Maths › Polar coordinates › Areas with polar coordinates
Areas with polar coordinates
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.
Builds on Polar curves and Compound angles and the harmonic form.
IN THIS TOPIC
- Apply A = ½∫r² dθ with limits that trace the region exactly once.
- Square and simplify r with double angle identities before integrating.
- Validate answers against circles and sectors of known area.
WHAT YOU PROBABLY THINK
Area in polar coordinates is the integral of r dθ, by analogy with the area under y = f(x).
Sectors, not strips
Between θ and θ + dθ the radius sweeps a thin sector, near enough a triangle of area ½r × r dθ. Integrating the sectors gives the polar area formula:
The r is squared, which is what the opening claim misses: a strip under a cartesian graph has area y dx, but a sector's area grows with the square of its radius. Sanity check on r = a over a full turn: ½ × a² × 2π = πa², the circle's area.
WORKED EXAMPLE
A quarter of a circle
Find the area swept by r = 2 as θ runs from 0 to π/2.
A = ½∫4 dθ from 0 to π/2 = 2 × π/2 = π.
The full circle has area 4π, and a quarter of it is π: the formula agrees with geometry before any hard curve is attempted.
Cardioids and petals
Real questions square a trig expression, so the double angle identity cos²θ = (1 + cos 2θ)/2 does the heavy lifting. Keep limits that trace the region once, and use symmetry to halve the work where the curve allows.
WORKED EXAMPLE
The area inside a cardioid
Find the area enclosed by r = 1 + cos θ.
A = ½∫(1 + cos θ)² dθ over 0 to 2π = ½∫(1 + 2 cos θ + cos²θ) dθ.
The cos θ term integrates to zero over a full turn; cos²θ contributes π via the double angle.
A = ½(2π + 0 + π) = 3π/2.
TRY IT UNSEEN
One petal of a rose
Find the area of one loop of r = cos 2θ, using the loop traced for −π/4 ≤ θ ≤ π/4.
Show the working
A = ½∫cos²2θ dθ = ½∫(1 + cos 4θ)/2 dθ over the loop.
The cos 4θ part integrates to zero across the symmetric limits, leaving ½ × ½ × π/2.
A = π/8. All four petals together cover π/2, half the unit circle's area.
THE EXAM BIT
- Square r before integrating, then reach for the double angle identities immediately.
- Choose limits by finding where r = 0; a loop runs between consecutive zeros.
- Over a full turn, plain cos θ and sin θ terms vanish; say so rather than integrating them longhand.
- Answers are exact multiples of π more often than not; a decimal is usually a wrong turn.
CHECK YOURSELF
Find the area swept by r = 3 as θ runs from 0 to 2π/3.
Show a hint
½∫9 dθ over the sector.
Show the answer
A
=
½
×
9
×
2
π
/
3
=
3
π
.
A
s
a
c
h
e
c
k
,
t
h
a
t
i
s
o
n
e
t
h
i
r
d
o
f
t
h
e
f
u
l
l
c
i
r
c
l
e
'
s
9
π
,
m
a
t
c
h
i
n
g
t
h
e
o
n
e
-
t
h
i
r
d
t
u
r
n
.
Polar area is ½∫r² dθ: thin sectors, so the radius comes in squared.
Square out, deploy cos²θ = (1 + cos 2θ)/2, and set limits between zeros of r.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.
- Apply A = ½∫r² dθ with limits that trace the region exactly once.
- Square and simplify r with double angle identities before integrating.
- Validate answers against circles and sectors of known area.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.