Explanation Given: The region lies inside r = 3 sin θ and outside circle of radius, r = 1 + sin θ . Formula used: Area of a region R in polar form, A = ∬ R d A = ∫ α β ∫ g 1 ( θ ) g 2 ( θ ) r d r d θ Graph: Area of the given region lies inside r = 3 s i n θ and outside circle of radius, r = 1 + sin θ ⇒there are two graphs, one is of circle with center ( 0 , 0 ) and radius 4. Another graph is of cardiod r = 3 + 2 cos θ Plot the two graphs: Calculation: Clearly from the graph, r varies from r = 1 + sin θ to r = 3 sin θ . For limits of θ , consider the intersection of the two curves r = 1 + sin θ and r = 3 sin θ . ∴, from 1 + sin θ = 3 sin θ , θ varies from θ = π 6 to θ = 5 π 6 . ∵ of symmetry of the region, we can consider it from θ = π 6 to θ = π 2 and take twice the value of integral obtained to get area of the region

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ISBN: 9781337604789

Author: Larson

Publisher: Cengage

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Chapter 14, Problem 32RE

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To calculate: Thearea of the given region which lies inside r=3sinθ and outside circle of r=1+sinθ using a double integral and sketch the region.

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Chapter 14, Problem 31RE

Chapter 14, Problem 33RE

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Thearea of the given region which lies inside r = 3 sin θ and outside circle of r = 1 + sin θ using a double integral and sketch the region.

Solution Summary: The author calculates the area of the given region which lies inside r=3mathrmsintheta and outside circle of

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To calculate: Thearea of the given region which lies inside r=3sinθ and outside circle of r=1+sinθ using a double integral and sketch the region.

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Chapter 14 Solutions