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  1. #1
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    drilled sphere

    A ball of radius 14 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.

    I keep getting 10086.6068 but its not right any help would be nice thanks
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  2. #2
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    Quote Originally Posted by waite3 View Post
    A ball of radius 14 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.

    I keep getting 10086.6068 but its not right any help would be nice thanks
    Since the volume of a sphere is (4/3)Πr^3, if you find the volume of the larger sphere [I get (10976/3)Π] and then subtract the volume of the smaller sphere [I get (256/3)Π], then you have found the volume of the resulting solid. [I get (10720/3)Π which is ≈ 11, 225.96]
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  3. #3
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    Quote Originally Posted by waite3 View Post
    A ball of radius 14 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.

    I keep getting 10086.6068 but its not right any help would be nice thanks
    Volume(Sphere 1) = \frac{10976\pi}{3}

    Volume (Sphere 2) = \frac{256\pi}{3}

    Volume of Resulting Solid = \frac{10720\pi}{3}
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  4. #4
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    Quote Originally Posted by waite3 View Post
    A ball of radius 14 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.

    I keep getting 10086.6068 but its not right any help would be nice thanks
    Quote Originally Posted by TheHolly View Post
    Since the volume of a sphere is (4/3)Πr^3, if you find the volume of the larger sphere [I get (10976/3)Π] and then subtract the volume of the smaller sphere [I get (256/3)Π], then you have found the volume of the resulting solid. [I get (10720/3)Π which is ≈ 11, 225.96]
    But the volume that has been removed isn't a sphere, is it? I would think it's a cylindrical solid with rounded end caps.

    -Dan
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  5. #5
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    ya the shape that is removed would be cylindrical with rounder caps but im not sure how to find the volume of that.
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  6. #6
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    I read it as a round whole not a spherical hole. So think a drillbit of radius 4 straight through the center of the sphere. The sphere has a volume of \frac{10976\pi}{3}.

    The cylindrical portion of the hole has a radius of 4 but what is the height? Draw a circle of radius 14 and inscribe in it a rectangle whose top and bottom have length 8 (a crossection of the sphere through a great circle "parallel" to the hole). Draw a radius to a point on the rectangle and and the circle (the corner of the rectanlge), and draw a vertical radius. Now you have a right trangle whose hypotenuse is length 14 and whose base is length 4.

    Therefore the height of the cylinder is h = 2\sqrt{180}.

    So the volume of this cylinder is V_c = 32 \pi \sqrt{180}

    What tools do you have for calculating volume. Do you have integral calculus?
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  7. #7
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    Quote Originally Posted by waite3 View Post
    A ball of radius 14 has a round hole of radius 4 drilled through its center. Find the volume of the resulting solid.

    I keep getting 10086.6068 but its not right any help would be nice thanks
    The solid you are looking for is a symmetric spherical layer with the central cylinder missing.

    The volume of the layer is:

    V_{layer}=\frac{\pi h}{6} \cdot \left(6r^2+h^2\right)

    The volume of the central cylinder is

    V_{cyl}=\pi r^2 \cdot h

    Thus the volume of the solid is:

    V_{solid}=\frac{\pi h}{6} \cdot \left(6r^2+h^2\right) - \pi r^2 \cdot h = \frac16 \pi h^3

    r = 2 and I've got h = 6 \sqrt{5}
    Attached Thumbnails Attached Thumbnails drilled sphere-kugelschicht.gif  
    Last edited by earboth; March 26th 2008 at 08:37 AM.
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  8. #8
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    Hello, waite3

    I must assume that this is a Calculus problem.


    A ball of radius 14 has a round hole of radius 4 drilled through its center.
    Find the volume of the resulting solid.
    Code:
                  14|
                  * * *
              *:::::|:::::*
            *:::::::|:::::::*
           *--------+--------*
                   4|
          *         |         * 
      - - * - - - - + - - - - * - -
          *         |         *
                    | 
           *- - - - + - - - -*
            *       |       *
              *     |     *
                  * * *
                    |

    We have a circle: . x^2+y^2 \:=\:14^2\quad\Rightarrow\quad y^2 \:=\:196 - x^2

    We have the region bounded by the upper semicircle and y = 4
    Revolve it about the x-axis to generate the desired solid.

    The circle and horizontal line intersect at: x \:=\:\pm6\sqrt{5}

    The volume of the solid is: . V \;=\;2 \times \pi\int^{6\sqrt{5}}_0\bigg[(196 - x^2) - 4^2\bigg]\,dx

    And we have: . V \;=\;2\pi\int^{6\sqrt{5}}_0\left(180-x^2\right)\,dx


    I get: . 1440\pi\sqrt{5} \:\approx\:10,115.73

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  9. #9
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    nice thanks a lot the answer was 10,115.73
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