Polar Coordinates – Area and Arc Length

In the previous lesson we saw that it is sometimes more convenient to represent curves using polar coordinates as oppose to rectangular coordinates. We are also aware from previous lessons that it is sometimes necessary to find the area under a curve or the arc length of a curve, and we have learned techniques for computing these features. However, the techniques we learned applied to curves that were represented using rectangular coordinates. In this lesson we learn how to compute these features for curves represented in polar coordinates.

Area in Polar Coordinates When computing the area under a curve in rectangular coordinates we used rectangles with infinitesimal width, 푑푥, as shown in the figure below. The use of rectangles is facilitated by the grid lines associated with the rectangular coordinate system.

y f(x)

퐵 퐴 = ∫ 푓(푥)푑푥 퐴

x A dx B

We purposely italicized the word “under” in the above description to call attention to the fact that for polar coordinates we do not compute the area under the curve but rather the area of a sector bounded by the curve. Because of the grid lines associated with polar coordinates a more suitable shape for computing the area are narrow sectors as shown in the figure below.

f(θ)

dθ

θB θA x

Therefore, instead of summing infinitesimal rectangles to find the area under a curve as we did in the rectangular coordinate system, we will sum infinitesimally narrow sectors to find the area bounded by a curve. To write the integral for computing this quantity we’ll first derive the formula for the area of a sector of a circle. We can do this using a ratio as shown below.

퐴푟푒푎 표푓 푎 푆푒푐푡표푟 푟푎푑𝑖푎푛푠 𝑖푛 푎 푠푒푐푡표푟 = 퐴푟푒푎 표푓 퐴 퐶𝑖푟푐푙푒 푟푎푑𝑖푎푛푠 𝑖푛 푐𝑖푟푐푙푒 θ 퐴 휃 = r 휋푟2 2휋 1 퐴 = 푟2휃 2

To apply this formula to the infinitesimally narrow sectors from above we need only make the following substitutions: 푟 = 푓(휃) 휃 = 푑휃

Therefore, the area of the infinitesimal sector, 푑퐴, is given as 1 푑퐴 = 푓2(휃)푑휃 2

Integrating both sides over [휃퐴, 휃퐵], we have

1 휃퐵 퐴 = ∫ 푓2(휃)푑휃 2 휃퐴

The result is stated formally below.

Area in Polar Coordinates The area bounded by the curve, 푟 = 푓(휃), and the rays 휃 = 휃퐴 and 휃 = 휃퐵, with 휃퐴 < 휃퐵, is equal to

1 휃퐵 퐴 = ∫ 푓2(휃)푑휃 2 휃퐴

Arc Length in Polar Coordinates When computing the arc length for a curve in rectangular coordinates we created an infinitesimal right triangle and used the Pythagorean theorem as shown below. 푑푥2 푑푠2 = (푑푥2 + 푑푦2) 푑푥2 푑푦 2 y f(x) 푑푠2 = (1 + ( ) ) 푑푥2 푑푥 dy dx 푑푦 2 푑푠 = (√1 + ( ) ) 푑푥 푑푥

2 푑푠 = (√1 + (푓′(푥)) ) 푑푥

x 푏 a b 2 푠 = ∫ (√1 + (푓′(푥)) ) 푑푥 푎

We can derive an expression for the arc length of a curve in polar coordinates using a similar method. However, in this case we will make use of the formula for the arc length of a circle, which we derive below using a ratio as we did for the area of a sector above.

푆푒푐푡표푟 퐴푟푐 퐿푒푛푔푡ℎ 푟푎푑𝑖푎푛푠 𝑖푛 푎 푠푒푐푡표푟 = s 퐶𝑖푟푢푚푓푒푟푒푛푐푒 푟푎푑𝑖푎푛푠 𝑖푛 푐𝑖푟푐푙푒 θ 푠 휃 r = 2휋푟 2휋 푠 = 푟휃

We can now derive the arc length of a curve using the figure below.

dθ

θ B θA x

The infinitesimal arc length is shown as the hypotenuse of a right triangle with side lengths of 푑푟 and 푟푑휃. Therefore, from the Pythagorean theorem we can write the following. 푑푠2 = 푑푟2 + 푟2푑휃2 Using a similar “trick” as we did for the rectangular coordinate case, we have 푑휃2 푑푠2 = (푑푟2 + 푟2푑휃2) 푑휃2 푑푟 2 푑푠2 = (( ) + 푟2) 푑휃2 푑휃

푑푟 2 푑푠 = √ (( ) + 푟2) 푑휃 푑휃

And since 푟 = 푓(휃) we can write the infinitesimal arc length as

푑푠 = √ 푓′(휃)2 + 푓(휃)2푑휃 Finally, to find the total arc length we integrate.

휃퐵 푠 = ∫ (√푓′(휃)2 + 푓(휃)2) 푑휃 휃퐴

The result is stated formally below.

Arc Length in Polar Coordinates The arc length of a curve, 푟 = 푓(휃), between the rays 휃 = 휃퐴 and 휃 = 휃퐵, with 휃퐴 < 휃퐵, is equal to

휃퐵 푠 = ∫ (√푓′(휃)2 + 푓(휃)2) 푑휃 휃퐴

Let’s do some examples utilizing our newly derived formulas. Example 1: Using the polar representation of circle derive an equation for the area of a circle of radius 푅. Solution: The figure below is described by the following polar equation. 푓(휃) = 2푅 푠𝑖푛(휃)

Next, we note that the circle is traced out in only a half cycle of the sine wave, therefore our integration limits are from 0 to 휋. 1 휋 퐴 = ∫ 푓2(휃)푑휃 2 0 1 휋 = ∫ (2푅)2 푠𝑖푛2(휃) 푑휃 2 0 휋 1 − 푐표푠(2휃) = 2푅2 ∫ ( ) 푑휃 0 2 휋 = ∫ (1 − 푐표푠(2휃))푑휃 0 1 휋 = 푅2 (휃 − 푠𝑖푛(2휃))| 2 0 1 1 = 푅2 ((휋 − 푠𝑖푛(2휋)) − (0 − 푠𝑖푛(0))) 2 2 = 푅2((휋 − 0) − (0 − 0)) = 휋푅2

Which agrees with area of a circle formula we are already familiar with.

Example 2: Find the area of the top half of the cardioid shown in the figure below and given by the following equation. 푓(휃) = 3 + 3푐표푠(휃)

Solution: The top half of the cardioid is traced out as 휃 varies from 0 to 휋. 1 휋 퐴 = ∫ 푓2(휃)푑휃 2 0 1 휋 = ∫ 32(1 + 푐표푠(휃))2푑휃 2 0 9 휋 = ∫ (1 + 2 푐표푠(휃) + 푐표푠2(휃))푑휃 2 0 9 휋 1 1 = ∫ (1 + 2 푐표푠(휃) + − 푠𝑖푛(2휃)) 푑휃 2 0 2 2 9 1 1 휋 = (휃 + 2 푠𝑖푛(휃) + 휃 + 푐표푠(2휃))| 2 2 4 0 9 1 1 1 1 = (휋 + 2 푠𝑖푛(휋) + 휋 + 푐표푠(2휋)) − (0 + 2 푠𝑖푛(0) + 0 + 푐표푠(0)) 2 2 4 2 4 9 1 1 1 = (휋 + 0 + 휋 + ∙ 1) − (0 + 0 + 0 + ∙ 1) 2 2 4 4 9 3 27 = ( 휋) = 휋 2 2 4

Example 3: Find the area that lies inside 푓(휃) = 2푐표푠(휃) and outside 푓(휃) = 1

Solution: The first task is to find where the two circles intersect by equating the two equations. 1 2푐표푠(휃) = 1 → 푐표푠(휃) = 2 Which gives us the following intersection points. 휋 휋 휃 = − , 3 3 The shaded area is found by computing the area bounded by 푓(휃) = 2푐표푠(휃) and then subtracting the area bounded by 푓(휃) = 1. The figure below illustrates the procedure.

x =

y y

Θ = π/3

x - x

Θ = -π/3 1 휋⁄3 1 휋⁄3 퐴 = ( ∫ 4푐표푠2(휃) 푑휃) − ( ∫ 1푑휃) 2 −휋⁄3 2 −휋⁄3 휋⁄3 1 휋⁄3 = ∫ 1 + 푐표푠(2휃) 푑휃 − ∫ 1푑휃 −휋⁄3 2 휋⁄3 휋⁄3 1 = ∫ + 푐표푠(2휃) 푑휃 −휋⁄3 2 1 1 휋⁄3 = ( 휃 + 푠𝑖푛(2휃))| 2 2 −휋⁄3 1 휋⁄3 = (휃 + 푠𝑖푛(2휃))| 2 −휋⁄3 1 휋 휋 = ((휋⁄3 + 푠𝑖푛 (2 )) − (− 휋⁄3 + 푠𝑖푛 (−2 ))) 2 3 3 1 2휋 휋 = ( + 2 푠𝑖푛 (2 )) 2 3 3

1 2휋 √3 = (( + 2 )) 2 3 2

휋 √3 = + ≅ 1.91 3 2 Example 4: Find the area of the shaded region in the figure below enclosed by 푓(휃) = 1⁄2 and 푓(휃) = 푐표푠(3휃).

f(θ)= 1/2

f(θ)= cos(3θ)

Solution: For this case we need to compute the area for the entire pedal and subtract the area of the pedal that lies outside the circle. The figure below illustrates this.

A1 - A2

To compute 퐴1 we need to determine the angle through which the given rose pedal is traced. For this we can use techniques from the previous lesson. Using 푟 = 푐표푠(휑) we find that the top half of the pedal is traced out as 휑 varies from 0 to 휋⁄2. The bottom half is traced out as 휑 varies from 휋⁄2 to 휋, or similarly from − 휋⁄2 to 0. For the current problem, where 휑 = 3휃, the top trace becomes 0 to 휋⁄6 , and the bottom trace becomes 휋⁄6 to 휋⁄3 or − 휋⁄6 to 0, respectively. Integrating using the positive angles region we have

휋⁄3 1 2 퐴1 = ( ∫ 푐표푠 (3휃) 푑휃) 2 0 1 휋⁄3 = ( ∫ (1 + 푐표푠(6휃))푑휃) 4 0 1 1 휋⁄3 = (휃 + 푠𝑖푛(6휃))| 4 6 0 1 휋 1 = ( + (푠𝑖푛(2휋) − 푠𝑖푛(0))) 4 3 6 1 휋 1 = ( + (0 − 0)) 4 3 6 휋 = 12

For the second region we need first to find when circle intersects with the first pedal of the rose function. To find this we set the two functions equal as follows: 푐표푠(3휃) = 1⁄2 The cosine function evaluates to 1⁄2, i.e. ± 휋⁄3. Therefore, the points of intersection, and hence the integration limits are computed as follows: 휋 휋 3휃 = ± → 휃 = ± 3 9

Now to compute 퐴2 we need to subtract two regions as shown in the figure below.

A A3 - 4

퐴2 = (퐴3) − (퐴4) 1 휋⁄9 1 휋⁄9 1 2 = ( ∫ 푐표푠2(3휃) 푑휃) − ( ∫ ( ) 푑휃) 2 −휋⁄9 2 −휋⁄9 2 1 휋⁄9 1 2휋 = ( ∫ (1 + 푐표푠(6휃))푑휃) − ( ∙ ) 4 −휋⁄9 8 9 1 2휋 1 2휋 2휋 휋 = ( + (푠𝑖푛 ( ) − 푠𝑖푛 (− ))) − ( ) 4 9 6 3 3 36 1 2휋 1 2휋 휋 = ( + (2 푠𝑖푛 ( ))) − ( ) 4 9 6 3 36

1 2휋 √3 휋 = ( + ) − ( ) 4 9 6 36

휋 √3 = + 36 24

Finally, the desired area is given as

퐴 = 퐴1 − 퐴2 휋 휋 √3 = − ( + ) ≅ 0.1024 12 36 24

Example 5: Find the length of the spiral curve 푓(휃) = 휃2 for [0 ≤ 휃 ≤ 휋]. Solution: Using the arc length formula from above we have

휋 휋 휋 푠 = ∫ (√푓′(휃)2 + 푓(휃)2) 푑휃 = ∫ (√4휃2 + 휃4) 푑휃 = ∫ 휃 (√4 + 휃2) 푑휃 0 0 0

To solve this integral we use the following substitution: 푢 = 4 + 휃2 푑푢 = 2휃푑휃 → 휃푑휃 = 푑푢⁄2

2 1 4+휋 = ∫ √푢푑푢 2 4 2 1 2 4+휋 1 = ( 푢3⁄2)| = ((4 + 휋2)3⁄2 − 8) ≅ 14.55 2 3 4 3

Example 7: Find the length of the cardioid given by the equation, 푓(휃) = 푎 − 푎 푐표푠(휃). Solution: The cardioid is traced out in the interval [0 ≤ 휃 ≤ 2휋].

2휋 푠 = ∫ (√푓′(휃)2 + 푓(휃)2) 푑휃 0 2휋 = ∫ (√푎2 푠𝑖푛2(휃) + 푎2(1 − 푐표푠(휃))2) 푑휃 0 2휋 = 푎 ∫ (√푠𝑖푛2(휃) + 1 − 2 푐표푠(휃) + 푐표푠2(휃)) 푑휃 0 2휋 = 푎 ∫ (√2 − 2 푐표푠(휃)) 푑휃 0 2휋 = √2푎 ∫ (√1 − 푐표푠(휃)) 푑휃 0

2휋 휃 = √2푎 ∫ (√2 푠𝑖푛2 ( )) 푑휃 0 2

2휋 휃 = 2푎 ∫ (푠𝑖푛 ( )) 푑휃 0 2 = 2푎(−2 푐표푠(휋) − (−2푐표푠(0))) = 8푎

Final Summary for Polar Coordinates – Area and Arc Length

Area in Polar Coordinates The area bounded by the curve, 푟 = 푓(휃), and the rays 휃 = 휃퐴 and 휃 = 휃퐵, with 휃퐴 < 휃퐵, is equal to

1 휃퐵 퐴 = ∫ 푓2(휃)푑휃 2 휃퐴 Arc Length in Polar Coordinates

The arc length of a curve, 푟 = 푓(휃), between the rays 휃 = 휃퐴 and 휃 = 휃퐵, with 휃퐴 < 휃퐵, is equal to

휃퐵 푠 = ∫ (√푓′(휃)2 + 푓(휃)2) 푑휃 휃퐴

By: ferrantetutoring