Practise › Questions › Areas with polar coordinates
Areas with polar coordinates questions
Sweep a radius round a polar curve and it brushes out area in thin sectors. Half r squared, integrated over the angle, measures any region a polar equation encloses.
6 original questions · 25 marks · the areas with polar coordinates notes · Polar coordinates
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Write down the formula for the area swept out by a polar curve r = f(θ) between θ = α and θ = β, and explain the ½.
Worked answer
A = ½∫r² dθ from α to β. Each thin wedge is nearly a triangle with base r dθ and height r, and a triangle's half shows up in front of the r² dθ. B1 for the formula, B1 for the explanation.Find the area swept by r = 4 as θ runs from 0 to π/3.
Worked answer
A = ½∫16 dθ = 8 × π/3 = 8π/3. M1 for the integral with the right limits, A1 for the area. Check against geometry. The full circle has area 16π, and a sixth of it is 8π/3, matching the sixth of a turn.Find the total area enclosed by the cardioid r = 2(1 + cos θ).
Worked answer
A = ½∫4(1 + cos θ)² dθ over 0 to 2π = 2∫(1 + 2 cos θ + cos²θ) dθ. The cos θ term dies over a full turn and cos²θ contributes π, so A = 2(2π + π) = 6π. M1 for the polar area integral, A1 for the expanded integrand, M1 for the double angle on cos²θ, M1 for the limits, A1 for the area. Squaring first and reaching for the double angle is the whole method.Find the area of one petal of the rose r = 3 cos 2θ, using the petal traced for −π/4 ≤ θ ≤ π/4.
Worked answer
A = ½∫9 cos²2θ dθ = (9/4)∫(1 + cos 4θ) dθ over the petal. The cos 4θ part vanishes across the symmetric limits, leaving (9/4)(π/2) = 9π/8. M1 for the polar area integral, M1 for the double angle, A1 for the reduced integrand, M1 for the limits, A1 for the area. The limits come from r = 0. A single loop is bounded by consecutive zeros of cos 2θ, and taking any wider interval sweeps part of a neighbouring petal as well.The curve r = 6 sin θ is a circle of radius 3. Verify its area using the polar formula with 0 ≤ θ ≤ π.
Worked answer
A = ½∫36 sin²θ dθ = 9∫(1 − cos 2θ) dθ over 0 to π = 9π, which is π × 3², the circle's area. M1 for the polar area integral, M1 for the double angle, A1 for 9π, B1 for identifying it with the area of a circle of radius 3. The circle is traced once by θ from 0 to π; running to 2π would count it twice.The circle r = 3 and the cardioid r = 2(1 + cos θ) intersect at two points. Find the exact area of the region lying inside both curves.
Worked answer
Setting 2(1 + cos θ) = 3 gives cos θ = 1/2, so the curves meet at θ = ±π/3. For |θ| < π/3 the cardioid is outside the circle, so the circle supplies the boundary; beyond π/3 the cardioid does. Using symmetry in the initial line, the area is 2[½∫9 dθ from 0 to π/3 + ½∫4(1 + cos θ)² dθ from π/3 to π]. The first piece is 3π. For the second, expanding gives 2∫(3/2 + 2 cos θ + ½cos 2θ) dθ = 2[3θ/2 + 2 sin θ + (sin 2θ)/4], which runs from π/2 + 9√3/8 at π/3 to 3π/2 at π, giving 2π − 9√3/4 before doubling. Total 7π − 9√3/2, about 14.20. M1 A1 for the intersections at θ = ±π/3, B1 for deciding which curve bounds on each side, M1 for the circular sector, M1 A1 for the cardioid piece, A1 for the exact total. Deciding which curve is nearer the pole on each side of the intersection is the whole question; a single integral of either curve is worth nothing.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise areas with polar coordinates one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.