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Area inside r = 2cos θ

Step by step, with the reason for every step · Area in polar coordinates · Calculus 2

Step by step

The question

Question: find the area inside the curve r = 2cos θ.

r = 2cos θ

1 Draw it

A polar grid: circles around the center, and a line for each angle.

For each angle θ, go out a distance r.

The curve closes up.

2 One trip around

This circle is traced once as θ goes from 0 to π.

Going to 2π would go round it twice.

θ from 0 to π

3 Write the integral

Cut the area into thin slices from the center.

Each is almost a thin triangle: one half r squared dθ, like a slice of a circle.

slice ≈ 12r2 dθ

Add up the slices: A is one half of the integral of r squared.

A = 12∫ from 0 to π of (2cos θ)2 dθ

Careful!

Do not forget the one half!

A slice is half of r squared dθ.

Careful: not A = ∫ r2 dθ ✗

4 Square it

Square r.

A = 12∫ from 0 to π of (4cos2 θ) dθ

5 Use the half-angle rule

The half-angle rule: cos squared is 1 plus cos of double the angle, over 2.

cos2 θ = 1 + cos 2θ2

Use it, then add the plain numbers: 2.

= 12∫ from 0 to π of (2 + 2 cos 2θ) dθ

6 Integrate

Integrate: a number gives the number times θ.

cos gives sin, divided by the number in front of θ.

A = 12[ 2θ + sin 2θ ] from 0 to π

7 Top minus bottom

Every sin part is 0 at these limits: sin of a whole number of half turns is 0.

sin 0 = sin π = sin 2π = 0

Only the θ part is left: 2 times the length of the interval, π.

= 12 × 2 × (π − 0)

So the area is π square units, about 3.14.

A = π ≈ 3.14

8 Check

Let's check: the curve is a circle with diameter 2, radius 1.

π r squared gives π.

The same.

Right!

circle, radius 1: π × (1)² = π

Let's look back.

First: draw it.

Then: one trip around.

Then: write the integral.

Then: square it.

Then: use the half-angle rule.

Then: integrate.

Then: top minus bottom.

Last: check.

Answer
A = π ≈ 3.14

Your turn

Try this one yourself, then check:

Find the area inside the curve r = 2sin θ
Show the answer
A = π

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