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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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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Tagged in
Math
Calculus

Integration