Area Between Polar Curves

The area between polar curves is A = ½∫ (R² − r²) dθ, where R is the outer curve, r is the inner curve, and the limits are the angles where the curves meet. Find those angles by setting the two equations equal. Square each radius before you subtract.

Before this lesson: Area in Polar Coordinates, Solving Trig Equations and Area Between Curves

The idea: outer wedge minus inner wedge

Each thin wedge of the region starts at the inner curve and ends at the outer curve. Its area is the big wedge, ½R²Δθ, minus the small wedge, ½r²Δθ.

Adding those pieces gives ½∫ (R² − r²) dθ. It works like the washer method: you subtract squares, not the radii first. The limits are where the outer and inner curves cross.

Area between polar curves formula

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

1. Sketch both curves and shade the region.
2. Find where they meet: set the equations equal and solve for θ. Also check the pole (r = 0), which this can miss.
3. On each interval, decide which curve is outer (farther from the origin).
4. If the outer and inner curves switch, split the integral at the crossing.
5. Square each radius, subtract, integrate, and multiply by ½. Use symmetry to save work.

Area between polar curves examples

Example 1: Inside a cardioid and outside a circle

Find the area inside r = 2 + 2 cos θ and outside r = 2.

2 + 2\cos\theta = 2

\cos\theta = 0 \Rightarrow \theta = \pm\dfrac{\pi}{2}

A = \dfrac{1}{2}\int_{-\pi/2}^{\pi/2} \left[(2 + 2\cos\theta)^2 - 2^2\right]d\theta

(2 + 2\cos\theta)^2 - 4 = 8\cos\theta + 4\cos^2\theta

\int_{-\pi/2}^{\pi/2} 8\cos\theta\,d\theta = \left[8\sin\theta\right]_{-\pi/2}^{\pi/2} = 16

\int_{-\pi/2}^{\pi/2} 4\cos^2\theta\,d\theta = \left[2\theta + \sin 2\theta\right]_{-\pi/2}^{\pi/2} = 2\pi

A = \dfrac{1}{2}(16 + 2\pi) = 8 + \pi

Answer: 8 + π ≈ 11.14.

Example 2: Inside one circle, outside a cardioid

Find the area inside r = 3 sin θ and outside r = 1 + sin θ.

3\sin\theta = 1 + \sin\theta

\sin\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}

A = \dfrac{1}{2}\int_{\pi/6}^{5\pi/6} \left[9\sin^2\theta - (1 + \sin\theta)^2\right]d\theta

9\sin^2\theta - (1 + 2\sin\theta + \sin^2\theta) = 8\sin^2\theta - 2\sin\theta - 1

8\sin^2\theta - 2\sin\theta - 1 = 3 - 4\cos 2\theta - 2\sin\theta

\int_{\pi/6}^{5\pi/6} (3 - 4\cos 2\theta - 2\sin\theta)\,d\theta = \left[3\theta - 2\sin 2\theta + 2\cos\theta\right]_{\pi/6}^{5\pi/6}

\left(\dfrac{5\pi}{2} + \sqrt{3} - \sqrt{3}\right) - \left(\dfrac{\pi}{2} - \sqrt{3} + \sqrt{3}\right) = 2\pi

A = \dfrac{1}{2}(2\pi) = \pi

Answer: π ≈ 3.14. The square roots cancel, which is a good sign but not something to expect every time.

Example 3: Inside both curves

Find the area inside both r = 3 cos θ and r = 1 + cos θ.

3\cos\theta = 1 + \cos\theta \Rightarrow \cos\theta = \dfrac{1}{2}

\int_0^{\pi/3} (1 + \cos\theta)^2\,d\theta = \left[\dfrac{3\theta}{2} + 2\sin\theta + \dfrac{\sin 2\theta}{4}\right]_0^{\pi/3}

\dfrac{\pi}{2} + \sqrt{3} + \dfrac{\sqrt{3}}{8} = \dfrac{\pi}{2} + \dfrac{9\sqrt{3}}{8}

\int_{\pi/3}^{\pi/2} 9\cos^2\theta\,d\theta = \left[\dfrac{9\theta}{2} + \dfrac{9\sin 2\theta}{4}\right]_{\pi/3}^{\pi/2}

\dfrac{9\pi}{4} - \left(\dfrac{3\pi}{2} + \dfrac{9\sqrt{3}}{8}\right) = \dfrac{3\pi}{4} - \dfrac{9\sqrt{3}}{8}

A = 2 \cdot \dfrac{1}{2}\left(\dfrac{\pi}{2} + \dfrac{3\pi}{4}\right)

A = \dfrac{5\pi}{4}

Answer: 5π/4 ≈ 3.93. When the boundary switches from one curve to the other, split the integral at the crossing.

Example 4: Between the loops of a limaçon

The limaçon r = 1 + 2 cos θ has an inner loop. Find the area between the inner and outer loops.

1 + 2\cos\theta = 0 \Rightarrow \theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}

\int (1 + 2\cos\theta)^2\,d\theta = 3\theta + 4\sin\theta + \sin 2\theta

A_{\text{inner}} = \dfrac{1}{2}\left[\left(4\pi - 2\sqrt{3} + \tfrac{\sqrt{3}}{2}\right) - \left(2\pi + 2\sqrt{3} - \tfrac{\sqrt{3}}{2}\right)\right]

A_{\text{inner}} = \dfrac{1}{2}\left(2\pi - 3\sqrt{3}\right) = \pi - \dfrac{3\sqrt{3}}{2}

A_{\text{outer}} = \dfrac{1}{2}\left[3\theta + 4\sin\theta + \sin 2\theta\right]_{-2\pi/3}^{2\pi/3} = 2\pi + \dfrac{3\sqrt{3}}{2}

A = A_{\text{outer}} - A_{\text{inner}} = \pi + 3\sqrt{3}

Answer: π + 3√3 ≈ 8.34. The inner loop alone has area π − 3√3/2 ≈ 0.544.

Example 5: A spotlight on a stage (word problem)

A round stage is the region inside r = 6, in meters. A spotlight lights the region inside the cardioid r = 4 + 4 cos θ. How much of the stage is lit?

4 + 4\cos\theta = 6 \Rightarrow \cos\theta = \dfrac{1}{2}

\dfrac{1}{2}\int_0^{\pi/3} 36\,d\theta = 6\pi

\int_{\pi/3}^{\pi} (4 + 4\cos\theta)^2\,d\theta = \left[24\theta + 32\sin\theta + 4\sin 2\theta\right]_{\pi/3}^{\pi}

24\pi - \left(8\pi + 16\sqrt{3} + 2\sqrt{3}\right) = 16\pi - 18\sqrt{3}

\dfrac{1}{2}\left(16\pi - 18\sqrt{3}\right) = 8\pi - 9\sqrt{3}

A = 2(6\pi + 8\pi - 9\sqrt{3}) = 28\pi - 18\sqrt{3}

A \approx 87.96 - 31.18 = 56.79

Answer: about 56.8 m² of the 36π ≈ 113.1 m² stage is lit, almost exactly half.

Common mistakes

Related methods

Area inside one polar curve

With no inner curve, r = 0 and the formula becomes ½∫ R² dθ from area in polar coordinates.

Rectangular area and washers

Area between curves subtracts heights, top minus bottom. The polar version subtracts squared radii, like the washer method subtracts squared radii for volume.

Arc length of polar curves

To find the length of a polar boundary instead of the area it encloses, use the arc length integral, the next lesson.

Practice problems

1. Find the area inside r = 2 and outside r = 1.
2. Find the area inside r = 4 cos θ and outside r = 2.
3. Find the area inside r = 1 + cos θ and outside r = 1.
4. Find the area inside r = 1 and outside r = 1 + cos θ.
5. Find all intersection points of r = sin θ and r = cos θ.
6. Find the area inside both r = sin θ and r = cos θ.
7. Find the area inside r = 2 sin θ and outside r = 1.
8. Find the area of the inner loop of r = 1 − 2 cos θ.
9. Find the area inside r = 3 + 2 sin θ and outside r = 2.
10. A round pond is the region inside r = 4, in meters. A fountain sprays water over the region inside r = 4 + 4 sin θ. How much area does the spray cover outside the pond?

A = \dfrac{1}{2}\int_0^{2\pi} (4 - 1)\,d\theta = 3\pi

4\cos\theta = 2 \Rightarrow \theta = \pm\dfrac{\pi}{3}

A = \dfrac{1}{2}\int_{-\pi/3}^{\pi/3} (16\cos^2\theta - 4)\,d\theta

A = \dfrac{1}{2}\left[4\theta + 4\sin 2\theta\right]_{-\pi/3}^{\pi/3}

A = \dfrac{4\pi}{3} + 2\sqrt{3} \approx 7.65

A = \dfrac{1}{2}\int_{-\pi/2}^{\pi/2} (2\cos\theta + \cos^2\theta)\,d\theta

A = \dfrac{1}{2}\left(4 + \dfrac{\pi}{2}\right) = 2 + \dfrac{\pi}{4}

A = \dfrac{1}{2}\int_{\pi/2}^{3\pi/2} (-2\cos\theta - \cos^2\theta)\,d\theta

A = \dfrac{1}{2}\left(4 - \dfrac{\pi}{2}\right) = 2 - \dfrac{\pi}{4}

\sin\theta = \cos\theta \Rightarrow \theta = \dfrac{\pi}{4}

r = \dfrac{\sqrt{2}}{2} \Rightarrow (x, y) = \left(\tfrac{1}{2}, \tfrac{1}{2}\right)

r = 0 \text{ at } \theta = 0 \text{ and } \theta = \dfrac{\pi}{2}

A = 2 \cdot \dfrac{1}{2}\int_0^{\pi/4} \sin^2\theta\,d\theta

A = \left[\dfrac{\theta}{2} - \dfrac{\sin 2\theta}{4}\right]_0^{\pi/4} = \dfrac{\pi}{8} - \dfrac{1}{4}

2\sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}

A = \dfrac{1}{2}\int_{\pi/6}^{5\pi/6} (1 - 2\cos 2\theta)\,d\theta

A = \dfrac{1}{2}\left[\theta - \sin 2\theta\right]_{\pi/6}^{5\pi/6} = \dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2}

1 - 2\cos\theta = 0 \Rightarrow \theta = \pm\dfrac{\pi}{3}

A = \dfrac{1}{2}\left[3\theta - 4\sin\theta + \sin 2\theta\right]_{-\pi/3}^{\pi/3}

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

3 + 2\sin\theta = 2 \Rightarrow \theta = -\dfrac{\pi}{6}, \dfrac{7\pi}{6}

A = \dfrac{1}{2}\int_{-\pi/6}^{7\pi/6} (5 + 12\sin\theta + 4\sin^2\theta)\,d\theta

A = \dfrac{1}{2}\left[7\theta - 12\cos\theta - \sin 2\theta\right]_{-\pi/6}^{7\pi/6}

A = \dfrac{14\pi}{3} + \dfrac{11\sqrt{3}}{2} \approx 24.19

4 + 4\sin\theta = 4 \Rightarrow \theta = 0, \pi

A = \dfrac{1}{2}\int_0^{\pi} (32\sin\theta + 16\sin^2\theta)\,d\theta

A = \dfrac{1}{2}(64 + 8\pi) = 32 + 4\pi \approx 44.57

Area between polar curves FAQ

How do you find the area between two polar curves?

Find where the curves meet, decide which is outer on each interval, and compute ½∫ (R² − r²) dθ between those angles.

How do you find the limits of integration?

Set the two equations equal and solve for θ. Also check whether both curves pass through the pole, because that intersection may not show up when you set them equal.

Why isn’t the formula ½∫ (R − r)² dθ?

Each piece is a big wedge minus a small wedge, ½R²Δθ − ½r²Δθ. Squaring the difference would give the wrong shape.

How do you find the area inside both curves?

Split the region at the crossing points. On each piece, integrate ½r² for whichever curve is closer to the origin, then add.

Summary

• Area between polar curves: A = ½∫ (R² − r²) dθ.
• Limits come from setting the equations equal; check the pole separately.
• R is the outer curve on that interval; split if the curves switch.
• Square each radius, subtract, then integrate.
• Use symmetry to integrate half the region and double.

Previous lesson: Area in Polar Coordinates

Next lesson: Arc Length of Polar Curves

Back to the Calculus 2 Index