Assume that [a, b] is a closed interval contained in the open interval (-1, 1). Let f(x) = V1 – x², and let S' be the area of the surface obtained by revolving the graph of f on [a, b] about the x axis (refer to Figure 6.32). Show that S= 27(b – a) (Thus the surface area depends only on the width of the interval [a, b] and not on its location within (-1, 1).)

Assume that [a, b] is a closed interval contained in the open interval (-1, 1). Let f(x) = V1 – x², and let S' be the area of the surface obtained by revolving the graph of f on [a, b] about the x axis (refer to Figure 6.32). Show that S= 27(b – a) (Thus the surface area depends only on the width of the interval [a, b] and not on its location within (-1, 1).)

Name: I want an answer of question number 10. Assume that [a, b] is a closed
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Description: I want an answer of question number 10. Assume that [a, b] is a closed interval contained in theopen interval (-1, 1). Let f(x) = V1 – x², and let S' be thearea of the surface obtained by revolving the graph of f on[a, b] about the x axis (refer to F

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Chapter1: Variables, Expressions, And Integers

Section1.4: Comparing And Ordering Integers

Problem 10E

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