Explanation Calculation: Consider the given equation and draw the figure as shown: r = 1 + 2 sin θ The value of ? can be calculated by making r = 0 , 1 + 2 sin θ = 0 2 sin θ = − 1 sin θ = − 1 2 θ = 7 π 6 , 11 π 6 Now, look for the region which is outside the small loop and inside the larger loop of the limacon. Area bounded = ∫ θ 1 θ 2 ∫ r 1 r 2 r d r d θ ∫ 0 2 π ∫ 0 1 + 2 sin θ r d r d θ = ∫ 0 2 π [ r 2 2 ] 0 1 + 2 sin θ d θ = ∫ 0 2 π [ ( 1 + 2 sin θ ) 2 2 ] d θ (Since we are looking for the region which is outside of small loop and inside of large loop, the Area = 2 times of minus ?= 3 π 2 ) ∫ 0 2 π ∫ 0 1 + 2 sin θ r d r d θ = 2 [ 1 2 ∫ π 2 7 π 6 [ 1 + 2 sin θ ] 2 d θ − 1 2 ∫ 7 π 6 3 π 2 ( 1 + 2 sin θ ) 2 d θ ] Now simplify ( 1 + 2 sin θ ) 2 ( 1 + 2 sin θ ) 2 = 1 + 4 sin 2 θ + 4 sin θ = 1 + 4 ( 1 − cos 2 θ 2 ) + 4 sin θ = 1 + 2 ( 1 − cos 2 θ ) + 4 sin θ = 1 + 2 − 2 cos 2 θ + 4 sin θ = 3 − 2 cos 2 θ + 4 sin θ ( 1 + 2 sin θ ) 2 = 2 [ 1 2 ∫ π 2 7 π 6 [ 1 + 2 sin θ ] 2 d θ − 1 2 ∫ 7 π 6 3 π 2 ( 1 + 2 sin θ ) 2 d θ ] = [ ∫ π 2 7 π 6 ( 3 − 2 cos 2 θ + 4 sin θ ) d θ − ∫ 7 π 6 3 π 2 ( 3 − 2 cos 2 θ + 4 sin θ ) d θ ] − − − − − − > 1 ∫ π 2 7 π 6 ( 3 − 2 cos 2 θ + 4 sin θ ) d θ = [ 3 θ − 2 sin 2 θ 2 − 4 cos θ ] π 2 7 π 6 = [ 3 θ − 2 sin θ cos θ − 4 cos θ ] π 2 7 π 6 = [ 7 π 2 − 2 sin

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ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 12, Problem 55SP

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To calculate: the area outside the small loop and inside the larger loop of the limacon r=1+2sinθ by using double integral.

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Chapter 12, Problem 54SP

Chapter 12, Problem 56SP

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Chapter 12 Solutions

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the area outside the small loop and inside the larger loop of the limacon r = 1 + 2 sin θ by using double integral.

Solution Summary: The author calculates the area outside the small loop and inside the larger loop of the limacon by using a double integral.

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To calculate: the area outside the small loop and inside the larger loop of the limacon r=1+2sinθ by using double integral.

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Chapter 12 Solutions