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Volume of cylindrical hole removed from a sphere

Given a solid sphere of radius 2r, what is the volume of material from a hole of radius r through the center of the sphere.


First, the volume of a sphere of radius 2r is given by

    \[ V_S = \frac{4}{3} \pi (2r)^3 = \frac{32}{3} \pi r^3. \]

Then, the volume of a sphere with a hole drilled in it is the volume of the solid of revolution generated by the region between f(x) = \sqrt{4r^2 - x^2} and g(x) = r from -\sqrt{3} r to \sqrt{3} r. Denoting this volume by V_T we then have,

    \begin{align*}  V_T &= \pi \int_{-\sqrt{3} r}^{\sqrt{3} r} (4r^2 - x^2 - r^2) \, dx \\      &= \pi \left( \int_{-\sqrt{3} r}^{\sqrt{3}r} (3r^2 - x^2) \, dx \right) \\      &= 6 \sqrt{3} \pi r^3 - 2 \sqrt{3} \pi r^3 \\      &= 4 \sqrt{3} \pi r^3. \end{align*}

Thus, the volume of the material removed from the sphere by drilling a hole in it is given by

    \[ \frac{32}{3} \pi r^3 - 4\sqrt{3} \pi r^3 = \left( \frac{32}{3} - 4 \sqrt{3} \right) \pi r^3. \]

9 comments

    • AudioRebel says:

      The trick is that the hole/tube does not end at the x-axis where f(x) = 0 but where f(x) = r.

      From the middle inside the hole/tube it goes r units up and r units down. thus set the equation of the semi-sphere f(x) = r and solve for x. This will give you the limits of integration.

      • Anonymous says:

        How is this correct though? If we follow Rori’s solution, it assumes that our hole never penetrated the surface, since there should a hole from -2 to 2. I believe we would have to remove all the part from -2 to

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        and then remove the cylinder of radius r from

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        to

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        and then from the entire sphere from from

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        to 2. That’s because the first and last interval would have revolutions smaller than r, so they would be lost.

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