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Calculus Volume 2

7.4 Area and Arc Length in Polar Coordinates

Calculus Volume 27.4 Area and Arc Length in Polar Coordinates
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 7.4.1. Apply the formula for area of a region in polar coordinates.
  • 7.4.2. Determine the arc length of a polar curve.

In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function y=f(x)y=f(x) defined from x=ax=a to x=bx=b where f(x)>0f(x)>0 on this interval, the area between the curve and the x-axis is given by A=abf(x)dx.A=abf(x)dx. This fact, along with the formula for evaluating this integral, is summarized in the Fundamental Theorem of Calculus. Similarly, the arc length of this curve is given by L=ab1+(f(x))2dx.L=ab1+(f(x))2dx. In this section, we study analogous formulas for area and arc length in the polar coordinate system.

Areas of Regions Bounded by Polar Curves

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

Consider a curve defined by the function r=f(θ),r=f(θ), where αθβ.αθβ. Our first step is to partition the interval [α,β][α,β] into n equal-width subintervals. The width of each subinterval is given by the formula Δθ=(βα)/n,Δθ=(βα)/n, and the ith partition point θiθi is given by the formula θi=α+iΔθ.θi=α+iΔθ. Each partition point θ=θiθ=θi defines a line with slope tanθitanθi passing through the pole as shown in the following graph.

On the polar coordinate plane, a curve is drawn in the first quadrant, and there are rays from the origin that intersect this curve at a regular interval. Every time one of these rays intersects the curve, a perpendicular line is made from the ray to the next ray. The first instance of a ray-curve intersection is labeled θ = α; the last instance is labeled θ = β. The intervening ones are marked θ1, θ2, …, θn−1.
Figure 7.39 A partition of a typical curve in polar coordinates.

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

A circle is drawn with radius r and a sector of angle θ. It is noted that A = (1/2) θ r2.
Figure 7.40 The area of a sector of a circle is given by A=12θr2.A=12θr2.

Recall that the area of a circle is A=πr2.A=πr2. When measuring angles in radians, 360 degrees is equal to 2π2π radians. Therefore a fraction of a circle can be measured by the central angle θ.θ. The fraction of the circle is given by θ2π,θ2π, so the area of the sector is this fraction multiplied by the total area:

A=(θ2π)πr2=12θr2.A=(θ2π)πr2=12θr2.

Since the radius of a typical sector in Figure 7.39 is given by ri=f(θi),ri=f(θi), the area of the ith sector is given by

Ai=12(Δθ)(f(θi))2.Ai=12(Δθ)(f(θi))2.

Therefore a Riemann sum that approximates the area is given by

An=i=1nAii=1n12(Δθ)(f(θi))2.An=i=1nAii=1n12(Δθ)(f(θi))2.

We take the limit as nn to get the exact area:

A=limnAn=12αβ(f(θ))2dθ.A=limnAn=12αβ(f(θ))2dθ.

This gives the following theorem.

Theorem 7.6

Area of a Region Bounded by a Polar Curve

Suppose ff is continuous and nonnegative on the interval αθβαθβ with 0<βα2π.0<βα2π. The area of the region bounded by the graph of r=f(θ)r=f(θ) between the radial lines θ=αθ=α and θ=βθ=β is

A=12αβ[f(θ)]2dθ=12αβr2dθ.A=12αβ[f(θ)]2dθ=12αβr2dθ.
7.9

Example 7.16

Finding an Area of a Polar Region

Find the area of one petal of the rose defined by the equation r=3sin(2θ).r=3sin(2θ).

Solution

The graph of r=3sin(2θ)r=3sin(2θ) follows.

A four-petaled rose with furthest extent 3 from the origin at π/4, 3π/4, 5π/4, and 7π/4.
Figure 7.41 The graph of r=3sin(2θ).r=3sin(2θ).

When θ=0θ=0 we have r=3sin(2(0))=0.r=3sin(2(0))=0. The next value for which r=0r=0 is θ=π/2.θ=π/2. This can be seen by solving the equation 3sin(2θ)=03sin(2θ)=0 for θ.θ. Therefore the values θ=0θ=0 to θ=π/2θ=π/2 trace out the first petal of the rose. To find the area inside this petal, use Equation 7.9 with f(θ)=3sin(2θ),f(θ)=3sin(2θ), α=0,α=0, and β=π/2:β=π/2:

A=12αβ[f(θ)]2dθ=120π/2[3sin(2θ)]2dθ=120π/29sin2(2θ)dθ.A=12αβ[f(θ)]2dθ=120π/2[3sin(2θ)]2dθ=120π/29sin2(2θ)dθ.

To evaluate this integral, use the formula sin2α=(1cos(2α))/2sin2α=(1cos(2α))/2 with α=2θ:α=2θ:

A=120π/29sin2(2θ)dθ=920π/2(1cos(4θ))2dθ=94(0π/21cos(4θ)dθ)=94(θsin(4θ)4)0π/2=94(π2sin2π4)94(0sin4(0)4)=9π8.A=120π/29sin2(2θ)dθ=920π/2(1cos(4θ))2dθ=94(0π/21cos(4θ)dθ)=94(θsin(4θ)4)0π/2=94(π2sin2π4)94(0sin4(0)4)=9π8.
Checkpoint 7.15

Find the area inside the cardioid defined by the equation r=1cosθ.r=1cosθ.

Example 7.16 involved finding the area inside one curve. We can also use Area of a Region Bounded by a Polar Curve to find the area between two polar curves. However, we often need to find the points of intersection of the curves and determine which function defines the outer curve or the inner curve between these two points.

Example 7.17

Finding the Area between Two Polar Curves

Find the area outside the cardioid r=2+2sinθr=2+2sinθ and inside the circle r=6sinθ.r=6sinθ.

Solution

First draw a graph containing both curves as shown.

A cardioid with equation r = 2 + 2 sinθ is shown, so it has its upper heart part at the origin and the rest of the cardioid is pointed up. There is a circle with radius 6 centered at (3, π/2). The area above the cardioid but below the circle is shaded orange.
Figure 7.42 The region between the curves r=2+2sinθr=2+2sinθ and r=6sinθ.r=6sinθ.

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for θ:θ:

6sinθ=2+2sinθ4sinθ=2sinθ=12.6sinθ=2+2sinθ4sinθ=2sinθ=12.

This gives the solutions θ=π6θ=π6 and θ=5π6,θ=5π6, which are the limits of integration. The circle r=3sinθr=3sinθ is the red graph, which is the outer function, and the cardioid r=2+2sinθr=2+2sinθ is the blue graph, which is the inner function. To calculate the area between the curves, start with the area inside the circle between θ=π6θ=π6 and θ=5π6,θ=5π6, then subtract the area inside the cardioid between θ=π6θ=π6 and θ=5π6:θ=5π6:

A=circlecardioid=12π/65π/6[6sinθ]2dθ12π/65π/6[2+2sinθ]2dθ=12π/65π/636sin2θdθ12π/65π/64+8sinθ+4sin2θdθ=18π/65π/61cos(2θ)2dθ2π/65π/61+2sinθ+1cos(2θ)2dθ=9[θsin(2θ)2]π/65π/62[3θ22cosθsin(2θ)4]π/65π/6=9(5π6sin2(5π/6)2)9(π6sin2(π/6)2)(3(5π6)4cos5π6sin2(5π/6)2)+(3(π6)4cosπ6sin2(π/6)2)=4π.A=circlecardioid=12π/65π/6[6sinθ]2dθ12π/65π/6[2+2sinθ]2dθ=12π/65π/636sin2θdθ12π/65π/64+8sinθ+4sin2θdθ=18π/65π/61cos(2θ)2dθ2π/65π/61+2sinθ+1cos(2θ)2dθ=9[θsin(2θ)2]π/65π/62[3θ22cosθsin(2θ)4]π/65π/6=9(5π6sin2(5π/6)2)9(π6sin2(π/6)2)(3(5π6)4cos5π6sin2(5π/6)2)+(3(π6)4cosπ6sin2(π/6)2)=4π.
Checkpoint 7.16

Find the area inside the circle r=4cosθr=4cosθ and outside the circle r=2.r=2.

In Example 7.17 we found the area inside the circle and outside the cardioid by first finding their intersection points. Notice that solving the equation directly for θθ yielded two solutions: θ=π6θ=π6 and θ=5π6.θ=5π6. However, in the graph there are three intersection points. The third intersection point is the origin. The reason why this point did not show up as a solution is because the origin is on both graphs but for different values of θ.θ. For example, for the cardioid we get

2+2sinθ=0sinθ=−1,2+2sinθ=0sinθ=−1,

so the values for θθ that solve this equation are θ=3π2+2nπ,θ=3π2+2nπ, where n is any integer. For the circle we get

6sinθ=0.6sinθ=0.

The solutions to this equation are of the form θ=nπθ=nπ for any integer value of n. These two solution sets have no points in common. Regardless of this fact, the curves intersect at the origin. This case must always be taken into consideration.

Arc Length in Polar Curves

Here we derive a formula for the arc length of a curve defined in polar coordinates.

In rectangular coordinates, the arc length of a parameterized curve (x(t),y(t))(x(t),y(t)) for atbatb is given by

L=ab(dxdt)2+(dydt)2dt.L=ab(dxdt)2+(dydt)2dt.

In polar coordinates we define the curve by the equation r=f(θ),r=f(θ), where αθβ.αθβ. In order to adapt the arc length formula for a polar curve, we use the equations

x=rcosθ=f(θ)cosθandy=rsinθ=f(θ)sinθ,x=rcosθ=f(θ)cosθandy=rsinθ=f(θ)sinθ,

and we replace the parameter t by θ.θ. Then

dxdθ=f(θ)cosθf(θ)sinθdydθ=f(θ)sinθ+f(θ)cosθ.dxdθ=f(θ)cosθf(θ)sinθdydθ=f(θ)sinθ+f(θ)cosθ.

We replace dtdt by dθ,dθ, and the lower and upper limits of integration are αα and β,β, respectively. Then the arc length formula becomes

L=ab(dxdt)2+(dydt)2dt=αβ(dxdθ)2+(dydθ)2dθ=αβ(f(θ)cosθf(θ)sinθ)2+(f(θ)sinθ+f(θ)cosθ)2dθ=αβ(f(θ))2(cos2θ+sin2θ)+(f(θ))2(cos2θ+sin2θ)dθ=αβ(f(θ))2+(f(θ))2dθ=αβr2+(drdθ)2dθ.L=ab(dxdt)2+(dydt)2dt=αβ(dxdθ)2+(dydθ)2dθ=αβ(f(θ)cosθf(θ)sinθ)2+(f(θ)sinθ+f(θ)cosθ)2dθ=αβ(f(θ))2(cos2θ+sin2θ)+(f(θ))2(cos2θ+sin2θ)dθ=αβ(f(θ))2+(f(θ))2dθ=αβr2+(drdθ)2dθ.

This gives us the following theorem.

Theorem 7.7

Arc Length of a Curve Defined by a Polar Function

Let ff be a function whose derivative is continuous on an interval αθβ.αθβ. The length of the graph of r=f(θ)r=f(θ) from θ=αθ=α to θ=βθ=β is

L=αβ[f(θ)]2+[f(θ)]2dθ=αβr2+(drdθ)2dθ.L=αβ[f(θ)]2+[f(θ)]2dθ=αβr2+(drdθ)2dθ.
7.10

Example 7.18

Finding the Arc Length of a Polar Curve

Find the arc length of the cardioid r=2+2cosθ. r=2+2cosθ.

Solution

When θ=0,r=2+2cos0=4.θ=0,r=2+2cos0=4. Furthermore, as θθ goes from 00 to 2π,2π, the cardioid is traced out exactly once. Therefore these are the limits of integration. Using f(θ)=2+2cosθ,f(θ)=2+2cosθ, α=0,α=0, and β=2π,β=2π, Equation 7.10 becomes

L=αβ[ f(θ) ]2+[ f(θ) ]2dθ=02π[ 2+2cosθ ]2+[ 2sinθ ]2dθ=02π4+8cosθ+4cos2θ+4sin2θdθ=02π4+8cosθ+4(cos2θ+sin2θ)dθ=02π8+8cosθdθ=202π2+2cosθdθ.L=αβ[ f(θ) ]2+[ f(θ) ]2dθ=02π[ 2+2cosθ ]2+[ 2sinθ ]2dθ=02π4+8cosθ+4cos2θ+4sin2θdθ=02π4+8cosθ+4(cos2θ+sin2θ)dθ=02π8+8cosθdθ=202π2+2cosθdθ.

Next, using the identity cos(2α)=2cos2α1,cos(2α)=2cos2α1, add 1 to both sides and multiply by 2. This gives 2+2cos(2α)=4cos2α.2+2cos(2α)=4cos2α. Substituting α=θ/2α=θ/2 gives 2+2cosθ=4cos2(θ/2),2+2cosθ=4cos2(θ/2), so the integral becomes

L=202π2+2cosθdθ=202π4cos2(θ2)dθ=202π2|cos(θ2)|dθ.L=202π2+2cosθdθ=202π4cos2(θ2)dθ=202π2|cos(θ2)|dθ.

The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue, change the limits from 00 to ππ and double the answer. This strategy works because cosine is positive between 00 and π2.π2. Thus,

L=402π|cos(θ2)|dθ=80πcos(θ2)dθ=8(2sin(θ2))0π=16.L=402π|cos(θ2)|dθ=80πcos(θ2)dθ=8(2sin(θ2))0π=16.
Checkpoint 7.17

Find the total arc length of r=3sinθ.r=3sinθ.

Section 7.4 Exercises

For the following exercises, determine a definite integral that represents the area.

188.

Region enclosed by r=4r=4

189.

Region enclosed by r=3sinθr=3sinθ

190.

Region in the first quadrant within the cardioid r=1+sinθr=1+sinθ

191.

Region enclosed by one petal of r=8sin(2θ)r=8sin(2θ)

192.

Region enclosed by one petal of r=cos(3θ)r=cos(3θ)

193.

Region below the polar axis and enclosed by r=1sinθr=1sinθ

194.

Region in the first quadrant enclosed by r=2cosθr=2cosθ

195.

Region enclosed by the inner loop of r=23sinθr=23sinθ

196.

Region enclosed by the inner loop of r=34cosθr=34cosθ

197.

Region enclosed by r=12cosθr=12cosθ and outside the inner loop

198.

Region common to r=3sinθandr=2sinθr=3sinθandr=2sinθ

199.

Region common to r=2andr=4cosθr=2andr=4cosθ

200.

Region common to r=3cosθandr=3sinθr=3cosθandr=3sinθ

For the following exercises, find the area of the described region.

201.

Enclosed by r=6sinθr=6sinθ

202.

Above the polar axis enclosed by r=2+sinθr=2+sinθ

203.

Below the polar axis and enclosed by r=2cosθr=2cosθ

204.

Enclosed by one petal of r=4cos(3θ)r=4cos(3θ)

205.

Enclosed by one petal of r=3cos(2θ)r=3cos(2θ)

206.

Enclosed by r=1+sinθr=1+sinθ

207.

Enclosed by the inner loop of r=3+6cosθr=3+6cosθ

208.

Enclosed by r=2+4cosθr=2+4cosθ and outside the inner loop

209.

Common interior of r=4sin(2θ)andr=2r=4sin(2θ)andr=2

210.

Common interior of r=32sinθandr=−3+2sinθr=32sinθandr=−3+2sinθ

211.

Common interior of r=6sinθandr=3r=6sinθandr=3

212.

Inside r=1+cosθr=1+cosθ and outside r=cosθr=cosθ

213.

Common interior of r=2+2cosθandr=2sinθr=2+2cosθandr=2sinθ

For the following exercises, find a definite integral that represents the arc length.

214.

r=4cosθon the interval0θπ2r=4cosθon the interval0θπ2

215.

r=1+sinθr=1+sinθ on the interval 0θ2π0θ2π

216.

r=2secθon the interval0θπ3r=2secθon the interval0θπ3

217.

r=eθon the interval0θ1r=eθon the interval0θ1

For the following exercises, find the length of the curve over the given interval.

218.

r=6on the interval0θπ2r=6on the interval0θπ2

219.

r=e3θon the interval0θ2r=e3θon the interval0θ2

220.

r=6cosθon the interval0θπ2r=6cosθon the interval0θπ2

221.

r=8+8cosθon the interval0θπr=8+8cosθon the interval0θπ

222.

r=1sinθon the interval0θ2πr=1sinθon the interval0θ2π

For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.

223.

[T] r=3θon the interval0θπ2r=3θon the interval0θπ2

224.

[T] r=2θon the intervalπθ2πr=2θon the intervalπθ2π

225.

[T] r=sin2(θ2)on the interval0θπr=sin2(θ2)on the interval0θπ

226.

[T] r=2θ2on the interval0θπr=2θ2on the interval0θπ

227.

[T] r=sin(3cosθ)on the interval0θπr=sin(3cosθ)on the interval0θπ

For the following exercises, use the familiar formula from geometry to find the area of the region described and then confirm by using the definite integral.

228.

r=3sinθon the interval0θπr=3sinθon the interval0θπ

229.

r=sinθ+cosθon the interval0θπr=sinθ+cosθon the interval0θπ

230.

r=6sinθ+8cosθon the interval0θπr=6sinθ+8cosθon the interval0θπ

For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.

231.

r=3sinθon the interval0θπr=3sinθon the interval0θπ

232.

r=sinθ+cosθon the interval0θπr=sinθ+cosθon the interval0θπ

233.

r=6sinθ+8cosθon the interval0θπr=6sinθ+8cosθon the interval0θπ

234.

Verify that if y=rsinθ=f(θ)sinθy=rsinθ=f(θ)sinθ then dydθ=f(θ)sinθ+f(θ)cosθ.dydθ=f(θ)sinθ+f(θ)cosθ.

For the following exercises, find the slope of a tangent line to a polar curve r=f(θ).r=f(θ). Let x=rcosθ=f(θ)cosθx=rcosθ=f(θ)cosθ and y=rsinθ=f(θ)sinθ,y=rsinθ=f(θ)sinθ, so the polar equation r=f(θ)r=f(θ) is now written in parametric form.

235.

Use the definition of the derivative dydx=dy/dθdx/dθdydx=dy/dθdx/dθ and the product rule to derive the derivative of a polar equation.

236.

r=1sinθ;r=1sinθ; (12,π6)(12,π6)

237.

r=4cosθ;r=4cosθ; (2,π3)(2,π3)

238.

r=8sinθ;r=8sinθ; (4,5π6)(4,5π6)

239.

r=4+sinθ;r=4+sinθ; (3,3π2)(3,3π2)

240.

r=6+3cosθ;r=6+3cosθ; (3,π)(3,π)

241.

r=4cos(2θ);r=4cos(2θ); tips of the leaves

242.

r=2sin(3θ);r=2sin(3θ); tips of the leaves

243.

r=2θ;r=2θ; (π2,π4)(π2,π4)

244.

Find the points on the interval πθππθπ at which the cardioid r=1cosθr=1cosθ has a vertical or horizontal tangent line.

245.

For the cardioid r=1+sinθ,r=1+sinθ, find the slope of the tangent line when θ=π3.θ=π3.

For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of θ.θ.

246.

r=3cosθ,θ=π3r=3cosθ,θ=π3

247.

r=θ,r=θ, θ=π2θ=π2

248.

r=lnθ,r=lnθ, θ=eθ=e

249.

[T] Use technology: r=2+4cosθr=2+4cosθ at θ=π6θ=π6

For the following exercises, find the points at which the following polar curves have a horizontal or vertical tangent line.

250.

r=4cosθr=4cosθ

251.

r2=4cos(2θ)r2=4cos(2θ)

252.

r=2sin(2θ)r=2sin(2θ)

253.

The cardioid r=1+sinθr=1+sinθ

254.

Show that the curve r=sinθtanθr=sinθtanθ (called a cissoid of Diocles) has the line x=1x=1 as a vertical asymptote.

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