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Math Help - Integration via shell method

  1. #1
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    Integration via shell method

    Reviewing for my final and got one (of many) problems I keep getting a weird answer for. The problem reads:

    Use the shell method to integrate. Find the volume generated by revolving the region bounded by the graphs of  y = \sqrt{x-1} and  y = x - 1 about the line  x = 3.

    Using the shell method I set it up as  \int{(3 - x)(\sqrt{x - 1} - x - 1) from [1,2]. However, I get a negative volume. Any help would be greatly appreciated.

    Thanks
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Poptimus View Post
    Reviewing for my final and got one (of many) problems I keep getting a weird answer for. The problem reads:

    Use the shell method to integrate. Find the volume generated by revolving the region bounded by the graphs of  y = \sqrt{x-1} and  y = x - 1 about the line  x = 3.

    Using the shell method I set it up as  \int{(3 - x)(\sqrt{x - 1} - \Big(x - 1\Big)) from [1,2]. However, I get a negative volume. Any help would be greatly appreciated.

    Thanks
    Well, you have forgotten to put parentheses around (x-1), that's why you get a negative result. Also, you have left out a factor of \scriptstyle 2\pi, I believe.
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  3. #3
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    Well, one possible reason you're getting negative numbers is that you should have

    V=2\pi\int_{1}^{2}(3 - x)(\sqrt{x - 1} - x + 1)\,dx.

    Sign error in there will do it every time. I think you also forgot the 2\pi factor out front. Don't forget: with the shell method, you could slice up one shell, flatten it out, and you'd have a rectangular prism whose volume is quite easy to compute: dV=2\pi\,r\,h\,dx. The radius r=3-x, like you had.

    [EDIT]: Failure failed to allow my post to go first.
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  4. #4
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    Appreciate it guys, such a simple mistake. I can't believe I overlooked it.
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