What is area enclosed by a polar curve?

This comes from summing thin circular sectors of angle dθ, each of area (1)/(2)r^2\,dθ.

What is wrong limits for a single loop?

A petal or loop is traced between consecutive zeros of r; integrating over 0 to 2π for a single petal overcounts.

CCEA CCEA A2 2020-style practice: Find the area enclosed by the cardioid r = 2(1 + cosθ) for 0 ≤ θ ≤ 2π.

The polar area is A = 12_0^2π r^2\,dθ with r = 2(1 + cosθ): r^2 = 4(1 + cosθ)^2 = 4(1 + 2cosθ + cos^2θ). Use cos^2θ = (1)/(2)(1 + cos 2θ): r^2 = 4(1 + 2cosθ + 12 + 12cos 2θ) = 6 + 8cosθ + 2cos 2θ. Integrate over 0 to 2π: the cosθ and cos 2θ terms integrate to zero over a full period, leaving A = 12[6θ]_0^2π = 12(12π) = 6π. Markers reward the polar-area formula, the double-angle substitution, and the value 6π.

Northern IrelandFurther MathsSyllabus dot point

How do polar coordinates describe curves, and how do you find the area they enclose?

Polar coordinates and their relationship to Cartesian coordinates, sketching curves given in polar form, and the area enclosed by a polar curve using the half r squared integral.

A CCEA A2 Further Maths answer on polar coordinates and their conversion to Cartesian form, sketching polar curves such as circles, cardioids and roses, and finding the area enclosed by a polar curve using the half integral of r squared.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

CCEA wants you to use polar coordinates $(r, \theta)$ , convert between polar and Cartesian form, sketch curves given in polar form (circles, cardioids, spirals, roses), and find the area enclosed by a polar curve using the $\frac{1}{2}\int r^2\,d\theta$ formula.

The answer

Polar coordinates

Converting between polar and Cartesian

Sketching polar curves

Area enclosed by a polar curve

This comes from summing thin circular sectors of angle $d\theta$ , each of area $\frac{1}{2}r^2\,d\theta$ .

Area of one petal of a rose

Find the area of one petal of $r = 3\sin 2\theta$ .

step Find the limits for one petal

One petal is traced as $r$ goes from $0$ up and back to $0$ , that is from $\theta = 0$ to $\theta = \frac{\pi}{2}$ (where $\sin 2\theta = 0$ at both ends).

step Set up the area integral

A = \frac{1}{2}\int_{0}^{\pi/2} (3\sin 2\theta)^2\,d\theta = \frac{1}{2}\int_{0}^{\pi/2} 9\sin^2 2\theta\,d\theta.

step Use the double-angle form and integrate

$\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)$ :

A = \frac{9}{2}\int_{0}^{\pi/2}\frac{1}{2}(1 - \cos 4\theta)\,d\theta = \frac{9}{4}\left[\theta - \frac{\sin 4\theta}{4}\right]_{0}^{\pi/2} = \frac{9}{4}\cdot\frac{\pi}{2} = \frac{9\pi}{8}.

Converting a Cartesian equation to polar form

Express the circle $x^2 + y^2 = 6y$ in polar form, and state where it passes through the pole.

step Substitute the polar relationships

Replace $x^2 + y^2$ by $r^2$ and $y$ by $r\sin\theta$ :

r^2 = 6r\sin\theta.

step Simplify

Divide by $r$ (valid for $r \neq 0$ ):

r = 6\sin\theta.

step Interpret

This is a circle through the pole: at $\theta = 0$ , $r = 0$ , so the curve passes through the origin, and its maximum $r = 6$ occurs at $\theta = \frac{\pi}{2}$ , giving a circle of radius $3$ centred at $(0, 3)$ in Cartesian terms. Converting between the two forms is a routine exam step, and the $r = a\sin\theta$ and $r = a\cos\theta$ families are always circles through the pole.

Examples in context

Example 1. Radar and antenna patterns. The strength of a signal from a directional antenna is naturally a function of angle, $r = f(\theta)$ , so its coverage is a polar curve. The lobes of the pattern are exactly the petals of a rose-type polar graph.

Example 2. Planetary orbits. Kepler's first law puts a planet on an ellipse with the Sun at a focus, written compactly in polar form $r = \frac{l}{1 + e\cos\theta}$ . Polar coordinates, centred on the focus, make orbital mechanics far simpler than Cartesian.

Try this

Q1. Convert the point with Cartesian coordinates $(0, 5)$ to polar form. [1 mark]

Cue. $r = 5$ , $\theta = \frac{\pi}{2}$ .

Q2. State the polar area formula. [1 mark]

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

Q3. What curve is $r = 6$ in polar coordinates? [1 mark]

Cue. A circle of radius $6$ centred at the pole.

Exam-style practice questions

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

CCEA A2 20206 marksFind the area enclosed by the cardioid

r = 2(1 + \cos\theta)

for

0 \leq \theta \leq 2\pi

Show worked answer →

The polar area is $A = \dfrac{1}{2}\displaystyle\int_{0}^{2\pi} r^2\,d\theta$ with $r = 2(1 + \cos\theta)$ :

$r^2 = 4(1 + \cos\theta)^2 = 4(1 + 2\cos\theta + \cos^2\theta).$

Use $\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)$ :

$r^2 = 4\left(1 + 2\cos\theta + \tfrac{1}{2} + \tfrac{1}{2}\cos 2\theta\right) = 6 + 8\cos\theta + 2\cos 2\theta.$

Integrate over $0$ to $2\pi$ : the $\cos\theta$ and $\cos 2\theta$ terms integrate to zero over a full period, leaving

$A = \dfrac{1}{2}\left[6\theta\right]_{0}^{2\pi} = \dfrac{1}{2}(12\pi) = 6\pi.$

Markers reward the polar-area formula, the double-angle substitution, and the value $6\pi$ .

CCEA A2 20185 marksConvert the polar equation

r = 4\cos\theta

to Cartesian form, and describe the curve.

Show worked answer →

Multiply both sides by $r$ : $r^2 = 4r\cos\theta$ .

Use $r^2 = x^2 + y^2$ and $x = r\cos\theta$ :

$x^2 + y^2 = 4x.$

Complete the square in $x$ : $x^2 - 4x + y^2 = 0 \Rightarrow (x - 2)^2 + y^2 = 4.$

This is a circle of radius $2$ centred at $(2, 0)$ , passing through the origin.

Markers reward multiplying by $r$ , the substitutions for $r^2$ and $r\cos\theta$ , completing the square, and identifying the circle.

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)