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anonymous
 5 years ago
how do you find the volume of a solid made by rotating y=2(x1)^(1/2) and y=x1 about the line x=1 using the shell method?
anonymous
 5 years ago
how do you find the volume of a solid made by rotating y=2(x1)^(1/2) and y=x1 about the line x=1 using the shell method?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can use an online whiteboard, try twiddla.com

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've tried it.. I got the height, but I can't get the right radius

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I tried it using disc and got it, but can't figure it out using shell

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What was your integral? was it \[\int\limits_{1}^{5} 2\pi (x+1)[2\sqrt{x1}  (x  1)] dx?\] The radius is (x + 1). Normally when you revolve around the yaxis (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.
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