Topic 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve: compute the area swept by a polar curve r = f(theta) using the one-half r-squared integral (BC).

What this topic is asking

The College Board (Topic 9.8, BC only) computes the area enclosed by a polar curve $r = f(\theta)$. Because polar regions are swept out by a rotating radius rather than bounded by vertical strips, the area integral looks different from the Cartesian one: it sums tiny circular sectors, giving the $\frac{1}{2}r^2$ formula.

The sector formula and why one-half r-squared

Choosing the limits: the real difficulty

Most polar-area errors are limit errors, not integration errors. The integral $\frac{1}{2}\int r^2\,d\theta$ is only correct if the chosen angle interval sweeps the region exactly once. For a curve that passes through the origin, the natural limits are the consecutive angles where $r = 0$, which mark the start and end of a single loop or petal. For the four-petal rose $r = 2\sin(2\theta)$, $r = 0$ at $\theta = 0$ and $\theta = \frac{\pi}{2}$, so one petal is traced on $\left[0, \frac{\pi}{2}\right]$, as in the worked free-response question. Integrating over $[0, 2\pi]$ for that rose would quadruple-count or double-count depending on the petal structure, giving the wrong area. Always determine the $\theta$-range for one sweep first, then multiply by the number of identical pieces if you want the whole figure.

A worked area inside one loop

Power-reduction: the standard no-calculator tool

When $r$ involves $\sin\theta$ or $\cos\theta$, squaring it produces $\sin^2$ or $\cos^2$, which you cannot integrate directly. The fix is the power-reduction identities $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ and $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$, which turn the square into a constant plus a single cosine that integrates immediately. This is the workhorse for no-calculator polar-area questions, as both worked examples show. On calculator-active questions you can skip the identity and integrate $\frac{1}{2}r^2$ numerically, but recognizing the power-reduction step is essential whenever an exact answer is required.

Symmetry as a shortcut

Many polar curves are symmetric about an axis, which lets you integrate over half the region and double. The cardioid $r = 1 + \cos\theta$ is symmetric about the $x$-axis, so its area is $2\cdot\frac{1}{2}\int_0^{\pi}(1 + \cos\theta)^2\,d\theta$, often simpler than the full $[0, 2\pi]$ integral. Symmetry must be justified (the curve genuinely repeats), and you must double consistently. Used carefully, symmetry both reduces algebra and guards against the limit errors above, because the half-range is easier to pin to a single sweep.