Area in Polar Coordinates

Quick answer

A polar region is swept out by a turning radius, so it is sliced into thin sectors rather than rectangles. A sector of radius r and angle Δθ has area ½r²Δθ, and adding them gives A = ½∫ r² dθ between the starting and ending angles. For the region between two polar curves, subtract: A = ½∫ (R² − r²) dθ, with R the outer curve. The limits are angles, often found by setting the two equations equal, and a curve must be traced only once between them.

What you'll learn

Derive the polar area formula from thin sectors
Find the area enclosed by a polar curve
Find the area between two polar curves
Choose angle limits that trace a region exactly once

Slicing by angle

Rectangles fit regions under graphs $y = f(x)$ . A polar region is swept out by a radius turning from angle $\alpha$ to angle $\beta$ , so slice it into thin sectors, like pieces of a pie. A sector of radius $r$ and angle $\Delta\theta$ is the fraction $\tfrac{\Delta\theta}{2\pi}$ of a disk of area $\pi r^2$ , so its area is $\tfrac{1}{2}r^2\,\Delta\theta$ . Adding the sectors and taking the limit:

A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta

r = 1 + cos θ

The cardioid r = 1 + cos θ, sliced into sectors

Why the formula has a half

A thin sector is nearly a triangle: its two long sides have length $r$ , and its short side, an arc, has length about $r\,\Delta\theta$ . The area of a triangle is half the base times the height, and here that is $\tfrac{1}{2} \cdot r\,\Delta\theta \cdot r$ . Sectors are triangles, not rectangles, so their area carries the half from the triangle formula. Forgetting it doubles every polar area.

The region between two curves

For the region between an outer curve $R(\theta)$ and an inner curve $r(\theta)$ , subtract the inner sectors from the outer ones:

A = \frac{1}{2}\int_{\alpha}^{\beta} \left(R^2 - r^2\right)d\theta

The limits are usually the angles where the curves meet. Set the two equations equal to find them, and sketch to see which curve is outside.

Worked examples

Example 3 — Inside the cardioid, outside the circle

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

r = 1
r = 1 + cos θ

Inside the cardioid and outside the circle r = 1

The curves meet where $1 + \cos\theta = 1$ , so $\cos\theta = 0$ : $\theta = \pm\tfrac{\pi}{2}$ . Between those angles the cardioid is outside:

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)

That is $2 + \tfrac{\pi}{4} \approx 2.79$ .

Common mistakes

Practice problems

Find the area inside $r = 3$ with the polar area formula.

Answer

$9\pi$

Full solution

$\tfrac{1}{2}\int_0^{2\pi} 9\,d\theta = 9\pi$ .
Find the area inside $r = 2\cos\theta$ .

Answer

$\pi$

Full solution

The circle is traced once for $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$ . $\tfrac{1}{2}\int_{-\pi/2}^{\pi/2} 4\cos^2\theta\,d\theta = 2 \cdot \tfrac{\pi}{2} = \pi$ : a circle of radius $1$ .
Find the area of one petal of $r = \sin 3\theta$ .

Answer

$\tfrac{\pi}{12}$

Full solution

One petal runs from $\theta = 0$ to $\tfrac{\pi}{3}$ . $\tfrac{1}{2}\int_0^{\pi/3} \sin^2 3\theta\,d\theta = \tfrac{1}{2} \cdot \tfrac{\pi}{6}$ .
Find the area inside $r = 1 + \sin\theta$ .

Answer

$\tfrac{3\pi}{2}$

Full solution

As in Example 1: $\tfrac{1}{2}\int_0^{2\pi} \left(1 + 2\sin\theta + \sin^2\theta\right)d\theta = \tfrac{1}{2}(2\pi + 0 + \pi)$ . It is the cardioid turned upward.
Find the area swept by the spiral $r = \theta$ as $\theta$ runs from $0$ to $\pi$ .

Answer

$\tfrac{\pi^3}{6} \approx 5.17$

Full solution

$\tfrac{1}{2}\int_0^{\pi} \theta^2\,d\theta = \tfrac{1}{2} \cdot \tfrac{\pi^3}{3}$ .
Find the area inside $r = 3\sin\theta$ and outside $r = 1 + \sin\theta$ .

Answer

$\pi$

Full solution

The curves meet where $3\sin\theta = 1 + \sin\theta$ , so $\sin\theta = \tfrac{1}{2}$ : $\theta = \tfrac{\pi}{6}$ and $\tfrac{5\pi}{6}$ . Between them, $\tfrac{1}{2}\int \left(9\sin^2\theta - (1 + \sin\theta)^2\right)d\theta = \tfrac{1}{2}\int_{\pi/6}^{5\pi/6} \left(8\sin^2\theta - 2\sin\theta - 1\right)d\theta$ .

The three pieces integrate to $8\left(\tfrac{\pi}{3} + \tfrac{\sqrt{3}}{4}\right)$ , $-2\sqrt{3}$ and $-\tfrac{2\pi}{3}$ , which add to $2\pi$ . Half of that is $\pi$ .
At which angles do $r = 1$ and $r = 2\cos\theta$ meet?

Answer

$\theta = \pm\tfrac{\pi}{3}$

Full solution

$2\cos\theta = 1$ gives $\cos\theta = \tfrac{1}{2}$ .
Find the area inside both $r = 1$ and $r = 2\cos\theta$ .

Answer

$\tfrac{2\pi}{3} - \tfrac{\sqrt{3}}{2} \approx 1.23$

Full solution

By symmetry, double the upper half. For $0 \le \theta \le \tfrac{\pi}{3}$ the boundary is the circle $r = 1$ ; for $\tfrac{\pi}{3} \le \theta \le \tfrac{\pi}{2}$ it is $r = 2\cos\theta$ .

$2\left(\tfrac{1}{2}\int_0^{\pi/3} 1\,d\theta + \tfrac{1}{2}\int_{\pi/3}^{\pi/2} 4\cos^2\theta\,d\theta\right) = \tfrac{\pi}{3} + \big[2\theta + \sin 2\theta\big]_{\pi/3}^{\pi/2} = \tfrac{\pi}{3} + \pi - \tfrac{2\pi}{3} - \tfrac{\sqrt{3}}{2}$ .
Find the total area of the four petals of $r = \cos 2\theta$ .

Answer

$\tfrac{\pi}{2}$

Full solution

Four petals of area $\tfrac{\pi}{8}$ each, from Example 2. Directly, $\tfrac{1}{2}\int_0^{2\pi} \cos^2 2\theta\,d\theta = \tfrac{1}{2} \cdot \pi$ .
A student finds the area inside $r = \cos 2\theta$ as $\tfrac{1}{2}\int_0^{2\pi} \cos 2\theta\,d\theta = 0$ . What went wrong?

Hint

What is the area of a sector of radius $r$ ?

Answer

The student forgot to square $r$ . The area is $\tfrac{\pi}{2}$ .

Full solution

The integrand is $r^2 = \cos^2 2\theta$ , never negative, so the area cannot come out $0$ .

$\tfrac{1}{2}\int_0^{2\pi} \cos^2 2\theta\,d\theta = \tfrac{\pi}{2}$ , as in problem 9.

Frequently asked questions

What is the formula for area in polar coordinates?

A = ½ ∫ from α to β of r² dθ, where the radius sweeps from angle α to angle β and traces the region once.

Why is there a ½ in the polar area formula?

The region is built from thin sectors, and a sector of radius r and angle Δθ has area ½r²Δθ, not r²Δθ.

How do I find the area between two polar curves?

Use ½ ∫ (R² − r²) dθ, where R is the outer curve and r the inner one, between the angles where the curves meet.

How do I find the limits of integration?

Find where the curve returns to the origin (r = 0) or where two curves meet (set them equal), and check that the region is traced once between the limits.

Can I subtract the radii before squaring?

No. ½∫(R − r)² dθ is not the area between the curves. Square each radius, then subtract.

What to learn next

Sequences, Series and Geometric SeriesWhat it means for an infinite sum to converge: sequences and their limits, partial sums, geometric series, and the nth-term test for divergence.

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