   Chapter 10.4, Problem 16E

Chapter
Section
Textbook Problem

Graph the curve and find the area that it encloses.16. r = 1 + 5 sin 6θ

To determine

To find: The area of the region that the polar equation encloses.

Explanation

Given:

The polar equation is r=1+5sin6θ .

Assume θ=0

Calculate the value of r.

r=1+5sin6θ=1+5sin6(0×π180)=1.00

Calculate the value of x.

x=rcosθ

Substitute 1 for r and 0 for θ .

x=rcosθ=1×cos(0×π180)=1

Calculate the value of y.

y=rsinθ

Substitute 1 for r and 0 for θ .

y=1×sin(0×π180)=0

Similarly calculate the values of x and y using the value of θ from 0 to 360 .

Tabulate the values of x and y in the table (1).

 θ r=1+cos2(5θ) x=rcosθ y=rsinθ 0.00 1.00 1.00 0.00 10.00 5.33 5.25 0.93 20.00 5.33 5.01 1.82 30.00 1.00 0.87 0.50 40.00 -3.33 -2.55 -2.14 50.00 -3.33 -2.14 -2.55 60.00 1.00 0.50 0.87 70.00 5.33 1.82 5.01 80.00 5.33 0.93 5.25 90.00 1.00 0.00 1.00 100.00 -3.33 0.58 -3.28 110.00 -3.33 1.14 -3.13 120.00 1.00 -0.50 0.87 130.00 5.33 -3.43 4.08 140.00 5.33 -4.08 3.43 150.00 1.00 -0.87 0.50 160.00 -3.33 3.13 -1.14 170.00 -3.33 3.28 -0.58 180.00 1.00 -1.00 0.00 190.00 5.33 -5.25 -0.93 200.00 5.33 -5.01 -1.82 210.00 1.00 -0.87 -0.50 220.00 -3.33 2.55 2.14 230.00 -3.33 2.14 2.55 240.00 1.00 -0.50 -0.87 250.00 5.33 -1.82 -5.01 260.00 5.33 -0.93 -5.25 270.00 1.00 0.00 -1.00 280.00 -3.33 -0.58 3.28 290.00 -3.33 -1.14 3.13 300.00 1.00 0.50 -0.87 310.00 5.33 3.43 -4.08 320.00 5.33 4.08 -3.43 330.00 1.00 0.87 -0.50 340.00 -3.33 -3.13 1.14 350.00 -3.33 -3.28 0.58 360.00 1.00 1.00 0.00

Graph:

The graph is plotted for x and y as shown in figure (1).

From the figure (1), the curve lies on all the quadrant, thus the limit tends to be 0θ2π

Calculate the area of the region using the polar area formula.

A=12abr2dθ (1)

Substitute (1+5sin6θ) for r in the equation (1).

A=12abr2dθ=1202π(1+5sin6θ)2dθ=1202π(12+(5sin6θ)2+(2)(1)(5sin6θ))dθ=1202π(1+25sin2(6θ)+10sin6θ)dθ

Substitute u for 6θ in the above equation

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