Area in Polar Coordinates

The single most elegant result in the polar strand of AQA Further Mathematics is the formula for the area swept out by a radius vector: $A=\tfrac12\displaystyle\int_\alpha^\beta r^2\,\mathrm d\theta$ . It looks nothing like the Cartesian $\int y\,\mathrm dx$ , and the factor of $\tfrac12$ and the squared $r$ are precisely where marks are won and lost. This lesson derives the formula from the sector-area idea, applies it to circles, cardioids, roses, spirals and lemniscates, and then tackles the harder "area between two curves" problems where finding the correct limits is everything.

1 Spec mapping

The polar area integral is compulsory pure content for Papers 1 and 2 of AQA 7367. A single area question blends all three assessment objectives: AO1 to carry out the integration, AO2 to justify the limits and use symmetry, and AO3 when the question is set in an unfamiliar "inside one curve, outside another" configuration. Because the integrand is almost always a squared trig expression, success depends on fluent use of the double-angle identities from A-Level Maths — the algebra of the integrand, not the calculus, is what separates the bands. It is also one of the highest-value topics to revise, because the same formula $\tfrac12\int r^2\,\mathrm d\theta$ answers every question in the strand; there is no menagerie of separate methods to memorise, just one formula applied with care over the right limits.

2 Core theory — deriving the area formula

Take a curve $r=f(\theta)$ and a thin angular slice between $\theta$ and $\theta+\delta\theta$ . Over so small an angle the bounding arc is almost a circular arc of radius $r$ , so the slice is almost a circular sector of radius $r$ and angle $\delta\theta$ . The area of a sector of radius $r$ and angle $\delta\theta$ (radians) is $\tfrac12 r^2\,\delta\theta$ , so

$\delta A \approx \tfrac12\,r^2\,\delta\theta.$

Summing the slices from $\theta=\alpha$ to $\theta=\beta$ and letting $\delta\theta\to0$ turns the sum into an integral:

$A = \lim_{\delta\theta\to0}\sum \tfrac12 r^2\,\delta\theta = \boxed{\ \frac12\int_\alpha^\beta r^2\,\mathrm d\theta\ } = \frac12\int_\alpha^\beta \big[f(\theta)\big]^2\,\mathrm d\theta.$

Key idea. This is the area of the region between the curve and the pole, swept by the radius vector as $\theta$ runs from $\alpha$ to $\beta$ . The $\tfrac12$ is inherited directly from the sector-area formula $\tfrac12 r^2\theta$ ; never omit it, and never forget to square $r$ .

It is worth pausing on why the sector area is $\tfrac12 r^2\,\delta\theta$ , because that single fact carries the whole formula. A full circle of radius $r$ has area $\pi r^2$ and corresponds to an angle of $2\pi$ ; a sector subtending angle $\delta\theta$ is the fraction $\tfrac{\delta\theta}{2\pi}$ of the disc, so its area is $\tfrac{\delta\theta}{2\pi}\cdot\pi r^2=\tfrac12 r^2\,\delta\theta$ . The approximation in $\delta A\approx\tfrac12 r^2\,\delta\theta$ is only that $r$ varies slightly across the slice, and that error vanishes in the limit — which is exactly what the integral sign records. This is the same "sum of infinitely many infinitesimal pieces" logic that underlies every integral you have met; the only novelty is that the pieces are wedges rather than rectangles.

The integrand $r^2=[f(\theta)]^2$ is almost always a squared cosine or sine, so the very first move after writing the integral is to apply a double-angle identity:

$\cos^2\theta = \tfrac12\big(1+\cos 2\theta\big), \qquad \sin^2\theta = \tfrac12\big(1-\cos 2\theta\big).$

The reason this step is unavoidable is that there is no elementary antiderivative of $\cos^2\theta$ or $\sin^2\theta$ written in those forms; you must first linearise them into a constant plus a single cosine of the double angle, which integrate term by term. When the curve involves $\cos n\theta$ , the identity produces $\cos 2n\theta$ instead, but the principle is identical. Train yourself to reach for the double-angle identity the instant you see a squared trig term inside an integral — it is the workhorse of this entire lesson.

3 Worked examples with mark scheme

Worked Example 1 — area of a circle (sanity check)

Find the area enclosed by $r=a$ .

$A = \frac12\int_0^{2\pi} a^2\,\mathrm d\theta = \frac{a^2}{2}\Big[\theta\Big]_0^{2\pi} = \frac{a^2}{2}\cdot 2\pi = \pi a^2.$

This recovers the familiar area of a disc — a reassuring check that the formula and its $\tfrac12$ are right. It is a good habit to test any new integral formula against a case whose answer you already know; if the polar formula had not produced $\pi a^2$ for the circle, you would know at once that you had dropped the $\tfrac12$ or mishandled the limits. The whole circle is swept once as $\theta$ runs over $[0,2\pi]$ because $r=a$ is constant — there is no over-tracing to worry about here.

Mark scheme: M1 apply $\tfrac12\int r^2\,\mathrm d\theta$ with $r=a$ , limits $0$ to $2\pi$ ; A1 $\pi a^2$ .

Worked Example 2 — area of a cardioid

Find the area enclosed by $r=a(1+\cos\theta)$ .

Two preliminary decisions fix the integral. First, the tracing interval: a cardioid is traced exactly once as $\theta$ runs over $[0,2\pi]$ , so those are the natural limits. Second, symmetry: because the equation depends only on $\cos\theta$ the curve is symmetric about the initial line, so the area below the axis equals the area above it. We may therefore integrate over $[0,\pi]$ and double, which both halves the arithmetic and makes the boundary terms cleaner. With those decisions made:

$\begin{aligned} A &= 2\cdot\frac12\int_0^{\pi} a^2(1+\cos\theta)^2\,\mathrm d\theta = a^2\int_0^{\pi}\big(1+2\cos\theta+\cos^2\theta\big)\,\mathrm d\theta \\ &= a^2\int_0^{\pi}\left(1+2\cos\theta+\tfrac12+\tfrac12\cos 2\theta\right)\mathrm d\theta \qquad\big(\cos^2\theta=\tfrac12(1+\cos 2\theta)\big)\\ &= a^2\int_0^{\pi}\left(\tfrac32+2\cos\theta+\tfrac12\cos 2\theta\right)\mathrm d\theta \\ &= a^2\left[\tfrac32\theta+2\sin\theta+\tfrac14\sin 2\theta\right]_0^{\pi} = a^2\left(\tfrac{3\pi}{2}+0+0\right) = \frac{3\pi a^2}{2}. \end{aligned}$

Notice how cleanly the boundary terms vanished: at $\theta=\pi$ both $\sin\theta$ and $\sin 2\theta$ are zero, and at $\theta=0$ every term is zero, so only the $\tfrac32\theta$ term survives. This is typical of cardioid and rose areas — the oscillatory parts integrate to zero over a symmetric interval, leaving a clean multiple of $\pi$ . The result $\tfrac{3\pi a^2}{2}$ is one you should memorise, since the cardioid recurs constantly in two-curve problems.

Mark scheme: M1 set up $\tfrac12\int r^2$ with correct limits (symmetry doubling acceptable); M1 expand and apply $\cos^2\theta=\tfrac12(1+\cos 2\theta)$ ; A1 correct integration; A1 $\tfrac{3\pi a^2}{2}$ . The identity-substitution mark is the discriminator.

Worked Example 3 — one petal of a rose

Find the area of one petal of $r=a\cos 2\theta$ , and the total area of the rose.

Finding the right limits is the only subtlety. A single petal is swept out between two consecutive directions where $r=0$ ; for $r=a\cos 2\theta$ these are where $\cos 2\theta=0$ , i.e. $2\theta=\pm\tfrac{\pi}{2}$ , giving $\theta=\pm\tfrac{\pi}{4}$ . Between those angles $\cos 2\theta\ge0$ , so $r\ge0$ and we are tracing one complete petal pointing along the initial line. Hence:

$\begin{aligned} A_{\text{petal}} &= \frac12\int_{-\pi/4}^{\pi/4} a^2\cos^2 2\theta\,\mathrm d\theta = \frac{a^2}{2}\int_{-\pi/4}^{\pi/4}\tfrac12\big(1+\cos 4\theta\big)\,\mathrm d\theta \\ &= \frac{a^2}{4}\left[\theta+\tfrac14\sin 4\theta\right]_{-\pi/4}^{\pi/4} = \frac{a^2}{4}\left[\left(\tfrac{\pi}{4}+0\right)-\left(-\tfrac{\pi}{4}+0\right)\right] = \frac{a^2}{4}\cdot\frac{\pi}{2} = \frac{\pi a^2}{8}. \end{aligned}$

The rose $r=a\cos 2\theta$ has $2n=4$ petals, so the total area is $4\cdot\dfrac{\pi a^2}{8}=\dfrac{\pi a^2}{2}$ . It is tempting to compute the total in one go as $\tfrac12\int_0^{2\pi}a^2\cos^2 2\theta\,\mathrm d\theta$ , and indeed that gives $\tfrac{\pi a^2}{2}$ too — but only because, over a full turn, the four petals are each traced exactly once (the negative- $r$ arcs filling the gaps). For a rose with an odd number of petals this shortcut would double-count, so the safe method is always "one petal, then multiply by the petal count".

Mark scheme: M1 correct petal limits $\pm\tfrac{\pi}{4}$ ; M1 $\cos^2 2\theta=\tfrac12(1+\cos 4\theta)$ ; A1 $\tfrac{\pi a^2}{8}$ per petal; A1 $\tfrac{\pi a^2}{2}$ total (with the petal count justified).

Worked Example 4 — area swept by a spiral

Find the area swept by the Archimedean spiral $r=\theta$ from $\theta=0$ to $\theta=2\pi$ .

This is the area enclosed between the first full turn of the spiral and the initial line — a snail-shell-shaped region. Here $r=\theta$ , so $r^2=\theta^2$ is a simple power, and no trig identity is needed at all:

$A = \frac12\int_0^{2\pi}\theta^2\,\mathrm d\theta = \frac12\left[\frac{\theta^3}{3}\right]_0^{2\pi} = \frac12\cdot\frac{8\pi^3}{3} = \frac{4\pi^3}{3}.$

Mark scheme: M1 $\tfrac12\int\theta^2\,\mathrm d\theta$ ; A1 $\tfrac{4\pi^3}{3}$ . No trig identity here — a useful reminder that the method is universal.

4 Area between two polar curves

Many exam questions ask not for the area enclosed by a single curve but for the area of a region bounded by two curves — "inside this and outside that", or "common to both". The principle is simply subtraction: the area swept by the outer curve, minus the area swept by the inner curve, leaves the region between them. If $f(\theta)\ge g(\theta)\ge0$ on $[\alpha,\beta]$ , the area between the outer curve $r=f(\theta)$ and the inner curve $r=g(\theta)$ is therefore the difference of the two swept areas:

$A = \frac12\int_\alpha^\beta\Big(\big[f(\theta)\big]^2-\big[g(\theta)\big]^2\Big)\,\mathrm d\theta.$

Crucial. It is $f^2-g^2$ , not $(f-g)^2$ . Subtract the squares of the radii, never the radii first. Writing $(f-g)^2$ is one of the most heavily penalised errors in the whole topic because it gives a plausible-looking but wrong number.

To see why it must be $f^2-g^2$ , go back to the derivation: the area swept by the outer curve alone is $\tfrac12\int f^2\,\mathrm d\theta$ and the area swept by the inner curve alone is $\tfrac12\int g^2\,\mathrm d\theta$ ; the annular region between them is the difference of these two areas, which is $\tfrac12\int(f^2-g^2)\,\mathrm d\theta$ . The quantity $(f-g)$ is the gap between the radii, which is not itself an area at all — squaring it has no geometric meaning here. A numerical sanity check makes the point: for two concentric circles $f=3,\ g=1$ the true annulus area over a full turn is $\tfrac12\int_0^{2\pi}(9-1)\,\mathrm d\theta=8\pi=\pi(3^2-1^2)$ , the familiar $\pi(R^2-r^2)$ ; the wrong formula $\tfrac12\int(3-1)^2=4\pi$ is exactly half of it, and plainly nonsense.

Worked Example 5 — inside one curve, outside another

Find the area inside $r=2$ and outside the cardioid $r=2(1-\cos\theta)$ .

The region we want is the part of the disc $r\le2$ that lies outside the cardioid. To set up the difference integral we need the angles at which the boundary switches from one curve to the other, which are the intersections. Find them by equating the radii:

$2 = 2(1-\cos\theta)\ \implies\ \cos\theta=0\ \implies\ \theta=\pm\tfrac{\pi}{2}.$

For $-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$ the cardioid radius $2(1-\cos\theta)$ is smaller than $2$ , so the circle $r=2$ is the outer curve and the cardioid is the inner one — exactly the configuration the formula needs. The region is symmetric about the initial line, so integrate over $[0,\tfrac{\pi}{2}]$ and double:

Area in Polar Coordinates

Area in Polar Coordinates

1 Spec mapping

2 Core theory — deriving the area formula

3 Worked examples with mark scheme

Worked Example 1 — area of a circle (sanity check)

Worked Example 2 — area of a cardioid

Worked Example 3 — one petal of a rose

Worked Example 4 — area swept by a spiral

4 Area between two polar curves

Worked Example 5 — inside one curve, outside another

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