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Math Help - Shell Method

  1. #1
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    Shell Method

    So you have:
    x=1+ (y-2)^2 and you are rotating around x=2
    For shell method the equation is 2 pi times the integral from 1 to 3 of x=1+ (y-2)^2
    The answer is 16/3 but I keep getting 32/3 pi. What am i doing wrong? It also has to be done by shels
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  2. #2
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    "For shell method the equation is 2 pi times the integral from 1 to 3 of x=1+ (y-2)^2"

    No, it is not. You are rotating around y= 2, not the x-axis. And it makes no sense to integrate a function of y from x= 1 to x= 3. Nor does it make sense to integrate from y= 1 when you are rotating around y= 2. You must mean to integrate from y= 2 to y= 3.

    You don't give a right boundary but x= 2 is implied. For that I get neither 16/3 nor (32/3)pi. Doing it by both by shells and by disks, I get a volume of pi/4.
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  3. #3
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    Quote Originally Posted by andrewho93 View Post
    So you have:
    x=1+ (y-2)^2 and you are rotating around x=2
    For shell method the equation is 2 pi times the integral from 1 to 3 of x=1+ (y-2)^2
    The answer is 16/3 but I keep getting 32/3 pi. What am i doing wrong? It also has to be done by shels
    The answer cannot be 16/3, since the volume must be less than the volume of a sphere with radius 1 unit.

    The radius of each cylinder is (2-x)

    The height of each cylinder is h=2|y-2|=2\sqrt{x-1}

    Surface area of each cylinder is 2{\pi}rh

    Therefore the volume of revolution is 2{\pi}\displaystyle\int_{1}^2rh}dx=4{\pi}\int_{1}^  2(2-x)\sqrt{x-1}}dx

    which comes out at \displaystyle\frac{16}{15}{\pi}

    The volume of revolution lies between the volumes of the purple spheres in the 2nd attachment.
    Attached Thumbnails Attached Thumbnails Shell Method-shells1.jpg   Shell Method-shells2.jpg  
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