Explanation Given: The provided equations are as shown below: y = x 16 − x 2 , y = 0 , x = 0 and x = 4 . Formula used: ∫ a 2 − x 2 d x = 1 2 a [ ( 2 x 2 − a 2 ) ( a 2 − x 2 ) + a sin − 1 ( x a ) ] Volume of the bounded region is: V = 2 π ∫ a b x y d x Calculation: The region bounded by the equations isshown in the text book. Now, find the volume of the region bounded by the equations y = x 16 − x 2 , y = 0 , x = 0 and x = 4 . For this note that the volume of the bounded region is V is, V = 2 π ∫ a b x y d x Now, substitute the values and then find the volume of the region as calculated below, V = 2 π ∫ 0 4 x ( x 16 − x 2 ) d x = 2 π ∫ 0 4

Calculus: Early Transcendental Functions (MindTap Course List)

6th Edition

ISBN: 9781285774770

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 8.6, Problem 71E

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To calculate: The volume of the region bounded by the following equations y=x16−x2,y=0,x=0 and x=4.

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Chapter 8.6, Problem 70E

Chapter 8.6, Problem 72E

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Chapter 8 Solutions

Calculus: Early Transcendental Functions (MindTap Course List)

Ch. 8.1 - Choosing an Antiderivative In Exercises 3 and 4,...Ch. 8.1 - Choosing an Antiderivative In Exercises 14, select...Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, select the...Ch. 8.1 - Choosing a Formula In Exercises 5-14, selectthe...

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The volume of the region bounded by the following equations y = x 16 − x 2 , y = 0 , x = 0 and x = 4 .

Solution Summary: The author calculates the volume of the region bounded by the equations shown in the book.

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To calculate: The volume of the region bounded by the following equations y=x16−x2,y=0,x=0 and x=4.

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Chapter 8 Solutions