Explanation Given: The region D is a sphere of radius a . Formula used: If f is a polar rectangle R given by 0 ≤ a ≤ r ≤ b , α ≤ θ ≤ β , where 0 ≤ β − α ≤ 2 π , then, ∬ R f ( x , y ) d A = ∫ α β ∫ a b f ( r cos θ , r sin θ ) r d r d θ (1) If g ( x ) is the function of x and h ( y ) is the function of y then, ∫ a b ∫ c d g ( x ) h ( y ) d y d x = ∫ a b g ( x ) d x ∫ c d h ( y ) d y (2) ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x (3) Calculation: From the given region D , let the equation of the sphere be x 2 + y 2 + z 2 = a 2 . Therefore, z = a 2 − x 2 − y 2 . Thus, r varies from -a to a and θ varies from 0 to 2 π . Substitute x = r cos θ and y = r sin θ in the equation (1) and obtain the required volume. ∬ D z d A = ∫ 0 2 π ∫ − a a a 2 − r 2 ( r ) d r d θ = ∫ 0 2 π ∫ − a a r a 2 − r 2 d r d θ Integrate the function with respect to r and θ by using the equation (2). ∫ 0 2 π ∫ − a a r a 2 − r 2 d r d θ = ∫ 0 2 π d θ ∫ − a a r a 2 − r 2 d r = 2 ∫ 0 2 π d θ ∫ 0 a r a 2 − r 2 d r [ by equation (3) ] Let t = r 2

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ISBN: 9781337772020

Author: Stewart

Publisher: VST

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Chapter 12.3, Problem 15E

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To find: The volume of the given solid by using polar coordinates.

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Chapter 12.3, Problem 16E

Chapter 12.3, Problem 17E

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Chapter 12 Solutions

EBK WEBASSIGN FOR STEWART'S ESSENTIAL C

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The volume of the given solid by using polar coordinates.

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Chapter 12 Solutions