Area in Polar Coordinates

For $y = f(x)$, you found area by stacking thin rectangles. Polar curves sweep out from the pole like the hands of a clock, so the natural slices are thin pie wedges instead. That gives a new area formula. Once you have it, the main skill is choosing the right $\theta$-limits, which means knowing exactly how the curve is traced (see polar coordinates and derivatives).

Key ideas

Where the formula comes from

A sector (pie wedge) of a circle with radius $r$ and angle $\Delta\theta$ (in radians) has area $\dfrac{1}{2}r^2\Delta\theta$. For a polar curve $r = f(\theta)$, the radius changes with $\theta$, but over a tiny angle it’s nearly constant. Add up the thin wedges and let $\Delta\theta \to 0$:

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

This is the area swept out by the segment from the pole to the curve as $\theta$ goes from $\alpha$ to $\beta$. The formula needs radians, and the region should be swept out exactly once (no overlaps).

Choosing the limits

• For a whole closed curve, use one full trip around it: often $0$ to $2\pi$, but a circle like $r = 2a\sin\theta$ and a rose with an odd number of petals are each traced once on $[0, \pi]$.
• For one petal or loop, find the angles where $r = 0$ on either side of it. The petal starts and ends at the pole.
• Use symmetry when it helps: find half (or one petal) and multiply.

Squares of sine and cosine

Squaring $r$ often produces $\sin^2$ or $\cos^2$. To integrate them, use the power-reducing identities:

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

One petal of $r = 2\sin 3\theta$ (Example 2) is swept out as $\theta$ goes from $0$ to $\frac{\pi}{3}$, starting and ending at the pole.

Area between two polar curves

If $R(\theta) \ge r(\theta) \ge 0$ for $\alpha \le \theta \le \beta$, the area between them (outside $r$, inside $R$) is

$$A = \frac{1}{2}\int_\alpha^\beta \Big( \big( R(\theta) \big)^2 - \big( r(\theta) \big)^2 \Big)\, d\theta$$

That’s outer area minus inner area. Note it is $R^2 - r^2$, not $(R - r)^2$.

Finding intersections, including the pole

To find where two polar curves cross, solve $f(\theta) = g(\theta)$. But that can miss points! The pole is a common point whenever each curve passes through it, even if they reach it at different angles (for example, $r = 2\sin\theta$ is at the pole when $\theta = 0$, but $r = 2\cos\theta$ is there when $\theta = \frac{\pi}{2}$). Sketch the curves to see which intersections matter.

On the AP exam

Polar area usually appears on the calculator-active section. Write the integral with its limits and the $\frac{1}{2}$, then evaluate numerically and give $3$ decimal places. If the intersection angles aren’t nice, find them with your calculator and store them (don’t round to $3$ decimals before integrating).

Worked examples

Example 1: A circle, as a check

Find the area enclosed by $r = 4\sin\theta$.

Solution. This circle has radius $2$, so we expect $\pi(2)^2 = 4\pi$. It is traced once for $0 \le \theta \le \pi$ (then $r$ goes negative and the circle is traced again).

$$A = \frac{1}{2}\int_0^{\pi} 16\sin^2\theta\, d\theta = 8\int_0^{\pi} \frac{1 - \cos(2\theta)}{2}\, d\theta = 4\left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\pi} = 4\pi$$

Using $0$ to $2\pi$ would give $8\pi$, counting the circle twice.

Example 2: One petal of a rose

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

Solution. $r = 0$ when $\sin(3\theta) = 0$: at $\theta = 0$ and next at $\theta = \dfrac{\pi}{3}$. Between them $r \gt 0$, so one petal is traced for $0 \le \theta \le \dfrac{\pi}{3}$ (see the figure).

$$\begin{aligned} A &= \frac{1}{2}\int_0^{\pi/3} 4\sin^2(3\theta)\, d\theta = 2\int_0^{\pi/3} \frac{1 - \cos(6\theta)}{2}\, d\theta \\ &= \left[ \theta - \frac{\sin(6\theta)}{6} \right]_0^{\pi/3} = \frac{\pi}{3} - \frac{\sin(2\pi)}{6} = \frac{\pi}{3} \end{aligned}$$

The whole rose has $3$ petals, so its total area is $\pi$.

Example 3: Inside one curve, outside another

Find the area of the region inside $r = 4\cos\theta$ and outside $r = 2$.

Solution. Intersections: $4\cos\theta = 2$ gives $\cos\theta = \dfrac{1}{2}$, so $\theta = \pm\dfrac{\pi}{3}$. Between these angles, the circle $r = 4\cos\theta$ is the outer curve.

The region inside $r = 4\cos\theta$ and outside $r = 2$, between $\theta = -\frac{\pi}{3}$ and $\theta = \frac{\pi}{3}$.

The region is symmetric about the $x$-axis, so double the integral from $0$:

$$\begin{aligned} A &= 2 \cdot \frac{1}{2}\int_0^{\pi/3} \left( 16\cos^2\theta - 4 \right) d\theta = \int_0^{\pi/3} \big( 8 + 8\cos(2\theta) - 4 \big)\, d\theta \\ &= \Big[ 4\theta + 4\sin(2\theta) \Big]_0^{\pi/3} = \frac{4\pi}{3} + 4 \cdot \frac{\sqrt{3}}{2} = \frac{4\pi}{3} + 2\sqrt{3} \approx 7.653 \end{aligned}$$

Example 4: Inside both curves (don’t forget the pole)

Find the area of the region inside both $r = 2\sin\theta$ and $r = 2\cos\theta$.

Solution. Both are circles of radius $1$, one centred at $(0, 1)$ and one at $(1, 0)$. Solving $2\sin\theta = 2\cos\theta$ gives $\theta = \dfrac{\pi}{4}$, the point $(1, 1)$. They also both pass through the pole: that’s the other corner of the overlap.

The overlap is a lens. For $0 \le \theta \le \dfrac{\pi}{4}$ its boundary is $r = 2\sin\theta$ (the smaller $r$); for $\dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{2}$ it’s $r = 2\cos\theta$. By symmetry the two halves are equal:

$$\begin{aligned} A &= 2 \cdot \frac{1}{2}\int_0^{\pi/4} 4\sin^2\theta\, d\theta = 2\int_0^{\pi/4} \big( 1 - \cos(2\theta) \big)\, d\theta \\ &= 2\left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\pi/4} = 2\left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{\pi}{2} - 1 \approx 0.571 \end{aligned}$$

Common mistakes

Forgetting the one-half. The formula is $\frac{1}{2}\int r^2\, d\theta$. Without the $\frac{1}{2}$, every answer is doubled.

Forgetting to square r. $\frac{1}{2}\int r\, d\theta$ is not an area. Square the whole expression: $(1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta$, not $1 + \cos^2\theta$.

Using (R − r)² for the region between curves. The area between is $\frac{1}{2}\int \left( R^2 - r^2 \right) d\theta$: outer area minus inner area. Squaring the difference gives a different (wrong) number.

Going around twice. Circles through the pole and odd-petal roses are traced once on an interval of length $\pi$. Integrating over $[0, 2\pi]$ counts them twice. Check a sketch or a table of values.

Missing the pole as an intersection. Solving $f(\theta) = g(\theta)$ only finds crossings at the same angle. If both curves pass through the pole, it’s also an intersection, as in Example 4.

Degree mode. The formula assumes radians. In degree mode the calculator’s integral is meaningless.

Practice

1. (Warm-up) Use the polar area formula to find the area inside $r = 3$. Check with the circle area formula.

Solution

$$A = \frac{1}{2}\int_0^{2\pi} 9\, d\theta = \frac{9}{2}(2\pi) = 9\pi$$

Check: $\pi(3)^2 = 9\pi$.

2. (Warm-up) Find the area swept out by the spiral $r = \theta$ for $0 \le \theta \le \pi$.

Solution

$$A = \frac{1}{2}\int_0^{\pi} \theta^2\, d\theta = \frac{1}{2}\left[ \frac{\theta^3}{3} \right]_0^{\pi} = \frac{\pi^3}{6} \approx 5.168$$

3. (Warm-up) One petal of $r = \cos(2\theta)$ is traced for $-\dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{4}$. Write the integral for its area, then evaluate it.

Solution

$$A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \cos^2(2\theta)\, d\theta = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \frac{1 + \cos(4\theta)}{2}\, d\theta = \frac{1}{4}\left[ \theta + \frac{\sin(4\theta)}{4} \right]_{-\pi/4}^{\pi/4} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}$$

4. (Core) Find the area enclosed by the cardioid $r = 2 + 2\cos\theta$.

Solution

The cardioid is traced once for $0 \le \theta \le 2\pi$.

$$(2 + 2\cos\theta)^2 = 4 + 8\cos\theta + 4\cos^2\theta = 4 + 8\cos\theta + 2 + 2\cos(2\theta)$$$$A = \frac{1}{2}\int_0^{2\pi} \big( 6 + 8\cos\theta + 2\cos(2\theta) \big)\, d\theta = \frac{1}{2}\Big[ 6\theta + 8\sin\theta + \sin(2\theta) \Big]_0^{2\pi} = \frac{1}{2}(12\pi) = 6\pi$$

5. (Core) Find the area of one petal of $r = 3\cos(2\theta)$, and the total area of the rose.

Solution

$r = 0$ at $\theta = \pm\dfrac{\pi}{4}$, so one petal is traced for $-\dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{4}$:

$$A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} 9\cos^2(2\theta)\, d\theta = \frac{9}{4}\left[ \theta + \frac{\sin(4\theta)}{4} \right]_{-\pi/4}^{\pi/4} = \frac{9}{4} \cdot \frac{\pi}{2} = \frac{9\pi}{8}$$

With $n = 2$ there are $4$ petals, so the total area is $4 \cdot \dfrac{9\pi}{8} = \dfrac{9\pi}{2}$.

6. (Core) Find the area of the region inside the cardioid $r = 1 + \cos\theta$ and outside the circle $r = 1$.

Solution

Intersections: $1 + \cos\theta = 1$ gives $\cos\theta = 0$, so $\theta = \pm\dfrac{\pi}{2}$. For $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$, $\cos\theta \ge 0$ and the cardioid is outside the circle.

$$\begin{aligned} A &= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \Big( (1 + \cos\theta)^2 - 1 \Big)\, d\theta = \frac{1}{2}\int_{-\pi/2}^{\pi/2} \left( 2\cos\theta + \cos^2\theta \right) d\theta \\ &= \frac{1}{2}\left[ 2\sin\theta + \frac{\theta}{2} + \frac{\sin(2\theta)}{4} \right]_{-\pi/2}^{\pi/2} = \frac{1}{2}\left( 4 + \frac{\pi}{2} \right) = 2 + \frac{\pi}{4} \approx 2.785 \end{aligned}$$

7. (Core) (Calculator active.) Let $R$ be the region in the first quadrant that is inside the circle $r = 4\cos\theta$ and outside the spiral $r = 1 + \theta$ (for $\theta \ge 0$).

• (a) Find the value of $\theta$ where the curves intersect in the first quadrant.
• (b) Find the area of $R$.

Solution

(a) Solve $4\cos\theta = 1 + \theta$ with a calculator (radian mode): $\theta = a \approx 1.037$. Store the full value.

(b) At $\theta = 0$ the circle ($r = 4$) is outside the spiral ($r = 1$), and it stays outside until $\theta = a$. So

$$A = \frac{1}{2}\int_0^{a} \Big( (4\cos\theta)^2 - (1 + \theta)^2 \Big)\, d\theta \approx 4.658$$

8. (Challenge) The limaçon $r = 1 + 2\cos\theta$ has an inner loop. Find the exact area inside the inner loop.

Solution

The inner loop is traced while $r \le 0$, between the angles where $r = 0$: $\cos\theta = -\dfrac{1}{2}$, so $\theta = \dfrac{2\pi}{3}$ to $\dfrac{4\pi}{3}$. (Squaring makes the negative $r$ values harmless.)

$$(1 + 2\cos\theta)^2 = 1 + 4\cos\theta + 4\cos^2\theta = 3 + 4\cos\theta + 2\cos(2\theta)$$$$A = \frac{1}{2}\Big[ 3\theta + 4\sin\theta + \sin(2\theta) \Big]_{2\pi/3}^{4\pi/3}$$

At $\dfrac{4\pi}{3}$: $4\pi - 2\sqrt{3} + \dfrac{\sqrt{3}}{2}$. At $\dfrac{2\pi}{3}$: $2\pi + 2\sqrt{3} - \dfrac{\sqrt{3}}{2}$. The difference is $2\pi - 3\sqrt{3}$, so

$$A = \pi - \frac{3\sqrt{3}}{2} \approx 0.544$$

9. (Challenge) Find the area of the region inside both the circle $r = 6\cos\theta$ and the cardioid $r = 2 + 2\cos\theta$.

Solution

Intersections: $6\cos\theta = 2 + 2\cos\theta$ gives $\cos\theta = \dfrac{1}{2}$, so $\theta = \pm\dfrac{\pi}{3}$. Both curves also pass through the pole.

Use the upper half and double it. For $0 \le \theta \le \dfrac{\pi}{3}$, the cardioid is the inner curve (at $\theta = 0$: $4 \lt 6$). For $\dfrac{\pi}{3} \le \theta \le \dfrac{\pi}{2}$, the circle is inner (it reaches the pole at $\theta = \frac{\pi}{2}$).

$$A = 2\left[ \frac{1}{2}\int_0^{\pi/3} (2 + 2\cos\theta)^2\, d\theta + \frac{1}{2}\int_{\pi/3}^{\pi/2} 36\cos^2\theta\, d\theta \right]$$

First integral: $(2 + 2\cos\theta)^2 = 6 + 8\cos\theta + 2\cos(2\theta)$, so

$$\int_0^{\pi/3} (2 + 2\cos\theta)^2\, d\theta = \Big[ 6\theta + 8\sin\theta + \sin(2\theta) \Big]_0^{\pi/3} = 2\pi + 4\sqrt{3} + \frac{\sqrt{3}}{2} = 2\pi + \frac{9\sqrt{3}}{2}$$

Second integral:

$$\int_{\pi/3}^{\pi/2} 36\cos^2\theta\, d\theta = 18\left[ \theta + \frac{\sin(2\theta)}{2} \right]_{\pi/3}^{\pi/2} = 18\left( \frac{\pi}{2} - \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) = 3\pi - \frac{9\sqrt{3}}{2}$$

Adding (the factor $2 \cdot \frac{1}{2} = 1$): $A = 2\pi + \dfrac{9\sqrt{3}}{2} + 3\pi - \dfrac{9\sqrt{3}}{2} = 5\pi \approx 15.708$.