Area Inside a Polar Curve

In rectangular coordinates, area is built from thin vertical rectangles: $\int_a^b f(x)\,dx$. A polar region fans out from the pole, so it is built from thin wedges instead. This is the topic where the most points were lost on the 2025 exam, and the losses had three causes: $r$ not squared, the $\frac{1}{2}$ dropped, and wrong limits.

Where $\frac{1}{2}\int r^2\,d\theta$ comes from

A slice of a circle of radius $r$ with angle $\Delta\theta$ is the fraction $\frac{\Delta\theta}{2\pi}$ of the whole disk, so its area is

$$\frac{\Delta\theta}{2\pi} \cdot \pi r^2 = \frac{1}{2}r^2\,\Delta\theta.$$

Now take the region swept by the radius to $r = f(\theta)$ as $\theta$ goes from $\alpha$ to $\beta$. Cut it into thin wedges. In each one, $r$ barely changes, so the wedge is nearly a slice of a circle. Add the slices and you have a Riemann sum; let $\Delta\theta \to 0$:

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

The figure shows the three steps on the cardioid $r = 2 + 2\cos\theta$. Because the integrand is a square, it is never negative, so a polar area integral is never negative either.

A whole closed curve

The identities $\cos^2 u$$= \frac{1 + \cos(2u)}{2}$ and $\sin^2 u$$= \frac{1 - \cos(2u)}{2}$ do most of the work on no-calculator polar areas. Over a full period, $\int \cos(ku)$ and $\int \sin(ku)$ are 0.

Choosing the limits

The limits must sweep the region exactly once. Three habits help:

• A petal or a loop starts and ends at the pole, so its limits are consecutive zeros of $r$. For $r = 3\sin(2\theta)$, $r = 0$ at $\theta = 0$ and $\frac{\pi}{2}$, so one petal is $\frac{1}{2}\int_0^{\pi/2} 9\sin^2(2\theta)\,d\theta$$= \frac{9\pi}{8}$.
• A region bounded by rays uses those rays' angles: the region swept from $\theta = 0$ to $\theta = \frac{\pi}{2}$ by the cardioid is $\frac{1}{2}\int_0^{\pi/2} (2 + 2\cos\theta)^2\,d\theta$$= \frac{3\pi + 8}{2}$.
• Watch for retracing. The circle $r = 4\sin\theta$ is traced once for $0 \le \theta \le \pi$. On $[\pi, 2\pi]$, $r < 0$ and the same circle is drawn again, so $\frac{1}{2}\int_0^{2\pi}$ gives twice the area.

Calculator questions and tables

On a calculator question, write the setup first, exactly as you will enter it, for example $\frac{1}{2}\int_0^{\pi} \left(1 + \theta\sin\theta\right)^2\,d\theta$$\approx 6.904$, in radian mode. A bald number earns nothing. If $r$ is given only by a table, approximate the integral with a Riemann or trapezoidal sum of the values $\frac{1}{2}r^2$, not $r$.