Area in Polar Coordinates

The area in polar coordinates inside a curve r = f(θ), from θ = α to θ = β, is A = ½∫ r² dθ with limits α and β. The formula adds up thin wedges, each shaped like a sector of a circle with area ½r²Δθ. Choose the limits so the curve is traced exactly once.

Before this lesson: Area of a Sector, Graphing Polar Equations and Derivatives of Polar Functions

The idea: add up thin wedges

In rectangular coordinates, you find area with thin rectangles. In polar coordinates the natural pieces are thin wedges that fan out from the origin, like slices of pizza.

A sector of a circle with radius r and angle Δθ (in radians) has area ½r²Δθ. When r changes with θ, each thin wedge is almost a sector, so its area is about ½(f(θ))²Δθ. Adding the wedges and letting Δθ shrink gives an integral.

Polar area formula

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

1. Sketch the curve, or picture its shape.
2. Find limits that trace the region once. Angles where r = 0 often mark where a loop or petal starts and ends.
3. Square r, expanding carefully.
4. Replace sin² and cos² with the power-reducing identities: cos² u = (1 + cos 2u)/2 and sin² u = (1 − cos 2u)/2.
5. Integrate, then multiply by ½.

Symmetry saves work: find the area of one petal or one half and multiply.

Area in polar coordinates examples

Example 1: A circle

Use the formula to find the area inside r = 3.

A = \dfrac{1}{2}\int_0^{2\pi} 3^2\,d\theta

A = \dfrac{9}{2}(2\pi - 0) = 9\pi

Answer: 9π, matching πr² for radius 3.

Example 2: A cardioid

Find the area inside the cardioid r = 1 + cos θ.

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

(1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta

\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}

\int_0^{2\pi} 1\,d\theta = 2\pi

\int_0^{2\pi} 2\cos\theta\,d\theta = \left[2\sin\theta\right]_0^{2\pi} = 0

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

A = \dfrac{1}{2}(2\pi + 0 + \pi) = \dfrac{3\pi}{2}

Answer: 3π/2 ≈ 4.71.

Example 3: A circle that isn’t centered at the origin

Find the area inside r = 4 sin θ.

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

A = 8\int_0^{\pi} \dfrac{1 - \cos 2\theta}{2}\,d\theta

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

A = 4(\pi - 0) = 4\pi

Answer: 4π. The curve is a circle of radius 2, and π(2)² = 4π agrees.

Example 4: One petal of a rose

Find the area of one petal of r = 2 cos 3θ, and the area of the whole rose.

2\cos 3\theta = 0 \Rightarrow \theta = \pm\dfrac{\pi}{6}

A = \dfrac{1}{2}\int_{-\pi/6}^{\pi/6} 4\cos^2 3\theta\,d\theta

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

A = \left[\theta + \dfrac{\sin 6\theta}{6}\right]_{-\pi/6}^{\pi/6}

A = \dfrac{\pi}{6} - \left(-\dfrac{\pi}{6}\right) = \dfrac{\pi}{3}

3 \cdot \dfrac{\pi}{3} = \pi

Answer: one petal has area π/3 ≈ 1.047, and the whole rose has area π.

Example 5: Area a sprinkler covers (word problem)

A lawn sprinkler waters the region inside r = 4 + 2 sin θ, in meters. How many square meters does it water?

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

(4 + 2\sin\theta)^2 = 16 + 16\sin\theta + 4\sin^2\theta

\int_0^{2\pi} 16\,d\theta = 32\pi, \quad \int_0^{2\pi} 16\sin\theta\,d\theta = 0

\int_0^{2\pi} 4\sin^2\theta\,d\theta = \int_0^{2\pi} (2 - 2\cos 2\theta)\,d\theta = 4\pi

A = \dfrac{1}{2}(32\pi + 0 + 4\pi) = 18\pi \approx 56.55

Answer: about 56.5 square meters. A full circle with the average radius, 4 m, would be 16π ≈ 50.3 m², so the bulge adds about 6 m².

Common mistakes

Related methods

Area between curves

For the region between two polar curves, subtract: ½∫ (R² − r²) dθ. That is the next lesson, Area Between Polar Curves. The rectangular version is area between curves.

Trig integrals

Squared sines and cosines appear in almost every polar area problem. Trigonometric integrals covers higher powers.

Arc length of polar curves

The length of the boundary, rather than the area inside it, uses a different integral. That lesson comes at the end of this unit.

Practice problems

1. Use the polar area formula to find the area inside r = 5.
2. Find the area inside r = 2 cos θ.
3. Find the area inside the cardioid r = 2 + 2 sin θ.
4. Find the area of one petal of r = sin 2θ, and of the whole four-petal rose.
5. Find the area swept out by the spiral r = θ for 0 ≤ θ ≤ π.
6. Find the area inside the limaçon r = 3 + cos θ.
7. Find the area swept out by r = e^θ for 0 ≤ θ ≤ 1.
8. Find the area inside r = 4 for 0 ≤ θ ≤ π/3. Check it with the sector formula.
9. Find the total area inside the lemniscate r² = 4 cos 2θ.
10. A garden bed is the region inside r = 2 + cos θ, in meters. Soil costs $8 per square meter. What does it cost to fill the bed?

A = \dfrac{1}{2}\int_0^{2\pi} 25\,d\theta = 25\pi

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

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

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

(2 + 2\sin\theta)^2 = 4 + 8\sin\theta + 4\sin^2\theta

\int_0^{2\pi} (4 + 8\sin\theta + 4\sin^2\theta)\,d\theta = 8\pi + 0 + 4\pi

A = \dfrac{1}{2}(12\pi) = 6\pi

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

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

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

4 \cdot \dfrac{\pi}{8} = \dfrac{\pi}{2}

A = \dfrac{1}{2}\int_0^{\pi} \theta^2\,d\theta = \dfrac{1}{2}\cdot\dfrac{\pi^3}{3}

A = \dfrac{\pi^3}{6} \approx 5.168

(3 + \cos\theta)^2 = 9 + 6\cos\theta + \cos^2\theta

\int_0^{2\pi} (9 + 6\cos\theta + \cos^2\theta)\,d\theta = 18\pi + 0 + \pi

A = \dfrac{19\pi}{2} \approx 29.85

A = \dfrac{1}{2}\int_0^{1} e^{2\theta}\,d\theta

A = \dfrac{1}{2}\left[\dfrac{e^{2\theta}}{2}\right]_0^1 = \dfrac{e^2 - 1}{4} \approx 1.597

A = \dfrac{1}{2}\int_0^{\pi/3} 16\,d\theta = 8 \cdot \dfrac{\pi}{3} = \dfrac{8\pi}{3}

\dfrac{1}{2}(4)^2\left(\dfrac{\pi}{3}\right) = \dfrac{8\pi}{3}

A_{\text{loop}} = \dfrac{1}{2}\int_{-\pi/4}^{\pi/4} 4\cos 2\theta\,d\theta

A_{\text{loop}} = \left[\sin 2\theta\right]_{-\pi/4}^{\pi/4} = 1 - (-1) = 2

2 \cdot 2 = 4

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

A = \dfrac{1}{2}(8\pi + 0 + \pi) = \dfrac{9\pi}{2} \approx 14.14

8 \dfrac{\text{dollars}}{\text{m}^2} \times 14.137\,\text{m}^2 \approx 113.10

Polar area FAQ

What is the formula for area in polar coordinates?

A = ½∫ r² dθ, from θ = α to θ = β, where the curve r = f(θ) is traced once on that interval.

Why is there a ½ in the polar area formula?

The pieces are wedges, not rectangles. A wedge with radius r and angle Δθ has area ½r²Δθ, the same as a sector of a circle.

How do you find the limits for a polar area?

Find the angles that trace the region exactly once. For loops and petals, the limits are usually consecutive angles where r = 0.

Why is my polar area twice the right answer?

The interval probably traced the curve twice. For example, r = 4 sin θ and r = 2 cos 3θ are each traced once on an interval of length π, not 2π.

Summary

• Polar area: A = ½∫ r² dθ between the limits α and β.
• The pieces are wedges with area ½r²Δθ.
• Choose limits that trace the region once; petals run between zeros of r.
• Expand r² fully and use power-reducing identities for sin² and cos².
• Use symmetry: find one petal or half, then multiply.

Previous lesson: Derivatives of Polar Functions

Next lesson: Area Between Polar Curves

Back to the Calculus 2 Index