Difference of squares, in polar form BC only

At each angle the region between two polar curves is the gap between two sectors sharing a vertex at the pole. Subtracting their areas gives

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

with $R$ the outer radius and $r$ the inner one. This is two of 9.8's integrals subtracted, exactly as 8.11's washer was two disc integrals subtracted.

And it is never $\frac{1}{2}\int (R - r)^{2}\,d\theta$. The same numbers make the case as in 8.11: with $R = 5$ and $r = 3$, $R^{2} - r^{2} = 16$ while $(R - r)^{2} = 4$. They agree only when $r = 0$, which is the case where the inner curve collapses to the pole and the region is a plain sector again.

The geometric reason is worth restating in polar terms. $R - r$ is the radial thickness of the region at that angle. A thin band far from the pole covers far more area than a small sector of that thickness at the pole, because it is stretched around a longer arc. Subtracting the radii throws that stretching away.

The limits are the angles where the two curves cross, and they have to be solved for before anything is integrated.

1. Set the radii equal and solve for $\theta$.
2. Test an angle strictly between two consecutive solutions to see which curve is outer there.
3. Integrate over that interval with the roles assigned.

Take $r = 3\cos\theta$ and $r = 1 + \cos\theta$. Setting them equal gives $3\cos\theta = 1 + \cos\theta$, so $\cos\theta = \frac{1}{2}$ and $\theta = \pm\frac{\pi}{3}$. Testing $\theta = 0$ gives $3$ for the circle and $2$ for the cardioid, so the circle is outer on $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$, and

$$A = \frac{1}{2}\int_{-\pi/3}^{\pi/3}\left(9\cos^{2}\theta - (1 + \cos\theta)^{2}\right)d\theta = \pi.$$

One warning specific to polar coordinates. Setting the radii equal can miss intersections, because a point has many polar names and the two curves may pass through it at different angles. The pole is the usual offender: both of these curves reach it, the circle at $\theta = \frac{\pi}{2}$ and the cardioid at $\theta = \pi$, and no value of $\theta$ satisfies the equation there. Always check the pole separately, by asking whether each curve reaches it at all.

Against the true $\pi \approx 3.142$ for that region:

1. $\frac{1}{2}\int (R - r)^{2}\,d\theta \approx 0.544$. The radii subtracted before squaring. Badly too small, and it is smaller rather than larger, which surprises people.
2. $\frac{1}{2}\int R^{2}\,d\theta \approx 6.661$. The inner curve never subtracted, so the region includes everything the circle sweeps over that interval.
3. Integrating over $[0, 2\pi] \approx 9.425$. The integrand is right and the interval has nothing to do with the region, which exists only between the intersection angles.

The third is the one with no analogue in Unit 8, and it is CA11 returning. In rectangular coordinates the limits are forced by the picture; here an interval can be written down that is perfectly legal, evaluates cleanly, and describes no region at all.

A check that catches all three: the answer must be smaller than the outer curve's own area over the same interval and larger than zero. Here that window is $0 < A < 6.661$, which rules out the third answer immediately and makes the first look implausibly small for a region that is clearly a substantial fraction of the circle.

Which curve is outer can change partway around, and then the interval splits, exactly as an area between rectangular curves splits at a crossing in 8.6.

For $r = 3\cos\theta$ and $r = 1 + \cos\theta$ over a wider range, the circle is outer on $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$ and the cardioid is outer beyond those angles until each curve reaches the pole. A region described as "inside one and outside the other" therefore has to be read carefully: the phrase names which curve is $R$ and which is $r$, and it holds only on the interval where that assignment is true.

Three habits close the topic, and all three are inherited:

1. Solve for the intersection angles, and check the pole separately. From this section.
2. Test an angle inside each subinterval to assign $R$ and $r$. This is 8.4's test point, in polar form.
3. Split wherever the assignment changes. This is 8.6, in polar form.

Nothing in this topic is new except the coordinate system. The integrand is 8.11's, the interval logic is 8.4's and 8.6's, and the retracing caution is 9.8's. That is a reasonable place for the unit to end.