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).)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I want an answer of question number 10.
![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).)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1974f639-b7c8-43df-92b8-417d565eba23%2Faca825ee-9ef4-448c-8828-13c96c78c183%2F8coyl8t.png&w=3840&q=75)
Transcribed Image Text: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).)
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