HookRadar never spoke Cartesian
In the summer of 1940, the RAF's Chain Home stations could detect incoming aircraft over a hundred miles out — and not one of those stations reported an \(x\) or a \(y\). A radar pulse travels out along a bearing and bounces back after a measured delay, so what the operator reads off is a distance and an angle: range 90 miles, bearing 120 degrees. The instrument speaks in a distance from a fixed point and an angle from a fixed direction, because that is what the physics hands it. Converting those readings onto the map grid — the Cartesian picture the fighter controllers actually plotted — was a calculation performed thousands of times a day, by hand, under fire.
That pair \((r,\theta)\) is a polar coordinate, and this section teaches the three skills AQA attaches to it: translating fluently between polar and Cartesian descriptions, sketching curves written as \(r=f(\theta)\) — a family including circles, spirals and the heart-shaped cardioid that Cartesian equations describe only clumsily — and integrating to find the area a polar curve encloses. The recurring theme is that some shapes are simply native to polar form: anything built from rotation, sweep or distance-from-a-centre collapses from a messy Cartesian equation into one clean line of \(r=f(\theta)\). The skill being bought is choosing the coordinate system that makes the problem small.
ModelTwo numbers, a different pair
Fix a point \(O\) — the pole — and a ray from it, the initial line, conventionally along the positive \(x\)-axis. Any point of the plane is then addressed by \(r\), its distance from the pole, and \(\theta\), the anticlockwise angle from the initial line. AQA works with \(r\ge 0\), and \(\theta\) taken in a standard interval such as \(-\pi<\theta\le\pi\) or \(0\le\theta<2\pi\); questions state the window in force.
The conversion dictionary is four lines. Polar to Cartesian is unconditional: \(x=r\cos\theta\), \(y=r\sin\theta\). Cartesian to polar needs more care: \(r=\sqrt{x^2+y^2}\) always, but \(\theta\) is not blindly \(\tan^{-1}\frac{y}{x}\) — the calculator's arctangent only answers between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), so any point in the second or third quadrant comes back with the wrong angle unless you adjust by \(\pi\). Plot the point, see its quadrant, then commit. The same dictionary converts whole equations. The circle \(x^2+y^2=a^2\) becomes simply \(r=a\); the line \(y=x\) becomes \(\theta=\frac{\pi}{4}\); and running the dictionary in reverse, \(r=2a\cos\theta\) multiplies up to \(r^2=2ar\cos\theta\), i.e. \(x^2+y^2=2ax\), a circle of radius \(a\) through the pole centred at \((a,0)\). That multiply-both-sides-by-\(r\) move — engineering the appearances of \(r^2\), \(r\cos\theta\) and \(r\sin\theta\) that the dictionary knows how to translate — is the workhorse of every conversion question.
Convert the Cartesian point \((-1,\sqrt{3})\) to polar form with \(-\pi<\theta\le\pi\).
Radius: \(r=\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{4}=2\). Angle: the calculator offers \(\tan^{-1}\big(\frac{\sqrt{3}}{-1}\big)=-\frac{\pi}{3}\) — but that angle points into the fourth quadrant, and \((-1,\sqrt{3})\) sits in the second (negative \(x\), positive \(y\)). Correct by adding \(\pi\): \(\theta=-\frac{\pi}{3}+\pi=\frac{2\pi}{3}\). So the point is \(\big(2,\frac{2\pi}{3}\big)\). Check forwards: \(x=2\cos\frac{2\pi}{3}=-1\), \(y=2\sin\frac{2\pi}{3}=\sqrt{3}\). \(\checkmark\) The five-second reverse check catches the quadrant error that costs more polar marks than any other single slip.
MechanismSketching r = f(θ): read the function like a flight log
A polar sketch is a story told in sweeps: as \(\theta\) advances, the point sits at distance \(f(\theta)\) from the pole, so the curve breathes in and out as the bearing rotates. The method is a disciplined table of values at the compass angles \(0\), \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\) plus wherever \(f\) peaks or vanishes, joined smoothly with three structural observations. Symmetry: if \(f(-\theta)=f(\theta)\) — any \(r=f(\cos\theta)\) qualifies — the curve is symmetric about the initial line, so you sketch the top half and mirror it. Maxima: the curve's furthest reach is at the maximum of \(f\). Zeros: where \(f(\theta)=0\) the curve arrives at the pole, and it does so travelling along the direction \(\theta\) — the solutions of \(f(\theta)=0\) are the tangent directions at the pole, which is what gives cardioids their pointed cusp and petals their sharp bases.
Under the \(r\ge 0\) convention, wherever \(f(\theta)<0\) the curve simply does not exist — you skip those \(\theta\) intervals entirely rather than plotting backwards. The gallery worth knowing cold: \(r=a\) is a circle about the pole; \(r=2a\cos\theta\) a circle through the pole (drawn for \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\) only, where \(\cos\theta\ge 0\)); \(r=a(1+\cos\theta)\) the cardioid, with cusp at the pole at \(\theta=\pi\); \(r=a+b\cos\theta\) the limaçon family, convex or dimpled according to how \(a\) compares with \(b\); \(r=a\cos 2\theta\) and \(r=a\sin 2\theta\) four-petalled roses, each petal living in an interval where the trig factor is non-negative; and \(r=a\theta\) the Archimedean spiral, winding outwards at a constant rate. Every one of them is painful in Cartesian form — the cardioid is the quartic \((x^2+y^2-ax)^2=a^2(x^2+y^2)\) — which is the entire argument for the polar language.
ModelArea: half the integral of r squared
Cartesian area is built from rectangle strips; polar area is built from pie slices. A thin wedge of angular width \(\delta\theta\) at radius \(r\) is approximately a circular sector, and a sector of angle \(\delta\theta\) in a circle of radius \(r\) has area \(\frac{\delta\theta}{2\pi}\times\pi r^2=\frac{1}{2}r^2\,\delta\theta\). Summing wedges and taking the limit gives the section's headline formula: the area swept between \(\theta=\alpha\) and \(\theta=\beta\) is \[A=\frac{1}{2}\int_{\alpha}^{\beta}r^2\,d\theta.\] Both peculiarities of the formula are doing real work: the \(\frac{1}{2}\) is the sector's half, and the square means a curve twice as far out sweeps four times the area — this is not a length integral.
The limits are where the thinking lives. They come from the geometry of the curve, not from habit: a full closed curve like the cardioid needs \(0\) to \(2\pi\); a single petal runs between consecutive solutions of \(r=0\); a region shared between two curves splits at their intersection angles. And since \(r^2\) almost always lands you on \(\cos^2\) or \(\sin^2\), the double-angle identities \(\cos^2\theta=\frac{1+\cos 2\theta}{2}\) and \(\sin^2\theta=\frac{1-\cos 2\theta}{2}\) are effectively part of the formula — expect to use one in every area question on the paper.
Find the area enclosed by the cardioid \(r=a(1+\cos\theta)\).
Set up: the curve closes over \(0\le\theta\le 2\pi\), so \[A=\frac{1}{2}\int_0^{2\pi}a^2(1+\cos\theta)^2\,d\theta=\frac{a^2}{2}\int_0^{2\pi}\big(1+2\cos\theta+\cos^2\theta\big)\,d\theta.\] Handle the square: \(\cos^2\theta=\frac{1+\cos 2\theta}{2}\), so the integrand is \(\frac{3}{2}+2\cos\theta+\frac{1}{2}\cos 2\theta\). Integrate: \[A=\frac{a^2}{2}\Big[\tfrac{3}{2}\theta+2\sin\theta+\tfrac{1}{4}\sin 2\theta\Big]_0^{2\pi}=\frac{a^2}{2}\cdot 3\pi=\frac{3\pi a^2}{2}.\] Both sine terms vanish at \(0\) and \(2\pi\) — over a full revolution the oscillating terms always die, leaving only the constant term's contribution. Sense check: the cardioid fits inside the circle \(r=2a\) (area \(4\pi a^2\)) and contains the circle \(r=a\) for a wide sweep of angles; \(\frac{3\pi a^2}{2}\approx 4.71a^2\) sits comfortably between.
CasePetals and shared regions: where the limits earn the marks
The rose \(r=4\sin 2\theta\) is the standard test of whether you choose limits by geometry. Under \(r\ge 0\) the curve exists only where \(\sin 2\theta\ge 0\); the first petal grows out of the pole at \(\theta=0\), reaches its maximum length \(4\) at \(\theta=\frac{\pi}{4}\), and returns to the pole at \(\theta=\frac{\pi}{2}\) — the two consecutive roots of \(r=0\) bracket the petal. Its area is \(\frac{1}{2}\int_0^{\pi/2}16\sin^2 2\theta\,d\theta=8\int_0^{\pi/2}\frac{1-\cos 4\theta}{2}\,d\theta=4\Big[\theta-\frac{\sin 4\theta}{4}\Big]_0^{\pi/2}=2\pi\). Integrating \(0\) to \(2\pi\) instead would sweep the pole-hugging gaps where the curve does not exist and silently double-count — the classic way this question is failed.
Regions bounded by two curves stack the same logic. First find the intersection angles by solving \(f(\theta)=g(\theta)\) — remembering the pole is a possible extra meeting point that the equation misses, since each curve can reach \(r=0\) at different values of \(\theta\). Then assemble the area as a sum of sweeps, using whichever curve is the inner boundary on each angular interval: the region inside both the circle \(r=3a\cos\theta\) and the cardioid \(r=a(1+\cos\theta)\), say, takes the cardioid's \(r\) where the cardioid is closer to the pole and the circle's where the circle is. A quick sketch first is not decoration; it is the instrument that tells you which curve rules which interval, and examiners' reports repeatedly attribute lost area marks to candidates who integrated the right formula between the wrong angles.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Polar questions on 7367 follow a stable script — convert, sketch, find an area — and the marks concentrate at two pressure points. The first is limits. Before integrating anything, solve \(r=0\) and state the interval your region occupies: 'petal from \(\theta=0\) to \(\theta=\frac{\pi}{2}\), where \(\sin 2\theta\ge 0\)' is one sentence that earns a method mark and prevents the doubled or halved answers that come from sweeping angles where the curve does not exist. Symmetry is the legitimate shortcut: integrating a symmetric half and doubling is faster and safer, but SAY you are doing it — a stray factor of 2 with no justification reads as an error.
The second pressure point is the squaring step. Expand \(r^2\) fully before integrating — \((1+\cos\theta)^2\) has a cross term that vanishing-under-symmetry intuition tempts candidates to drop — and bring in the double-angle identity the moment a \(\cos^2\) or \(\sin^2\) appears, since no antiderivative of a squared trig function exists without it. Keep exact values throughout: polar answers are almost always multiples of \(\pi\), and a decimal approximation forfeits the accuracy mark. In sketches, three labels buy the communication marks: the maximum value of \(r\) with the angle where it occurs, the tangent directions at the pole, and the intercepts on the initial line and the line \(\theta=\frac{\pi}{2}\). And when a conversion stalls, multiply both sides by \(r\) — manufacturing \(r^2\) and \(r\cos\theta\) is the move the question is nearly always waiting for.