Finding the Area of the Region Bounded by Two Polar Curves BC only

At each angle the region between two polar curves is the gap between two sectors, so $A = \frac{1}{2}\int_{\alpha}^{\beta}(R^{2} - r^{2})\,d\theta$: two of 9.8's integrals subtracted, exactly as 8.11's washer was two disc integrals subtracted. It is never $\frac{1}{2}\int (R-r)^{2}\,d\theta$, and the same numbers make the case, since $R = 5$ and $r = 3$ give $16$ against $4$. The limits are the angles where the radii are equal: $3\cos\theta = 1 + \cos\theta$ gives $\theta = \pm\frac{\pi}{3}$, testing $\theta = 0$ puts the circle outside, and the area comes to exactly $\pi$.

Setting the radii equal can miss intersections, because a point has many polar names and the pole in particular is reached by each curve at its own angle, so it must be checked separately. Three failures against that $\pi$: subtracting the radii before squaring gives about $0.544$, omitting the inner curve gives about $6.661$, and integrating over $[0, 2\pi]$ gives about $9.425$ while describing no region at all. Requiring the answer to lie between zero and the outer curve's own area over the interval rules out the last immediately. Where the outer and inner roles change, the interval splits, which is 8.6 in polar form.