How Do You Calculate the Volume of a Solid Revolved Around y=3?

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SUMMARY

The volume of the solid generated by revolving the region enclosed by the curves y=x^2+2, y=2x+1, and the y-axis around the line y=3 can be calculated using the washer method. The correct volume element is expressed as ΔV = π[(2x + 1 - 3)² - (x² + 2 - 3)²]Δx, integrated from 0 to 1. The initial setup using the disk/washer method was incorrect due to misplacement of the function values. Proper simplification and integration will yield the desired volume.

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Homework Statement


Find the volume of the solid generated by the revolution of the region enclosed by y=x^2+2, y=2x+1 and the y-axis about the line y=3



Homework Equations


Disk/Washer method, shell method


The Attempt at a Solution


This one is stumping me no matter how I go about it. What I have done so far is use the disk/washer method by inputting x+3 in for f(x) and x^2-2x+1 (obtained from subtracting the two functions given) in for g(x):
V=pi[tex]\int[/tex]((x+3)^2-(x^2-2x+1)^2)dx over [0,1]. I think that because of the location of the region on a graph, however, the x+3 might become x+2, and that I might have made a mistake in the way I set the formula up, but I do not know if/where I made this mistake. If anyone can point me in the right direction and point out any mistakes I made in setting up this question it would be greatly appreciated, thanks in advance.
 
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What you want for your typical volume element is this:
[tex]\Delta V = \pi [(2x + 1 - 3)^2 - (x^2 + 2 - 3)^2]\Delta x[/tex]
Simplify that a bit, and integrate between 0 and 1, and you should be good.
 

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