac{1}{2}\left(R^2 - r^2\right)\Delta\theta.$

Adding and taking the limit:

$A = \frac{1}{2}\int_{\alpha}^{\beta} \left(R(\theta)^2 - r(\theta)^2\right)d\theta.$

Square each radius, then subtract. Writing $\frac{1}{2}\int (R - r)^2\,d\theta$ is the most common setup error in this topic: $(R - r)^2$ is not the area of anything here. For $R = 3$ and $r = 2$ on a quarter turn, the correct area is $\frac{1}{2} \cdot (9 - 4) \cdot \frac{\pi}{2} = \frac{5\pi}{4}$ , while the wrong one gives $\frac{\pi}{4}$ .

The three-step setup

The cardioid r = 2 + 2 cos θ in emerald and the circle r = 3 in blue on a polar grid. They meet at (3, π/3) and (3, −π/3), marked with dashed rays from the pole. The region inside the cardioid and outside the circle, a crescent on the right, is shaded. A side panel gives the steps: set 2 + 2 cos θ = 3 to get θ = ±π/3; on that interval the cardioid is the outer curve; the area is one half the integral from −π/3 to π/3 of (2 + 2 cos θ) squared minus 3 squared, which is (9√3 − 2π)/2, about 4.653. — **Figure 1.** The region inside the cardioid $r = 2 + 2\cos\theta$ and outside the circle $r = 3$ . LevlPrep original diagram.

Where do the curves meet? Set the $r$ -values equal and solve for $\theta$ . Then check the pole on its own: each curve reaches $r = 0$ at its own angle, and the equation will not find that crossing.
Which curve is outer? Pick a test angle between the crossings and compare the two $r$ -values. The larger is $R$ .
Integrate $\frac{1}{2}\left(R^2 - r^2\right)$ between the crossing angles.

Worked example. Find the area inside $r = 2 + 2\cos\theta$ and outside $r = 3$ (the figure).

Step 1: $2 + 2\cos\theta = 3$ , so $\cos\theta = \frac{1}{2}$ and $\theta = -\frac{\pi}{3}$ or $\frac{\pi}{3}$ .

Step 2: at $\theta = 0$ , the cardioid has $r = 4$ and the circle $r = 3$ , so the cardioid is outer.

Step 3: $(2 + 2\cos\theta)^2 - 9$ $= 4$ $+ 8\cos\theta$ $+ 4\cos^2\theta - 9$ $= -3$ $+ 8\cos\theta$ $+ 2\cos(2\theta)$ .

$A$ $= \frac{1}{2}\int_{-\pi/3}^{\pi/3} \left(-3 + 8\cos\theta + 2\cos(2\theta)\right)d\theta$ $= \frac{1}{2}\Big[-3\theta + 8\sin\theta + \sin(2\theta)\Big]_{-\pi/3}^{\pi/3}$ .

$= \frac{1}{2}\left(-2\pi + 8\sqrt{3} + \sqrt{3}\right)$ $= \frac{9\sqrt{3} - 2\pi}{2}$ $\approx 4.653$ .

Symmetry saves work: this region is symmetric about the $x$ -axis, so $A$ $= 2 \cdot \frac{1}{2}\int_0^{\pi/3} \left(R^2 - r^2\right)d\theta$ gives the same answer with an easier lower limit.

Regions inside both curves

A region inside both curves is bounded by one curve for some angles and by the other for the rest, so it needs two integrals, or area arithmetic. Take $r = 4\sin\theta$ and $r = 2$ , which meet at $\theta = \frac{\pi}{6}$ and $\frac{5\pi}{6}$ . The crescent inside $r = 4\sin\theta$ and outside $r = 2$ has area $\frac{1}{2}\int_{\pi/6}^{5\pi/6} \left(16\sin^2\theta - 4\right)d\theta$ $= \frac{4\pi}{3}$ $+ 2\sqrt{3}$ . The whole disk inside $r = 4\sin\theta$ has area $4\pi$ , so the part inside both is

$4\pi - \left(\frac{4\pi}{3} + 2\sqrt{3}\right) = \frac{8\pi}{3} - 2\sqrt{3}.$

Region	Setup
inside $R$ , outside $r$	$\frac{1}{2}\int \left(R^2 - r^2\right)d\theta$ between the crossings
inside both	split at the crossing and use each curve where it is the boundary, or subtract from a known area
between two loops of one curve	outer loop's area minus inner loop's area

On the exam

Readers look for the $\frac{1}{2}$ , both squares and the right limits in the setup. On a calculator question, write the integral with its limits, then the value to three decimals. If the crossing angle needs a calculator, store it and use the stored value as a limit rather than a rounded one.