Topic 9.9 Finding the Area of the Region Bounded by Two Polar Curves: find the area between two polar curves by subtracting one-half r-squared integrals over the correct angle interval, after finding intersections (BC).

What this topic is asking

The College Board (Topic 9.9, BC only) extends polar area to the region between two polar curves $r = f(\theta)$ (outer) and $r = g(\theta)$ (inner). As in Cartesian "area between curves," you subtract, but here you subtract the $\frac{1}{2}r^2$ integrals, and the crucial first step is finding where the curves intersect so you integrate over the right angle range.

The subtraction-of-squares formula

Finding intersections (and the origin trap)

The limits come from intersection angles, so solving $f(\theta) = g(\theta)$ is the first task. For $r = 2$ and $r = 2 + 2\cos\theta$, setting them equal gives $\cos\theta = 0$, so $\theta = \pm\frac{\pi}{2}$. A subtlety unique to polar coordinates is that two curves can intersect at the origin at different $\theta$-values, so the algebraic equation $f(\theta) = g(\theta)$ may miss the origin crossing. The origin is on a curve whenever $r = 0$ for some $\theta$, regardless of which $\theta$. On the AP exam you usually confirm intersections by sketching or, on calculator sections, by graphing both curves and reading the crossing angles, which catches origin intersections that pure algebra can overlook.

A worked area between curves

When the curves swap outer and inner

The single integral $\frac{1}{2}\int(f^2 - g^2)\,d\theta$ is valid only where the same curve stays outer. If the curves cross and the outer/inner roles swap, you must split the angle range at the crossing and use the correct ordering on each piece, exactly as in Cartesian area-between-curves problems where the top function changes. A common harder version asks for the area inside both curves (their intersection region), which is typically the inner curve on part of the range and the other curve on the rest, requiring a piecewise setup. Sketching the two curves and shading the target region is the surest way to decide where to split and which curve bounds each piece.

Subtracting squares, not curves

The most damaging error in this topic is computing $\frac{1}{2}\int(f - g)^2\,d\theta$ instead of $\frac{1}{2}\int(f^2 - g^2)\,d\theta$. These are different: the correct formula subtracts the areas of the outer and inner sectors, which means subtracting $f^2$ and $g^2$ separately, then halving. Writing $(f - g)^2$ subtracts the radii first and squares, which has no geometric meaning here and gives a wrong answer. Keep the structure "half of (outer squared minus inner squared)" fixed in mind, and expand $f^2$ and $g^2$ as separate terms rather than combining $f - g$ before squaring.