Paper 1 · Further Pure 1 ✦

Polar Coordinates

Stop describing a point by how far across and up. Describe it by how far out and which way round — and curves that were ugly in $x$ and $y$ become one clean line of algebra.

Syllabus 1.5Convention r ⩾ 0Area = ½∫r²dθ (memorise)

What the examiner expects you to do

Convert between Cartesian and polar coordinates and equations of curves, both directions.
Sketch simple polar curves over $0 \leq θ < 2 π$ or $- π < θ \leq π$ (or a subset), showing the significant features: symmetry, intersections with the initial line, the form of the curve at the pole, and the least / greatest values of $r$ .
Recall and use $A = \frac{1}{2} \int r^{2} d θ$ for the area of a sector.

Detailed plotting is never required — a sketch with the right features beats a laboured table of values every time.

§1 Cartesian ⇄ polar

◆ The dictionary

A point is $(r, θ)$ : $r$ is the distance from the pole (the origin), $θ$ the angle anticlockwise from the initial line (the positive $x$ -axis).

x = r \cos θ, y = r \sin θ, r^{2} = x^{2} + y^{2}, \tan θ = \frac{y}{x} .

CAIE uses the convention $r \geq 0$ . That single line has real consequences: any $θ$ making $r$ negative is simply not part of the curve.

⚠

\tan θ = y / x

does not by itself give

θ

$\tan$ repeats every $π$ , so it cannot tell $(- 1, \sqrt{3})$ from $(1, - \sqrt{3})$ . Always look at which quadrant the point is in before writing down $θ$ , or check the signs of $\cos θ$ and $\sin θ$ .

✎ Worked example — a point

Write $(- 1, \sqrt{3})$ in polar form with $r \geq 0$ and $0 \leq θ < 2 π$ .

$r = \sqrt{(- 1)^{2} + (\sqrt{3})^{2}} = \sqrt{4} = 2$ . Now $\cos θ = - \frac{1}{2}$ and $\sin θ = \frac{\sqrt{3}}{2}$ , which is the second quadrant, so $θ = \frac{2 π}{3}$ :

(- 1, \sqrt{3}) = (2, \frac{2 π}{3}) .

Taking $\arctan (\sqrt{3} / (- 1)) = - \frac{π}{3}$ blindly would have put the point in the fourth quadrant — the wrong half of the plane.

✎ Worked example — Cartesian ⟶ polar

Convert $x^{2} + y^{2} = 4 x$ . Substitute $x^{2} + y^{2} = r^{2}$ and $x = r \cos θ$ :

r^{2} = 4 r \cos θ ⟹ r = 4 \cos θ (or r = 0) .

Dividing by $r$ is safe here because $r = 0$ is recovered at $θ = \frac{π}{2}$ . The Cartesian form $(x - 2)^{2} + y^{2} = 4$ shows it is a circle of radius $2$ through the pole, centre $(2,0)$ .

✎ Worked example — polar ⟶ Cartesian

Convert $r = \frac{2}{1 - \cos θ}$ . Clear the fraction and isolate the surd:

r - r \cos θ = 2 ⟹ \sqrt{x^{2} + y^{2}} = 2 + x .

Square both sides: $x^{2} + y^{2} = 4 + 4 x + x^{2}$ , so

y^{2} = 4 (x + 1)

— a parabola. (Squaring is safe: on this curve $x \geq - 1$ , so $2 + x \geq 1 > 0$ and no spurious branch is created.)

★ A* tip — the three moves that solve every conversion

Multiply through by $r$ to create $r^{2}$ , $r \cos θ$ or $r \sin θ$ ; replace them by $x^{2} + y^{2}$ , $x$ , $y$ ; isolate any remaining lone $r$ and square. Those three moves cover essentially every conversion CAIE sets.

§2 Sketching — the four features that score

◆ The checklist the mark scheme is looking for

Symmetry. If replacing $θ$ by $- θ$ leaves $r$ unchanged, the curve is symmetric about the initial line (true for anything in $\cos θ$ ). If replacing $θ$ by $π - θ$ leaves $r$ unchanged, it is symmetric about $θ = \frac{π}{2}$ (true for anything in $\sin θ$ ).
Intersections with the initial line — evaluate $r$ at $θ = 0$ and $θ = π$ .
The form at the pole. Solve $r = 0$ ; each solution $θ_{0}$ is the direction along which the curve arrives at the pole, so the line $θ = θ_{0}$ is the tangent there.
Least and greatest $r$ , and the $θ$ at which they occur.

⚠ The

r \geq 0

convention decides the shape

Where the formula would give $r < 0$ , the curve simply stops existing. Two consequences worth memorising:

$r = a (1 + 2 \cos θ)$ has $r < 0$ for $\frac{2 π}{3} < θ < \frac{4 π}{3}$ , so under this convention you draw the outer loop only — no inner loop.
$r = a \cos 2 θ$ gives $r \geq 0$ on only half of $[0,2 π)$ , so you get two petals, not four.

Boards that allow $r < 0$ draw those extra loops. CAIE does not. Use the visualiser below to see it.

◇ The standard curves worth recognising on sight

Equation	Curve	Key features
$r = a$	Circle, centre the pole	radius $a$ ; no $θ$ dependence
$r = 2 a \cos θ$	Circle through the pole	centre $(a, 0)$ , radius $a$ ; exists for $- \frac{π}{2} \leq θ \leq \frac{π}{2}$
$r = a (1 + \cos θ)$	Cardioid	max $r = 2 a$ at $θ = 0$ ; reaches the pole at $θ = π$ ; symmetric about the initial line
$r = a (1 + 2 \cos θ)$	Limaçon	max $3 a$ at $θ = 0$ ; $r = 0$ at $θ = \frac{2 π}{3}$
$r = a \cos 2 θ$ , $r = a \sin 3 θ$	Rose curves	petals; each petal spans the $θ$ -interval where $r \geq 0$
$r = a θ$	Spiral	$r$ grows steadily with $θ$ ; starts at the pole

✎ Worked example — sketching

r = a (1 + \cos θ)

0 \leq θ < 2 π

Symmetry: $\cos (- θ) = \cos θ$ , so $r$ is unchanged — symmetric about the initial line. Sketch $0 \leq θ \leq π$ and reflect.
Initial line: $θ = 0 \Rightarrow r = 2 a$ ; $θ = π \Rightarrow r = 0$ .
At the pole: $r = 0$ only at $θ = π$ , so the curve reaches the pole along the direction $θ = π$ and that line is the tangent there.
Extremes: greatest $r = 2 a$ at $θ = 0$ ; least $r = 0$ at $θ = π$ . Also $r = a$ at $θ = \frac{π}{2}$ .

That is a heart-shaped curve — the cardioid — and those five facts are the whole sketch.

§3 The area of a sector

◆ The formula — not on MF19, so learn it

A = \frac{1}{2} \int_{α}^{β} r^{2} d θ

the area swept out by the radius as $θ$ runs from $α$ to $β$ — bounded by the curve and the two half-lines $θ = α$ , $θ = β$ .

Where the ½ and the square come from

A circular sector of radius $r$ and angle $δ θ$ has area $\frac{1}{2} r^{2} δ θ$ . Between $θ$ and $θ + δ θ$ the polar curve is almost a circular arc, so that thin slice has area

δ A \approx \frac{1}{2} r^{2} δ θ .

Adding the slices and letting $δ θ \to 0$ turns the sum into an integral, giving $A = \frac{1}{2} \int_{α}^{β} r^{2} d θ$ . $■$

★ A* method — you will always need a double angle

Squaring $r$ nearly always produces $\cos^{2} θ$ or $\sin^{2} θ$ , which you cannot integrate directly. Convert before integrating:

\cos^{2} θ = \frac{1}{2} (1 + \cos 2 θ), \sin^{2} θ = \frac{1}{2} (1 - \cos 2 θ) .

✎ Worked example — the area inside the cardioid

r = a (1 + \cos θ)

The whole curve is traced once as $θ$ goes from $0$ to $2 π$ , so

A = \frac{1}{2} \int_{0}^{2 π} a^{2} (1 + \cos θ)^{2} d θ = \frac{a^{2}}{2} \int_{0}^{2 π} (1 + 2 \cos θ + \cos^{2} θ) d θ .

Replace $\cos^{2} θ$ by $\frac{1}{2} (1 + \cos 2 θ)$ :

= \frac{a^{2}}{2} \int_{0}^{2 π} (\frac{3}{2} + 2 \cos θ + \frac{1}{2} \cos 2 θ) d θ = \frac{a^{2}}{2} [\frac{3}{2} θ + 2 \sin θ + \frac{1}{4} \sin 2 θ]_{0}^{2 π} .

Both sine terms vanish at $0$ and $2 π$ , leaving

A = \frac{a^{2}}{2} \cdot 3 π = \frac{3 π a^{2}}{2} .

✎ Worked example — a genuine sector, not a whole curve

Find the exact area of the region bounded by $r = 2 + \cos θ$ and the half-lines $θ = 0$ and $θ = \frac{π}{2}$ .

A = \frac{1}{2} \int_{0}^{π / 2} (2 + \cos θ)^{2} d θ = \frac{1}{2} \int_{0}^{π / 2} (4 + 4 \cos θ + \cos^{2} θ) d θ .

Using $\cos^{2} θ = \frac{1}{2} (1 + \cos 2 θ)$ the integrand becomes $\frac{9}{2} + 4 \cos θ + \frac{1}{2} \cos 2 θ$ , so

A = \frac{1}{2} [\frac{9}{2} θ + 4 \sin θ + \frac{1}{4} \sin 2 θ]_{0}^{π / 2} = \frac{1}{2} (\frac{9 π}{4} + 4 + 0) = \frac{9 π}{8} + 2 .

Numerically $\approx 5.534$ — worth a quick check on the calculator against $\frac{1}{2} \int_{0}^{π / 2} (2 + \cos θ)^{2} d θ$ .

★ A* tip — use the symmetry

For a curve symmetric about the initial line, integrate over $0 \leq θ \leq π$ and double it. Half the algebra, and the examiner expects to see you say why.

§4 Interactive: trace the curve, sweep the area

Pick a curve, then drag the two half-lines $θ = α$ and $θ = β$ . The shaded region is exactly what $\frac{1}{2} \int_{α}^{β} r^{2} d θ$ measures — and the panel checks the exact integral against a numerical one, so you can see the calculus is right.

Scale

a

Start

α

0°

End

β

180°

the shaded slice is ½∫r²dθ ✎

§5 Examiner traps & A* checklist

⚠ The seven most common mark-losers

Writing $\int r d θ$ or dropping the $\frac{1}{2}$ . The formula is $\frac{1}{2} \int r^{2} d θ$ , and it is not on MF19.
Integrating $\cos^{2} θ$ as if it were $\cos θ$ — convert with the double angle first.
Plotting parts of a curve where $r < 0$ . CAIE uses $r \geq 0$ , so those $θ$ are excluded.
Getting $θ$ from $\arctan (y / x)$ without checking the quadrant.
Integrating over the wrong $θ$ -range: use the range that traces the region once. For a single rose petal that is the interval where $r \geq 0$ , not $0$ to $2 π$ .
A sketch with no labelled features. Mark the greatest $r$ , the intersections with the initial line, and the tangent direction at the pole.
Losing the $r = 0$ solution when you divide an equation through by $r$ .

★ Before you turn the page

Is my answer for an area positive, and does it look the right size compared with a circle of the same rough radius?
Did I use symmetry to halve the work, and say so?
Are my $θ$ limits in radians?

§6 Video explainers (curated)

▶Converting between Cartesian and polar
Curated search — points and equations, both directions.↗ ▶Sketching polar curves
Curated search — cardioids, limaçons and rose curves.↗ ▶Area with ½∫r²dθ
Curated search — sector areas, worked with double angles.↗

§7 Check yourself

Score 0 / 5

Q1 · a point

In polar form with $r \geq 0$ and $0 \leq θ < 2 π$ , the point $(- 1, \sqrt{3})$ is:

Solution

$r = \sqrt{1 + 3} = 2$ . The point is in the second quadrant ( $x < 0, y > 0$ ), and $\cos θ = - \frac{1}{2}, \sin θ = \frac{\sqrt{3}}{2}$ give $θ = \frac{2 π}{3}$ . The trap answer $- \frac{π}{3}$ is what $\arctan (y / x)$ returns if you ignore the quadrant.

Q2 · conversion

The Cartesian equation of $r = 4 \cos θ$ is:

Solution

Multiply by $r$ : $r^{2} = 4 r \cos θ$ , so $x^{2} + y^{2} = 4 x$ . Completing the square gives $(x - 2)^{2} + y^{2} = 4$ — a circle of radius $2$ , centre $(2,0)$ , passing through the pole.

Q3 · area

The area enclosed by $r = 2 a \cos θ$ (taking $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ ) is:

Solution

$A = \frac{1}{2} \int_{- π / 2}^{π / 2} 4 a^{2} \cos^{2} θ d θ = a^{2} \int_{- π / 2}^{π / 2} (1 + \cos 2 θ) d θ = a^{2} [θ + \frac{1}{2} \sin 2 θ]_{- π / 2}^{π / 2} = π a^{2}$ .

Sanity check: the curve is a circle of radius $a$ , and $π a^{2}$ is exactly its area. ✓

Q4 · the r ⩾ 0 convention

For $0 \leq θ < 2 π$ , the curve $r = 1 + 2 \cos θ$ exists for:

Solution

We need $1 + 2 \cos θ \geq 0$ , i.e. $\cos θ \geq - \frac{1}{2}$ . On $[0,2 π)$ that holds for $0 \leq θ \leq \frac{2 π}{3}$ and $\frac{4 π}{3} \leq θ < 2 π$ . In between, the formula gives $r < 0$ , so under the CAIE convention there is no inner loop.

Q5 · full method — try it on paper first

The curve $C$ has polar equation $r = 1 + \cos θ$ for $0 \leq θ < 2 π$ .

(i) Show that $C$ is symmetric about the initial line. (ii) State the greatest and least values of $r$ and where they occur, and describe the curve at the pole. (iii) Find the exact area enclosed by $C$ .

Solution

(i) Replacing $θ$ by $- θ$ gives $1 + \cos (- θ) = 1 + \cos θ$ , which is unchanged. So the point $(r, - θ)$ is on $C$ whenever $(r, θ)$ is, and $C$ is symmetric about the initial line.

(ii) $\cos θ$ runs from $- 1$ to $1$ , so the greatest value is $r = 2$ at $θ = 0$ and the least is $r = 0$ at $θ = π$ . Since $r = 0$ only at $θ = π$ , the curve reaches the pole along that single direction, so the line $θ = π$ is the tangent at the pole. (Also $r = 1$ at $θ = \frac{π}{2}$ .)

(iii) The whole curve is traced once for $0 \leq θ < 2 π$ :

A = \frac{1}{2} \int_{0}^{2 π} (1 + \cos θ)^{2} d θ = \frac{1}{2} \int_{0}^{2 π} (1 + 2 \cos θ + \cos^{2} θ) d θ .

Using $\cos^{2} θ = \frac{1}{2} (1 + \cos 2 θ)$ , the integrand is $\frac{3}{2} + 2 \cos θ + \frac{1}{2} \cos 2 θ$ , so

A = \frac{1}{2} [\frac{3}{2} θ + 2 \sin θ + \frac{1}{4} \sin 2 θ]_{0}^{2 π} = \frac{1}{2} (3 π + 0 + 0) = \frac{3 π}{2} .

Check: the curve sits inside a circle of radius $2$ (area $4 π \approx 12.57$ ) and outside one of radius $1$ ; $\frac{3 π}{2} \approx 4.71$ is comfortably between. ✓ $■$

FP1 · Topic 5 · Polar Coordinates