What is wrong quadrant for the angle?

When converting from Cartesian, tanθ = (y)/(x) alone does not fix the quadrant; check the signs of x and y.

AQA AQA 2019-style practice: Find the exact area enclosed by the cardioid r = 2(1 + cosθ).

The area is A = (1)/(2)_α^β r^2\,dθ. The whole cardioid is traced as θ runs from 0 to 2π. r^2 = 4(1 + cosθ)^2 = 4(1 + 2cosθ + cos^2θ). Replace cos^2θ = (1)/(2)(1 + cos 2θ): r^2 = 4(1 + 2cosθ + (1)/(2) + (1)/(2)cos 2θ) = 6 + 8cosθ + 2cos 2θ. So A = (1)/(2)_0^2π(6 + 8cosθ + 2cos 2θ)\,dθ = (1)/(2)[6θ + 8sinθ + sin 2θ]_0^2π. The sine terms vanish at both limits, leaving A = (1)/(2)(12π) = 6π. Markers reward the area formula, the double-angle substitution to integrate cos^2θ, and the correct limits over a full revolution.

EnglandFurther MathsSyllabus dot point

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

Polar coordinates and the relationship with Cartesian coordinates, sketching polar curves, and finding areas enclosed by polar curves using integration.

A focused answer to the AQA A-Level Further Mathematics polar coordinates content, covering the relationship between polar and Cartesian coordinates, sketching polar curves such as cardioids and spirals, and finding areas enclosed by polar curves using integration.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
Polar and Cartesian coordinates
Sketching polar curves
Area enclosed by a polar curve

What this dot point is asking

AQA wants you to convert between polar and Cartesian coordinates, sketch polar curves such as circles, cardioids and spirals, identify where curves cut the initial line or have maximum distance from the pole, and find the area enclosed by a polar curve using integration.

Polar and Cartesian coordinates

In the polar system a point is located by its distance $r$ from the pole (origin) and the angle $\theta$ that the radius makes with the initial line (the positive $x$ axis). The same point has infinitely many polar names because adding $2\pi$ to $\theta$ returns to it, so AQA usually restricts $\theta$ to $[0, 2\pi)$ or $(-\pi, \pi]$ and takes $r \geq 0$ . Converting an equation between systems is a routine first step: to go from polar to Cartesian you often multiply through by $r$ so that the convertible groups $r^2$ , $r\cos\theta$ and $r\sin\theta$ appear, then substitute.

Sketching polar curves

For a curve $r = f(\theta)$ you build a table of $r$ values, note where $r = 0$ (the curve passes through the pole) and where $r$ is greatest. Common shapes are the cardioid $r = a(1 + \cos\theta)$ , the limacon $r = a + b\cos\theta$ (which has an inner loop when $b > a$ ), and spirals like $r = a\theta$ . Symmetry shortcuts save time: a curve in $\cos\theta$ is symmetric about the initial line, and one in $\sin\theta$ is symmetric about the line $\theta = \frac{\pi}{2}$ . The maximum value of $r$ marks the furthest point from the pole, found where $\frac{dr}{d\theta} = 0$ , and the tangent at the pole occurs at the values of $\theta$ for which $r = 0$ .

Because the equation is built from $\cos\theta$ , you only need to evaluate $r$ for $\theta$ from $0$ to $\pi$ and reflect the result in the initial line, which halves the plotting work.

Area enclosed by a polar curve

Area inside a polar curve with a squared term

Find the area enclosed by one loop of $r = 2\sin 2\theta$ , which exists between $\theta = 0$ and $\theta = \dfrac{\pi}{2}$ .

Step 1: Set up the polar area integral

The polar area formula is $A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta$ . The limits are the values of $\theta$ where the loop begins and ends, i.e. where $r = 0$ . For $r = 2\sin 2\theta$ , the first loop is swept as $\theta$ goes from $0$ to $\frac{\pi}{2}$ . We square the expression for $r$ before integrating.

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

Step 2: Apply the double-angle identity to make the integral tractable

$\sin^2 2\theta$ has no direct antiderivative in that form, so we use the identity $\sin^2 u = \frac{1}{2}(1 - \cos 2u)$ with $u = 2\theta$ . This converts the squared sine into a linear combination of cosines, which integrates straightforwardly.

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

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

Step 3: Integrate and evaluate the limits

Integrate term by term and substitute the limits. At both $\theta = 0$ and $\theta = \frac{\pi}{2}$ the sine term $\sin 4\theta$ is zero, so the cosine integral contributes nothing at either limit.

A = \left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}

Final answer: The area of one loop is $\dfrac{\pi}{2}$ .

When the squared integrand contains $\cos^2\theta$ or $\sin^2\theta$ , always reduce it with a double-angle identity before integrating, since $\int\cos^2\theta\,d\theta$ has no elementary antiderivative in that raw form. To find the area between two polar curves, integrate the difference of their $r^2$ values over the range where one lies outside the other, and to find the area of one loop of a multi-loop curve, set $r = 0$ to locate the limits that bound a single loop.

Exam-style practice questions

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

AQA 20196 marksFind the exact area enclosed by the cardioid

r = 2(1 + \cos\theta)

Show worked answer →

The area is $A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$ . The whole cardioid is traced as $\theta$ runs from $0$ to $2\pi$ .

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

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

So $A = \frac{1}{2}\int_0^{2\pi}(6 + 8\cos\theta + 2\cos 2\theta)\,d\theta = \frac{1}{2}\left[6\theta + 8\sin\theta + \sin 2\theta\right]_0^{2\pi}$ .

The sine terms vanish at both limits, leaving $A = \frac{1}{2}(12\pi) = 6\pi$ .

Markers reward the area formula, the double-angle substitution to integrate $\cos^2\theta$ , and the correct limits over a full revolution.

AQA 20214 marksThe curve

C

has polar equation

r = 4\cos\theta

. Convert this to a Cartesian equation 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: $x^2 - 4x + y^2 = 0$ , so $(x - 2)^2 + y^2 = 4$ .

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

Markers reward multiplying by $r$ to create convertible terms, the substitutions, completing the square, and identifying the circle with its centre and radius.

Related dot points

Sources & how we know this

AQA A-level Further Mathematics (7367) specification — AQA (2017)