What are wrong loop limits?

A single loop runs between consecutive zeros of r; using 0 to 2π for a multi-loop curve counts several loops (or cancels overlaps) and gives a wrong area.

What is not linearising the square?

sin^2θ and cos^2θ must be rewritten with double-angle identities before integrating; integrating them directly is not valid.

Write the integral for the area enclosed by r = 1 + cosθ for 0 θ 2π. [2 marks]

OCR OCR 2018-style practice: Find the area enclosed by the circle r = 3sinθ for 0 θ π.

Use the polar area formula A = 12 r^2\,dθ (M1): A = 12_0^π (3sinθ)^2\,dθ = 92_0^πsin^2θ\,dθ (A1). Use sin^2θ = 12(1 - cos 2θ) (M1): A = 92·12_0^π(1 - cos 2θ)\,dθ = 94[θ - 12sin 2θ]_0^π (A1). Evaluate (A1): = 94[(π - 0) - 0] = 9π4. Markers reward the area formula, squaring r, the double-angle substitution, and the exact area 9π4.

OCR OCR 2022-style practice: Find the area of one loop of the curve r = 2cos 2θ.

The curve r = 2cos 2θ has loops; one loop is traced as r goes from 0 up to its maximum and back to 0. Set r = 0 (M1): cos 2θ = 0, so 2θ = π2, that is θ = π4; the loop spans -π4 θ π4 (A1). Apply the area formula (M1): A = 12_-π/4^π/4(2cos 2θ)^2\,dθ = 12_-π/4^π/4 4cos^2 2θ\,dθ (A1). Use cos^2 2θ = 12(1 + cos 4θ) (M1): A = 2·12_-π/4^π/4(1 + cos 4θ)\,dθ = [θ + 14sin 4θ]_-π/4^π/4 = π2 (A1). Markers reward finding the loop limits from r = 0, the area formula, the double-angle reduction, and the exact area π2.

EnglandFurther MathsSyllabus dot point

How do you find the area enclosed by a polar curve?

The area enclosed by a polar curve using the formula one half the integral of r squared with respect to theta, including areas between two curves and the area of one loop.

A focused answer to the OCR A-Level Further Mathematics A content on the area enclosed by a polar curve, covering the formula one half the integral of r squared with respect to theta, choosing the correct limits, finding the area of a single loop, and finding the area between two polar curves.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The area swept out by the radius of a polar curve $r = f(\theta)$ between $\theta = \alpha$ and $\theta = \beta$ is $A = \tfrac{1}{2}\displaystyle\int_{\alpha}^{\beta} r^2\,d\theta$ , summing thin circular sectors of area $\tfrac{1}{2}r^2\,d\theta$ . The limits matter: for a full closed curve like a circle use the range that traces it once, and for a single loop find the two consecutive values of $\theta$ where $r = 0$ . The integrand $r^2$ usually contains $\sin^2$ or $\cos^2$ , so reduce with the double-angle identities $\sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)$ and $\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)$ before integrating. The area between two curves is the difference of two such integrals over the overlapping range.

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What this dot point is asking
The polar area formula
Choosing the limits
Reducing the integrand
Area between two curves
Try this

What this dot point is asking

OCR wants you to find the area enclosed by a polar curve using $A = \tfrac{1}{2}\int r^2\,d\theta$ , to choose the correct $\theta$ -limits (especially for a single loop, found from where $r = 0$ ), and to find the area between two polar curves by subtracting the appropriate integrals.

The polar area formula

The area is built from thin sectors, like slices of a pie, each subtending a small angle $d\theta$ at the pole. A sector of radius $r$ and angle $d\theta$ has area $\tfrac{1}{2}r^2\,d\theta$ , and integrating sums them.

Choosing the limits

The hardest part is usually the limits, not the integration. For a full closed curve, integrate over the range that traces it exactly once (for example $0$ to $\pi$ for a circle $r = a\sin\theta$ ). For one loop of a multi-loop curve, the loop begins and ends where $r = 0$ , so solve $f(\theta) = 0$ for the two consecutive angles bounding the loop.

Reducing the integrand

Because $r$ is usually a sine or cosine, $r^2$ contains $\sin^2$ or $\cos^2$ , which you cannot integrate directly. Use the double-angle identities to linearise first.

Area between two curves

When two polar curves overlap, the area of the region between them is the larger area minus the smaller, both via the polar area formula over the range where the region lies. Find the intersection angles by setting the two $r$ -expressions equal and solving for $\theta$ ; these angles become the limits of the integrals. Sketch both curves first so you can see which curve is the outer one over each part of the range, because the outer curve can switch between the two as $\theta$ varies, and a clear sketch prevents you from subtracting the areas in the wrong order or over the wrong interval.

The polar area formula reuses the double-angle identities and the integration skills from the core, and parallels the area-under-a-curve idea in Cartesian coordinates.

Try this

Q1. Write the integral for the area enclosed by $r = 1 + \cos\theta$ for $0 \le \theta \le 2\pi$ . [2 marks]

Cue. $A = \tfrac{1}{2}\displaystyle\int_0^{2\pi}(1 + \cos\theta)^2\,d\theta$ .

Q2. State the double-angle identity you would use to integrate $\cos^2\theta$ . [1 mark]

Cue. $\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20185 marksFind the area enclosed by the circle

r = 3\sin\theta

for

0 \le \theta \le \pi

Show worked answer →

Use the polar area formula $A = \tfrac{1}{2}\displaystyle\int r^2\,d\theta$ (M1): $A = \tfrac{1}{2}\displaystyle\int_0^{\pi} (3\sin\theta)^2\,d\theta = \tfrac{9}{2}\displaystyle\int_0^{\pi}\sin^2\theta\,d\theta$ (A1).

Use $\sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)$ (M1): $A = \tfrac{9}{2}\cdot\tfrac{1}{2}\displaystyle\int_0^{\pi}(1 - \cos 2\theta)\,d\theta = \tfrac{9}{4}\left[\theta - \tfrac{1}{2}\sin 2\theta\right]_0^{\pi}$ (A1).

Evaluate (A1): $= \tfrac{9}{4}\left[(\pi - 0) - 0\right] = \tfrac{9\pi}{4}$ .

Markers reward the area formula, squaring $r$ , the double-angle substitution, and the exact area $\tfrac{9\pi}{4}$ .

OCR 20226 marksFind the area of one loop of the curve

r = 2\cos 2\theta

Show worked answer →

The curve $r = 2\cos 2\theta$ has loops; one loop is traced as $r$ goes from $0$ up to its maximum and back to $0$ . Set $r = 0$ (M1): $\cos 2\theta = 0$ , so $2\theta = \pm\tfrac{\pi}{2}$ , that is $\theta = \pm\tfrac{\pi}{4}$ ; the loop spans $-\tfrac{\pi}{4} \le \theta \le \tfrac{\pi}{4}$ (A1).

Apply the area formula (M1): $A = \tfrac{1}{2}\displaystyle\int_{-\pi/4}^{\pi/4}(2\cos 2\theta)^2\,d\theta = \tfrac{1}{2}\displaystyle\int_{-\pi/4}^{\pi/4} 4\cos^2 2\theta\,d\theta$ (A1).

Use $\cos^2 2\theta = \tfrac{1}{2}(1 + \cos 4\theta)$ (M1): $A = 2\cdot\tfrac{1}{2}\displaystyle\int_{-\pi/4}^{\pi/4}(1 + \cos 4\theta)\,d\theta = \left[\theta + \tfrac{1}{4}\sin 4\theta\right]_{-\pi/4}^{\pi/4} = \tfrac{\pi}{2}$ (A1).

Markers reward finding the loop limits from $r = 0$ , the area formula, the double-angle reduction, and the exact area $\tfrac{\pi}{2}$ .

Related dot points

Sources & how we know this

OCR A Level Further Mathematics A (H245) specification — OCR (2017)