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# Volume of cylindrical hole removed from a sphere

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

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

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 and from to . Denoting this volume by we then have,

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

1. Anonymous says:

How did you find the interval of integration to be negative root 3 to root 3?

• Mihajlo says:

It is the height of the equilateral triangle, because the radius of the base is r.

• 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 to

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and then remove the cylinder of radius from

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-3^0.5

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to

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3^0.5

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and then from the entire sphere from from

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-3^0.5

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

• Anonymous says:

Nevermind, we already exclude the excess volume when taking the interval from -3^0.5 to 3^0.5.

• Anonymous says:

Nevermind.

2. Ejnota Rom says:

I can’t undrrstand from where root of 3 come

3. Sue Harris says:

The third step of your simplification of V(T) has an extra π (Pi) coefficient in the second term.

• RoRi says:

Thanks! Fixed.