Explanation Let y = f : [ a , b ] → R be continuous and f ( x ) ≥ 0 . Consider the curve C given by the graph of the function f . Let S be the surface generated by revolving this curve about the x -axis. We will define the surface area of S in terms of an integral expression. Let us find the area of the surface generated by revolving the curve y = 1 2 ( x 2 + 1 ) ; 0 ≤ x ≤ 1 about the x -axis between the interval [a, b]. Here the function y is increasing hence it is one-one and onto. Hence, the volume of object formed after rotation is given by V = ∫ a b 2 π f ( x ) 1 + { f ′ ( x ) } 2 d x = ∫ a b 2 π { 1 2 ( x 2 + 1 ) } 1 + x 2 d x = ∫ a b π ( x 2 + 1 ) 1 + x 2 d x Also, we can write x in terms of as x = 2 y − 1 ...

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ISBN: 9781319055844

Author: FRANZOSA

Publisher: VST

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Chapter 6.4, Problem 61E

To determine

The function of which the curve when rotated about the x-axis, the region R under the graph of given function for given interval equals the volume obtained by rotating the same region R about the y-axis.

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Chapter 6.4, Problem 60E

Chapter 6.4, Problem 62E

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Chapter 6 Solutions

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The function of which the curve when rotated about the x -axis, the region R under the graph of given function for given interval equals the volume obtained by rotating the same region R about the y -axis.

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The function of which the curve when rotated about the x-axis, the region R under the graph of given function for given interval equals the volume obtained by rotating the same region R about the y-axis.

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Chapter 6 Solutions