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Math Help - shells

  1. #1
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    shells

    A hole of diameter a is bored through the centre of a sphere of radius a. Show that the remaining volume is root3/2.pi.a(cubed).

    please help!!
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  2. #2
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    Lexington, MA (USA)
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    Hello, dannyshox!

    A hole of diameter a is bored through the centre of a sphere of radius a.
    Show that the remaining volume is: \frac{\sqrt{3}}{2}\pi a^3
    Code:
                    |
                  * * *
              *:::::|:::::*
          - * - - - + - - - * -   y = a/2
           *|       |       |*
            |       |       |
          * |       |       | *
      ----*-+-------+-------+-*----
          * |       |       | *a
            |       |       |
           *|       |       |*
            * - - - | - - - *
              *     |     *
                  * * *
                    |

    The equation of the circle is: . x^2+y^2\:=\:a^2 \quad\Rightarrow\quad y^2 \:=\:a^2-x^2

    The equation of the horizontal line is: . y \:=\:\frac{a}{2}

    Revolve the shaded region about the x-axis.
    That is the desired volume.

    The graphs intersect for: . \begin{array}{cc}x^2+y^2 \:=\:a^2 & [1] \\ y \:=\:\frac{a}{2} & [2]\end{array}

    . . Substitute [2] into [1]: . x^2 + \left(\frac{a}{2}\right)^2 \:=\:a^2 \quad\Rightarrow\quad x^2 \:=\:\frac{3a^2}{4} \quad\Rightarrow\quad x \:=\:\pm\frac{\sqrt{3}}{2}a


    Formula: . V \;=\;\pi\int^b_a\bigg(y_2^2 - y_1^2\bigg)\,dx

    Due to symmetry: . V \;=\;2 \times \pi\int^{\frac{\sqrt{3}}{2}a} _0 \bigg[(a^2-x^2) - \left(\frac{a}{2}\right)^2\bigg]\,dx . = \;2\pi\int^{\frac{\sqrt{3}}{2}a}_0\bigg[\frac{3}{4}a^2 - x^2\bigg]\,dx

    Got it?

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  3. #3
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    Oct 2009
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    got it

    thanks!
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