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 − x 2 ; 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 = ∫ 0 1 2 π f ( x ) 1 + { f ′ ( x ) } 2 d x = ∫ 0 1 2 π 1 − x 2 1 1 − x 2 d x = ∫ 0 1 2 π d x = 2 π Also, we can write x in terms of as x = 1 − y 2

Loose-leaf Version For Calculus: Early Transcendentals

4th Edition

ISBN: 9781319055912

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 6.4, Problem 63E

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 62E

Chapter 6.4, Problem 64E

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Chapter 6 Solutions

Loose-leaf Version For Calculus: Early Transcendentals

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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