Explanation Given information: The given function is f ( r , θ , z ) = r . The region is bounded by the cylinder r = 1 between the planes z = − 1 and z = 1 . Calculation: The volume bounded by the surface in the form of triple integral is expressed as, V = ∭ S d V For cylindrical coordinates, x = r cos θ , y = r sin θ , z = z d V = d z r d r d θ and x 2 + y 2 = r 2 . The limits in cylindrical coordinates of z is from − 1 to 1 , limits of r is from 0 to 1, and limits of θ is from 0 to 2 π . Therefore, the volume integral in cylindrical coordinates is V = ∫ 0 2 π ∫ 0 1 ∫ − 1 1 d z r d r d θ = ∫ 0 2 π ∫ 0 1 [ z ] − 1 1 r d r d θ = ∫ 0 2 π ∫ 0 1 [ 1 + 1 ] r d r d θ = 2 ∫ 0 2 π ∫ 0 1 r d r d θ V = 2 ∫ 0 2 π [ r 2 2 ] 0 1 d θ = 2 ∫ 0 2 π [ 1 2 ] d θ = ∫ 0 2 π d θ = [ θ ] 0 2 π V = 2 π The average value of the function f over the region D in space is defined by the expression as shown below: Average value of f over D = 1 volume of D ∭ D f ( r , θ , z ) d V Substitute 2 π for volume of D and r for f ( r , θ , z )

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 14.7, Problem 85E

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Calculate the average value of the given function f over the region bounded.

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Chapter 14.7, Problem 84E

Chapter 14.7, Problem 86E

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Chapter 14 Solutions

University Calculus: Early Transcendentals (4th Edition)

Ch. 14.1 - In Exercises 1-14. evaluate the iterated...Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - In Exercises 1-14, evaluate the iterated...Ch. 14.1 - In Exercises 1-14, evaluate the iterated...Ch. 14.1 - In Exercises 1-14, evaluate the iterated...Ch. 14.1 - In Exercises 1-14, evaluate the iterated integral....Ch. 14.1 - In Exercises 1-14, evaluate the iterated...Ch. 14.1 - In Exercises 1-14, evaluate the iterated integral....Ch. 14.1 - Prob. 10E

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