Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

Areas and Lengths in Polar Coordinates

In this section we develop the formula for the area of a region whose boundary is given by a polar equation.We need to use the formula for the area of a sector of a circle

A =1

2r2θ (1)

where r is the radius and θ is the radian measure of the central angle. Formula 1 follows from the factthat the area of a sector is proportional to its central angle:

A =θ

2π· πr2 =

1

2r2θ

Let R be the region bounded by the polar curve r = f(θ) and by the rays θ = a and θ = b, where f is apositive continuous function and where 0 < b − a ≤ 2π.

We divide the interval [a, b] into subintervals with endpoints θ0, θ1, θ2, . . . , θn and equal width ∆θ. Therays θ = θi then divide R into n smaller regions with central angle ∆θ = θi − θi−1. If we choose θ∗i in theith subinterval [θi−1, θi], then the area ∆Ai of the ith region is approximated by the area of the sector ofa circle with central angle ∆θ and radius f(θ∗i ). Thus from Formula 1 we have

∆Ai ≈1

2[f(θ∗i )]

2∆θ (2)

and so an approximation to the total area A of R is A ≈n

∑

i=1

1

2[f(θ∗i )]

2∆θ. One can see that the approxima-

tion in (2) improves as n → ∞. But the sums in (2) are Riemann sums for the function g(θ) =1

2[f(θ)]2,

so

limn→∞

n∑

i=1

1

2[f(θ∗i )]

2∆θ =

∫ b

a

1

2[f(θ)]2dθ

It therefore appears plausible (and can in fact be proved) that the formula for the area A of the polarregion R is

A =

∫ b

a

1

2[f(θ)]2dθ (3)

This formula is often written as

A =

∫ b

a

1

2r2dθ (4)

1

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find the area of each of the following regions:

(a) (b) (c) (d)

Solution:

(a) We have

A =

∫ 3π/4

π/4

1

2· 12dθ =

1

2

∫ 3π/4

π/4

dθ =1

2

(

3π

4− π

4

)

=1

2

(

2π

4

)

=π

4

(b) We have

A =

∫ 2π+π/4

3π/4

1

2· 12dθ =

1

2

∫ 2π+π/4

3π/4

dθ =1

2

(

2π +π

4− 3π

4

)

=1

2

(

2π − 2π

4

)

= π − π

4=

3π

4

(c) We have

A =

∫ 7π/4

5π/4

1

2· 12dθ =

1

2

∫ 7π/4

5π/4

dθ =1

2

(

7π

4− 5π

4

)

=1

2

(

2π

4

)

=π

4

(d) We have

A =

∫ 2π+π/4

7π/4

1

2· 12dθ =

1

2

∫ 2π+π/4

7π/4

dθ =1

2

(

2π +π

4− 7π

4

)

=1

2

(

2π − 6π

4

)

= π − 3π

4=

π

4

or

A =

∫ π/4

−π/4

1

2· 12dθ =

1

2

∫ π/4

−π/4

dθ =1

2

(π

4−

(

−π

4

))

=1

2

(π

4+

π

4

)

=π

4

EXAMPLE: Find the area of the inner loop of r = 2 + 4 cos θ.

2

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find the area of the inner loop of r = 2 + 4 cos θ.

Solution: We first find a and b:

2 + 4 cos θ = 0 =⇒ cos θ = −1

2=⇒ θ =

2π

3,

4π

3

Therefore the area is

A =

∫ 4π/3

2π/3

1

2(2 + 4 cos θ)2dθ =

∫ 4π/3

2π/3

1

2(4 + 16 cos θ + 16 cos2 θ)dθ

=

∫ 4π/3

2π/3

(2 + 8 cos θ + 8 cos2 θ)dθ =

∫ 4π/3

2π/3

(

2 + 8 cos θ + 8 · 1 + cos 2θ

2

)

dθ

=

∫ 4π/3

2π/3

(2 + 8 cos θ + 4(1 + cos 2θ)dθ =

∫ 4π/3

2π/3

(6 + 8 cos θ + 4 cos 2θ)dθ

=[

6θ + 8 sin θ + 2 sin 2θ]4π/3

2π/3= 4π − 6

√3 ≈ 2.174

EXAMPLE: Find the area enclosed by one loop of the four-leaved rose r = cos 2θ.

3

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find the area enclosed by one loop of the four-leaved rose r = cos 2θ.

Solution: Notice that the region enclosed by the right loop is swept out by a ray that rotates from θ = −π/4to θ = π/4. Therefore, Formula 4 gives

A =

∫ π/4

−π/4

1

2r2dθ =

1

2

∫ π/4

−π/4

cos2 2θdθ =

∫ π/4

0

cos2 2θdθ

=

∫ π/4

0

1

2(1 + cos 4θ)dθ =

1

2

[

θ +1

4sin 4θ

]π/4

0

=π

8

Let R be the region bounded by curves with polar equations r = f(θ), r = g(θ), θ = a, and θ = b, wheref(θ) ≥ g(θ) ≥ 0 and 0 < b − a ≤ 2π. Then the area A of R is

A =

∫ b

a

1

2

(

[f(θ)]2 − [g(θ)]2)

dθ

EXAMPLE: Find the area that lies inside r = 3 + 2 sin θ and outside r = 2.

4

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find the area that lies inside r = 3 + 2 sin θ and outside r = 2.

Solution: We first find a and b:

3 + 2 sin θ = 2 =⇒ sin θ = −1

2=⇒ θ =

7π

6, −π

6

(

11π

6

)

Therefore the area is

A =

∫ 7π/6

−π/6

1

2

[

(3 + 2 sin θ)2 − 22]

dθ =

∫ 7π/6

−π/6

1

2(5 + 12 sin θ + 4 sin2 θ)dθ

=

∫ 7π/6

−π/6

1

2

(

5 + 12 sin θ + 4 · 1 − cos 2θ

2

)

dθ =

∫ 7π/6

−π/6

1

2(5 + 12 sin θ + 2(1 − cos 2θ))dθ

=

∫ 7π/6

−π/6

1

2(7 + 12 sin θ − 2 cos 2θ)dθ =

1

2

[

7θ − 12 cos θ − sin 2θ]7π/6

−π/6

=11√

3

2+

14π

3≈ 24.187

EXAMPLE: Find the area of the region outside r = 3 + 2 sin θ and inside r = 2.

5

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find the area of the region outside r = 3 + 2 sin θ and inside r = 2.

Solution: We have

A =

∫ 11π/6

7π/6

1

2

[

22 − (3 + 2 sin θ)2]

dθ

=

∫ 11π/6

7π/6

1

2(−5 − 12 sin θ − 4 sin2 θ)dθ

=

∫ 11π/6

7π/6

1

2(−7 − 12 sin θ + 2 cos 2θ)dθ =

1

2

[

− 7θ + 12 cos θ + sin 2θ]11π/6

7π/6=

11√

3

2− 7π

3≈ 2.196

EXAMPLE: Find all points of intersection of the curves r = cos 2θ and r =1

2.

Solution: If we solve the equations r = cos 2θ and r =1

2, we get cos 2θ =

1

2and, therefore,

2θ = π/3, 5π/3, 7π/3, 11π/3

Thus the values of θ between 0 and 2π that satisfy both equations are

θ = π/6, 5π/6, 7π/6, 11π/6

We have found four points of intersection:(

1

2, π/6

)

,

(

1

2, 5π/6

)

,

(

1

2, 7π/6

)

, and

(

1

2, 11π/6

)

However, you can see from the above figure that the curves have four other points of intersection — namely,(

1

2, π/3

)

,

(

1

2, 2π/3

)

,

(

1

2, 4π/3

)

, and

(

1

2, 5π/3

)

These can be found using symmetry or by noticing that another equation of the circle is r = −1

2and then

solving the equations r = cos 2θ and r = −1

2.

6

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

Arc Length

To find the length of a polar curve r = f(θ), a ≤ θ ≤ b, we regard θ as a parameter and write theparametric equations of the curve as

x = r cos θ = f(θ) cos θ y = r sin θ = f(θ) sin θ

Using the Product Rule and differentiating with respect to θ, we obtain

dx

dθ=

dr

dθcos θ − r sin θ

dy

dθ=

dr

dθsin θ + r cos θ

So, using cos2 θ + sin2 θ = 1, we have

(

dx

dθ

)2

+

(

dy

dθ

)2

=

(

dr

dθ

)2

cos2 θ − 2rdr

dθcos θ sin θ + r2 sin2 θ

+

(

dr

dθ

)2

sin2 θ + 2rdr

dθsin θ cos θ + r2 cos2 θ =

(

dr

dθ

)2

+ r2

Assuming that f ′ is continuous, we can use one of the formulas from Section 9.2 to write the arc length as

L =

∫ b

a

√

(

dx

dθ

)2

+

(

dy

dθ

)2

dθ

Therefore, the length of a curve with polar equation r = f(θ), a ≤ θ ≤ b, is

L =

∫ b

a

√

r2 +

(

dr

dθ

)2

dθ (5)

EXAMPLE: Find the length of the curve r = θ, 0 ≤ θ ≤ 1.

7

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find the length of the curve r = θ, 0 ≤ θ ≤ 1.

Solution: We have

L =

∫ 1

0

√θ2 + 1dθ =

θ = tanx =⇒√

θ2 + 1 =√

tan2 x + 1 =√

sec2 x = | sec x| = sec x

dθ = d tanx

dθ = sec2 xdx

=

∫ π/4

0

sec3 xdx =

[

1

2(sec x tan x + ln | sec x + tanx|)

]π/4

0

=1

2(√

2 + ln(1 +√

2))

EXAMPLE: Find the length of the cardioid r = 1 − cos θ.

8

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find the length of the cardioid r = 1 − cos θ.

Solution: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5 gives

L =

∫ 2π

0

√

r2 +

(

dr

dθ

)2

dθ =

∫ 2π

0

√

(1 − cos θ)2 + sin2 θdθ

=

∫ 2π

0

√

1 − 2 cos θ + cos2 θ + sin2 θdθ

=

∫ 2π

0

√2 − 2 cos θdθ

=

∫ 2π

0

√

4 sin2 θ

2dθ

=

∫ 2π

0

2

∣

∣

∣

∣

sinθ

2

∣

∣

∣

∣

dθ

=

∫ 2π

0

2 sinθ

2dθ

= −4 cosθ

2

]2π

0

= 4 + 4 = 8

EXAMPLE: Find the length of the cardioid r = 1 + sin θ.

9

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

EXAMPLE: Find the length of the cardioid r = 1 + sin θ.

Solution 1: Note thatr = 1 + sin θ = 1 − cos

(

θ +π

2

)

Therefore the graph of r = 1 + sin θ is the rotation of the graph of r = 1 − cos θ. Hence the length of thecardioid r = 1 + sin θ is 8 by the previous Example.

Solution 2: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5gives

L =

∫ 2π

0

√

r2 +

(

dr

dθ

)2

dθ

=

∫ 2π

0

√

(1 + sin θ)2 + cos2 θdθ

=

∫ 2π

0

√

1 + 2 sin θ + sin2 θ + cos2 θdθ

=

∫ 2π

0

√2 + 2 sin θdθ

=

∫ 2π

0

√

2 − 2 cos(

θ +π

2

)

dθ =

θ +π

2= u

d(

θ +π

2

)

= du

dθ = du

=

∫ 5π/2

π/2

√2 − 2 cos udu

=

∫ 5π/2

π/2

√

4 sin2 u

2du =

∫ 5π/2

π/2

2∣

∣

∣sin

u

2

∣

∣

∣du =

∫ 2π

π/2

2∣

∣

∣sin

u

2

∣

∣

∣du +

∫ 5π/2

2π

2∣

∣

∣sin

u

2

∣

∣

∣du

=

∫ 2π

π/2

2 sinu

2du −

∫ 5π/2

2π

2 sinu

2du

= −4 cosu

2

]2π

π/2+ 4 cos

u

2

]5π/2

2π

=(

−4 cos π + 4 cosπ

4

)

+

(

4 cos5π

4− 4 cos π

)

= (4 + 2√

2) + (−2√

2 + 4) = 8

10

Section 9.4 Areas and Lengths in Polar Coordinates 2010 Kiryl Tsishchanka

Solution 3: The full length of the cardioid is given by the parameter interval 0 ≤ θ ≤ 2π, so Formula 5gives

L =

∫ 2π

0

√

r2 +

(

dr

dθ

)2

dθ =

∫ 2π

0

√

(1 + sin θ)2 + cos2 θdθ =

∫ 2π

0

√

1 + 2 sin θ + sin2 θ + cos2 θdθ

=

∫ 2π

0

√2 + 2 sin θdθ =

√2

∫ 2π

0

√1 + sin θdθ =

√2

∫ 2π

0

√1 + sin θ

√1 − sin θ√

1 − sin θdθ

=√

2

∫ 2π

0

√

1 − sin2 θ√1 − sin θ

dθ =√

2

∫ 2π

0

√cos2 θ√

1 − sin θdθ =

√2

∫ 2π

0

| cos θ|√1 − sin θ

dθ

=√

2

∫ π/2

0

cos θ√1 − sin θ

dθ −√

2

∫ 3π/2

π/2

cos θ√1 − sin θ

dθ +√

2

∫ 2π

3π/2

cos θ√1 − sin θ

dθ

Note that

∫

cos θ√1 − sin θ

dθ =

1 − sin θ = u

d(1 − sin θ) = du

− cos θdθ = du

cos θdθ = −du

= −∫

du√u

= −∫

u−1/2du = − u−1/2+1

−1/2 + 1+ C

= −2√

u + C

= −2√

1 − sin θ + C

Therefore

L = −2√

2√

1 − sin θ]π/2

0+ 2

√2√

1 − sin θ]3π/2

π/2− 2

√2√

1 − sin θ]2π

3π/2

= −2√

2(

√

1 − sin(π/2) −√

1 − sin 0)

+2√

2(

√

1 − sin(3π/2) −√

1 − sin(π/2))

−2√

2(

√

1 − sin(2π) −√

1 − sin(3π/2))

= −2√

2 (0 − 1) + 2√

2(√

2 − 0)

− 2√

2(

1 −√

2)

= 2√

2 + 4 − 2√

2 + 4 = 8

11