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Math Help - Volume

  1. #1
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    Volume

    A cone is attached to a hemisphere of radius 4 cm. If the total height of the object is 10 cm, find its volume.


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  2. #2
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    Re: Volume

    Let's calculate the volume of the cone and the volume of the hemisphere seperatly and then add them. I will be accurate to 2 points after the decimal point.

    Hemisphere: V = (2/3)*pi*r^3 = 128*pi/3 = 134.04

    If the height of the entire object is 10, and the radius of the hemisphere is 4, then the height of the cone is 10 - 4 = 6.

    Cone: V = (1/3)*pi*r^2*h = 32*pi = 100.53

    Vcone + Vhemisphere = 134.04 + 100.53 = 234.57
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  3. #3
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    Re: Volume

    Hello, Farisco!

    A cone is attached to a hemisphere of radius 4 cm.
    If the total height of the object is 10 cm, find its volume.

    Code:
        -         * * *
        :     *     :     *
        :   *       :       *
        :  *        :4       *
        :           :
        : * - - - - * - - - - *
        :  \   4    :    4   /
       10   \       :       /
        :    \      :      /
        :     \     :6    /
        :      \    :    /
        :       \   :   /
        :        \  :  /
        :         \ : /
        :          \:/
        -           *
    We have a hemisphere with radius 4.
    We have cone with radius 4 and height 6.

    You should be able to find the total volume
    . . without a calculator and without rounded-off decimals.


    A sphere has volume: . V \:=\:\tfrac{4}{3}\pi r^3, where r is the radius.

    A half-sphere with radius 4 has volume: . V \:=\:\tfrac{1}{2} \times \tfrac{4}{3}\pi(4^3) \:=\:\frac{128\pi}{3}


    A circular cone has volume: . V \:=\:\tfrac{\pi}{3}r^2h\;\;(r = \text{radius, }\:h = \text{height})

    A cone with r = 4,\,h=6 has volume: . V \;=\;\tfrac{\pi}{3}(4^2)(6) \:=\:32\pi


    The total volume is: . \frac{128\pi}{3} + 32\pi \;=\;\boxed{\frac{224\pi}{3}\text{ cm}^3}


    If you want a decimal, now is the time to crank it out:

    . . . . \boxed{2}\,\boxed{2}\,\boxed{4}\;\;\boxed{\times} \;\; \boxed{\pi}\;\;\boxed{\div}\;\;\boxed{3}\;\;\boxed  {=}

    and we get: . \boxed{234.5722515}
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