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Area in polar coordinatesEdexcel International A Level Further Maths: Revision notes

Section 1

The area formula

A polar curve r=f(θ)r=f(\theta) is described by the distance rr from the pole OO and the angle θ\theta from the initial line. A thin sector of angle δθ\delta\theta and radius rr has area ≈12r2 δθ\approx\frac12r^2\,\delta\theta. Adding the sectors and letting δθ→0\delta\theta\to0 gives the area bounded by the curve and the half-lines θ=α\theta=\alpha and θ=β\theta=\beta: A=12∫αβr2 dθ.A=\frac12\int_{\alpha}^{\beta}r^2\,d\theta. Substitute r2r^2 as a function of θ\theta before integrating. The Edexcel convention is r≥0r\ge0, so each value of θ\theta gives one point of the curve, and the formula applies for α<β\alpha<\beta over a range where the curve is traced once.

Key termspolar curvepoleinitial linesector
Common mistake

Leaving out the factor 12\frac12, or integrating rr instead of r2r^2.

Exam tip

Write r2r^2 out in full first, e.g. (3+2cos⁡θ)2(3+2\cos\theta)^2, then expand.

Section 2

Choosing the limits

The limits α\alpha and β\beta are the values of θ\theta at the two bounding half-lines. If the region is enclosed by the curve alone, find the range of θ\theta for which r≥0r\ge0 and the curve is traced once. For example r=4cos⁡θr=4\cos\theta with r≥0r\ge0 runs from θ=−π2\theta=-\frac{\pi}{2} to π2\frac{\pi}{2}, and r=3+2cos⁡θr=3+2\cos\theta is traced once as θ\theta runs from 00 to 2π2\pi. Where r=0r=0 the curve meets the pole, and these values of θ\theta often give the limits. Symmetry about the initial line lets you double the area from 00 to π\pi (or from 00 to π2\frac{\pi}{2} for a curve that is also symmetrical about θ=π2\theta=\frac{\pi}{2}), but only if the curve really is symmetrical.

Key termslimitssymmetry
Common mistake

Taking a range of θ\theta where r<0r<0, or going round the curve twice, so the area is counted again.

Section 3

Integrating: the cos⁡2θ\cos^2\theta identity

Most area integrals contain cos⁡2θ\cos^2\theta or sin⁡2θ\sin^2\theta. Use cos⁡2θ=12(1+cos⁡2θ),sin⁡2θ=12(1−cos⁡2θ).\cos^2\theta=\tfrac12(1+\cos2\theta),\qquad\sin^2\theta=\tfrac12(1-\cos2\theta). Squares of brackets such as (a+bcos⁡θ)2=a2+2abcos⁡θ+b2cos⁡2θ(a+b\cos\theta)^2=a^2+2ab\cos\theta+b^2\cos^2\theta must be expanded first. Remember ∫cos⁡2θ dθ=12sin⁡2θ\int\cos2\theta\,d\theta=\frac12\sin2\theta.

Key termsdouble-angle identity
Common mistake

Writing ∫cos⁡2θ dθ=sin⁡2θ\int\cos2\theta\,d\theta=\sin2\theta and losing the 12\frac12.

Section 4

Worked example

Find the area enclosed by r=3+2cos⁡θr=3+2\cos\theta, 0≤θ≤2π0\le\theta\le2\pi. A=12∫02π(3+2cos⁡θ)2 dθ=12∫02π(11+12cos⁡θ+2cos⁡2θ)dθ=12[11θ+12sin⁡θ+sin⁡2θ]02π=11π.A=\frac12\int_0^{2\pi}(3+2\cos\theta)^2\,d\theta=\frac12\int_0^{2\pi}\left(11+12\cos\theta+2\cos2\theta\right)d\theta=\frac12\left[11\theta+12\sin\theta+\sin2\theta\right]_0^{2\pi}=11\pi. For a sector of the spiral r=2θr=2\theta from θ=π\theta=\pi to 2π2\pi: A=12∫π2π4θ2 dθ=23[θ3]π2π=14π33A=\frac12\int_\pi^{2\pi}4\theta^2\,d\theta=\frac23\left[\theta^3\right]_\pi^{2\pi}=\frac{14\pi^3}{3}.

Exam tip

Check with a known shape: r=4cos⁡θr=4\cos\theta is a circle of radius 22, so the integral must give 4π4\pi.

Section 5

Areas between curves

To find the area inside two curves, first find where they meet by equating the rr values. For r=3cos⁡θr=3\cos\theta and r=1+cos⁡θr=1+\cos\theta, cos⁡θ=12\cos\theta=\frac12, so θ=±π3\theta=\pm\frac{\pi}{3}. The boundary of the common region switches from one curve to the other at the intersection, so split the integral there, using for each range of θ\theta the curve with the smaller rr. For a region inside one curve but outside another, subtract the common area from the whole area (or subtract 12∫r22 dθ\frac12\int r_2^2\,d\theta from 12∫r12 dθ\frac12\int r_1^2\,d\theta over the same range).

Key termsintersectioncommon region
Exam tip

Sketch both curves roughly first, so you can see which curve is nearer the pole in each range of θ\theta.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Area in polar coordinates

  1. The curve CC has polar equation r=4cos⁡θr=4\cos\theta, for −π2≤θ≤π2-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}.
    Find the exact area of the region bounded by CC and the half-lines θ=0\theta=0 and θ=π4\theta=\frac{\pi}{4}.2 marks
  2. The spiral SS has polar equation r=2θr=2\theta, for θ≥0\theta\ge0.
    Find the exact area of the region bounded by SS and the half-lines θ=π\theta=\pi and θ=2π\theta=2\pi.2 marks
  3. The curve CC has polar equation r=3+2cos⁡θr=3+2\cos\theta, for 0≤θ≤2π0\le\theta\le2\pi.
    Show that the area enclosed by CC is 11π11\pi.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).