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how do you find the volume of a solid made by rotating y=2(x-1)^(1/2) and y=x-1 about the line x=-1 using the shell method?

Mathematics
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Ok well I'm sure you have the formula for the shell method. So all you need is the radius, height, and the boundary points for your cylinders. The radius should be x+1, the boundary for x is 1 to 5. The height is going to be the upper graph minus the lower. Can you do the rest?
It'd be tough to explain through open study since drawings are necessary to understand it. I'd recommend going to this http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx to help
you can use an online whiteboard, try twiddla.com

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I've tried it.. I got the height, but I can't get the right radius
I tried it using disc and got it, but can't figure it out using shell
What was your integral? was it \[\int\limits_{1}^{5} 2\pi (x+1)[2\sqrt{x-1} - (x - 1)] dx?\] The radius is (x + 1). Normally when you revolve around the y-axis (the line x = 0) your radius is just x. But now you've shifted the axis left one to get x = -1, so the radius increased by one. I think example 3 in the link I gave you is close to your problem.
oh that makes sense.

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