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Math Help - Volume of sphere with hole

  1. #1
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    Volume of sphere with hole

    A hole of diameter d is drilled through a sphere of radius r in such a way that the axis of the hole passes through the centre of the sphere. Find the volume of the solid that remains.

    I drew a picture of a sphere centred on the cartesian plane, with a hole going through its centre (in the shape of a cylinder).

    I already knew that the volume of the sphere was \frac {4}{3}\pi r^{3}. I then worked out the volume of the hole (cylinder):
    <br />
\int_{-r}^{r} \frac{1}{4}\pi d^{2}dy
    = \frac{1}{2}\pi d^{2} r<br />

    Volume of solid remaining = \frac {4}{3}\pi r^{3} - \frac{1}{2}\pi d^{2} r

    but the correct answer was \frac{1}{6}\pi (4r^{2} - d^{2})^{\frac{3}{2}}

    I cant see where I went wrong cause surely the objective of this problem is to minus the hole (cylinder) from a sphere to get the remaining volume?
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  2. #2
    A Plied Mathematician
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    The volume subtracted out by drilling the hole is not precisely a cylinder, because a cylinder usually has flat ends. The material removed has curved ends (matches up with the sphere). So what do you get when you change this?
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  3. #3
    Newbie Topher's Avatar
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    Very intresting problem. It sounds like Ackbeet is right in the fact that the figure removed is not
    Spoiler:
    only
    a cylinder.



    Good Luck!
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  4. #4
    MHF Contributor ebaines's Avatar
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    Rather than subtracting the round-ended cylinder from the volume of the sphere, I think it's probably easier to set up the integral for the volume. You can use cylindrical shell elements of width  dx , circumference 2 \pi x, and height  2  \sqrt {R^2 - x^2} , and integrate from x = d/2 to x = R.
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  5. #5
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    I got the integral V=-2\pi\int_{r^{2}-\frac{d^{2}}{4}}^{0} \sqrt{u} du after making the substitution of u=r^{2} - x^{2} then my final answer was:

    \frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}. Did I do something wrong or is there any way to simplify it to \frac{1}{6}\pi(4r^{2} -d^{2)^\frac{3}{2} ?
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  6. #6
    A Plied Mathematician
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    They are the same answer. Proof:

    \displaystyle{\frac{4}{3}\pi(r^{2}-\frac{d^{2}}{4})^{\frac{3}{2}}}

    \displaystyle{=\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}\left(\frac{1}{4}\right)^{\fra  c{3}{2}}}

    \displaystyle{=\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}\,\frac{1}{4^{3/2}}}

    \displaystyle{=\frac{1}{8}\,\frac{4}{3}\pi(4r^{2}-d^{2})^{\frac{3}{2}}}

    \displaystyle{=\frac{1}{6}\pi(4r^{2}-d^{2})^{\frac{3}{2}}.}
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