Polar coordinates

Further Maths · Further Pure Mathematics 1. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Polar coordinates fix a point by its distance r from the pole (the origin) and the angle θ, in radians, measured anticlockwise from the initial line (the positive x-axis). The syllabus uses r ≥ 0.

You need to change points and equations between the Cartesian and polar forms, and to sketch a curve r = f(θ) by noting its symmetry, where it meets the initial line, its greatest and least values of r, and the directions in which it passes through the pole (where r = 0).

The area swept out by the line from the pole to the curve, between two values of θ, is found by integrating ½r².

Rules to remember

x = r cos θ and y = r sin θ
r² = x² + y² and tan θ = y/x (use a sketch to choose the correct quadrant for θ)
Area of a sector = ½ ∫ r² dθ between θ = α and θ = β
To integrate squares: cos² θ = ½(1 + cos 2θ) and sin² θ = ½(1 − cos 2θ)
If r depends on cos θ only, the curve is symmetrical about the initial line; if on sin θ only, about the line θ = π/2
The curve passes through the pole when r = 0; those values of θ give the tangents at the pole
r = a is a circle with centre the pole; θ = α is a half-line from the pole; r = 2a cos θ is a circle of diameter 2a through the pole

Common mistake

Students forget the ½ in the area formula, or integrate r instead of r². Square r first, expand, then integrate and halve.

Worked example

Find the area enclosed by the curve r = 3 + sin θ, 0 ≤ θ < 2π.

r² = 9 + 6 sin θ + sin² θ = 9 + 6 sin θ + ½(1 − cos 2θ).
Integrate from 0 to 2π: 9 gives 18π, 6 sin θ gives 0, ½ gives π and −½ cos 2θ gives 0. Total = 19π.
Area = ½ × 19π = 19π/2.

Answer: 19π/2 (about 29.8)

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Polar Coordinates

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