Math Core

For regions under $y = f(x)$ , you add up thin rectangles. Polar regions are shaped like fans spreading out from the origin, so rectangles are the wrong building block. Instead you add up thin sectors, pie slices with their point at the origin, and that leads to a new area formula.

From sectors to an integral

A sector of a circle with radius $r$ and central angle $\Delta\theta$ (in radians) has area

\frac{1}{2}r^2\,\Delta\theta.

(A full circle has $\Delta\theta = 2\pi$ , giving $\pi r^2$ .)

Now take a polar curve $r = f(\theta)$ between the rays $\theta = \alpha$ and $\theta = \beta$ . Split the angle into thin wedges of width $\Delta\theta$ . Over one thin wedge, $r$ barely changes, so the region is almost a sector of radius $f(\theta)$ with area about $\dfrac{1}{2}f(\theta)^2\,\Delta\theta$ . Add them up and let $\Delta\theta \to 0$ .

Area of a polar region

The area of the region bounded by $r = f(\theta)$ and the rays $\theta = \alpha$ and $\theta = \beta$ (with $\alpha \lt \beta$ and the region swept out exactly once) is

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

Two things to notice. First, the $\dfrac{1}{2}$ comes from the sector formula; don't drop it. Second, the integrand is $r^2$ , which is never negative, so this formula never produces a negative area even when $r$ itself is negative.

The trig identities you'll need

Squaring $r$ usually produces $\sin^2$ or $\cos^2$ terms. Integrate them with the power-reducing identities:

\cos^2 u = \frac{1 + \cos(2u)}{2}, \qquad \sin^2 u = \frac{1 - \cos(2u)}{2}.

A useful consequence: over any interval that covers a whole number of periods of $\cos(2u)$ , both $\cos^2 u$ and $\sin^2 u$ average to $\dfrac{1}{2}$ .

Worked example: Area inside a cardioid

Find the area enclosed by $r = 1 + \cos\theta$ .

The cardioid is traced once as $\theta$ goes from $0$ to $2\pi$ .

\begin{aligned} A &= \frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2\, d\theta = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \cos^2\theta\right) d\theta \\ &= \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \frac{1 + \cos 2\theta}{2}\right) d\theta \\ &= \frac{1}{2}\left(2\pi + 0 + \pi + 0\right) = \frac{3\pi}{2}. \end{aligned}

The $\cos\theta$ and $\cos 2\theta$ terms integrate to $0$ over a full period.

Choosing the limits

The hardest part of most polar area problems is finding $\alpha$ and $\beta$ . The limits are the angles where the region starts and stops being swept out. Often that's where the curve passes through the origin, found by solving $r = 0$ .

Worked example: One petal of a rose

Find the area of one petal of $r = 3\sin(2\theta)$ .

The curve is at the origin when $\sin(2\theta) = 0$ , that is at $\theta = 0, \dfrac{\pi}{2}, \pi, \ldots$ Between $\theta = 0$ and $\theta = \dfrac{\pi}{2}$ , $r \ge 0$ and the curve traces one full petal in the first quadrant.

A = \frac{1}{2}\int_0^{\pi/2} 9\sin^2(2\theta)\, d\theta = \frac{9}{2}\int_0^{\pi/2} \frac{1 - \cos 4\theta}{2}\, d\theta = \frac{9}{4}\left[\theta - \frac{\sin 4\theta}{4}\right]_0^{\pi/2} = \frac{9}{4}\cdot\frac{\pi}{2} = \frac{9\pi}{8}.

The four-petal rose r = 3 sin(2θ). The first-quadrant petal is traced for 0 ≤ θ ≤ π/2 and has area 9π/8.Open in grapher →

Worked example: The inner loop of a limaçon

Find the area inside the inner loop of $r = 1 + 2\cos\theta$ .

Solve $r = 0$ : $\cos\theta = -\dfrac{1}{2}$ , so $\theta = \dfrac{2\pi}{3}$ and $\theta = \dfrac{4\pi}{3}$ . Between those angles $r \le 0$ , and the curve traces the small inner loop (on the right side of the pole, because $r$ is negative while $\theta$ points left).

\begin{aligned} A &= \frac{1}{2}\int_{2\pi/3}^{4\pi/3} (1 + 2\cos\theta)^2\, d\theta = \frac{1}{2}\int_{2\pi/3}^{4\pi/3} \left(3 + 4\cos\theta + 2\cos 2\theta\right) d\theta \\ &= \frac{1}{2}\Big[3\theta + 4\sin\theta + \sin 2\theta\Big]_{2\pi/3}^{4\pi/3} = \frac{1}{2}\left(2\pi - 4\sqrt{3} + \sqrt{3}\right) = \pi - \frac{3\sqrt{3}}{2}. \end{aligned}

That's about $0.543$ . Here $1 + 4\cos\theta + 4\cos^2\theta$ was rewritten as $3 + 4\cos\theta + 2\cos 2\theta$ using $4\cos^2\theta = 2 + 2\cos 2\theta$ .

The limaçon r = 1 + 2 cos θ. The small inner loop is traced for 2π/3 ≤ θ ≤ 4π/3.Open in grapher →

Common mistake

Integrating over $0$ to $2\pi$ isn't always right. For $r = 3\sin(2\theta)$ , the interval $0$ to $2\pi$ covers all four petals, and for curves like $r = 2\cos\theta$ (a circle), $0$ to $2\pi$ traces the circle twice and doubles the area. Sketch the curve, or check where $r = 0$ , to find an interval that sweeps the region exactly once.

Tip

Use symmetry to simplify limits. For the cardioid above, you could compute $2 \cdot \dfrac{1}{2}\displaystyle\int_0^{\pi}(1 + \cos\theta)^2\, d\theta$ , since the top and bottom halves are mirror images.