Math Core

Many AP polar questions ask for the area of a region bounded by two curves: inside one and outside the other, or inside both. The formula is a small step from the single-curve formula. The real work is finding where the curves meet and deciding which curve is farther from the origin.

Outer minus inner

Picture a thin wedge from the origin at angle $\theta$ . If the region lies between an inner curve $r_{\text{in}}$ and an outer curve $r_{\text{out}}$ , the wedge's area is the big sector minus the small sector:

\frac{1}{2}r_{\text{out}}^2\,\Delta\theta - \frac{1}{2}r_{\text{in}}^2\,\Delta\theta.

Adding up the wedges gives the formula.

Area between two polar curves

If $0 \le r_{\text{in}}(\theta) \le r_{\text{out}}(\theta)$ for $\alpha \le \theta \le \beta$ , the area of the region between them is

A = \frac{1}{2}\int_{\alpha}^{\beta} \left(r_{\text{out}}^2 - r_{\text{in}}^2\right) d\theta.

Common mistake

Square each radius separately. $\dfrac{1}{2}\displaystyle\int (r_{\text{out}} - r_{\text{in}})^2\, d\theta$ is wrong: the region is a difference of two sectors, and $(a - b)^2 \ne a^2 - b^2$ .

The procedure

Sketch both curves and identify the region.
Find the intersection angles by setting the two $r$ expressions equal and solving for $\theta$ . Check the origin separately, since two curves can both pass through the pole at different angles.
Decide which curve is outer on each part of the interval. Test an angle in between.
Integrate $\dfrac{1}{2}(r_{\text{out}}^2 - r_{\text{in}}^2)$ , splitting the interval where the roles change. Use symmetry when you can.

Worked example: Inside a cardioid, outside a circle

Find the area of the region inside $r = 1 + \cos\theta$ and outside $r = 1$ .

Intersections: $1 + \cos\theta = 1$ gives $\cos\theta = 0$ , so $\theta = \pm\dfrac{\pi}{2}$ .

Outer curve: for $-\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2}$ , $\cos\theta > 0$ so the cardioid is outside the circle. That's the region.

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

Here $\displaystyle\int_{-\pi/2}^{\pi/2} 2\cos\theta\, d\theta = 4$ and $\displaystyle\int_{-\pi/2}^{\pi/2} \cos^2\theta\, d\theta = \dfrac{\pi}{2}$ .

The cardioid r = 1 + cos θ and the unit circle r = 1 meet at θ = ±π/2, the points (0, ±1). The region inside the cardioid and outside the circle is the crescent on the right.Open in grapher →

Regions inside both curves

When the region is inside both curves, the boundary switches from one curve to the other at the intersection. You don't subtract; you add two single-curve areas, each using whichever curve is the boundary on its own interval.

Worked example: The overlap of two circles

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

Intersections: $2\sin\theta = 2\cos\theta$ gives $\theta = \dfrac{\pi}{4}$ . Both circles also pass through the origin.

Boundary: from $\theta = 0$ to $\dfrac{\pi}{4}$ , the edge of the overlap farthest from the origin is on $r = 2\sin\theta$ (the smaller value). From $\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$ , it's on $r = 2\cos\theta$ . The two halves are mirror images across the line $y = x$ , so

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

The circles r = 2 sin θ and r = 2 cos θ overlap in a lens between the origin and the point (1, 1).Open in grapher →

Tip

A quick way to decide the boundary for "inside both": at each angle, the region extends from the origin out to the nearer curve, the smaller of the two $r$ values. For "inside one, outside the other," the region runs from the curve you're outside of to the curve you're inside of.

Worked example: Inside a circle, outside a cardioid

Find the area inside $r = 3\cos\theta$ and outside $r = 1 + \cos\theta$ .

Intersections: $3\cos\theta = 1 + \cos\theta$ gives $\cos\theta = \dfrac{1}{2}$ , so $\theta = \pm\dfrac{\pi}{3}$ . At $\theta = 0$ , $3 > 2$ , so the circle is outer between the intersections.

A = \frac{1}{2}\int_{-\pi/3}^{\pi/3} \left[9\cos^2\theta - (1 + \cos\theta)^2\right] d\theta = \frac{1}{2}\int_{-\pi/3}^{\pi/3} \left(8\cos^2\theta - 2\cos\theta - 1\right) d\theta.

Using $8\cos^2\theta = 4 + 4\cos 2\theta$ , the integrand becomes $3 + 4\cos 2\theta - 2\cos\theta$ , so

A = \frac{1}{2}\Big[3\theta + 2\sin 2\theta - 2\sin\theta\Big]_{-\pi/3}^{\pi/3} = \frac{1}{2}\left(2\pi + 2\sqrt{3} - 2\sqrt{3}\right) = \pi.

Calculator-active area problems

When the intersection angles aren't nice, the AP exam lets you use a calculator. Find the intersections numerically and store them rather than retyping rounded values. Your written work should still show the integral with its limits and integrand; the calculator only supplies the final number.

Worked example: Intersections found numerically

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

Intersections: solve $4\sin\theta = 2 + \cos\theta$ on a calculator. In $0 \le \theta \le \pi$ the solutions are $\theta = a \approx 0.7514$ and $\theta = b \approx 2.8801$ .

Outer curve: at $\theta = \dfrac{\pi}{2}$ , $4\sin\theta = 4$ and $2 + \cos\theta = 2$ , so the circle is outer between $a$ and $b$ .

A = \frac{1}{2}\int_a^b \left[16\sin^2\theta - (2 + \cos\theta)^2\right] d\theta \approx 7.755.

Practice

Practice 1

Find the area of the region inside $r = 3$ and outside $r = 2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

The curves $r = 1$ and $r = 2\sin\theta$ intersect at two points with $0 \le \theta \le \pi$ . Find both values of $\theta$ .

Separate answers with commas, e.g. 2, -5

Practice 3

Find the area of the region inside $r = 2\sin\theta$ and outside $r = 1$ . Give an exact answer or a decimal to three places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Which expression gives the area of the region inside $r = 2$ and outside $r = 2 - 2\cos\theta$ ?

Practice 5

Find the area of the region inside $r = 4\cos\theta$ and outside $r = 2$ . Give an exact answer or a decimal to three places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Find the area of the region inside both $r = 1 + \cos\theta$ and $r = 1 - \cos\theta$ . Give an exact answer or a decimal to three places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.