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

*Related*

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

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

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.

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

and then remove the cylinder of radius from

to

and then from the entire sphere from from

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

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

Nevermind.

I can’t undrrstand from where root of 3 come

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

Thanks! Fixed.