Lecture 13Section 9.5 Area in Polar Coordinates
Jiwen He
1
1.1
Area of a Polar Region
Basic Polar Area
Area of a Polar Region
1
2
The area of the polar region Γ generated by
3
r = ρ(θ),
α≤θ≤β
Z
is
β
2
1
ρ(θ) dθ
2
A=
α
Proof
Let P = {θ0 , θ1 , · · · , θn } be a partition of [α, β]. Set ri = min ρ(θ) and Ri =
α≤θ≤β
1 2
1
r ∆θi ≤ Ai ≤ Ri2 ∆θi
2 i
2
max ρ(θ). Then
α≤θ≤β
2
1
Lf (P ) ≤ A ≤ Uf (P ) with f (θ) = ρ(θ)
2
Z β
Z β
2
1
A=
ρ(θ) dθ
f (θ) dθ =
α
α 2
Summing from i = 1 to i = n yields
Since P is arbitrary, we conclude
1.2
Circles
Area of a Circle of Radius a: A = πa2
Circle in Polar Coordinates
r = a, 0 ≤ θ ≤ 2π
r = −a, 0 ≤ θ ≤ 2π
r = 2a sin θ, 0 ≤ θ ≤ π
π
3π
r = −2a cos θ,
≤θ≤
2
2
Z 2π
Z β
2
1 2
1
1
ρ(θ) dθ =
a dθ = a2 · 2π = πa2
A=
2
2
0
α 2
Z
β
A=
α
2
1
ρ(θ) dθ =
2
Z
0
2π
1 2
1
−a dθ = a2 · 2π = πa2
2
2
Z π
Z π
2
2
1
1
ρ(θ) dθ =
2a sin θ dθ = 2a2 sin2 θdθ
0 2
0
α 2
Z π
Z π
1 1
1
π
2
2
− cos 2θ dθ = 2a
dθ = 2a2 · = πa2
= 2a
2
2
2
2
0
0
Z
β
A=
Z 3π
Z 3π
2 1
2
2
2
1
ρ(θ) dθ =
−2a cos θ dθ = 2a2
cos2 θdθ
π
π
2
2
α
2
2
Z 3π
Z 3π
2
2
1 1
1
π
= 2a2
+ cos 2θ dθ = 2a2
dθ = 2a2 · = πa2
π
π
2
2
2
2
2
2
Z
β
A=
4
1.3
Ribbons
Area of a Lemniscate (Ribbon): A = a2
Ribbon
3π
2
Sketch r2 = a2 cos 2θ in 4 stages: [0, π4 ], [ π4 , π2 ], [ π2 , 3π
4 ], [ 4 , π] Sketch r =
π
π π
π 3π
3π
2
a sin 2θ in 4 stages: [0, 4 ], [ 4 , 2 ], [ 2 , 4 ], [ 4 , π]
π4
Z π4
Z β
2
1 2
1
2 1
ρ(θ) dθ = 4
a cos 2θ dθ = 2a
= a2
sin 2θ
A=4
2
0 2
α 2
0
π4
Z β
Z π4
2
1
1 2
1
2
A=4
ρ(θ) dθ = 4
a sin 2θ dθ= 2a − cos 2θ
= a2
2
α 2
0 2
0
Quiz
Quiz
1.4
1. r = 2a sin θ is a
(a) line,
(b) circle,
(c) lemniscate.
2. r2 = a2 sin 2θ is a
(a) line,
(b) circle,
(c) lemniscate.
Flowers
Area of a Flower: A =
π π
4, 2
5
Flower
5π
5π
Sketch r = sin 3θ in 6 stages: [0, π6 ], [ π6 , π3 ], · · · , [ 2π
3 , 6 ], [ 6 , π] Sketch r = cos 4θ
π
π π
15π
in 16 stages: [0, 8 ], [ 8 , 4 ], · · · , [ 8 , 2π]
Z π6
Z β
2
2
1
1
ρ(θ) dθ = 6
sin 3θ dθ
A=6
0 2
α 2
π6
Z π6
1 1
3
1
π
=3
− cos 6θ dθ =
θ − sin 6θ
=
2
2
2
6
4
0
0
Z β
Z π8
2
2
1
1
A = 16
ρ(θ) dθ = 16
cos 4θ dθ
2
2
α
0
π8
Z π8
1 1
π
1
=
=8
+ cos 8θ dθ = 4 θ + sin 8θ
2
8
2
0 2
0
1.5
Limaçons
Limaçons (Snails): r = a + cos θ, a ≥ 1
Flower
Sketch r = a + cos θ, a ≥ 1, in 2 stages: [0, π], [π, 2π]
Z π
Z π
2
1
A=2
a + cos θ dθ =
a2 + 2a cos θ + cos2 θ dθ
0 2
0
π
Z π
1
1
1
1
2
2
=
a + 2a cos θ + + cos 2θ dθ = a + θ = a2 + π
2 2
2
2
0
0
Limaçon (Snail): r = 1 − 2 cos θ
6
Area between the Inner and Outer Loops
• Sketch in 4 stages: [0, 13 π], [ 13 π, π], [π, 53 π], [ 53 π, 2π].
√ √ √ • A = Aouter − Ainner = 2π + 32 3 − Ainner = 2π + 32 3 − π − 32 3 =
√
π+3 3
Area Within Outer Loop: Aouter
Z π
Z π
2
1
Aouter = 2
1 − 2 cos θ dθ =
1 − 4 cos θ + 4 cos2 θ dθ
π 2
π
3
3
π
Z π
1 − 4 cos θ + 2 + 2 cos 2θ dθ = 3θ − 4 sin θ + sin 2θ = · · ·
=
π
3
π
3
Area Within Inner Loop: Ainner
Ainner
Z
=2
0
Z
=
π
3
π
3
2
1
1 − 2 cos θ dθ =
2
Z
π
3
1 − 4 cos θ + 4 cos2 θ dθ
0
π3
1 − 4 cos θ + 2 + 2 cos 2θ dθ = 3θ − 4 sin θ + sin 2θ = · · ·
0
2
2.1
0
Area between Polar Curves
Between Polar Curves
Area between Polar Curves
7
Area between r = ρ1 (θ) and r = ρ2 (θ)
Z β2
2
1
area of Ω =
ρ2 (θ) dθ
α2 2
Z β1
2
1
−
ρ1 (θ) dθ
α1 2
Z β 2 2 1 =
ρ2 (θ) − ρ1 (θ)
dθ
α 2
if α1 = α2 = α, β1 = β2 = β.
Remark
Extra care is needed to determine the intervals of θ values (e.g, [α1 , β1 ] and
[α2 , β2 ]) over which the outer and inner boundaries of the region are traced out.
8
2.2
Between Circles
Area between Circles: r = 2 cos θ and r = 1
Area between r = 2 cos θ and r = 1
• The two intersection points: 2 cos θ = 1
⇒ [α, β] = [− π3 , π3 ]
9
⇒
cos θ =
1
2
⇒
π 5π
3, 3
• By symmetry,
area of Ω
Z β 2 2 1 dθ
=
ρ2 (θ) − ρ1 (θ)
α 2
Z π3 2 2 1 dθ
=2
2 cos θ − 1
2
0
Z π3
1 + 2 cos 2θ dθ
=
0
√
π3
π
3
= θ + sin 2θ
= +
3
2
0
Quiz
Quiz
3. area of r = 2a sin θ is :
(a) πa2 ,
(b)
1 2
πa ,
2
(c) a2 .
1 2
πa ,
2
(c) a2 .
4. area of r2 = a2 sin 2θ is :
(a) πa2 ,
2.3
(b)
Between Circle and Flower
Area outside Circle r = 2 sin θ and inside flower r = 2 sin 2θ
10
• The three intersection
points: 2 sin 2θ = 2 sin θ ⇒ 2 sin θ cos θ = sin θ
⇒ sin θ 2 cos θ − 1 = 0 ⇒ sin θ = 0 or cos θ = 12 ⇒ 0, or π3 , 5π
3 ⇒
[α, β] = [0, π3 ] and [α, β] = [ 5π
,
π]
3
• By symmetry,
Z
β
2 2 1 ρ2 (θ) − ρ1 (θ)
dθ
α 2
Z π3 2 2 1 2 sin 2θ − 2 sin θ
dθ
=
2
0
Z π3
=
cos 2θ − cos 4θ dθ
0
√
π3
1
3 3
1
=
=
sin 2θ − sin 4θ
2
4
8
0
A1 =
11
2.4
Between Circle and Limaçon
Area between Circle r = 2 sin θ and Limaçon r =
12
3
2
− sin θ
13
3
2
• The two intersection points: 2 sin θ =
1
2
− sin θ ⇒ sin θ =
π 5π
6, 6
⇒
• The area can be represented Zasπfollows:
6 1
2
A=
2 sin θ dθ
2
0
Z 5π
6 13
2
+
− sin θ dθ
π
2 2
Z 6π
2
1
2 sin θ dθ
+
5π 2
6
5
15 √
= ··· = π −
3
4
8
Outline
Contents
1 Area of a Polar Region
1.1 Basic Polar Area . .
1.2 Circles . . . . . . . .
1.3 Ribbons . . . . . . .
1.4 Flowers . . . . . . .
1.5 Limaçons . . . . . .
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1
1
4
5
5
6
2 Area between Polar Curves
2.1 Between Polar Curves . .
2.2 Between Circles . . . . . .
2.3 Circle&Flower . . . . . . .
2.4 Circle&Limaçon . . . . . .
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7
7
9
10
12
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14