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anonymous

  • 5 years ago

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?

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  1. anonymous
    • 5 years ago
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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?

  2. anonymous
    • 5 years ago
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    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

  3. anonymous
    • 5 years ago
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    you can use an online whiteboard, try twiddla.com

  4. anonymous
    • 5 years ago
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    I've tried it.. I got the height, but I can't get the right radius

  5. anonymous
    • 5 years ago
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    I tried it using disc and got it, but can't figure it out using shell

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    oh that makes sense.

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