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Math Help - Cylindrical shell method to find volume

  1. #1
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    Cylindrical shell method to find volume

    Hi I have been asked to find the volume generated by rotating about the y axis the area in the first quadrant lying over the x-iterval [0,a] and under the graph of y=\sqrt{x}.

    the formula i used for this was the cylindircal shell method:
    <br />
v=\int^b_{a} (2\pi{x}*f{(x)})dx<br /> <br />


    So i evaluated the volume to be:
    <br />
v=\int^a_{0} (2\pi{x}*\sqrt{x})dx<br /> <br />

    <br />
v=\int^a_{0} (2\pi*{x}^{3/2})dx<br /> <br />
    <br />
v=\frac{4\pi*{x}^{5/2}}5<br /> <br />

    <br />
v=\frac{4\pi*{a}^{5/2}}5<br /> <br />


    But my friend got the result :


    <br />
\frac{\pi{a}^{5/2}}{5}<br />

    so where did i go wrong? Can i not use this method?


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  2. #2
    Super Member
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    Quote Originally Posted by olski1 View Post
    Hi I have been asked to find the volume generated by rotating about the y axis the area in the first quadrant lying over the x-iterval [0,a] and under the graph of y=\sqrt{x}.

    the formula i used for this was the cylindircal shell method:
    <br />
v=\int^b_{a} (2\pi{x}*f{(x)})dx <br /> <br />


    So i evaluated the volume to be:
    <br />
v=\int^a_{0} (2\pi{x}*\sqrt{x})dx <br /> <br />

    <br />
v=\int^a_{0} (2\pi*{x}^{3/2})dx <br /> <br />
    <br />
v=\frac{4\pi*{x}^{5/2}}5 <br /> <br />

    <br />
v=\frac{4\pi*{a}^{5/2}}5 <br /> <br />


    But my friend got the result :

    \frac{\pi{a}^{5/2}}{5}
    " alt="
    \frac{\pi{a}^{5/2}}{5}
    " />

    so where did i go wrong? Can i not use this method?



    You're correct. You can test it by simply choosing value of "a". If you plug \int_0^22\pi x\sqrt{x}dx into a calculator, you obtain 14.217. Now plug 2 into the equation you obtained, and you'll get the same result.
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