You know how to graph polar curves like cardioids and roses. To find the area they enclose, slice the region the way polar coordinates see it: not into vertical strips, but into thin wedges fanning out from the origin. Then add the wedges with a definite integral.
Why it works
A sector of a circle with radius $r$ and angle $\Delta\theta$ (in radians) is the fraction $\frac{\Delta\theta}{2\pi}$ of the disk, so its area is $\frac{\Delta\theta}{2\pi}\cdot\pi r^{2} = \frac{1}{2}r^{2}\,\Delta\theta$. Cut the region between $\theta = \alpha$ and $\theta = \beta$ into thin wedges; each is close to a sector with radius $r(\theta)$. Adding them:
$$A = \lim\sum \tfrac{1}{2}r(\theta_i)^{2}\,\Delta\theta = \frac{1}{2}\int_{\alpha}^{\beta} r(\theta)^{2}\,d\theta.$$
Key idea
$$A = \frac{1}{2}\int_{\alpha}^{\beta} r^{2}\,d\theta,$$
where the ray sweeps the region exactly once as $\theta$ goes from $\alpha$ to $\beta$. Between two curves: $A = \frac{1}{2}\int_{\alpha}^{\beta}\left(r_{\text{outer}}^{2} - r_{\text{inner}}^{2}\right)d\theta$.
Worked example: A cardioid
Find the area inside $r = 1 + \cos\theta$.
- The curve is traced once for $0 \le \theta \le 2\pi$.
- $r^{2} = 1 + 2\cos\theta + \cos^{2}\theta$, and $\cos^{2}\theta = \frac{1 + \cos 2\theta}{2}$.
- $$\begin{aligned} A &= \frac{1}{2}\int_0^{2\pi}\left(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\right)d\theta \\ &= \frac{1}{2}\cdot\frac{3}{2}\cdot 2\pi \\ &= \frac{3\pi}{2}. \end{aligned}$$
Answer: $\frac{3\pi}{2}$
Worked example: One petal
Find the area of one petal of $r = \cos 3\theta$.
- $r = 0$ when $3\theta = \pm\frac{\pi}{2}$, so one petal runs over $-\frac{\pi}{6} \le \theta \le \frac{\pi}{6}$.
- $$\begin{aligned} A &= \frac{1}{2}\int_{-\pi/6}^{\pi/6}\cos^{2}3\theta\,d\theta \\ &= \frac{1}{2}\int_{-\pi/6}^{\pi/6}\frac{1 + \cos 6\theta}{2}\,d\theta. \end{aligned}$$
- $= \frac{1}{4}\cdot\frac{\pi}{3} + 0 = \frac{\pi}{12}$.
Answer: $\frac{\pi}{12}$; the whole three-petal rose has area $\frac{\pi}{4}$.
Common mistake
Using limits that trace part of the curve twice. The circle $r = 2\cos\theta$ is traced completely for $0 \le \theta \le \pi$; from $\pi$ to $2\pi$ it is traced again (with $r < 0$). So $\frac{1}{2}\int_0^{2\pi}4\cos^{2}\theta\,d\theta = 2\pi$, double the true area $\pi$ of this circle of radius 1. Check: find where $r = 0$, sketch the curve, and compare with the area of a shape you know.
Worked example: Between two curves
Find the area inside $r = 3\sin\theta$ and outside $r = 1 + \sin\theta$.
- They meet where $3\sin\theta = 1 + \sin\theta$, so $\sin\theta = \frac{1}{2}$: $\theta = \frac{\pi}{6}$ and $\frac{5\pi}{6}$.
- $$\begin{aligned} A &= \tfrac{1}{2}\int_{\pi/6}^{5\pi/6}\left[9\sin^{2}\theta - (1 + \sin\theta)^{2}\right]d\theta \\ &= \tfrac{1}{2}\int_{\pi/6}^{5\pi/6}\left(8\sin^{2}\theta - 2\sin\theta - 1\right)d\theta. \end{aligned}$$
- The three pieces are $\frac{8\pi}{3} + 2\sqrt{3}$, $2\sqrt{3}$ and $\frac{2\pi}{3}$, so $$\begin{aligned} A &= \frac{1}{2}\left(\frac{8\pi}{3} + 2\sqrt{3} - 2\sqrt{3} - \frac{2\pi}{3}\right) \\ &= \pi. \end{aligned}$$
Answer: $\pi$