Worked examples
The question
Question: find the area inside the curve r = 3.
r = 3
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
A circle around the center: once around is θ from 0 to 2π.
θ from 0 to 2π
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 2π of (3)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 2π of (9) dθ
5 Integrate
Integrate: a number gives the number times θ.
cos gives sin, divided by the number in front of θ.
A = 12[ 9θ ] from 0 to 2π
6 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: 9 times the length of the interval, 2π.
= 12 × 9 × (2π − 0)
So the area is 9π square units, about 28.27.
A = 9π ≈ 28.27
7 Check
Let's check: the curve is a circle, radius 3.
π r squared gives 9π.
The same.
Right!
circle, radius 3: π × (3)² = 9π
Let's look back.
First: draw it.
Then: one trip around.
Then: write the integral.
Then: square it.
Then: integrate.
Then: top minus bottom.
Last: check.
▶ Watch it solved
The question
Question: find the area inside the curve r = 2 + 2cos θ.
r = 2 + 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
The curve goes around once as θ goes from 0 to 2π.
θ from 0 to 2π
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 2π of (2 + 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 the bracket: first squared, two times first times second, second squared.
A = 12∫ from 0 to 2π of (4 + 8cos θ + 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: 4 plus 2 is 6.
= 12∫ from 0 to 2π of (6 + 8cos θ + 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; sin gives minus cos.
A = 12[ 6θ + 8sin θ + sin 2θ ] from 0 to 2π
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: 6 times the length of the interval, 2π.
= 12 × 6 × (2π − 0)
So the area is 6π square units, about 18.85.
A = 6π ≈ 18.85
8 Check
Let's check with a calculator: it adds up 1000 thin slices.
That gives about 18.85, the same as our answer.
Right!
1000 thin slices: ≈ 18.85
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.
▶ Watch it solved
The question
Question: find the area inside the curve r = 2 + sin θ.
r = 2 + sin 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
The curve goes around once as θ goes from 0 to 2π.
θ from 0 to 2π
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 2π of (2 + sin 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 the bracket: first squared, two times first times second, second squared.
A = 12∫ from 0 to 2π of (4 + 4sin θ + sin2 θ) dθ
5 Use the half-angle rule
The half-angle rule: sin squared is 1 minus cos of double the angle, over 2.
sin2 θ = 1 − cos 2θ2
Use it, then add the plain numbers: 4 plus 1/2 is 9/2.
= 12∫ from 0 to 2π of (92 + 4sin θ − 12 cos 2θ) dθ
6 Integrate
Integrate: a number gives the number times θ.
cos gives sin, divided by the number in front of θ; sin gives minus cos.
A = 12[ 92θ − 4cos θ − sin 2θ4 ] from 0 to 2π
7 Top minus bottom
Every sin part is 0 at these limits: sin of a whole number of half turns is 0.
And cos 2π − cos 0 is 0.
sin 0 = sin π = sin 2π = 0
Only the θ part is left: 9/2 times the length of the interval, 2π.
= 12 × 92 × (2π − 0)
So the area is 9π over 2 square units, about 14.14.
A = 9π2 ≈ 14.14
8 Check
Let's check with a calculator: it adds up 1000 thin slices.
That gives about 14.14, the same as our answer.
Right!
1000 thin slices: ≈ 14.14
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.
▶ Watch it solved
