   Chapter 10.4, Problem 21E

Chapter
Section
Textbook Problem

Find the area of the region enclosed by one loop of the curve.21. r = 1 + 2 sin θ (inner loop)

To determine

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

Explanation

Formula used:

The area A of the polar region R is A=ab12r2dθ where r=f(θ).

Calculation:

The polar equation is r=1+2sinθ (1)

Assume the value of θ=0.

Calculate the value of r in the equation r=1+2sinθ,

Substitute 0 for θ in the equation (1),

r=1+2sin(0×π180)=0

Calculate the value of x in the equation x=rcosθ

Substitute 0 for r and 0 for θ in x=rcosθ,

x=rcosθ=0×cos(0×π180)=0

Calculate the value of y in y=rsinθ

Substitute 0 for r and 0 for θ,

y=0×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 table (1).

 θ r=1+2sinθ x=rcosθ y=rsinθ 0.00 1.00 1.00 0.00 10.00 1.35 1.33 0.23 20.00 1.68 1.58 0.58 30.00 2.00 1.73 1.00 40.00 2.29 1.75 1.47 50.00 2.53 1.63 1.94 60.00 2.73 1.37 2.37 70.00 2.88 0.98 2.71 80.00 2.97 0.52 2.92 90.00 3.00 0.00 3.00 100.00 2.97 –0.52 2.92 110.00 2.88 –0.98 2.71 120.00 2.73 –1.37 2.37 130.00 2.53 –1.63 1.94 140.00 2.29 –1.75 1.47 150.00 2.00 –1.73 1.00 160.00 1.68 –1.58 0.58 170.00 1.35 –1.33 0.23 180.00 1.00 –1.00 0.00 190.00 0.65 –0.64 –0.11 200.00 0.32 –0.30 –0.11 210.00 0.00 0.00 0.00 220.00 –0.29 0.22 0.18 230.00 –0.53 0.34 0.41 240.00 –0.73 0.37 0.63 250.00 –0.88 0.30 0.83 260.00 –0.97 0.17 0.95 270.00 –1.00 0.00 1.00 280.00 –0.97 –0.17 0.95 290.00 –0.88 –0.30 0.83 300.00 –0.73 –0.37 0.63 310.00 –0.53 –0.34 0.41 320.00 –0.29 –0.22 0.18 330.00 0.00 0.00 0.00 340.00 0.32 0.30 –0.11 350.00 0.65 0.64 –0.11 360.00 1.00 1.00 0.00

Graph:

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

To find the limit of integration, substitute r=0 in the polar equation r=1+2sinθ,

1+2sinθ=0sinθ=12θ=7π6and11π6

We can integrate from 7π6 to 11π6

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