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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?

Question

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?

check_circleAnswer
Step 1

For this question radius= (x-1)^2 from x=0 to x=2

Step 2

We will use disk method to find the volume of the region. 

Volume
SIf()Fdx
= [(x-1)]'dx
S(x-1)'dx
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Volume SIf()Fdx = [(x-1)]'dx S(x-1)'dx

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Step 3

To integrate the integral we have to substitute z=x-1

W...

Z=x-1
dz dx
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Z=x-1 dz dx

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Tagged in

Math

Calculus

Integration

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