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Math Help - volume by shell; find my error

  1. #1
    Junior Member Tclack's Avatar
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    volume by shell; find my error

    Find volume of shape between:  y= \sqrt{x}, y=0 , x=9
    rotated about line x=9
    So, since I'm going around vertically, I rearranged x=y^2

    \pi \int_0^3 (9-y^2)^2 dy
    =\pi \int_0^3 (81-18y^2+y^4) dy
    =\pi(81y-6y^2+\frac{y^5}{5})\mid_0^3
    =\frac{1188}{5}\pi

    the answer is supposed to be 648pi/5

    The similarities between my answer and their answer makes me think my method is correct. See attached picture. I'm trying to get the shape to be like a. But I suspect my equation is giving me b. Especially because my answer is larger, and b looks like it would have a larger volume.
    Can anyone clarify?
    Attached Thumbnails Attached Thumbnails volume by shell; find my error-disk-prob.jpg  
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Tclack View Post
    Find volume of shape between:  y= \sqrt{x}, y=0 , x=9
    rotated about line x=9
    So, since I'm going around vertically, I rearranged x=y^2
    So you are using disks not "shells".

    \pi \int_0^3 (9-y^2)^2 dy
    =\pi \int_0^3 (81-18y^2+y^4) dy
    =\pi(81y-6y^2+\frac{y^5}{5})\mid_0^3
    You have a wrong power:
    =\pi(81y-6y^3+\frac{y^5}{5})\mid_0^3

    =\frac{1188}{5}\pi

    the answer is supposed to be 648pi/5

    The similarities between my answer and their answer makes me think my method is correct. See attached picture. I'm trying to get the shape to be like a. But I suspect my equation is giving me b. Especially because my answer is larger, and b looks like it would have a larger volume.
    Can anyone clarify?
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  3. #3
    Junior Member Tclack's Avatar
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    Damn, thanks, I went over it 4 times and I didn't catch it. So I thought the error was in the setup of the integral.

    Do the volumes of A and B equal each other?
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