The volume for solid of revolution between curves is the volume of the object between a curve f(x) and a curve g(x) rotated around the x - axis on an interval [a,b] given by - AV = = Sπ (f(x))² – (g(x))² - a BV = v = f*((f(x))² - (g(x))²) a CV= 1 = [((f(x))² - (g(x))²) π a b DV= = ['n ((f(x))² - n(g(x))²) T a

The volume for solid of revolution between curves is the volume of the object between a curve f(x) and a curve g(x) rotated around the x - axis on an interval [a,b] given by - AV = = Sπ (f(x))² – (g(x))² - a BV = v = f*((f(x))² - (g(x))²) a CV= 1 = [((f(x))² - (g(x))²) π a b DV= = ['n ((f(x))² - n(g(x))²) T a

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

ChapterA: Appendix

SectionA.2: Geometric Constructions

Problem 10P: A soda can has a volume of 25 cubic inches. Let x denote its radius and h its height, both in...

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I found a textbook example which intergrates an area function to find the volume of a solid of revolution bounded by the curve f(x) = (x-1)2 about the x-axis and the lines x=0 and x=2 using the disk method. The example shows a final solution change of intergration interval to x=1 and x=3 to arrive at the answer of 242pi/3 instead of the original interval x=0 and x=2? Would you show me step-wise how this was done or why it was necessary?
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