1. The Polar Coordinate System
In polar coordinates, a point P is located by its distance r from the origin (called the pole) and the angle θ from the initial line (positive x-axis, measured anticlockwise).
⭐ Cambridge Convention — r ≥ 0 Always
The syllabus strictly uses r ≥ 0. When a curve equation gives a negative value of r for some θ, those points are simply not plotted — they are outside the curve. This is a key difference from some other treatments.
PointP = (r, θ) where r ≥ 0
θ range0 ≤ θ < 2π or −π < θ ≤ π (or a subset)
2. Converting Between Systems
The four fundamental relationships come directly from basic trigonometry applied to the right triangle formed by the point, pole, and the projection onto the initial line.
Polar → Cartesian
x= r cosθ
y= r sinθ
r²= x² + y²
Cartesian → Polar
r= √(x² + y²)
tanθ= y/x
r cosθ= x, r sinθ = y
The "Multiply by r" Trick
When converting polar to Cartesian, if the equation is hard to convert directly, multiply both sides by r. This creates the recognisable terms r², r cosθ, and r sinθ.
Example conversions
r = 4→ r² = 16 → x² + y² = 16 (circle, radius 4)
r = 2 cosθ→ r² = 2r cosθ → x²+y² = 2x → (x−1)²+y² = 1
θ = π/4→ tanθ = 1 → y/x = 1 → y = x (half-line, x≥0)
x² + y² = 9→ r² = 9 → r = 3
y = 3→ r sinθ = 3 → r = 3/sinθ = 3 cosecθ
3. Sketching Polar Curves — The 5-Step Method
Cambridge does not require detailed plotting, but expects all significant features to be shown clearly on a sketch.
- 1Check symmetry first — this can halve your work (see symmetry table below).
- 2Find where r = 0 — these are the points where the curve passes through the pole. Set the equation equal to zero and solve for θ.
- 3Find maximum and minimum r — maximum r gives the furthest point from the pole. Use calculus or known trig bounds.
- 4Find intersections with the initial line (θ=0, π) and with the line θ=π/2 — these give key anchor points for the sketch.
- 5Build a table of (θ, r) for key angles: 0, π/6, π/4, π/3, π/2, 2π/3, π, etc. Plot and connect smoothly.
Symmetry Rules
| Condition on equation | Symmetry about | Example |
|---|---|---|
| Replace θ with −θ gives same equation | Initial line (x-axis) | r = 1 + cosθ (since cos(−θ) = cosθ) |
| Replace θ with π−θ gives same equation | Line θ = π/2 (y-axis) | r = 1 + sinθ... actually r = 1 − cosθ |
| Replace θ with θ+π gives same r | Pole (origin) | r = sin(2θ) — symmetry about origin |
| r depends only on sin θ | Line θ = π/2 | r = 2 sinθ, r = a(1 + sinθ) |
| r depends only on cos θ | Initial line | r = a cosθ, r = a(1 + cosθ) |
4. Common Curve Types
5. Area Formula — A = ½∫r²dθ
The area enclosed by a polar curve r = f(θ) between the radii θ = α and θ = β is found by summing infinitesimally thin sectors. Each thin sector has area ½r²dθ.
⭐ Area Formula (in MF19)
A = ½ ∫αβ r² dθ
where r = f(θ), integrated between the two bounding radial lines θ = α and θ = β
🔑 Three Key Steps for Area Problems
- Identify the limits α and β carefully — use the θ-values where r=0 (curve passes through pole) for a complete loop, or the given radial lines for a sector.
- Expand r² using trig identities before integrating. Most commonly: cos²θ = ½(1+cos2θ), sin²θ = ½(1−cos2θ).
- Use symmetry to simplify — if the curve is symmetric about the initial line, integrate from 0 to π/2 and double, for example.
Area Between Two Polar Curves
Area between r = f(θ) (outer) and r = g(θ) (inner)
A= ½ ∫[f(θ)² − g(θ)²] dθsubtract inner from outer
Example 1
Coordinate Conversions
★☆☆ Standard
(a) Convert the polar equation r = 4sinθ to Cartesian form and identify the curve.
(b) Convert x² + y² − 6x = 0 to polar form.
(c) Find the Cartesian coordinates of the point (3, 2π/3) in polar coordinates.
1
Part (a) — Multiply by r
r = 4sinθ→ r² = 4r sinθ
x² + y²= 4y
x² + (y−2)²= 4complete the square
This is a circle with centre (0, 2) and radius 2.
2
Part (b) — Substitute r²=x²+y², x=r cosθ
r² − 6r cosθ= 0
r(r − 6cosθ)= 0
Polar form:r = 6cosθ (r = 0 is just the pole, included)
3
Part (c) — Polar (3, 2π/3) to Cartesian
x= 3 cos(2π/3) = 3 × (−½) = −3/2
y= 3 sin(2π/3) = 3 × (√3/2) = 3√3/2
Example 2
Sketching a Cardioid
★★☆ Challenging
Sketch the curve r = a(1 + cosθ) for 0 ≤ θ ≤ 2π, where a > 0. State the symmetry, the maximum value of r, the value of r at the pole, and where the curve crosses the initial line.
1
Symmetry
Replace θ with −θ: cos(−θ) = cosθ, so equation is unchanged. Symmetric about the initial line.
This means we only need to plot 0 ≤ θ ≤ π and reflect below.
2
Key values table
θ = 0r = a(1+1) = 2a (maximum, rightmost point)
θ = π/3r = a(1+½) = 3a/2
θ = π/2r = a(1+0) = a
θ = 2π/3r = a(1−½) = a/2
θ = πr = a(1−1) = 0 (curve passes through pole)
3
Key features to mark on sketch
- Maximum r = 2a at θ = 0 → point (2a, 0)
- Passes through pole at θ = π (curve cusps here)
- Crosses θ = π/2 line at r = a → point (0, a) in Cartesian
- Symmetric about initial line — sketch upper half then reflect
- The curve is a cardioid (heart-shaped) — smooth everywhere except the cusp at the pole
Example 3
Area Enclosed by a Polar Curve
★★☆ Challenging
Find the area enclosed by the cardioid r = a(1 + cosθ).
1
Set up the integral — use symmetry
The curve is symmetric about the initial line. Area of full curve = 2 × area for 0 ≤ θ ≤ π.
A= 2 × ½ ∫₀^π r² dθ = ∫₀^π a²(1+cosθ)² dθ
2
Expand (1 + cosθ)²
(1+cosθ)²= 1 + 2cosθ + cos²θ
cos²θ= ½(1 + cos2θ)double angle identity
Integrand= 1 + 2cosθ + ½ + ½cos2θ = 3/2 + 2cosθ + ½cos2θ
3
Integrate and evaluate
A= a²∫₀^π (3/2 + 2cosθ + ½cos2θ) dθ
= a²[3θ/2 + 2sinθ + sin2θ/4]₀^π
At θ=π:3π/2 + 2sin π + sin2π/4 = 3π/2 + 0 + 0 = 3π/2
At θ=0:0
A= 3πa²/2
Example 4
Rose Curve — Area of One Petal
★★★ A* Level
The curve C has equation r = 3sin(2θ) for 0 ≤ θ ≤ 2π. (a) State the number of petals and their positions. (b) Find the area enclosed by one petal. (c) Hence find the total area enclosed by C.
1
Part (a) — Number of petals
For r = a sin(nθ): if n is even → 2n petals; if n is odd → n petals. Here n = 2 (even), so 4 petals.
Petals occur when sin(2θ) > 0, i.e. for 0 < θ < π/2 (first petal, in first quadrant), π < θ < 3π/2 (third petal), etc.
r = 0 at θ = 0, π/2, π, 3π/2 — the curve passes through the pole 4 times.
2
Part (b) — Area of one petal (0 ≤ θ ≤ π/2)
A1= ½ ∫₀^{π/2} (3sin2θ)² dθ = ½ ∫₀^{π/2} 9sin²(2θ) dθ
sin²(2θ)= ½(1 − cos4θ)
A1= (9/4) ∫₀^{π/2} (1 − cos4θ) dθ
= (9/4)[θ − sin4θ/4]₀^{π/2}
= (9/4)[π/2 − sin(2π)/4 − 0] = (9/4)(π/2) = 9π/8
3
Part (c) — Total area = 4 petals
Atotal= 4 × 9π/8 = 9π/2
Alternatively: ½∫₀^{2π} r²dθ = 9π/2, but this double-counts petals that overlap — for a rose curve, integrate over one period and multiply. The safest approach is always to identify one petal's limits and multiply by the count.
Interactive Polar Curve Plotter
Type any polar equation in terms of θ, choose a θ range, and explore the curve. Area computation is automatic.
🌀
Polar Curve r = f(θ) — Live Plotter
—Max r
—Area (approx)
—Passes through pole?
—Initial line sym.?
Practice Questions
Question 1 — Conversions
Convert the following to the indicated form: (a) r = 5 to Cartesian, (b) r = 2/cosθ to Cartesian, (c) x² + y² = 4y to polar, (d) the Cartesian point (−1, √3) to polar form (r, θ) with r ≥ 0 and 0 ≤ θ < 2π.
(a) Square both sides: r²=25 → x²+y²=25. (b) r cosθ = 2 → x = 2. (c) Use r²=x²+y² and y=r sinθ. (d) r=√(1+3)=2; for the angle, (−1,√3) is in quadrant 2, so θ = π − arctan(√3/1) = π − π/3 = 2π/3.
✓ Solution
(a) r=5r²=25 → x²+y²=25 (circle, radius 5)
(b) r=2/cosθr cosθ = 2 → x = 2 (vertical line)
(c) x²+y²=4yr² = 4r sinθ → r = 4sinθ
(d) r= √(1+3) = 2
θ= π − arctan(√3) = π − π/3 = 2π/3
Polar form(2, 2π/3)
Question 2 — Sketching Features
For the curve r = 2 + 3cosθ (a limaçon): (a) state the symmetry, (b) find the maximum and minimum values of r and where they occur, (c) find where the curve crosses the initial line and the line θ = π/2, (d) determine whether the curve has an inner loop (find where r = 0 if it exists).
(a) cosθ is unchanged by θ→−θ. (b) Max when cosθ=1 (θ=0), min when cosθ=−1 (θ=π). (c) At θ=0: r=5; at θ=π: r=−1... but r≥0 convention! If r<0 for some θ, the curve has an inner loop. (d) r=0 when cosθ=−2/3.
✓ Solution
(a) SymmetryAbout initial line (cosθ unchanged by θ→−θ)
(b) r_max= 2+3 = 5 at θ=0
r_min= 2−3 = −1 at θ=π. Since r≥0, r actually passes through 0 before reaching −1, so there IS an inner loop.
(c) θ=0:r=5 → point (5,0) on initial line
θ=π/2:r=2+3(0)=2 → point (0,2) in Cartesian
(d) r=0:2+3cosθ=0 → cosθ=−2/3 → θ=arccos(−2/3)≈131.8°
Inner loopexists for arccos(−2/3) < θ < 2π−arccos(−2/3)
Question 3 — Area Calculation
Find the area of the region enclosed by the curve r² = 4cos(2θ) (a lemniscate). State the range of θ for which r is real, and use symmetry to simplify your calculation.
r² = 4cos(2θ) requires cos(2θ) ≥ 0, so 2θ ∈ [0,π/2]∪[3π/2,2π], giving θ ∈ [0,π/4]∪[3π/4,π]... wait, check carefully. The curve has two loops. One loop: −π/4 ≤ θ ≤ π/4. Area of one loop = ½∫r²dθ = ½∫4cos(2θ)dθ. Then double for both loops.
✓ Solution
r² = 4cos(2θ) ≥ 0 requires cos(2θ) ≥ 0, so one loop exists for −π/4 ≤ θ ≤ π/4 (and another for 3π/4 ≤ θ ≤ 5π/4).
Area of one loop= ½ ∫_{−π/4}^{π/4} 4cos(2θ) dθ
= 2[sin(2θ)/2]_{−π/4}^{π/4} = [sin(2θ)]_{−π/4}^{π/4}
= sin(π/2) − sin(−π/2) = 1 − (−1) = 2
Total area= 2 × 2 = 4
Question 4 — A* Area Between Curves
The curves C₁: r = 1 and C₂: r = 2cosθ intersect at two points. (a) Find the polar coordinates of the intersection points. (b) Find the area of the region that lies inside C₂ but outside C₁. (c) Find the total area that lies inside both curves.
(a) Set r=1=2cosθ → cosθ=½ → θ=±π/3. (b) Region inside C₂ but outside C₁: integrate ½(r₂²−r₁²) from −π/3 to π/3. (c) Region inside both: for |θ|≤π/3, the inner curve is r=1 (circle); for π/3≤|θ|≤π/2, inner curve is r=2cosθ (circle C₂). Add the two parts.
✓ Solution
(a) 2cosθ=1cosθ=½ → θ=π/3 and θ=−π/3
Points(1, π/3) and (1, −π/3)
(b) A outside= ½ ∫_{−π/3}^{π/3} [(2cosθ)²−1²] dθ
= ∫₀^{π/3} (4cos²θ−1) dθ (symmetry)
= ∫₀^{π/3} (2(1+cos2θ)−1) dθ = ∫₀^{π/3} (1+2cos2θ) dθ
= [θ+sin2θ]₀^{π/3} = π/3+sin(2π/3) = π/3+√3/2
A outside= π/3 + √3/2
(c) A both= Area(C₂) − A outside = π − (π/3+√3/2) = 2π/3 − √3/2
Note: Area of C₂ (r=2cosθ) = ½∫_{−π/2}^{π/2}4cos²θ dθ = π. This is a circle of radius 1 centred at (1,0), area = π(1²) = π ✓
Formula Reference Sheet
Complete reference for Polar Coordinates — Cambridge 9231 P1, Section 1.5.
Fundamental Relationships
x = r cosθ
y = r sinθ
r² = x² + y²
tanθ = y/x
Convention: r ≥ 0
The "Multiply by r" Trick
r · r = r²creates x²+y²
r · cosθ = x/r · r = x
r · sinθ = y
Use when equation has r, cosθ, sinθ
Area Formula (MF19)
A = ½ ∫_α^β r² dθ
Between curves: ½∫(r₁²−r₂²)dθ
Use symmetry to halve limits
cos²θ = ½(1+cos2θ)
sin²θ = ½(1−cos2θ)
Standard Curve Types
r = acircle, centre O, radius a
r = a cosθcircle, diam. on init. line
r = a(1 ± cosθ)cardioid
r = a ± b cosθlimaçon
r = a sin(nθ)rose (n or 2n petals)
r² = a² cos(2θ)lemniscate
Sketching Features to Show
Symmetry axis
Max and min r (with θ)
Intersections with initial line
Where curve meets pole (r=0)
Behaviour at pole (cusp/tangent)
Symmetry Quick-Test
θ → −θ unchanged: sym. init. line
θ → π−θ unchanged: sym. θ=π/2
cos only → initial line sym.
sin only → θ=π/2 sym.
📋 Cambridge Exam Strategy — Polar Coordinates
- Always state r ≥ 0 explicitly when converting or sketching — this is the Cambridge convention and affects which parts of curves exist.
- For conversions: if stuck, multiply both sides by r to create r², r cosθ, r sinθ — all of which convert cleanly.
- For sketching: identify symmetry first — it can halve your table of values. State the symmetry explicitly on your answer.
- For area integrals: always expand r² fully and use the double-angle identity before integrating. Never integrate without expanding first.
- For a rose curve r = a sin(nθ): area of one petal = ½∫₀^{π/n} a² sin²(nθ) dθ. Identify the correct limits by finding where r = 0.
- When the question says "area inside curve A but outside curve B": the integrand is
r_A² − r_B²(outer minus inner), with limits from the intersection points.