Questions on areas with polar coordinates

Question 1

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.

Accepted 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.

Question 2

Find the area swept by r = 4 as θ runs from 0 to π/3.

Accepted 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.

Question 3

A sector of the circle r = 4 has area 4π. Find the angle θ it subtends.

Accepted Answer

A = ½∫16 dθ = 8θ over the sector, so 8θ = 4π gives θ = π/2. M1 for rearranging the sector formula for θ, A1 for θ = π/2. A quarter turn of a circle of radius 4 has area a quarter of 16π, which is 4π, confirming the answer.

Question 4

A sector of a polar curve r = f(θ) has area A over some interval of θ. If the curve is scaled to r = 2f(θ) over the same interval, state the new area in terms of A, and explain why.

Accepted Answer

The new area is 4A. B1. Since A = ½∫r² dθ, replacing r by 2r replaces r² by 4r², so every value inside the integral is multiplied by 4 and so is the integral itself. B1 for the reasoning. A cartesian strip of area y dx behaves differently: doubling y only doubles the area.

Question 5

Find the total area enclosed by the cardioid r = 2(1 + cos θ).

Accepted 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.

Question 6

Find the area of one petal of the rose r = 3 cos 2θ, using the petal traced for −π/4 ≤ θ ≤ π/4.

Accepted Answer

A = ½∫9 cos²2θ dθ = (9/4)∫(1 + cos 4θ) dθ over the petal. The cos 4θ part vanishes because −π/4 to π/4 is one full period of cos 4θ, 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.

Question 7

The curve r = 6 sin θ is a circle of radius 3. Verify its area using the polar formula with 0 ≤ θ ≤ π.

Accepted 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.

Question 8

Sketch the spiral r = 2θ for 0 ≤ θ ≤ π, and find the exact area of the region swept out by the radius over that interval.

Accepted Answer

The curve starts at the pole and unwinds anticlockwise, r growing steadily with the angle until it ends at r = 2π on the half-line θ = π. A = ½∫4θ² dθ from 0 to π = [2θ³/3] = 2π³/3, about 20.7. B1 for a sketch starting at the pole and reaching 2π at θ = π, M1 for the polar area integral with the right limits, A1 for the antiderivative 2θ³/3, A1 for the area. No identity is needed: r² is a power of θ, so it integrates directly.

Question 9

A curve has equation r = sin θ + cos θ. Show that r² = 1 + sin 2θ, and hence find the exact area swept by the curve as θ runs from 0 to π/2.

Accepted Answer

r² = sin²θ + 2 sin θ cos θ + cos²θ = (sin²θ + cos²θ) + 2 sin θ cos θ = 1 + sin 2θ, using sin²θ + cos²θ = 1 and the double angle sin 2θ = 2 sin θ cos θ. M1 for expanding the square, A1 for r² = 1 + sin 2θ. The area is A = ½∫(1 + sin 2θ) dθ from 0 to π/2 = ½[θ − (cos 2θ)/2] from 0 to π/2. M1 for the polar area integral, A1 for the antiderivative. Substituting the limits, ½[(π/2 − (cos π)/2) − (0 − (cos 0)/2)] = ½[(π/2 + 1/2) − (−1/2)] = π/4 + 1/2. A1 for the exact area.

Question 10

Find the exact area swept by the spiral r = 3θ as θ runs from π to 2π.

Accepted Answer

A = ½∫9θ² dθ from π to 2π = (3/2)[θ³] from π to 2π = (3/2)(8π³ − π³) = (3/2)(7π³) = 21π³/2. M1 for the polar area integral, A1 for the antiderivative (3/2)θ³, M1 for substituting both limits, A1 for the area. Unlike the spiral swept from the pole, both limits here are nonzero, so both must be substituted and subtracted.

Question 11

The circle r = 4 and the cardioid r = 2 + 2cos θ touch at θ = 0, and the circle never lies inside the cardioid. Find the exact area of the region inside the circle and outside the cardioid.

Accepted Answer

A = ½∫[16 − (2 + 2cos θ)²] dθ over 0 to 2π. Expanding, (2 + 2cos θ)² = 4 + 8cos θ + 4cos²θ. M1 for the difference integral, A1 for the expansion. Over a full turn, ∫cos θ dθ = 0 and ∫cos²θ dθ = π using the double angle identity, so ∫(2 + 2cos θ)² dθ = 8π + 0 + 4π = 12π and ∫16 dθ = 32π. M1 for both integrals, using the double angle where needed. So A = ½(32π − 12π) = 10π. A1 for the area.

Question 12

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.

Accepted 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/2. 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.

Question 13

Sketch the limaçon r = 3 + 2 cos θ and the circle r = 2 on the same diagram, and find the exact area of the region inside the limaçon and outside the circle.

Accepted Answer

The limaçon runs from r = 5 at θ = 0 down to r = 1 at θ = π, never reaching the pole; the circle is centred on the pole with radius 2. Setting 3 + 2 cos θ = 2 gives cos θ = −1/2, so the curves meet at θ = ±2π/3, and between those angles the limaçon is the outer curve. A = ½∫[(3 + 2 cos θ)² − 4] dθ from −2π/3 to 2π/3, and by symmetry this is ∫(7 + 12 cos θ + 2 cos 2θ) dθ from 0 to 2π/3 = [7θ + 12 sin θ + sin 2θ] = 14π/3 + 6√3 − √3/2. Total 14π/3 + 11√3/2, about 24.2. B1 for the two sketches with the limaçon enclosing the circle near θ = 0, M1 for setting the curves equal, A1 for θ = ±2π/3, M1 for the difference integral ½∫(outer² − inner²) dθ with those limits, M1 for expanding and using the double angle, A1 for the exact area. Both curves bound the region over the same range of θ, so the subtraction happens inside one integral; contrast the circle-and-cardioid region, where each curve took its own range.

Question 14

The circles r = 2 cos θ and r = 2 sin θ intersect at the pole and at one further point. Explain why solving 2 cos θ = 2 sin θ finds only one of the two intersections, then find the exact area of the region lying inside both circles.

Accepted Answer

Solving gives tan θ = 1, so θ = π/4 and the shared point (√2, π/4). The pole is missed because each curve arrives there at a different angle: r = 2 sin θ vanishes at θ = 0 while r = 2 cos θ vanishes at θ = π/2, so no single θ satisfies both equations there, and the pole must be checked separately. For 0 ≤ θ ≤ π/4 the boundary of the common region is the nearer curve r = 2 sin θ; past π/4 it is r = 2 cos θ. By symmetry about θ = π/4, A = 2 × ½∫4 sin²θ dθ from 0 to π/4 = 2∫(1 − cos 2θ) dθ = 2[θ − ½ sin 2θ] = 2(π/4 − 1/2) = π/2 − 1, about 0.571. M1 for solving the equation, A1 for θ = π/4, B1 for the pole explanation, M1 for splitting the region with the correct inner curve on each side, M1 for the double angle and integration, A1 for the exact area.

Question 15

The limaçon r = 1 + 2cos θ has an inner loop, traced as θ runs from 2π/3 to 4π/3, while the outer boundary is traced as θ runs from −2π/3 to 2π/3. Show that r² = 3 + 4cos θ + 2cos 2θ, find the area enclosed by the outer boundary and the area of the inner loop, and hence find the exact area enclosed by the outer boundary but outside the inner loop.

Accepted Answer

r² = (1 + 2cos θ)² = 1 + 4cos θ + 4cos²θ. Using cos²θ = (1 + cos 2θ)/2, this is 1 + 4cos θ + 2(1 + cos 2θ) = 3 + 4cos θ + 2cos 2θ. M1 for expanding the square, A1 for r² = 3 + 4cos θ + 2cos 2θ.
Outer boundary: by symmetry, A_outer = ½∫[−2π/3, 2π/3](3 + 4cos θ + 2cos 2θ) dθ = ∫[0, 2π/3](3 + 4cos θ + 2cos 2θ) dθ = [3θ + 4sin θ + sin 2θ] from 0 to 2π/3 = 2π + 2√3 − √3/2 = 2π + 3√3/2. M1 for using the symmetry and the antiderivative, A1 for the value.
Inner loop: by the same symmetry about θ = π, A_inner = ∫[2π/3, π](3 + 4cos θ + 2cos 2θ) dθ = [3θ + 4sin θ + sin 2θ] from 2π/3 to π = 3π − (2π + 3√3/2) = π − 3√3/2. M1 for the same antiderivative over the new limits, A1 for the value.
The region enclosed by the outer boundary but outside the inner loop is A_outer − A_inner = (2π + 3√3/2) − (π − 3√3/2) = π + 3√3. A1 for the exact area. Integrating r² over the full turn 0 to 2π instead gives A_outer + A_inner = 3π, since squaring makes the inner loop's negative r contribute positively a second time: that total is neither the area of the outer boundary alone nor of the region excluding the loop, which is why the two pieces have to be found and combined separately.