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Thread: volume of solid of revolution further integration

  1. #1
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    volume of solid of revolution further integration

    Question: An area in the first quadrant is bounded by the ellipse $\displaystyle 4x^2+9y^2=36 $ and the axes. This area is rotated through four right angles about the x-axis. Find
    (a) the volume of the solid generated
    (b) the x-coordinate of the centre of gravity of this solid


    My attempt:
    a) $\displaystyle 4x^2+9y^2=36 $


    $\displaystyle \frac{x^2}{9}+ \frac{y^2}{4 }=1 $


    $\displaystyle y =\frac{\sqrt{4-\frac{4}{9}x^2}}{2} $
    element of volume = $\displaystyle \pi y^2 \partial x $
    v =$\displaystyle \int_{0}^{3} \pi \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x $


    v =$\displaystyle 2 $
    b) $\displaystyle \bar{x}\int_{0}^{3} \pi \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x = \int_{0}^{3} \pi x \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x $


    $\displaystyle \bar{x}\int_{0}^{3} \left ( \frac{{4-\frac{4}{9}x^2}}{4} \right ) \mathrm{d} x = \int_{0}^{3} x - \frac{x^3}{9} \mathrm{d} x $




    $\displaystyle \bar{x} = \frac{9}{8} $




    but the book gives the answer for a) $\displaystyle 8 \pi $ b) $\displaystyle \frac{4}{5} $


    please can i have some help with this question
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  2. #2
    Member Walagaster's Avatar
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    Re: volume of solid of revolution further integration

    Quote Originally Posted by bigmansouf View Post
    Question: An area in the first quadrant is bounded by the ellipse $\displaystyle 4x^2+9y^2=36 $ and the axes. This area is rotated through four right angles about the x-axis. Find
    (a) the volume of the solid generated
    (b) the x-coordinate of the centre of gravity of this solid


    My attempt:
    a) $\displaystyle 4x^2+9y^2=36 $


    $\displaystyle \frac{x^2}{9}+ \frac{y^2}{4 }=1 $


    $\displaystyle y =\frac{\sqrt{4-\frac{4}{9}x^2}}{2} $
    Check your algebra solving for $y$ here.

    element of volume = $\displaystyle \pi y^2 \partial x $
    v =$\displaystyle \int_{0}^{3} \pi \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x $


    v =$\displaystyle 2 $
    What happened to the $\pi$?


    b) $\displaystyle \bar{x}\int_{0}^{3} \pi \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x = \int_{0}^{3} \pi x \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x $


    $\displaystyle \bar{x}\int_{0}^{3} \left ( \frac{{4-\frac{4}{9}x^2}}{4} \right ) \mathrm{d} x = \int_{0}^{3} x - \frac{x^3}{9} \mathrm{d} x $




    $\displaystyle \bar{x} = \frac{9}{8} $




    but the book gives the answer for a) $\displaystyle 8 \pi $ b) $\displaystyle \frac{4}{5} $


    please can i have some help with this question
    Please don't use color and font commands throughout like that because it makes it very difficult to edit.
    Thanks from topsquark and bigmansouf
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  3. #3
    Junior Member Cervesa's Avatar
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    Re: volume of solid of revolution further integration

    $\displaystyle V = \pi \int_0^3 y^2 \, dx$

    $\displaystyle V = \pi \int_0^3 4\left(1 - \frac{x^2}{9} \right) \, dx$

    $\displaystyle V = 4\pi \bigg[x - \frac{x^3}{27} \bigg]_0^3$

    $\displaystyle V = 4\pi \bigg[\left(3 - \frac{3^3}{27}\right) - (0) \bigg] = 8\pi$


    $\displaystyle \bar{x} = \dfrac{\pi \int_0^3 x \cdot y^2 \, dx}{\pi \int_0^3 y^2 \, dx} = \dfrac{4\pi \int_0^3 x - \frac{x^3}{9} \, dx}{8\pi}$

    $\displaystyle \bar{x} = \dfrac{1}{2} \int_0^3 x - \frac{x^3}{9} \, dx$

    $\displaystyle \bar{x} = \dfrac{1}{2}\bigg[\frac{x^2}{2} - \frac{x^4}{36} \bigg]_0^3$

    $\displaystyle \bar{x} = \dfrac{1}{2}\bigg[\left(\frac{9}{2} - \frac{9}{4} \right)-(0) \bigg] = \dfrac{9}{8}$
    Thanks from bigmansouf
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    Re: volume of solid of revolution further integration

    Quote Originally Posted by Walagaster View Post
    Check your algebra solving for $y$ here.



    What happened to the $\pi$?



    Please don't use color and font commands throughout like that because it makes it very difficult to edit.
    sorry but cannot you explain what you mean by font commands and color
    thank you
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  5. #5
    Member Walagaster's Avatar
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    Re: volume of solid of revolution further integration

    Quote Originally Posted by bigmansouf View Post
    sorry but cannot you explain what you mean by font commands and color
    thank you
    Try "reply with quote" to your original post and I think you will see what I am talking about. (I think so, anyway).
    Thanks from topsquark
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  6. #6
    Forum Admin topsquark's Avatar
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    Re: volume of solid of revolution further integration

    Quote Originally Posted by Walagaster View Post
    Try "reply with quote" to your original post and I think you will see what I am talking about. (I think so, anyway).
    Wow.

    -Dan
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