Explanation Given: The polar equation is r = 3 − 2 cos 4 θ Assume θ = 0 ∘ Calculate the value of r . r = 3 − 2 cos 4 θ Substitute 0 ∘ for θ and r = 3 − 2 cos ( 4 ( 0 ∘ × π 180 ) ) = 1 Calculate the value of x . x = r cos θ Substitute 1 for r and 0 ∘ for θ . x = r cos θ = 1 cos ( 0 ∘ × π 180 ) = 1 Calculate the value of y . y = r sin θ Substitute 1 for r and 0 ∘ for θ . y = 1 sin ( 0 × π 180 ) = 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 = 3 − 2 cos 4 θ x = r cos θ y = r sin θ 0.00 1.00 1.00 0.00 10.00 1.47 1.45 0.25 20.00 2.65 2.49 0.91 30.00 4.00 3.46 2.00 40.00 4.88 3.74 3.14 50.00 4.88 3.14 3.74 60.00 4.00 2.00 3.46 70.00 2.65 0.91 2.49 80.00 1.47 0.25 1.45 90.00 1.00 0.00 1.00 100.00 1.47 -0.25 1.45 110.00 2.65 -0.91 2.49 120.00 4.00 -2.00 3.46 130.00 4.88 -3.14 3.74 140.00 4.88 -3.74 3.14 150.00 4.00 -3.46 2.00 160.00 2.65 -2.49 0.91 170.00 1.47 -1.45 0.25 180.00 1.00 -1.00 0.00 190.00 1.47 -1.45 -0.25 200.00 2.65 -2.49 -0.91 210.00 4.00 -3.46 -2.00 220.00 4.88 -3.74 -3.14 230.00 4.88 -3.14 -3.74 240.00 4.00 -2.00 -3.46 250.00 2.65 -0.91 -2.49 260.00 1.47 -0.25 -1.45 270.00 1.00 0.00 -1.00 280.00 1.47 0.25 -1.45 290.00 2.65 0.91 -2.49 300.00 4.00 2.00 -3.46 310.00 4.88 3.14 -3.74 320.00 4.88 3.74 -3.14 330.00 4.00 3.46 -2.00 340.00 2.65 2.49 -0.91 350.00 1.47 1.45 -0.25 360.00 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 0 ≤ θ ≤ 2 π Calculate the area of the region using the polar area formula. A = 1 2 ∫ a b r 2 d θ (1) Substitute 3 − 2 cos 4 θ for r in equation (1). A = 1 2 ∫ 0 2 π ( 3 − 2 cos 4 θ ) 2 d θ = 1 2 ∫ 0 2 π ( 3 2 + ( 2 cos 4 θ ) 2 − ( 2 ) ( 3 ) ( 2 cos 4 θ ) ) d θ = 1 2 ∫ 0 2 π ( 9 + 4 cos 2 ( 4 θ ) + 12 cos 4 θ ) d θ Substitute u for 4 θ in the above equation

8th Edition

ISBN: 9781305266643

Author: James Stewart

Publisher: Cengage Learning

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Chapter 10.4, Problem 14E

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

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Chapter 10.4, Problem 13E

Chapter 10.4, Problem 15E

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Chapter 10 Solutions

Multivariable Calculus

Ch. 10.1 - Sketch the curve by using the parametric equations...Ch. 10.1 - Sketch the curve by using the parametric equations...Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

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The area of the region that the polar equation encloses.

Solution Summary: The author explains how the polar equation encloses the region of underset_A=11pi.

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